\(\int \frac {(c+d x)^2 (a+b x^2)^{3/2} (A+B x+C x^2)}{x^{10}} \, dx\) [74]

Optimal result

Integrand size = 32, antiderivative size = 365 \[ \int \frac {(c+d x)^2 \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^{10}} \, dx=\frac {b (3 b c (B c+2 A d)-8 a d (2 c C+B d)) \sqrt {a+b x^2}}{64 a x^4}+\frac {b^2 (3 b c (B c+2 A d)-8 a d (2 c C+B d)) \sqrt {a+b x^2}}{128 a^2 x^2}+\frac {(3 b c (B c+2 A d)-8 a d (2 c C+B d)) \left (a+b x^2\right )^{3/2}}{48 a x^6}-\frac {A c^2 \left (a+b x^2\right )^{5/2}}{9 a x^9}-\frac {c (B c+2 A d) \left (a+b x^2\right )^{5/2}}{8 a x^8}-\frac {\left (9 a c (c C+2 B d)-A \left (4 b c^2-9 a d^2\right )\right ) \left (a+b x^2\right )^{5/2}}{63 a^2 x^7}-\frac {\left (2 A b \left (4 b c^2-9 a d^2\right )+9 a \left (7 a C d^2-2 b c (c C+2 B d)\right )\right ) \left (a+b x^2\right )^{5/2}}{315 a^3 x^5}-\frac {b^3 (3 b c (B c+2 A d)-8 a d (2 c C+B d)) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{128 a^{5/2}} \] Output:

1/64*b*(3*b*c*(2*A*d+B*c)-8*a*d*(B*d+2*C*c))*(b*x^2+a)^(1/2)/a/x^4+1/128*b 
^2*(3*b*c*(2*A*d+B*c)-8*a*d*(B*d+2*C*c))*(b*x^2+a)^(1/2)/a^2/x^2+1/48*(3*b 
*c*(2*A*d+B*c)-8*a*d*(B*d+2*C*c))*(b*x^2+a)^(3/2)/a/x^6-1/9*A*c^2*(b*x^2+a 
)^(5/2)/a/x^9-1/8*c*(2*A*d+B*c)*(b*x^2+a)^(5/2)/a/x^8-1/63*(9*a*c*(2*B*d+C 
*c)-A*(-9*a*d^2+4*b*c^2))*(b*x^2+a)^(5/2)/a^2/x^7-1/315*(2*A*b*(-9*a*d^2+4 
*b*c^2)+9*a*(7*a*C*d^2-2*b*c*(2*B*d+C*c)))*(b*x^2+a)^(5/2)/a^3/x^5-1/128*b 
^3*(3*b*c*(2*A*d+B*c)-8*a*d*(B*d+2*C*c))*arctanh((b*x^2+a)^(1/2)/a^(1/2))/ 
a^(5/2)
 

Mathematica [A] (verified)

Time = 4.51 (sec) , antiderivative size = 389, normalized size of antiderivative = 1.07 \[ \int \frac {(c+d x)^2 \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^{10}} \, dx=-\frac {\sqrt {a+b x^2} \left (1024 A b^4 c^2 x^8-a b^3 x^6 \left (9 c x (105 B c+256 c C x+512 B d x)+2 A \left (256 c^2+945 c d x+1152 d^2 x^2\right )\right )+16 a^4 \left (10 A \left (28 c^2+63 c d x+36 d^2 x^2\right )+3 x \left (8 C x \left (15 c^2+35 c d x+21 d^2 x^2\right )+5 B \left (21 c^2+48 c d x+28 d^2 x^2\right )\right )\right )+6 a^2 b^2 x^4 \left (2 A \left (32 c^2+105 c d x+96 d^2 x^2\right )+3 x \left (8 C x \left (8 c^2+35 c d x+56 d^2 x^2\right )+B \left (35 c^2+128 c d x+140 d^2 x^2\right )\right )\right )+8 a^3 b x^2 \left (2 A \left (400 c^2+945 c d x+576 d^2 x^2\right )+3 x \left (4 C x \left (96 c^2+245 c d x+168 d^2 x^2\right )+B \left (315 c^2+768 c d x+490 d^2 x^2\right )\right )\right )\right )}{40320 a^3 x^9}+\frac {b^3 (-3 b c (B c+2 A d)+8 a d (2 c C+B d)) \text {arctanh}\left (\frac {-\sqrt {b} x+\sqrt {a+b x^2}}{\sqrt {a}}\right )}{64 a^{5/2}} \] Input:

Integrate[((c + d*x)^2*(a + b*x^2)^(3/2)*(A + B*x + C*x^2))/x^10,x]
 

Output:

-1/40320*(Sqrt[a + b*x^2]*(1024*A*b^4*c^2*x^8 - a*b^3*x^6*(9*c*x*(105*B*c 
+ 256*c*C*x + 512*B*d*x) + 2*A*(256*c^2 + 945*c*d*x + 1152*d^2*x^2)) + 16* 
a^4*(10*A*(28*c^2 + 63*c*d*x + 36*d^2*x^2) + 3*x*(8*C*x*(15*c^2 + 35*c*d*x 
 + 21*d^2*x^2) + 5*B*(21*c^2 + 48*c*d*x + 28*d^2*x^2))) + 6*a^2*b^2*x^4*(2 
*A*(32*c^2 + 105*c*d*x + 96*d^2*x^2) + 3*x*(8*C*x*(8*c^2 + 35*c*d*x + 56*d 
^2*x^2) + B*(35*c^2 + 128*c*d*x + 140*d^2*x^2))) + 8*a^3*b*x^2*(2*A*(400*c 
^2 + 945*c*d*x + 576*d^2*x^2) + 3*x*(4*C*x*(96*c^2 + 245*c*d*x + 168*d^2*x 
^2) + B*(315*c^2 + 768*c*d*x + 490*d^2*x^2)))))/(a^3*x^9) + (b^3*(-3*b*c*( 
B*c + 2*A*d) + 8*a*d*(2*c*C + B*d))*ArcTanh[(-(Sqrt[b]*x) + Sqrt[a + b*x^2 
])/Sqrt[a]])/(64*a^(5/2))
 

Rubi [A] (verified)

Time = 1.85 (sec) , antiderivative size = 328, normalized size of antiderivative = 0.90, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {2338, 25, 2338, 25, 2338, 27, 539, 27, 534, 243, 51, 51, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (a+b x^2\right )^{3/2} (c+d x)^2 \left (A+B x+C x^2\right )}{x^{10}} \, dx\)

\(\Big \downarrow \) 2338

\(\displaystyle -\frac {\int -\frac {\left (b x^2+a\right )^{3/2} \left (9 a C d^2 x^3+9 a d (2 c C+B d) x^2+\left (9 a c (c C+2 B d)-A \left (4 b c^2-9 a d^2\right )\right ) x+9 a c (B c+2 A d)\right )}{x^9}dx}{9 a}-\frac {A c^2 \left (a+b x^2\right )^{5/2}}{9 a x^9}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\left (b x^2+a\right )^{3/2} \left (9 a C d^2 x^3+9 a d (2 c C+B d) x^2+\left (9 a c (c C+2 B d)-A \left (4 b c^2-9 a d^2\right )\right ) x+9 a c (B c+2 A d)\right )}{x^9}dx}{9 a}-\frac {A c^2 \left (a+b x^2\right )^{5/2}}{9 a x^9}\)

\(\Big \downarrow \) 2338

\(\displaystyle \frac {-\frac {\int -\frac {\left (b x^2+a\right )^{3/2} \left (72 a^2 C d^2 x^2-9 a (3 b c (B c+2 A d)-8 a d (2 c C+B d)) x+8 a \left (9 a c (c C+2 B d)-A \left (4 b c^2-9 a d^2\right )\right )\right )}{x^8}dx}{8 a}-\frac {9 c \left (a+b x^2\right )^{5/2} (2 A d+B c)}{8 x^8}}{9 a}-\frac {A c^2 \left (a+b x^2\right )^{5/2}}{9 a x^9}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {\left (b x^2+a\right )^{3/2} \left (72 a^2 C d^2 x^2-9 a (3 b c (B c+2 A d)-8 a d (2 c C+B d)) x+8 a \left (9 a c (c C+2 B d)-A \left (4 b c^2-9 a d^2\right )\right )\right )}{x^8}dx}{8 a}-\frac {9 c \left (a+b x^2\right )^{5/2} (2 A d+B c)}{8 x^8}}{9 a}-\frac {A c^2 \left (a+b x^2\right )^{5/2}}{9 a x^9}\)

\(\Big \downarrow \) 2338

\(\displaystyle \frac {\frac {-\frac {\int \frac {a \left (63 a (3 b c (B c+2 A d)-8 a d (2 c C+B d))-8 \left (2 A b \left (4 b c^2-9 a d^2\right )+9 a \left (7 a C d^2-2 b c (c C+2 B d)\right )\right ) x\right ) \left (b x^2+a\right )^{3/2}}{x^7}dx}{7 a}-\frac {8 \left (a+b x^2\right )^{5/2} \left (9 a c (2 B d+c C)-A \left (4 b c^2-9 a d^2\right )\right )}{7 x^7}}{8 a}-\frac {9 c \left (a+b x^2\right )^{5/2} (2 A d+B c)}{8 x^8}}{9 a}-\frac {A c^2 \left (a+b x^2\right )^{5/2}}{9 a x^9}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {1}{7} \int \frac {\left (63 a (3 b c (B c+2 A d)-8 a d (2 c C+B d))-8 \left (2 A b \left (4 b c^2-9 a d^2\right )+9 a \left (7 a C d^2-2 b c (c C+2 B d)\right )\right ) x\right ) \left (b x^2+a\right )^{3/2}}{x^7}dx-\frac {8 \left (a+b x^2\right )^{5/2} \left (9 a c (2 B d+c C)-A \left (4 b c^2-9 a d^2\right )\right )}{7 x^7}}{8 a}-\frac {9 c \left (a+b x^2\right )^{5/2} (2 A d+B c)}{8 x^8}}{9 a}-\frac {A c^2 \left (a+b x^2\right )^{5/2}}{9 a x^9}\)

\(\Big \downarrow \) 539

\(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {\int \frac {3 a \left (16 \left (2 A b \left (4 b c^2-9 a d^2\right )+9 a \left (7 a C d^2-2 b c (c C+2 B d)\right )\right )+21 b (3 b c (B c+2 A d)-8 a d (2 c C+B d)) x\right ) \left (b x^2+a\right )^{3/2}}{x^6}dx}{6 a}+\frac {21 \left (a+b x^2\right )^{5/2} (3 b c (2 A d+B c)-8 a d (B d+2 c C))}{2 x^6}\right )-\frac {8 \left (a+b x^2\right )^{5/2} \left (9 a c (2 B d+c C)-A \left (4 b c^2-9 a d^2\right )\right )}{7 x^7}}{8 a}-\frac {9 c \left (a+b x^2\right )^{5/2} (2 A d+B c)}{8 x^8}}{9 a}-\frac {A c^2 \left (a+b x^2\right )^{5/2}}{9 a x^9}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {1}{2} \int \frac {\left (16 \left (2 A b \left (4 b c^2-9 a d^2\right )+9 a \left (7 a C d^2-2 b c (c C+2 B d)\right )\right )+21 b (3 b c (B c+2 A d)-8 a d (2 c C+B d)) x\right ) \left (b x^2+a\right )^{3/2}}{x^6}dx+\frac {21 \left (a+b x^2\right )^{5/2} (3 b c (2 A d+B c)-8 a d (B d+2 c C))}{2 x^6}\right )-\frac {8 \left (a+b x^2\right )^{5/2} \left (9 a c (2 B d+c C)-A \left (4 b c^2-9 a d^2\right )\right )}{7 x^7}}{8 a}-\frac {9 c \left (a+b x^2\right )^{5/2} (2 A d+B c)}{8 x^8}}{9 a}-\frac {A c^2 \left (a+b x^2\right )^{5/2}}{9 a x^9}\)

\(\Big \downarrow \) 534

\(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {1}{2} \left (21 b (3 b c (2 A d+B c)-8 a d (B d+2 c C)) \int \frac {\left (b x^2+a\right )^{3/2}}{x^5}dx-\frac {16 \left (a+b x^2\right )^{5/2} \left (2 A b \left (4 b c^2-9 a d^2\right )+9 a \left (7 a C d^2-2 b c (2 B d+c C)\right )\right )}{5 a x^5}\right )+\frac {21 \left (a+b x^2\right )^{5/2} (3 b c (2 A d+B c)-8 a d (B d+2 c C))}{2 x^6}\right )-\frac {8 \left (a+b x^2\right )^{5/2} \left (9 a c (2 B d+c C)-A \left (4 b c^2-9 a d^2\right )\right )}{7 x^7}}{8 a}-\frac {9 c \left (a+b x^2\right )^{5/2} (2 A d+B c)}{8 x^8}}{9 a}-\frac {A c^2 \left (a+b x^2\right )^{5/2}}{9 a x^9}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {1}{2} \left (\frac {21}{2} b (3 b c (2 A d+B c)-8 a d (B d+2 c C)) \int \frac {\left (b x^2+a\right )^{3/2}}{x^6}dx^2-\frac {16 \left (a+b x^2\right )^{5/2} \left (2 A b \left (4 b c^2-9 a d^2\right )+9 a \left (7 a C d^2-2 b c (2 B d+c C)\right )\right )}{5 a x^5}\right )+\frac {21 \left (a+b x^2\right )^{5/2} (3 b c (2 A d+B c)-8 a d (B d+2 c C))}{2 x^6}\right )-\frac {8 \left (a+b x^2\right )^{5/2} \left (9 a c (2 B d+c C)-A \left (4 b c^2-9 a d^2\right )\right )}{7 x^7}}{8 a}-\frac {9 c \left (a+b x^2\right )^{5/2} (2 A d+B c)}{8 x^8}}{9 a}-\frac {A c^2 \left (a+b x^2\right )^{5/2}}{9 a x^9}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {1}{2} \left (\frac {21}{2} b (3 b c (2 A d+B c)-8 a d (B d+2 c C)) \left (\frac {3}{4} b \int \frac {\sqrt {b x^2+a}}{x^4}dx^2-\frac {\left (a+b x^2\right )^{3/2}}{2 x^4}\right )-\frac {16 \left (a+b x^2\right )^{5/2} \left (2 A b \left (4 b c^2-9 a d^2\right )+9 a \left (7 a C d^2-2 b c (2 B d+c C)\right )\right )}{5 a x^5}\right )+\frac {21 \left (a+b x^2\right )^{5/2} (3 b c (2 A d+B c)-8 a d (B d+2 c C))}{2 x^6}\right )-\frac {8 \left (a+b x^2\right )^{5/2} \left (9 a c (2 B d+c C)-A \left (4 b c^2-9 a d^2\right )\right )}{7 x^7}}{8 a}-\frac {9 c \left (a+b x^2\right )^{5/2} (2 A d+B c)}{8 x^8}}{9 a}-\frac {A c^2 \left (a+b x^2\right )^{5/2}}{9 a x^9}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {1}{2} \left (\frac {21}{2} b (3 b c (2 A d+B c)-8 a d (B d+2 c C)) \left (\frac {3}{4} b \left (\frac {1}{2} b \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2-\frac {\sqrt {a+b x^2}}{x^2}\right )-\frac {\left (a+b x^2\right )^{3/2}}{2 x^4}\right )-\frac {16 \left (a+b x^2\right )^{5/2} \left (2 A b \left (4 b c^2-9 a d^2\right )+9 a \left (7 a C d^2-2 b c (2 B d+c C)\right )\right )}{5 a x^5}\right )+\frac {21 \left (a+b x^2\right )^{5/2} (3 b c (2 A d+B c)-8 a d (B d+2 c C))}{2 x^6}\right )-\frac {8 \left (a+b x^2\right )^{5/2} \left (9 a c (2 B d+c C)-A \left (4 b c^2-9 a d^2\right )\right )}{7 x^7}}{8 a}-\frac {9 c \left (a+b x^2\right )^{5/2} (2 A d+B c)}{8 x^8}}{9 a}-\frac {A c^2 \left (a+b x^2\right )^{5/2}}{9 a x^9}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {1}{2} \left (\frac {21}{2} b (3 b c (2 A d+B c)-8 a d (B d+2 c C)) \left (\frac {3}{4} b \left (\int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}-\frac {\sqrt {a+b x^2}}{x^2}\right )-\frac {\left (a+b x^2\right )^{3/2}}{2 x^4}\right )-\frac {16 \left (a+b x^2\right )^{5/2} \left (2 A b \left (4 b c^2-9 a d^2\right )+9 a \left (7 a C d^2-2 b c (2 B d+c C)\right )\right )}{5 a x^5}\right )+\frac {21 \left (a+b x^2\right )^{5/2} (3 b c (2 A d+B c)-8 a d (B d+2 c C))}{2 x^6}\right )-\frac {8 \left (a+b x^2\right )^{5/2} \left (9 a c (2 B d+c C)-A \left (4 b c^2-9 a d^2\right )\right )}{7 x^7}}{8 a}-\frac {9 c \left (a+b x^2\right )^{5/2} (2 A d+B c)}{8 x^8}}{9 a}-\frac {A c^2 \left (a+b x^2\right )^{5/2}}{9 a x^9}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {1}{7} \left (\frac {1}{2} \left (\frac {21}{2} b \left (\frac {3}{4} b \left (-\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {\sqrt {a+b x^2}}{x^2}\right )-\frac {\left (a+b x^2\right )^{3/2}}{2 x^4}\right ) (3 b c (2 A d+B c)-8 a d (B d+2 c C))-\frac {16 \left (a+b x^2\right )^{5/2} \left (2 A b \left (4 b c^2-9 a d^2\right )+9 a \left (7 a C d^2-2 b c (2 B d+c C)\right )\right )}{5 a x^5}\right )+\frac {21 \left (a+b x^2\right )^{5/2} (3 b c (2 A d+B c)-8 a d (B d+2 c C))}{2 x^6}\right )-\frac {8 \left (a+b x^2\right )^{5/2} \left (9 a c (2 B d+c C)-A \left (4 b c^2-9 a d^2\right )\right )}{7 x^7}}{8 a}-\frac {9 c \left (a+b x^2\right )^{5/2} (2 A d+B c)}{8 x^8}}{9 a}-\frac {A c^2 \left (a+b x^2\right )^{5/2}}{9 a x^9}\)

Input:

Int[((c + d*x)^2*(a + b*x^2)^(3/2)*(A + B*x + C*x^2))/x^10,x]
 

Output:

-1/9*(A*c^2*(a + b*x^2)^(5/2))/(a*x^9) + ((-9*c*(B*c + 2*A*d)*(a + b*x^2)^ 
(5/2))/(8*x^8) + ((-8*(9*a*c*(c*C + 2*B*d) - A*(4*b*c^2 - 9*a*d^2))*(a + b 
*x^2)^(5/2))/(7*x^7) + ((21*(3*b*c*(B*c + 2*A*d) - 8*a*d*(2*c*C + B*d))*(a 
 + b*x^2)^(5/2))/(2*x^6) + ((-16*(2*A*b*(4*b*c^2 - 9*a*d^2) + 9*a*(7*a*C*d 
^2 - 2*b*c*(c*C + 2*B*d)))*(a + b*x^2)^(5/2))/(5*a*x^5) + (21*b*(3*b*c*(B* 
c + 2*A*d) - 8*a*d*(2*c*C + B*d))*(-1/2*(a + b*x^2)^(3/2)/x^4 + (3*b*(-(Sq 
rt[a + b*x^2]/x^2) - (b*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/Sqrt[a]))/4))/2) 
/2)/7)/(8*a))/(9*a)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 534

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d   Int[ 
x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 
0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
 

rule 539

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[c*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] + Simp[1/(a*(m + 1)) 
   Int[x^(m + 1)*(a + b*x^2)^p*(a*d*(m + 1) - b*c*(m + 2*p + 3)*x), x], x] 
/; FreeQ[{a, b, c, d, p}, x] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
 

rule 2338

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ 
Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S 
imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( 
m + 1))   Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( 
m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt 
Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
 

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 433, normalized size of antiderivative = 1.19

method	result	size
default	\(\left (A \,d^{2}+2 B c d +C \,c^{2}\right ) \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{7 a \,x^{7}}+\frac {2 b \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{35 a^{2} x^{5}}\right )+A \,c^{2} \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{9 a \,x^{9}}-\frac {4 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{7 a \,x^{7}}+\frac {2 b \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{35 a^{2} x^{5}}\right )}{9 a}\right )-\frac {C \,d^{2} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5 a \,x^{5}}+c \left (2 A d +B c \right ) \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 a \,x^{8}}-\frac {3 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 a \,x^{6}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{4 a \,x^{4}}+\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{2 a \,x^{2}}+\frac {3 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )}{2 a}\right )}{4 a}\right )}{6 a}\right )}{8 a}\right )+d \left (B d +2 C c \right ) \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 a \,x^{6}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{4 a \,x^{4}}+\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{2 a \,x^{2}}+\frac {3 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )}{2 a}\right )}{4 a}\right )}{6 a}\right )\)	\(433\)
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-2304 A a \,b^{3} d^{2} x^{8}+1024 A \,b^{4} c^{2} x^{8}-4608 B a \,b^{3} c d \,x^{8}+8064 C \,a^{2} b^{2} d^{2} x^{8}-2304 C a \,b^{3} c^{2} x^{8}-1890 A a \,b^{3} c d \,x^{7}+2520 B \,a^{2} b^{2} d^{2} x^{7}-945 B a \,b^{3} c^{2} x^{7}+5040 C \,a^{2} b^{2} c d \,x^{7}+1152 A \,a^{2} b^{2} d^{2} x^{6}-512 A a \,b^{3} c^{2} x^{6}+2304 B \,a^{2} b^{2} c d \,x^{6}+16128 C \,a^{3} b \,d^{2} x^{6}+1152 C \,a^{2} b^{2} c^{2} x^{6}+1260 A \,a^{2} b^{2} c d \,x^{5}+11760 B \,a^{3} b \,d^{2} x^{5}+630 B \,a^{2} b^{2} c^{2} x^{5}+23520 C \,a^{3} b c d \,x^{5}+9216 A \,a^{3} b \,d^{2} x^{4}+384 A \,a^{2} b^{2} c^{2} x^{4}+18432 B \,a^{3} b c d \,x^{4}+8064 C \,a^{4} d^{2} x^{4}+9216 C \,a^{3} b \,c^{2} x^{4}+15120 A \,a^{3} b c d \,x^{3}+6720 B \,a^{4} d^{2} x^{3}+7560 B \,a^{3} b \,c^{2} x^{3}+13440 C \,a^{4} c d \,x^{3}+5760 A \,a^{4} d^{2} x^{2}+6400 A \,a^{3} b \,c^{2} x^{2}+11520 B \,a^{4} c d \,x^{2}+5760 C \,a^{4} c^{2} x^{2}+10080 A \,a^{4} c d x +5040 B \,a^{4} c^{2} x +4480 A \,a^{4} c^{2}\right )}{40320 x^{9} a^{3}}-\frac {\left (6 A b c d -8 a B \,d^{2}+3 b B \,c^{2}-16 C a c d \right ) b^{3} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{128 a^{\frac {5}{2}}}\)	\(510\)

Input:

int((d*x+c)^2*(b*x^2+a)^(3/2)*(C*x^2+B*x+A)/x^10,x,method=_RETURNVERBOSE)
 

Output:

(A*d^2+2*B*c*d+C*c^2)*(-1/7/a/x^7*(b*x^2+a)^(5/2)+2/35*b/a^2/x^5*(b*x^2+a) 
^(5/2))+A*c^2*(-1/9/a/x^9*(b*x^2+a)^(5/2)-4/9*b/a*(-1/7/a/x^7*(b*x^2+a)^(5 
/2)+2/35*b/a^2/x^5*(b*x^2+a)^(5/2)))-1/5*C*d^2/a/x^5*(b*x^2+a)^(5/2)+c*(2* 
A*d+B*c)*(-1/8/a/x^8*(b*x^2+a)^(5/2)-3/8*b/a*(-1/6/a/x^6*(b*x^2+a)^(5/2)-1 
/6*b/a*(-1/4/a/x^4*(b*x^2+a)^(5/2)+1/4*b/a*(-1/2/a/x^2*(b*x^2+a)^(5/2)+3/2 
*b/a*(1/3*(b*x^2+a)^(3/2)+a*((b*x^2+a)^(1/2)-a^(1/2)*ln((2*a+2*a^(1/2)*(b* 
x^2+a)^(1/2))/x)))))))+d*(B*d+2*C*c)*(-1/6/a/x^6*(b*x^2+a)^(5/2)-1/6*b/a*( 
-1/4/a/x^4*(b*x^2+a)^(5/2)+1/4*b/a*(-1/2/a/x^2*(b*x^2+a)^(5/2)+3/2*b/a*(1/ 
3*(b*x^2+a)^(3/2)+a*((b*x^2+a)^(1/2)-a^(1/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^( 
1/2))/x))))))
 

Fricas [A] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 944, normalized size of antiderivative = 2.59 \[ \int \frac {(c+d x)^2 \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^{10}} \, dx =\text {Too large to display} \] Input:

integrate((d*x+c)^2*(b*x^2+a)^(3/2)*(C*x^2+B*x+A)/x^10,x, algorithm="frica 
s")
 

Output:

[-1/80640*(315*(3*B*b^4*c^2 - 8*B*a*b^3*d^2 - 2*(8*C*a*b^3 - 3*A*b^4)*c*d) 
*sqrt(a)*x^9*log(-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(128* 
(36*B*a*b^3*c*d + 2*(9*C*a*b^3 - 4*A*b^4)*c^2 - 9*(7*C*a^2*b^2 - 2*A*a*b^3 
)*d^2)*x^8 + 315*(3*B*a*b^3*c^2 - 8*B*a^2*b^2*d^2 - 2*(8*C*a^2*b^2 - 3*A*a 
*b^3)*c*d)*x^7 - 4480*A*a^4*c^2 - 128*(18*B*a^2*b^2*c*d + (9*C*a^2*b^2 - 4 
*A*a*b^3)*c^2 + 9*(14*C*a^3*b + A*a^2*b^2)*d^2)*x^6 - 210*(3*B*a^2*b^2*c^2 
 + 56*B*a^3*b*d^2 + 2*(56*C*a^3*b + 3*A*a^2*b^2)*c*d)*x^5 - 384*(48*B*a^3* 
b*c*d + (24*C*a^3*b + A*a^2*b^2)*c^2 + 3*(7*C*a^4 + 8*A*a^3*b)*d^2)*x^4 - 
840*(9*B*a^3*b*c^2 + 8*B*a^4*d^2 + 2*(8*C*a^4 + 9*A*a^3*b)*c*d)*x^3 - 640* 
(18*B*a^4*c*d + 9*A*a^4*d^2 + (9*C*a^4 + 10*A*a^3*b)*c^2)*x^2 - 5040*(B*a^ 
4*c^2 + 2*A*a^4*c*d)*x)*sqrt(b*x^2 + a))/(a^3*x^9), 1/40320*(315*(3*B*b^4* 
c^2 - 8*B*a*b^3*d^2 - 2*(8*C*a*b^3 - 3*A*b^4)*c*d)*sqrt(-a)*x^9*arctan(sqr 
t(b*x^2 + a)*sqrt(-a)/a) + (128*(36*B*a*b^3*c*d + 2*(9*C*a*b^3 - 4*A*b^4)* 
c^2 - 9*(7*C*a^2*b^2 - 2*A*a*b^3)*d^2)*x^8 + 315*(3*B*a*b^3*c^2 - 8*B*a^2* 
b^2*d^2 - 2*(8*C*a^2*b^2 - 3*A*a*b^3)*c*d)*x^7 - 4480*A*a^4*c^2 - 128*(18* 
B*a^2*b^2*c*d + (9*C*a^2*b^2 - 4*A*a*b^3)*c^2 + 9*(14*C*a^3*b + A*a^2*b^2) 
*d^2)*x^6 - 210*(3*B*a^2*b^2*c^2 + 56*B*a^3*b*d^2 + 2*(56*C*a^3*b + 3*A*a^ 
2*b^2)*c*d)*x^5 - 384*(48*B*a^3*b*c*d + (24*C*a^3*b + A*a^2*b^2)*c^2 + 3*( 
7*C*a^4 + 8*A*a^3*b)*d^2)*x^4 - 840*(9*B*a^3*b*c^2 + 8*B*a^4*d^2 + 2*(8*C* 
a^4 + 9*A*a^3*b)*c*d)*x^3 - 640*(18*B*a^4*c*d + 9*A*a^4*d^2 + (9*C*a^4 ...
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3181 vs. \(2 (350) = 700\).

Time = 64.34 (sec) , antiderivative size = 3181, normalized size of antiderivative = 8.72 \[ \int \frac {(c+d x)^2 \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^{10}} \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)**2*(b*x**2+a)**(3/2)*(C*x**2+B*x+A)/x**10,x)
 

Output:

-35*A*a**8*b**(19/2)*c**2*sqrt(a/(b*x**2) + 1)/(315*a**7*b**9*x**8 + 945*a 
**6*b**10*x**10 + 945*a**5*b**11*x**12 + 315*a**4*b**12*x**14) - 110*A*a** 
7*b**(21/2)*c**2*x**2*sqrt(a/(b*x**2) + 1)/(315*a**7*b**9*x**8 + 945*a**6* 
b**10*x**10 + 945*a**5*b**11*x**12 + 315*a**4*b**12*x**14) - 114*A*a**6*b* 
*(23/2)*c**2*x**4*sqrt(a/(b*x**2) + 1)/(315*a**7*b**9*x**8 + 945*a**6*b**1 
0*x**10 + 945*a**5*b**11*x**12 + 315*a**4*b**12*x**14) - 15*A*a**6*b**(9/2 
)*d**2*sqrt(a/(b*x**2) + 1)/(105*a**5*b**4*x**6 + 210*a**4*b**5*x**8 + 105 
*a**3*b**6*x**10) - 40*A*a**5*b**(25/2)*c**2*x**6*sqrt(a/(b*x**2) + 1)/(31 
5*a**7*b**9*x**8 + 945*a**6*b**10*x**10 + 945*a**5*b**11*x**12 + 315*a**4* 
b**12*x**14) - 15*A*a**5*b**(11/2)*c**2*sqrt(a/(b*x**2) + 1)/(105*a**5*b** 
4*x**6 + 210*a**4*b**5*x**8 + 105*a**3*b**6*x**10) - 33*A*a**5*b**(11/2)*d 
**2*x**2*sqrt(a/(b*x**2) + 1)/(105*a**5*b**4*x**6 + 210*a**4*b**5*x**8 + 1 
05*a**3*b**6*x**10) + 5*A*a**4*b**(27/2)*c**2*x**8*sqrt(a/(b*x**2) + 1)/(3 
15*a**7*b**9*x**8 + 945*a**6*b**10*x**10 + 945*a**5*b**11*x**12 + 315*a**4 
*b**12*x**14) - 33*A*a**4*b**(13/2)*c**2*x**2*sqrt(a/(b*x**2) + 1)/(105*a* 
*5*b**4*x**6 + 210*a**4*b**5*x**8 + 105*a**3*b**6*x**10) - 17*A*a**4*b**(1 
3/2)*d**2*x**4*sqrt(a/(b*x**2) + 1)/(105*a**5*b**4*x**6 + 210*a**4*b**5*x* 
*8 + 105*a**3*b**6*x**10) + 30*A*a**3*b**(29/2)*c**2*x**10*sqrt(a/(b*x**2) 
 + 1)/(315*a**7*b**9*x**8 + 945*a**6*b**10*x**10 + 945*a**5*b**11*x**12 + 
315*a**4*b**12*x**14) - 17*A*a**3*b**(15/2)*c**2*x**4*sqrt(a/(b*x**2) +...
                                                                                    
                                                                                    
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 537, normalized size of antiderivative = 1.47 \[ \int \frac {(c+d x)^2 \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^{10}} \, dx=\frac {{\left (2 \, C c d + B d^{2}\right )} b^{3} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{16 \, a^{\frac {3}{2}}} - \frac {3 \, {\left (B c^{2} + 2 \, A c d\right )} b^{4} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{128 \, a^{\frac {5}{2}}} - \frac {{\left (2 \, C c d + B d^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}}{48 \, a^{3}} - \frac {{\left (2 \, C c d + B d^{2}\right )} \sqrt {b x^{2} + a} b^{3}}{16 \, a^{2}} + \frac {{\left (B c^{2} + 2 \, A c d\right )} {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{4}}{128 \, a^{4}} + \frac {3 \, {\left (B c^{2} + 2 \, A c d\right )} \sqrt {b x^{2} + a} b^{4}}{128 \, a^{3}} + \frac {{\left (2 \, C c d + B d^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}}{48 \, a^{3} x^{2}} - \frac {{\left (B c^{2} + 2 \, A c d\right )} {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{3}}{128 \, a^{4} x^{2}} - \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{2} c^{2}}{315 \, a^{3} x^{5}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} C d^{2}}{5 \, a x^{5}} + \frac {{\left (2 \, C c d + B d^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {5}{2}} b}{24 \, a^{2} x^{4}} - \frac {{\left (B c^{2} + 2 \, A c d\right )} {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}}{64 \, a^{3} x^{4}} + \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b c^{2}}{63 \, a^{2} x^{7}} + \frac {2 \, {\left (C c^{2} + 2 \, B c d + A d^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {5}{2}} b}{35 \, a^{2} x^{5}} - \frac {{\left (2 \, C c d + B d^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {5}{2}}}{6 \, a x^{6}} + \frac {{\left (B c^{2} + 2 \, A c d\right )} {\left (b x^{2} + a\right )}^{\frac {5}{2}} b}{16 \, a^{2} x^{6}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A c^{2}}{9 \, a x^{9}} - \frac {{\left (C c^{2} + 2 \, B c d + A d^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {5}{2}}}{7 \, a x^{7}} - \frac {{\left (B c^{2} + 2 \, A c d\right )} {\left (b x^{2} + a\right )}^{\frac {5}{2}}}{8 \, a x^{8}} \] Input:

integrate((d*x+c)^2*(b*x^2+a)^(3/2)*(C*x^2+B*x+A)/x^10,x, algorithm="maxim 
a")
 

Output:

1/16*(2*C*c*d + B*d^2)*b^3*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(3/2) - 3/128*( 
B*c^2 + 2*A*c*d)*b^4*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(5/2) - 1/48*(2*C*c*d 
 + B*d^2)*(b*x^2 + a)^(3/2)*b^3/a^3 - 1/16*(2*C*c*d + B*d^2)*sqrt(b*x^2 + 
a)*b^3/a^2 + 1/128*(B*c^2 + 2*A*c*d)*(b*x^2 + a)^(3/2)*b^4/a^4 + 3/128*(B* 
c^2 + 2*A*c*d)*sqrt(b*x^2 + a)*b^4/a^3 + 1/48*(2*C*c*d + B*d^2)*(b*x^2 + a 
)^(5/2)*b^2/(a^3*x^2) - 1/128*(B*c^2 + 2*A*c*d)*(b*x^2 + a)^(5/2)*b^3/(a^4 
*x^2) - 8/315*(b*x^2 + a)^(5/2)*A*b^2*c^2/(a^3*x^5) - 1/5*(b*x^2 + a)^(5/2 
)*C*d^2/(a*x^5) + 1/24*(2*C*c*d + B*d^2)*(b*x^2 + a)^(5/2)*b/(a^2*x^4) - 1 
/64*(B*c^2 + 2*A*c*d)*(b*x^2 + a)^(5/2)*b^2/(a^3*x^4) + 4/63*(b*x^2 + a)^( 
5/2)*A*b*c^2/(a^2*x^7) + 2/35*(C*c^2 + 2*B*c*d + A*d^2)*(b*x^2 + a)^(5/2)* 
b/(a^2*x^5) - 1/6*(2*C*c*d + B*d^2)*(b*x^2 + a)^(5/2)/(a*x^6) + 1/16*(B*c^ 
2 + 2*A*c*d)*(b*x^2 + a)^(5/2)*b/(a^2*x^6) - 1/9*(b*x^2 + a)^(5/2)*A*c^2/( 
a*x^9) - 1/7*(C*c^2 + 2*B*c*d + A*d^2)*(b*x^2 + a)^(5/2)/(a*x^7) - 1/8*(B* 
c^2 + 2*A*c*d)*(b*x^2 + a)^(5/2)/(a*x^8)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2195 vs. \(2 (329) = 658\).

Time = 0.18 (sec) , antiderivative size = 2195, normalized size of antiderivative = 6.01 \[ \int \frac {(c+d x)^2 \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^{10}} \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^2*(b*x^2+a)^(3/2)*(C*x^2+B*x+A)/x^10,x, algorithm="giac" 
)
 

Output:

1/64*(3*B*b^4*c^2 - 16*C*a*b^3*c*d + 6*A*b^4*c*d - 8*B*a*b^3*d^2)*arctan(- 
(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^2) - 1/20160*(945*(sqr 
t(b)*x - sqrt(b*x^2 + a))^17*B*b^4*c^2 - 5040*(sqrt(b)*x - sqrt(b*x^2 + a) 
)^17*C*a*b^3*c*d + 1890*(sqrt(b)*x - sqrt(b*x^2 + a))^17*A*b^4*c*d - 2520* 
(sqrt(b)*x - sqrt(b*x^2 + a))^17*B*a*b^3*d^2 - 40320*(sqrt(b)*x - sqrt(b*x 
^2 + a))^16*C*a^2*b^(5/2)*d^2 - 8190*(sqrt(b)*x - sqrt(b*x^2 + a))^15*B*a* 
b^4*c^2 - 63840*(sqrt(b)*x - sqrt(b*x^2 + a))^15*C*a^2*b^3*c*d - 16380*(sq 
rt(b)*x - sqrt(b*x^2 + a))^15*A*a*b^4*c*d - 31920*(sqrt(b)*x - sqrt(b*x^2 
+ a))^15*B*a^2*b^3*d^2 - 80640*(sqrt(b)*x - sqrt(b*x^2 + a))^14*C*a^2*b^(7 
/2)*c^2 - 161280*(sqrt(b)*x - sqrt(b*x^2 + a))^14*B*a^2*b^(7/2)*c*d + 1612 
80*(sqrt(b)*x - sqrt(b*x^2 + a))^14*C*a^3*b^(5/2)*d^2 - 80640*(sqrt(b)*x - 
 sqrt(b*x^2 + a))^14*A*a^2*b^(7/2)*d^2 - 97650*(sqrt(b)*x - sqrt(b*x^2 + a 
))^13*B*a^2*b^4*c^2 + 90720*(sqrt(b)*x - sqrt(b*x^2 + a))^13*C*a^3*b^3*c*d 
 - 195300*(sqrt(b)*x - sqrt(b*x^2 + a))^13*A*a^2*b^4*c*d + 45360*(sqrt(b)* 
x - sqrt(b*x^2 + a))^13*B*a^3*b^3*d^2 + 80640*(sqrt(b)*x - sqrt(b*x^2 + a) 
)^12*C*a^3*b^(7/2)*c^2 - 215040*(sqrt(b)*x - sqrt(b*x^2 + a))^12*A*a^2*b^( 
9/2)*c^2 + 161280*(sqrt(b)*x - sqrt(b*x^2 + a))^12*B*a^3*b^(7/2)*c*d - 322 
560*(sqrt(b)*x - sqrt(b*x^2 + a))^12*C*a^4*b^(5/2)*d^2 + 80640*(sqrt(b)*x 
- sqrt(b*x^2 + a))^12*A*a^3*b^(7/2)*d^2 - 106470*(sqrt(b)*x - sqrt(b*x^2 + 
 a))^11*B*a^3*b^4*c^2 + 30240*(sqrt(b)*x - sqrt(b*x^2 + a))^11*C*a^4*b^...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^2 \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^{10}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{3/2}\,{\left (c+d\,x\right )}^2\,\left (C\,x^2+B\,x+A\right )}{x^{10}} \,d x \] Input:

int(((a + b*x^2)^(3/2)*(c + d*x)^2*(A + B*x + C*x^2))/x^10,x)
 

Output:

int(((a + b*x^2)^(3/2)*(c + d*x)^2*(A + B*x + C*x^2))/x^10, x)
 

Reduce [B] (verification not implemented)

Time = 23.69 (sec) , antiderivative size = 1064, normalized size of antiderivative = 2.92 \[ \int \frac {(c+d x)^2 \left (a+b x^2\right )^{3/2} \left (A+B x+C x^2\right )}{x^{10}} \, dx =\text {Too large to display} \] Input:

int((d*x+c)^2*(b*x^2+a)^(3/2)*(C*x^2+B*x+A)/x^10,x)
 

Output:

( - 4480*sqrt(a + b*x**2)*a**5*c**2 - 10080*sqrt(a + b*x**2)*a**5*c*d*x - 
5760*sqrt(a + b*x**2)*a**5*d**2*x**2 - 6400*sqrt(a + b*x**2)*a**4*b*c**2*x 
**2 - 5040*sqrt(a + b*x**2)*a**4*b*c**2*x - 15120*sqrt(a + b*x**2)*a**4*b* 
c*d*x**3 - 11520*sqrt(a + b*x**2)*a**4*b*c*d*x**2 - 9216*sqrt(a + b*x**2)* 
a**4*b*d**2*x**4 - 6720*sqrt(a + b*x**2)*a**4*b*d**2*x**3 - 5760*sqrt(a + 
b*x**2)*a**4*c**3*x**2 - 13440*sqrt(a + b*x**2)*a**4*c**2*d*x**3 - 8064*sq 
rt(a + b*x**2)*a**4*c*d**2*x**4 - 384*sqrt(a + b*x**2)*a**3*b**2*c**2*x**4 
 - 7560*sqrt(a + b*x**2)*a**3*b**2*c**2*x**3 - 1260*sqrt(a + b*x**2)*a**3* 
b**2*c*d*x**5 - 18432*sqrt(a + b*x**2)*a**3*b**2*c*d*x**4 - 1152*sqrt(a + 
b*x**2)*a**3*b**2*d**2*x**6 - 11760*sqrt(a + b*x**2)*a**3*b**2*d**2*x**5 - 
 9216*sqrt(a + b*x**2)*a**3*b*c**3*x**4 - 23520*sqrt(a + b*x**2)*a**3*b*c* 
*2*d*x**5 - 16128*sqrt(a + b*x**2)*a**3*b*c*d**2*x**6 + 512*sqrt(a + b*x** 
2)*a**2*b**3*c**2*x**6 - 630*sqrt(a + b*x**2)*a**2*b**3*c**2*x**5 + 1890*s 
qrt(a + b*x**2)*a**2*b**3*c*d*x**7 - 2304*sqrt(a + b*x**2)*a**2*b**3*c*d*x 
**6 + 2304*sqrt(a + b*x**2)*a**2*b**3*d**2*x**8 - 2520*sqrt(a + b*x**2)*a* 
*2*b**3*d**2*x**7 - 1152*sqrt(a + b*x**2)*a**2*b**2*c**3*x**6 - 5040*sqrt( 
a + b*x**2)*a**2*b**2*c**2*d*x**7 - 8064*sqrt(a + b*x**2)*a**2*b**2*c*d**2 
*x**8 - 1024*sqrt(a + b*x**2)*a*b**4*c**2*x**8 + 945*sqrt(a + b*x**2)*a*b* 
*4*c**2*x**7 + 4608*sqrt(a + b*x**2)*a*b**4*c*d*x**8 + 2304*sqrt(a + b*x** 
2)*a*b**3*c**3*x**8 + 1890*sqrt(a)*log((sqrt(a + b*x**2) - sqrt(a) + sq...