\(\int \frac {(a+b x)^{5/2} (A+B x+C x^2+D x^3)}{(c+d x)^{5/2}} \, dx\) [123]

Optimal result

Integrand size = 34, antiderivative size = 522 \[ \int \frac {(a+b x)^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=-\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) (a+b x)^{5/2}}{3 d^4 (c+d x)^{3/2}}+\frac {2 \left (3 a d \left (2 c C d-B d^2-3 c^2 D\right )-b \left (11 c^2 C d-8 B c d^2+5 A d^3-14 c^3 D\right )\right ) (a+b x)^{3/2}}{3 d^5 \sqrt {c+d x}}-\frac {5 \left (a^3 d^3 D-a^2 b d^2 (8 C d-21 c D)+a b^2 d \left (112 c C d-48 B d^2-189 c^2 D\right )-b^3 \left (168 c^2 C d-112 B c d^2+64 A d^3-231 c^3 D\right )\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b d^6}-\frac {\left (5 a^2 d^2 D-10 a b d (4 C d-11 c D)+b^2 \left (136 c C d-48 B d^2-259 c^2 D\right )\right ) (a+b x)^{3/2} \sqrt {c+d x}}{96 b d^5}+\frac {(8 b C d-23 b c D-a d D) (a+b x)^{5/2} \sqrt {c+d x}}{24 b d^4}+\frac {D (a+b x)^{7/2} \sqrt {c+d x}}{4 b d^3}+\frac {5 (b c-a d) \left (a^3 d^3 D-a^2 b d^2 (8 C d-21 c D)+a b^2 d \left (112 c C d-48 B d^2-189 c^2 D\right )-b^3 \left (168 c^2 C d-112 B c d^2+64 A d^3-231 c^3 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{3/2} d^{13/2}} \] Output:

-2/3*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(b*x+a)^(5/2)/d^4/(d*x+c)^(3/2)+2/3*(3* 
a*d*(-B*d^2+2*C*c*d-3*D*c^2)-b*(5*A*d^3-8*B*c*d^2+11*C*c^2*d-14*D*c^3))*(b 
*x+a)^(3/2)/d^5/(d*x+c)^(1/2)-5/64*(a^3*d^3*D-a^2*b*d^2*(8*C*d-21*D*c)+a*b 
^2*d*(-48*B*d^2+112*C*c*d-189*D*c^2)-b^3*(64*A*d^3-112*B*c*d^2+168*C*c^2*d 
-231*D*c^3))*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b/d^6-1/96*(5*a^2*d^2*D-10*a*b*d* 
(4*C*d-11*D*c)+b^2*(-48*B*d^2+136*C*c*d-259*D*c^2))*(b*x+a)^(3/2)*(d*x+c)^ 
(1/2)/b/d^5+1/24*(8*C*b*d-D*a*d-23*D*b*c)*(b*x+a)^(5/2)*(d*x+c)^(1/2)/b/d^ 
4+1/4*D*(b*x+a)^(7/2)*(d*x+c)^(1/2)/b/d^3+5/64*(-a*d+b*c)*(a^3*d^3*D-a^2*b 
*d^2*(8*C*d-21*D*c)+a*b^2*d*(-48*B*d^2+112*C*c*d-189*D*c^2)-b^3*(64*A*d^3- 
112*B*c*d^2+168*C*c^2*d-231*D*c^3))*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/ 
(d*x+c)^(1/2))/b^(3/2)/d^(13/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 12.54 (sec) , antiderivative size = 626, normalized size of antiderivative = 1.20 \[ \int \frac {(a+b x)^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\frac {\sqrt {\frac {b (c+d x)}{b c-a d}} \left (\frac {14 \sqrt {b c-a d} (C d-3 c D) \left (15 b^3 d (b c-a d)^{5/2} (a+b x) \sqrt {\frac {b (c+d x)}{b c-a d}}-10 b^3 d^2 (b c-a d)^{3/2} (a+b x)^2 \sqrt {\frac {b (c+d x)}{b c-a d}}+8 b^3 d^3 \sqrt {b c-a d} (a+b x)^3 \sqrt {\frac {b (c+d x)}{b c-a d}}-15 b^3 \sqrt {d} (b c-a d)^3 \sqrt {a+b x} \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )\right )}{3 b^4}+\frac {7 \sqrt {d} D \left (15 b \sqrt {d} (b c-a d)^3 (a+b x) (c+d x)-10 b d^{3/2} (b c-a d)^2 (a+b x)^2 (c+d x)+8 b d^{5/2} (b c-a d) (a+b x)^3 (c+d x)+48 b d^{7/2} (a+b x)^4 (c+d x)-15 (b c-a d)^{9/2} \sqrt {a+b x} \sqrt {\frac {b (c+d x)}{b c-a d}} \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )\right )}{12 b^2 \sqrt {\frac {b (c+d x)}{b c-a d}}}-\frac {32 d^4 \left (-2 c C d+B d^2+3 c^2 D\right ) (a+b x)^4 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {7}{2},\frac {9}{2},\frac {d (a+b x)}{-b c+a d}\right )}{-b c+a d}+\frac {32 b d^4 \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) (a+b x)^4 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {7}{2},\frac {9}{2},\frac {d (a+b x)}{-b c+a d}\right )}{(b c-a d)^2}\right )}{112 d^7 \sqrt {a+b x} \sqrt {c+d x}} \] Input:

Integrate[((a + b*x)^(5/2)*(A + B*x + C*x^2 + D*x^3))/(c + d*x)^(5/2),x]
 

Output:

(Sqrt[(b*(c + d*x))/(b*c - a*d)]*((14*Sqrt[b*c - a*d]*(C*d - 3*c*D)*(15*b^ 
3*d*(b*c - a*d)^(5/2)*(a + b*x)*Sqrt[(b*(c + d*x))/(b*c - a*d)] - 10*b^3*d 
^2*(b*c - a*d)^(3/2)*(a + b*x)^2*Sqrt[(b*(c + d*x))/(b*c - a*d)] + 8*b^3*d 
^3*Sqrt[b*c - a*d]*(a + b*x)^3*Sqrt[(b*(c + d*x))/(b*c - a*d)] - 15*b^3*Sq 
rt[d]*(b*c - a*d)^3*Sqrt[a + b*x]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c 
 - a*d]]))/(3*b^4) + (7*Sqrt[d]*D*(15*b*Sqrt[d]*(b*c - a*d)^3*(a + b*x)*(c 
 + d*x) - 10*b*d^(3/2)*(b*c - a*d)^2*(a + b*x)^2*(c + d*x) + 8*b*d^(5/2)*( 
b*c - a*d)*(a + b*x)^3*(c + d*x) + 48*b*d^(7/2)*(a + b*x)^4*(c + d*x) - 15 
*(b*c - a*d)^(9/2)*Sqrt[a + b*x]*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSinh[( 
Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]]))/(12*b^2*Sqrt[(b*(c + d*x))/(b*c 
- a*d)]) - (32*d^4*(-2*c*C*d + B*d^2 + 3*c^2*D)*(a + b*x)^4*Hypergeometric 
2F1[3/2, 7/2, 9/2, (d*(a + b*x))/(-(b*c) + a*d)])/(-(b*c) + a*d) + (32*b*d 
^4*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D)*(a + b*x)^4*Hypergeometric2F1[5/2, 
7/2, 9/2, (d*(a + b*x))/(-(b*c) + a*d)])/(b*c - a*d)^2))/(112*d^7*Sqrt[a + 
 b*x]*Sqrt[c + d*x])
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 455, normalized size of antiderivative = 0.87, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2124, 27, 1193, 27, 90, 60, 60, 60, 66, 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.

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

\(\Big \downarrow \) 2124

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1193

\(\displaystyle \frac {\frac {2 \int \frac {3 (a+b x)^{5/2} \left (\left (-28 D c^3+21 C d c^2-14 B d^2 c+8 A d^3\right ) b^2-2 a d \left (-11 D c^2+7 C d c-3 B d^2\right ) b+a^2 d^2 (C d-2 c D)+d (b c-a d)^2 D x\right )}{2 d^3 \sqrt {c+d x}}dx}{b c-a d}+\frac {2 (a+b x)^{7/2} \left (3 a d \left (-B d^2-3 c^2 D+2 c C d\right )-b \left (4 A d^3-7 B c d^2-13 c^3 D+10 c^2 C d\right )\right )}{d^3 \sqrt {c+d x} (b c-a d)}}{3 (b c-a d)}+\frac {2 (a+b x)^{7/2} \left (A+\frac {c \left (-B d^2+c^2 (-D)+c C d\right )}{d^3}\right )}{3 (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \int \frac {(a+b x)^{5/2} \left (\left (-28 D c^3+21 C d c^2-14 B d^2 c+8 A d^3\right ) b^2-2 a d \left (-11 D c^2+7 C d c-3 B d^2\right ) b+a^2 d^2 (C d-2 c D)+d (b c-a d)^2 D x\right )}{\sqrt {c+d x}}dx}{d^3 (b c-a d)}+\frac {2 (a+b x)^{7/2} \left (3 a d \left (-B d^2-3 c^2 D+2 c C d\right )-b \left (4 A d^3-7 B c d^2-13 c^3 D+10 c^2 C d\right )\right )}{d^3 \sqrt {c+d x} (b c-a d)}}{3 (b c-a d)}+\frac {2 (a+b x)^{7/2} \left (A+\frac {c \left (-B d^2+c^2 (-D)+c C d\right )}{d^3}\right )}{3 (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {\frac {3 \left (\frac {D (a+b x)^{7/2} \sqrt {c+d x} (b c-a d)^2}{4 b}-\frac {\left (a^3 d^3 D-a^2 b d^2 (8 C d-21 c D)+a b^2 d \left (-48 B d^2-189 c^2 D+112 c C d\right )-\left (b^3 \left (64 A d^3-112 B c d^2-231 c^3 D+168 c^2 C d\right )\right )\right ) \int \frac {(a+b x)^{5/2}}{\sqrt {c+d x}}dx}{8 b}\right )}{d^3 (b c-a d)}+\frac {2 (a+b x)^{7/2} \left (3 a d \left (-B d^2-3 c^2 D+2 c C d\right )-b \left (4 A d^3-7 B c d^2-13 c^3 D+10 c^2 C d\right )\right )}{d^3 \sqrt {c+d x} (b c-a d)}}{3 (b c-a d)}+\frac {2 (a+b x)^{7/2} \left (A+\frac {c \left (-B d^2+c^2 (-D)+c C d\right )}{d^3}\right )}{3 (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\frac {3 \left (\frac {D (a+b x)^{7/2} \sqrt {c+d x} (b c-a d)^2}{4 b}-\frac {\left (a^3 d^3 D-a^2 b d^2 (8 C d-21 c D)+a b^2 d \left (-48 B d^2-189 c^2 D+112 c C d\right )-\left (b^3 \left (64 A d^3-112 B c d^2-231 c^3 D+168 c^2 C d\right )\right )\right ) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}}dx}{6 d}\right )}{8 b}\right )}{d^3 (b c-a d)}+\frac {2 (a+b x)^{7/2} \left (3 a d \left (-B d^2-3 c^2 D+2 c C d\right )-b \left (4 A d^3-7 B c d^2-13 c^3 D+10 c^2 C d\right )\right )}{d^3 \sqrt {c+d x} (b c-a d)}}{3 (b c-a d)}+\frac {2 (a+b x)^{7/2} \left (A+\frac {c \left (-B d^2+c^2 (-D)+c C d\right )}{d^3}\right )}{3 (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\frac {3 \left (\frac {D (a+b x)^{7/2} \sqrt {c+d x} (b c-a d)^2}{4 b}-\frac {\left (a^3 d^3 D-a^2 b d^2 (8 C d-21 c D)+a b^2 d \left (-48 B d^2-189 c^2 D+112 c C d\right )-\left (b^3 \left (64 A d^3-112 B c d^2-231 c^3 D+168 c^2 C d\right )\right )\right ) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}}dx}{4 d}\right )}{6 d}\right )}{8 b}\right )}{d^3 (b c-a d)}+\frac {2 (a+b x)^{7/2} \left (3 a d \left (-B d^2-3 c^2 D+2 c C d\right )-b \left (4 A d^3-7 B c d^2-13 c^3 D+10 c^2 C d\right )\right )}{d^3 \sqrt {c+d x} (b c-a d)}}{3 (b c-a d)}+\frac {2 (a+b x)^{7/2} \left (A+\frac {c \left (-B d^2+c^2 (-D)+c C d\right )}{d^3}\right )}{3 (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\frac {3 \left (\frac {D (a+b x)^{7/2} \sqrt {c+d x} (b c-a d)^2}{4 b}-\frac {\left (a^3 d^3 D-a^2 b d^2 (8 C d-21 c D)+a b^2 d \left (-48 B d^2-189 c^2 D+112 c C d\right )-\left (b^3 \left (64 A d^3-112 B c d^2-231 c^3 D+168 c^2 C d\right )\right )\right ) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}\right )}{4 d}\right )}{6 d}\right )}{8 b}\right )}{d^3 (b c-a d)}+\frac {2 (a+b x)^{7/2} \left (3 a d \left (-B d^2-3 c^2 D+2 c C d\right )-b \left (4 A d^3-7 B c d^2-13 c^3 D+10 c^2 C d\right )\right )}{d^3 \sqrt {c+d x} (b c-a d)}}{3 (b c-a d)}+\frac {2 (a+b x)^{7/2} \left (A+\frac {c \left (-B d^2+c^2 (-D)+c C d\right )}{d^3}\right )}{3 (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {\frac {3 \left (\frac {D (a+b x)^{7/2} \sqrt {c+d x} (b c-a d)^2}{4 b}-\frac {\left (a^3 d^3 D-a^2 b d^2 (8 C d-21 c D)+a b^2 d \left (-48 B d^2-189 c^2 D+112 c C d\right )-\left (b^3 \left (64 A d^3-112 B c d^2-231 c^3 D+168 c^2 C d\right )\right )\right ) \left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{d}\right )}{4 d}\right )}{6 d}\right )}{8 b}\right )}{d^3 (b c-a d)}+\frac {2 (a+b x)^{7/2} \left (3 a d \left (-B d^2-3 c^2 D+2 c C d\right )-b \left (4 A d^3-7 B c d^2-13 c^3 D+10 c^2 C d\right )\right )}{d^3 \sqrt {c+d x} (b c-a d)}}{3 (b c-a d)}+\frac {2 (a+b x)^{7/2} \left (A+\frac {c \left (-B d^2+c^2 (-D)+c C d\right )}{d^3}\right )}{3 (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {3 \left (\frac {D (a+b x)^{7/2} \sqrt {c+d x} (b c-a d)^2}{4 b}-\frac {\left (\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 d}-\frac {5 (b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}\right )}{4 d}\right )}{6 d}\right ) \left (a^3 d^3 D-a^2 b d^2 (8 C d-21 c D)+a b^2 d \left (-48 B d^2-189 c^2 D+112 c C d\right )-\left (b^3 \left (64 A d^3-112 B c d^2-231 c^3 D+168 c^2 C d\right )\right )\right )}{8 b}\right )}{d^3 (b c-a d)}+\frac {2 (a+b x)^{7/2} \left (3 a d \left (-B d^2-3 c^2 D+2 c C d\right )-b \left (4 A d^3-7 B c d^2-13 c^3 D+10 c^2 C d\right )\right )}{d^3 \sqrt {c+d x} (b c-a d)}}{3 (b c-a d)}+\frac {2 (a+b x)^{7/2} \left (A+\frac {c \left (-B d^2+c^2 (-D)+c C d\right )}{d^3}\right )}{3 (c+d x)^{3/2} (b c-a d)}\)

Input:

Int[((a + b*x)^(5/2)*(A + B*x + C*x^2 + D*x^3))/(c + d*x)^(5/2),x]
 

Output:

(2*(A + (c*(c*C*d - B*d^2 - c^2*D))/d^3)*(a + b*x)^(7/2))/(3*(b*c - a*d)*( 
c + d*x)^(3/2)) + ((2*(3*a*d*(2*c*C*d - B*d^2 - 3*c^2*D) - b*(10*c^2*C*d - 
 7*B*c*d^2 + 4*A*d^3 - 13*c^3*D))*(a + b*x)^(7/2))/(d^3*(b*c - a*d)*Sqrt[c 
 + d*x]) + (3*(((b*c - a*d)^2*D*(a + b*x)^(7/2)*Sqrt[c + d*x])/(4*b) - ((a 
^3*d^3*D - a^2*b*d^2*(8*C*d - 21*c*D) + a*b^2*d*(112*c*C*d - 48*B*d^2 - 18 
9*c^2*D) - b^3*(168*c^2*C*d - 112*B*c*d^2 + 64*A*d^3 - 231*c^3*D))*(((a + 
b*x)^(5/2)*Sqrt[c + d*x])/(3*d) - (5*(b*c - a*d)*(((a + b*x)^(3/2)*Sqrt[c 
+ d*x])/(2*d) - (3*(b*c - a*d)*((Sqrt[a + b*x]*Sqrt[c + d*x])/d - ((b*c - 
a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(Sqrt[b]*d^ 
(3/2))))/(4*d)))/(6*d)))/(8*b)))/(d^3*(b*c - a*d)))/(3*(b*c - a*d))
 

Defintions of rubi rules used

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 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 66

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

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 1193

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x 
 + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + b*x + c*x^2)^p, d + 
e*x, x]}, Simp[R*(d + e*x)^(m + 1)*((f + g*x)^(n + 1)/((m + 1)*(e*f - d*g)) 
), x] + Simp[1/((m + 1)*(e*f - d*g))   Int[(d + e*x)^(m + 1)*(f + g*x)^n*Ex 
pandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), x], x], x]] /; FreeQ[{a 
, b, c, d, e, f, g, n}, x] && IGtQ[p, 0] && ILtQ[2*m, -2] &&  !IntegerQ[n] 
&&  !(EqQ[m, -2] && EqQ[p, 1] && EqQ[2*c*d - b*e, 0])
 

rule 2124

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px 
, a + b*x, x]}, Simp[R*(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((m + 1)*(b*c - 
 a*d))), x] + Simp[1/((m + 1)*(b*c - a*d))   Int[(a + b*x)^(m + 1)*(c + d*x 
)^n*ExpandToSum[(m + 1)*(b*c - a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr 
eeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && LtQ[m, -1] && (IntegerQ[m] ||  ! 
ILtQ[n, -1])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3663\) vs. \(2(472)=944\).

Time = 0.54 (sec) , antiderivative size = 3664, normalized size of antiderivative = 7.02

method	result	size
default	\(\text {Expression too large to display}\)	\(3664\)

Input:

int((b*x+a)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(5/2),x,method=_RETURNVERBOS 
E)
 

Output:

1/384*(b*x+a)^(1/2)*(-966*D*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*a^2*b*c*d^ 
4*x^2-3600*C*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c 
)/(d*b)^(1/2))*a^2*b^2*c^2*d^4*x+8400*C*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c) 
)^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a*b^3*c^3*d^3*x-1280*A*a*b^2*c*d 
^4*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)-512*B*a^2*b*c*d^4*((b*x+a)*(d*x+c)) 
^(1/2)*(d*b)^(1/2)-1792*A*a*b^2*d^5*x*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+ 
1680*B*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b 
)^(1/2))*b^4*c^2*d^4*x^2-2520*C*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)* 
(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*b^4*c^3*d^3*x^2+3360*B*ln(1/2*(2*b*d*x+2 
*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*b^4*c^3*d^3*x-9 
240*D*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^3*c^4*d*x+3680*B*(d*b)^(1/2)*( 
(b*x+a)*(d*x+c))^(1/2)*a*b^2*c^2*d^3-6720*C*(d*b)^(1/2)*((b*x+a)*(d*x+c))^ 
(1/2)*a*b^2*c^3*d^2-960*A*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^ 
(1/2)+a*d+b*c)/(d*b)^(1/2))*b^4*c^3*d^3+30*D*((b*x+a)*(d*x+c))^(1/2)*(d*b) 
^(1/2)*a^3*d^5*x^2-30*D*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1 
/2)+a*d+b*c)/(d*b)^(1/2))*a^4*c*d^5*x+120*C*ln(1/2*(2*b*d*x+2*((b*x+a)*(d* 
x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^3*b*c^2*d^4-2400*B*ln(1/2* 
(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a*b^3 
*c*d^5*x^2+2592*C*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)*a^2*b*c*d^4*x-1536*C 
*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b^2*c*d^4*x^2+2322*D*(d*b)^(1/2)...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1022 vs. \(2 (469) = 938\).

Time = 21.14 (sec) , antiderivative size = 2058, normalized size of antiderivative = 3.94 \[ \int \frac {(a+b x)^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate((b*x+a)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(5/2),x, algorithm="fr 
icas")
 

Output:

[1/768*(15*(231*D*b^4*c^6 - 84*(5*D*a*b^3 + 2*C*b^4)*c^5*d + 14*(15*D*a^2* 
b^2 + 20*C*a*b^3 + 8*B*b^4)*c^4*d^2 - 4*(5*D*a^3*b + 30*C*a^2*b^2 + 40*B*a 
*b^3 + 16*A*b^4)*c^3*d^3 - (D*a^4 - 8*C*a^3*b - 48*B*a^2*b^2 - 64*A*a*b^3) 
*c^2*d^4 + (231*D*b^4*c^4*d^2 - 84*(5*D*a*b^3 + 2*C*b^4)*c^3*d^3 + 14*(15* 
D*a^2*b^2 + 20*C*a*b^3 + 8*B*b^4)*c^2*d^4 - 4*(5*D*a^3*b + 30*C*a^2*b^2 + 
40*B*a*b^3 + 16*A*b^4)*c*d^5 - (D*a^4 - 8*C*a^3*b - 48*B*a^2*b^2 - 64*A*a* 
b^3)*d^6)*x^2 + 2*(231*D*b^4*c^5*d - 84*(5*D*a*b^3 + 2*C*b^4)*c^4*d^2 + 14 
*(15*D*a^2*b^2 + 20*C*a*b^3 + 8*B*b^4)*c^3*d^3 - 4*(5*D*a^3*b + 30*C*a^2*b 
^2 + 40*B*a*b^3 + 16*A*b^4)*c^2*d^4 - (D*a^4 - 8*C*a^3*b - 48*B*a^2*b^2 - 
64*A*a*b^3)*c*d^5)*x)*sqrt(b*d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + 
a^2*d^2 + 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 
8*(b^2*c*d + a*b*d^2)*x) + 4*(48*D*b^4*d^6*x^5 - 3465*D*b^4*c^5*d - 128*A* 
a^2*b^2*d^6 + 105*(49*D*a*b^3 + 24*C*b^4)*c^4*d^2 - 21*(83*D*a^2*b^2 + 160 
*C*a*b^3 + 80*B*b^4)*c^3*d^3 + (15*D*a^3*b + 904*C*a^2*b^2 + 1840*B*a*b^3 
+ 960*A*b^4)*c^2*d^4 - 128*(2*B*a^2*b^2 + 5*A*a*b^3)*c*d^5 - 8*(11*D*b^4*c 
*d^5 - (17*D*a*b^3 + 8*C*b^4)*d^6)*x^4 + 2*(99*D*b^4*c^2*d^4 - 2*(79*D*a*b 
^3 + 36*C*b^4)*c*d^5 + (59*D*a^2*b^2 + 104*C*a*b^3 + 48*B*b^4)*d^6)*x^3 - 
3*(231*D*b^4*c^3*d^3 - 3*(129*D*a*b^3 + 56*C*b^4)*c^2*d^4 + (161*D*a^2*b^2 
 + 256*C*a*b^3 + 112*B*b^4)*c*d^5 - (5*D*a^3*b + 88*C*a^2*b^2 + 144*B*a*b^ 
3 + 64*A*b^4)*d^6)*x^2 - 2*(2310*D*b^4*c^4*d^2 - 21*(167*D*a*b^3 + 80*C...
 

Sympy [F]

\[ \int \frac {(a+b x)^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\int \frac {\left (a + b x\right )^{\frac {5}{2}} \left (A + B x + C x^{2} + D x^{3}\right )}{\left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((b*x+a)**(5/2)*(D*x**3+C*x**2+B*x+A)/(d*x+c)**(5/2),x)
 

Output:

Integral((a + b*x)**(5/2)*(A + B*x + C*x**2 + D*x**3)/(c + d*x)**(5/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+b x)^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((b*x+a)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(5/2),x, algorithm="ma 
xima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m 
ore detail
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1449 vs. \(2 (469) = 938\).

Time = 0.39 (sec) , antiderivative size = 1449, normalized size of antiderivative = 2.78 \[ \int \frac {(a+b x)^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate((b*x+a)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(5/2),x, algorithm="gi 
ac")
 

Output:

1/192*(((2*(4*(b*x + a)*(6*(D*b^5*c*d^10*abs(b) - D*a*b^4*d^11*abs(b))*(b* 
x + a)/(b^6*c*d^11 - a*b^5*d^12) - (11*D*b^6*c^2*d^9*abs(b) + 2*D*a*b^5*c* 
d^10*abs(b) - 8*C*b^6*c*d^10*abs(b) - 13*D*a^2*b^4*d^11*abs(b) + 8*C*a*b^5 
*d^11*abs(b))/(b^6*c*d^11 - a*b^5*d^12)) + 3*(33*D*b^7*c^3*d^8*abs(b) - 27 
*D*a*b^6*c^2*d^9*abs(b) - 24*C*b^7*c^2*d^9*abs(b) + 3*D*a^2*b^5*c*d^10*abs 
(b) + 16*C*a*b^6*c*d^10*abs(b) + 16*B*b^7*c*d^10*abs(b) - 9*D*a^3*b^4*d^11 
*abs(b) + 8*C*a^2*b^5*d^11*abs(b) - 16*B*a*b^6*d^11*abs(b))/(b^6*c*d^11 - 
a*b^5*d^12))*(b*x + a) - 3*(231*D*b^8*c^4*d^7*abs(b) - 420*D*a*b^7*c^3*d^8 
*abs(b) - 168*C*b^8*c^3*d^8*abs(b) + 210*D*a^2*b^6*c^2*d^9*abs(b) + 280*C* 
a*b^7*c^2*d^9*abs(b) + 112*B*b^8*c^2*d^9*abs(b) - 20*D*a^3*b^5*c*d^10*abs( 
b) - 120*C*a^2*b^6*c*d^10*abs(b) - 160*B*a*b^7*c*d^10*abs(b) - 64*A*b^8*c* 
d^10*abs(b) - D*a^4*b^4*d^11*abs(b) + 8*C*a^3*b^5*d^11*abs(b) + 48*B*a^2*b 
^6*d^11*abs(b) + 64*A*a*b^7*d^11*abs(b))/(b^6*c*d^11 - a*b^5*d^12))*(b*x + 
 a) - 20*(231*D*b^9*c^5*d^6*abs(b) - 651*D*a*b^8*c^4*d^7*abs(b) - 168*C*b^ 
9*c^4*d^7*abs(b) + 630*D*a^2*b^7*c^3*d^8*abs(b) + 448*C*a*b^8*c^3*d^8*abs( 
b) + 112*B*b^9*c^3*d^8*abs(b) - 230*D*a^3*b^6*c^2*d^9*abs(b) - 400*C*a^2*b 
^7*c^2*d^9*abs(b) - 272*B*a*b^8*c^2*d^9*abs(b) - 64*A*b^9*c^2*d^9*abs(b) + 
 19*D*a^4*b^5*c*d^10*abs(b) + 128*C*a^3*b^6*c*d^10*abs(b) + 208*B*a^2*b^7* 
c*d^10*abs(b) + 128*A*a*b^8*c*d^10*abs(b) + D*a^5*b^4*d^11*abs(b) - 8*C*a^ 
4*b^5*d^11*abs(b) - 48*B*a^3*b^6*d^11*abs(b) - 64*A*a^2*b^7*d^11*abs(b)...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\int \frac {{\left (a+b\,x\right )}^{5/2}\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (c+d\,x\right )}^{5/2}} \,d x \] Input:

int(((a + b*x)^(5/2)*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^(5/2),x)
 

Output:

int(((a + b*x)^(5/2)*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^(5/2), x)
 

Reduce [F]

\[ \int \frac {(a+b x)^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\int \frac {\left (b x +a \right )^{\frac {5}{2}} \left (D x^{3}+C \,x^{2}+B x +A \right )}{\left (d x +c \right )^{\frac {5}{2}}}d x \] Input:

int((b*x+a)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(5/2),x)
 

Output:

int((b*x+a)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(5/2),x)