\(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^{19}} \, dx\) [183]

Optimal result

Integrand size = 26, antiderivative size = 207 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{19}} \, dx=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{15 b x^{20}}-\frac {(3 b B-2 A c) \left (b x^2+c x^4\right )^{5/2}}{39 b^2 x^{18}}+\frac {8 c (3 b B-2 A c) \left (b x^2+c x^4\right )^{5/2}}{429 b^3 x^{16}}-\frac {16 c^2 (3 b B-2 A c) \left (b x^2+c x^4\right )^{5/2}}{1287 b^4 x^{14}}+\frac {64 c^3 (3 b B-2 A c) \left (b x^2+c x^4\right )^{5/2}}{9009 b^5 x^{12}}-\frac {128 c^4 (3 b B-2 A c) \left (b x^2+c x^4\right )^{5/2}}{45045 b^6 x^{10}} \] Output:

-1/15*A*(c*x^4+b*x^2)^(5/2)/b/x^20-1/39*(-2*A*c+3*B*b)*(c*x^4+b*x^2)^(5/2) 
/b^2/x^18+8/429*c*(-2*A*c+3*B*b)*(c*x^4+b*x^2)^(5/2)/b^3/x^16-16/1287*c^2* 
(-2*A*c+3*B*b)*(c*x^4+b*x^2)^(5/2)/b^4/x^14+64/9009*c^3*(-2*A*c+3*B*b)*(c* 
x^4+b*x^2)^(5/2)/b^5/x^12-128/45045*c^4*(-2*A*c+3*B*b)*(c*x^4+b*x^2)^(5/2) 
/b^6/x^10
 

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.64 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{19}} \, dx=-\frac {\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (3 b B x^2 \left (1155 b^4-840 b^3 c x^2+560 b^2 c^2 x^4-320 b c^3 x^6+128 c^4 x^8\right )+A \left (3003 b^5-2310 b^4 c x^2+1680 b^3 c^2 x^4-1120 b^2 c^3 x^6+640 b c^4 x^8-256 c^5 x^{10}\right )\right )}{45045 b^6 x^{20}} \] Input:

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^19,x]
 

Output:

-1/45045*((x^2*(b + c*x^2))^(5/2)*(3*b*B*x^2*(1155*b^4 - 840*b^3*c*x^2 + 5 
60*b^2*c^2*x^4 - 320*b*c^3*x^6 + 128*c^4*x^8) + A*(3003*b^5 - 2310*b^4*c*x 
^2 + 1680*b^3*c^2*x^4 - 1120*b^2*c^3*x^6 + 640*b*c^4*x^8 - 256*c^5*x^10))) 
/(b^6*x^20)
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {1940, 1220, 1129, 1129, 1129, 1129, 1123}

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.

\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{19}} \, dx\)

\(\Big \downarrow \) 1940

\(\displaystyle \frac {1}{2} \int \frac {\left (B x^2+A\right ) \left (c x^4+b x^2\right )^{3/2}}{x^{20}}dx^2\)

\(\Big \downarrow \) 1220

\(\displaystyle \frac {1}{2} \left (\frac {(3 b B-2 A c) \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^{18}}dx^2}{3 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{15 b x^{20}}\right )\)

\(\Big \downarrow \) 1129

\(\displaystyle \frac {1}{2} \left (\frac {(3 b B-2 A c) \left (-\frac {8 c \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^{16}}dx^2}{13 b}-\frac {2 \left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}\right )}{3 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{15 b x^{20}}\right )\)

\(\Big \downarrow \) 1129

\(\displaystyle \frac {1}{2} \left (\frac {(3 b B-2 A c) \left (-\frac {8 c \left (-\frac {6 c \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^{14}}dx^2}{11 b}-\frac {2 \left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}\right )}{13 b}-\frac {2 \left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}\right )}{3 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{15 b x^{20}}\right )\)

\(\Big \downarrow \) 1129

\(\displaystyle \frac {1}{2} \left (\frac {(3 b B-2 A c) \left (-\frac {8 c \left (-\frac {6 c \left (-\frac {4 c \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^{12}}dx^2}{9 b}-\frac {2 \left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}}\right )}{11 b}-\frac {2 \left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}\right )}{13 b}-\frac {2 \left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}\right )}{3 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{15 b x^{20}}\right )\)

\(\Big \downarrow \) 1129

\(\displaystyle \frac {1}{2} \left (\frac {(3 b B-2 A c) \left (-\frac {8 c \left (-\frac {6 c \left (-\frac {4 c \left (-\frac {2 c \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^{10}}dx^2}{7 b}-\frac {2 \left (b x^2+c x^4\right )^{5/2}}{7 b x^{12}}\right )}{9 b}-\frac {2 \left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}}\right )}{11 b}-\frac {2 \left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}\right )}{13 b}-\frac {2 \left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}\right )}{3 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{15 b x^{20}}\right )\)

\(\Big \downarrow \) 1123

\(\displaystyle \frac {1}{2} \left (\frac {\left (-\frac {8 c \left (-\frac {6 c \left (-\frac {4 c \left (\frac {4 c \left (b x^2+c x^4\right )^{5/2}}{35 b^2 x^{10}}-\frac {2 \left (b x^2+c x^4\right )^{5/2}}{7 b x^{12}}\right )}{9 b}-\frac {2 \left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}}\right )}{11 b}-\frac {2 \left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}\right )}{13 b}-\frac {2 \left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}\right ) (3 b B-2 A c)}{3 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{15 b x^{20}}\right )\)

Input:

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^19,x]
 

Output:

((-2*A*(b*x^2 + c*x^4)^(5/2))/(15*b*x^20) + ((3*b*B - 2*A*c)*((-2*(b*x^2 + 
 c*x^4)^(5/2))/(13*b*x^18) - (8*c*((-2*(b*x^2 + c*x^4)^(5/2))/(11*b*x^16) 
- (6*c*((-2*(b*x^2 + c*x^4)^(5/2))/(9*b*x^14) - (4*c*((-2*(b*x^2 + c*x^4)^ 
(5/2))/(7*b*x^12) + (4*c*(b*x^2 + c*x^4)^(5/2))/(35*b^2*x^10)))/(9*b)))/(1 
1*b)))/(13*b)))/(3*b))/2
 

Defintions of rubi rules used

rule 1123

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

rule 1129

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

rule 1220

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x 
^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e 
*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1))   Int[(d + e*x 
)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, 
 x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0 
]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 
]
 

rule 1940

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_) 
^(n_))^(q_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1) 
*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x] /; 
FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && I 
ntegerQ[Simplify[j/n]] && IntegerQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 
 1)/n]] && NeQ[n^2, 1]
 

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.60

method	result	size
pseudoelliptic	\(-\frac {\left (c \,x^{2}+b \right )^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (\left (\frac {15 B \,x^{2}}{13}+A \right ) b^{5}-\frac {10 c \,x^{2} \left (\frac {12 B \,x^{2}}{11}+A \right ) b^{4}}{13}+\frac {80 c^{2} x^{4} \left (B \,x^{2}+A \right ) b^{3}}{143}-\frac {160 c^{3} x^{6} \left (\frac {6 B \,x^{2}}{7}+A \right ) b^{2}}{429}+\frac {640 \left (\frac {3 B \,x^{2}}{5}+A \right ) c^{4} x^{8} b}{3003}-\frac {256 A \,c^{5} x^{10}}{3003}\right )}{15 x^{16} b^{6}}\)	\(124\)
gosper	\(-\frac {\left (c \,x^{2}+b \right ) \left (-256 A \,c^{5} x^{10}+384 B b \,c^{4} x^{10}+640 A b \,c^{4} x^{8}-960 B \,b^{2} c^{3} x^{8}-1120 A \,b^{2} c^{3} x^{6}+1680 B \,b^{3} c^{2} x^{6}+1680 A \,b^{3} c^{2} x^{4}-2520 B \,b^{4} c \,x^{4}-2310 A \,b^{4} c \,x^{2}+3465 B \,b^{5} x^{2}+3003 A \,b^{5}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{45045 x^{18} b^{6}}\)	\(142\)
default	\(-\frac {\left (c \,x^{2}+b \right ) \left (-256 A \,c^{5} x^{10}+384 B b \,c^{4} x^{10}+640 A b \,c^{4} x^{8}-960 B \,b^{2} c^{3} x^{8}-1120 A \,b^{2} c^{3} x^{6}+1680 B \,b^{3} c^{2} x^{6}+1680 A \,b^{3} c^{2} x^{4}-2520 B \,b^{4} c \,x^{4}-2310 A \,b^{4} c \,x^{2}+3465 B \,b^{5} x^{2}+3003 A \,b^{5}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{45045 x^{18} b^{6}}\)	\(142\)
orering	\(-\frac {\left (c \,x^{2}+b \right ) \left (-256 A \,c^{5} x^{10}+384 B b \,c^{4} x^{10}+640 A b \,c^{4} x^{8}-960 B \,b^{2} c^{3} x^{8}-1120 A \,b^{2} c^{3} x^{6}+1680 B \,b^{3} c^{2} x^{6}+1680 A \,b^{3} c^{2} x^{4}-2520 B \,b^{4} c \,x^{4}-2310 A \,b^{4} c \,x^{2}+3465 B \,b^{5} x^{2}+3003 A \,b^{5}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{45045 x^{18} b^{6}}\)	\(142\)
trager	\(-\frac {\left (-256 A \,c^{7} x^{14}+384 B b \,c^{6} x^{14}+128 A b \,c^{6} x^{12}-192 B \,b^{2} c^{5} x^{12}-96 A \,b^{2} c^{5} x^{10}+144 B \,b^{3} c^{4} x^{10}+80 A \,b^{3} c^{4} x^{8}-120 B \,b^{4} c^{3} x^{8}-70 A \,b^{4} c^{3} x^{6}+105 B \,b^{5} c^{2} x^{6}+63 A \,b^{5} c^{2} x^{4}+4410 B \,b^{6} c \,x^{4}+3696 A \,b^{6} c \,x^{2}+3465 B \,b^{7} x^{2}+3003 A \,b^{7}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{45045 b^{6} x^{16}}\)	\(183\)
risch	\(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (-256 A \,c^{7} x^{14}+384 B b \,c^{6} x^{14}+128 A b \,c^{6} x^{12}-192 B \,b^{2} c^{5} x^{12}-96 A \,b^{2} c^{5} x^{10}+144 B \,b^{3} c^{4} x^{10}+80 A \,b^{3} c^{4} x^{8}-120 B \,b^{4} c^{3} x^{8}-70 A \,b^{4} c^{3} x^{6}+105 B \,b^{5} c^{2} x^{6}+63 A \,b^{5} c^{2} x^{4}+4410 B \,b^{6} c \,x^{4}+3696 A \,b^{6} c \,x^{2}+3465 B \,b^{7} x^{2}+3003 A \,b^{7}\right )}{45045 x^{16} b^{6}}\)	\(183\)

Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^19,x,method=_RETURNVERBOSE)
 

Output:

-1/15*(c*x^2+b)^2*(x^2*(c*x^2+b))^(1/2)*((15/13*B*x^2+A)*b^5-10/13*c*x^2*( 
12/11*B*x^2+A)*b^4+80/143*c^2*x^4*(B*x^2+A)*b^3-160/429*c^3*x^6*(6/7*B*x^2 
+A)*b^2+640/3003*(3/5*B*x^2+A)*c^4*x^8*b-256/3003*A*c^5*x^10)/x^16/b^6
 

Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.87 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{19}} \, dx=-\frac {{\left (128 \, {\left (3 \, B b c^{6} - 2 \, A c^{7}\right )} x^{14} - 64 \, {\left (3 \, B b^{2} c^{5} - 2 \, A b c^{6}\right )} x^{12} + 48 \, {\left (3 \, B b^{3} c^{4} - 2 \, A b^{2} c^{5}\right )} x^{10} - 40 \, {\left (3 \, B b^{4} c^{3} - 2 \, A b^{3} c^{4}\right )} x^{8} + 3003 \, A b^{7} + 35 \, {\left (3 \, B b^{5} c^{2} - 2 \, A b^{4} c^{3}\right )} x^{6} + 63 \, {\left (70 \, B b^{6} c + A b^{5} c^{2}\right )} x^{4} + 231 \, {\left (15 \, B b^{7} + 16 \, A b^{6} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{45045 \, b^{6} x^{16}} \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^19,x, algorithm="fricas")
 

Output:

-1/45045*(128*(3*B*b*c^6 - 2*A*c^7)*x^14 - 64*(3*B*b^2*c^5 - 2*A*b*c^6)*x^ 
12 + 48*(3*B*b^3*c^4 - 2*A*b^2*c^5)*x^10 - 40*(3*B*b^4*c^3 - 2*A*b^3*c^4)* 
x^8 + 3003*A*b^7 + 35*(3*B*b^5*c^2 - 2*A*b^4*c^3)*x^6 + 63*(70*B*b^6*c + A 
*b^5*c^2)*x^4 + 231*(15*B*b^7 + 16*A*b^6*c)*x^2)*sqrt(c*x^4 + b*x^2)/(b^6* 
x^16)
 

Sympy [F]

\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{19}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{19}}\, dx \] Input:

integrate((B*x**2+A)*(c*x**4+b*x**2)**(3/2)/x**19,x)
 

Output:

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**19, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 385 vs. \(2 (183) = 366\).

Time = 0.05 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.86 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{19}} \, dx=-\frac {1}{30030} \, B {\left (\frac {256 \, \sqrt {c x^{4} + b x^{2}} c^{6}}{b^{5} x^{2}} - \frac {128 \, \sqrt {c x^{4} + b x^{2}} c^{5}}{b^{4} x^{4}} + \frac {96 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{b^{3} x^{6}} - \frac {80 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{2} x^{8}} + \frac {70 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b x^{10}} - \frac {63 \, \sqrt {c x^{4} + b x^{2}} c}{x^{12}} - \frac {693 \, \sqrt {c x^{4} + b x^{2}} b}{x^{14}} + \frac {3003 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{16}}\right )} + \frac {1}{180180} \, A {\left (\frac {1024 \, \sqrt {c x^{4} + b x^{2}} c^{7}}{b^{6} x^{2}} - \frac {512 \, \sqrt {c x^{4} + b x^{2}} c^{6}}{b^{5} x^{4}} + \frac {384 \, \sqrt {c x^{4} + b x^{2}} c^{5}}{b^{4} x^{6}} - \frac {320 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{b^{3} x^{8}} + \frac {280 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{2} x^{10}} - \frac {252 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b x^{12}} + \frac {231 \, \sqrt {c x^{4} + b x^{2}} c}{x^{14}} + \frac {3003 \, \sqrt {c x^{4} + b x^{2}} b}{x^{16}} - \frac {15015 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{18}}\right )} \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^19,x, algorithm="maxima")
 

Output:

-1/30030*B*(256*sqrt(c*x^4 + b*x^2)*c^6/(b^5*x^2) - 128*sqrt(c*x^4 + b*x^2 
)*c^5/(b^4*x^4) + 96*sqrt(c*x^4 + b*x^2)*c^4/(b^3*x^6) - 80*sqrt(c*x^4 + b 
*x^2)*c^3/(b^2*x^8) + 70*sqrt(c*x^4 + b*x^2)*c^2/(b*x^10) - 63*sqrt(c*x^4 
+ b*x^2)*c/x^12 - 693*sqrt(c*x^4 + b*x^2)*b/x^14 + 3003*(c*x^4 + b*x^2)^(3 
/2)/x^16) + 1/180180*A*(1024*sqrt(c*x^4 + b*x^2)*c^7/(b^6*x^2) - 512*sqrt( 
c*x^4 + b*x^2)*c^6/(b^5*x^4) + 384*sqrt(c*x^4 + b*x^2)*c^5/(b^4*x^6) - 320 
*sqrt(c*x^4 + b*x^2)*c^4/(b^3*x^8) + 280*sqrt(c*x^4 + b*x^2)*c^3/(b^2*x^10 
) - 252*sqrt(c*x^4 + b*x^2)*c^2/(b*x^12) + 231*sqrt(c*x^4 + b*x^2)*c/x^14 
+ 3003*sqrt(c*x^4 + b*x^2)*b/x^16 - 15015*(c*x^4 + b*x^2)^(3/2)/x^18)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 582 vs. \(2 (183) = 366\).

Time = 2.11 (sec) , antiderivative size = 582, normalized size of antiderivative = 2.81 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{19}} \, dx =\text {Too large to display} \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^19,x, algorithm="giac")
 

Output:

256/45045*(18018*(sqrt(c)*x - sqrt(c*x^2 + b))^20*B*c^(13/2)*sgn(x) + 6006 
0*(sqrt(c)*x - sqrt(c*x^2 + b))^18*A*c^(15/2)*sgn(x) - 12870*(sqrt(c)*x - 
sqrt(c*x^2 + b))^16*B*b^2*c^(13/2)*sgn(x) + 128700*(sqrt(c)*x - sqrt(c*x^2 
 + b))^16*A*b*c^(15/2)*sgn(x) - 32175*(sqrt(c)*x - sqrt(c*x^2 + b))^14*B*b 
^3*c^(13/2)*sgn(x) + 141570*(sqrt(c)*x - sqrt(c*x^2 + b))^14*A*b^2*c^(15/2 
)*sgn(x) + 15015*(sqrt(c)*x - sqrt(c*x^2 + b))^12*B*b^4*c^(13/2)*sgn(x) + 
50050*(sqrt(c)*x - sqrt(c*x^2 + b))^12*A*b^3*c^(15/2)*sgn(x) + 9009*(sqrt( 
c)*x - sqrt(c*x^2 + b))^10*B*b^5*c^(13/2)*sgn(x) + 6006*(sqrt(c)*x - sqrt( 
c*x^2 + b))^10*A*b^4*c^(15/2)*sgn(x) + 4095*(sqrt(c)*x - sqrt(c*x^2 + b))^ 
8*B*b^6*c^(13/2)*sgn(x) - 2730*(sqrt(c)*x - sqrt(c*x^2 + b))^8*A*b^5*c^(15 
/2)*sgn(x) - 1365*(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*b^7*c^(13/2)*sgn(x) + 
910*(sqrt(c)*x - sqrt(c*x^2 + b))^6*A*b^6*c^(15/2)*sgn(x) + 315*(sqrt(c)*x 
 - sqrt(c*x^2 + b))^4*B*b^8*c^(13/2)*sgn(x) - 210*(sqrt(c)*x - sqrt(c*x^2 
+ b))^4*A*b^7*c^(15/2)*sgn(x) - 45*(sqrt(c)*x - sqrt(c*x^2 + b))^2*B*b^9*c 
^(13/2)*sgn(x) + 30*(sqrt(c)*x - sqrt(c*x^2 + b))^2*A*b^8*c^(15/2)*sgn(x) 
+ 3*B*b^10*c^(13/2)*sgn(x) - 2*A*b^9*c^(15/2)*sgn(x))/((sqrt(c)*x - sqrt(c 
*x^2 + b))^2 - b)^15
 

Mupad [B] (verification not implemented)

Time = 12.42 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.72 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{19}} \, dx=\frac {2\,A\,c^3\,\sqrt {c\,x^4+b\,x^2}}{1287\,b^2\,x^{10}}-\frac {16\,A\,c\,\sqrt {c\,x^4+b\,x^2}}{195\,x^{14}}-\frac {B\,b\,\sqrt {c\,x^4+b\,x^2}}{13\,x^{14}}-\frac {14\,B\,c\,\sqrt {c\,x^4+b\,x^2}}{143\,x^{12}}-\frac {A\,c^2\,\sqrt {c\,x^4+b\,x^2}}{715\,b\,x^{12}}-\frac {A\,b\,\sqrt {c\,x^4+b\,x^2}}{15\,x^{16}}-\frac {16\,A\,c^4\,\sqrt {c\,x^4+b\,x^2}}{9009\,b^3\,x^8}+\frac {32\,A\,c^5\,\sqrt {c\,x^4+b\,x^2}}{15015\,b^4\,x^6}-\frac {128\,A\,c^6\,\sqrt {c\,x^4+b\,x^2}}{45045\,b^5\,x^4}+\frac {256\,A\,c^7\,\sqrt {c\,x^4+b\,x^2}}{45045\,b^6\,x^2}-\frac {B\,c^2\,\sqrt {c\,x^4+b\,x^2}}{429\,b\,x^{10}}+\frac {8\,B\,c^3\,\sqrt {c\,x^4+b\,x^2}}{3003\,b^2\,x^8}-\frac {16\,B\,c^4\,\sqrt {c\,x^4+b\,x^2}}{5005\,b^3\,x^6}+\frac {64\,B\,c^5\,\sqrt {c\,x^4+b\,x^2}}{15015\,b^4\,x^4}-\frac {128\,B\,c^6\,\sqrt {c\,x^4+b\,x^2}}{15015\,b^5\,x^2} \] Input:

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^19,x)
 

Output:

(2*A*c^3*(b*x^2 + c*x^4)^(1/2))/(1287*b^2*x^10) - (16*A*c*(b*x^2 + c*x^4)^ 
(1/2))/(195*x^14) - (B*b*(b*x^2 + c*x^4)^(1/2))/(13*x^14) - (14*B*c*(b*x^2 
 + c*x^4)^(1/2))/(143*x^12) - (A*c^2*(b*x^2 + c*x^4)^(1/2))/(715*b*x^12) - 
 (A*b*(b*x^2 + c*x^4)^(1/2))/(15*x^16) - (16*A*c^4*(b*x^2 + c*x^4)^(1/2))/ 
(9009*b^3*x^8) + (32*A*c^5*(b*x^2 + c*x^4)^(1/2))/(15015*b^4*x^6) - (128*A 
*c^6*(b*x^2 + c*x^4)^(1/2))/(45045*b^5*x^4) + (256*A*c^7*(b*x^2 + c*x^4)^( 
1/2))/(45045*b^6*x^2) - (B*c^2*(b*x^2 + c*x^4)^(1/2))/(429*b*x^10) + (8*B* 
c^3*(b*x^2 + c*x^4)^(1/2))/(3003*b^2*x^8) - (16*B*c^4*(b*x^2 + c*x^4)^(1/2 
))/(5005*b^3*x^6) + (64*B*c^5*(b*x^2 + c*x^4)^(1/2))/(15015*b^4*x^4) - (12 
8*B*c^6*(b*x^2 + c*x^4)^(1/2))/(15015*b^5*x^2)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.49 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{19}} \, dx=\frac {-3003 \sqrt {c \,x^{2}+b}\, a \,b^{7}-3696 \sqrt {c \,x^{2}+b}\, a \,b^{6} c \,x^{2}-63 \sqrt {c \,x^{2}+b}\, a \,b^{5} c^{2} x^{4}+70 \sqrt {c \,x^{2}+b}\, a \,b^{4} c^{3} x^{6}-80 \sqrt {c \,x^{2}+b}\, a \,b^{3} c^{4} x^{8}+96 \sqrt {c \,x^{2}+b}\, a \,b^{2} c^{5} x^{10}-128 \sqrt {c \,x^{2}+b}\, a b \,c^{6} x^{12}+256 \sqrt {c \,x^{2}+b}\, a \,c^{7} x^{14}-3465 \sqrt {c \,x^{2}+b}\, b^{8} x^{2}-4410 \sqrt {c \,x^{2}+b}\, b^{7} c \,x^{4}-105 \sqrt {c \,x^{2}+b}\, b^{6} c^{2} x^{6}+120 \sqrt {c \,x^{2}+b}\, b^{5} c^{3} x^{8}-144 \sqrt {c \,x^{2}+b}\, b^{4} c^{4} x^{10}+192 \sqrt {c \,x^{2}+b}\, b^{3} c^{5} x^{12}-384 \sqrt {c \,x^{2}+b}\, b^{2} c^{6} x^{14}-256 \sqrt {c}\, a \,c^{7} x^{15}+384 \sqrt {c}\, b^{2} c^{6} x^{15}}{45045 b^{6} x^{15}} \] Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^19,x)
 

Output:

( - 3003*sqrt(b + c*x**2)*a*b**7 - 3696*sqrt(b + c*x**2)*a*b**6*c*x**2 - 6 
3*sqrt(b + c*x**2)*a*b**5*c**2*x**4 + 70*sqrt(b + c*x**2)*a*b**4*c**3*x**6 
 - 80*sqrt(b + c*x**2)*a*b**3*c**4*x**8 + 96*sqrt(b + c*x**2)*a*b**2*c**5* 
x**10 - 128*sqrt(b + c*x**2)*a*b*c**6*x**12 + 256*sqrt(b + c*x**2)*a*c**7* 
x**14 - 3465*sqrt(b + c*x**2)*b**8*x**2 - 4410*sqrt(b + c*x**2)*b**7*c*x** 
4 - 105*sqrt(b + c*x**2)*b**6*c**2*x**6 + 120*sqrt(b + c*x**2)*b**5*c**3*x 
**8 - 144*sqrt(b + c*x**2)*b**4*c**4*x**10 + 192*sqrt(b + c*x**2)*b**3*c** 
5*x**12 - 384*sqrt(b + c*x**2)*b**2*c**6*x**14 - 256*sqrt(c)*a*c**7*x**15 
+ 384*sqrt(c)*b**2*c**6*x**15)/(45045*b**6*x**15)