\(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^{15}} \, dx\) [181]

Optimal result

Integrand size = 26, antiderivative size = 133 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{15}} \, dx=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{11 b x^{16}}-\frac {(11 b B-6 A c) \left (b x^2+c x^4\right )^{5/2}}{99 b^2 x^{14}}+\frac {4 c (11 b B-6 A c) \left (b x^2+c x^4\right )^{5/2}}{693 b^3 x^{12}}-\frac {8 c^2 (11 b B-6 A c) \left (b x^2+c x^4\right )^{5/2}}{3465 b^4 x^{10}} \] Output:

-1/11*A*(c*x^4+b*x^2)^(5/2)/b/x^16-1/99*(-6*A*c+11*B*b)*(c*x^4+b*x^2)^(5/2 
)/b^2/x^14+4/693*c*(-6*A*c+11*B*b)*(c*x^4+b*x^2)^(5/2)/b^3/x^12-8/3465*c^2 
*(-6*A*c+11*B*b)*(c*x^4+b*x^2)^(5/2)/b^4/x^10
 

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.67 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{15}} \, dx=-\frac {\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (11 b B x^2 \left (35 b^2-20 b c x^2+8 c^2 x^4\right )+3 A \left (105 b^3-70 b^2 c x^2+40 b c^2 x^4-16 c^3 x^6\right )\right )}{3465 b^4 x^{16}} \] Input:

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^15,x]
 

Output:

-1/3465*((x^2*(b + c*x^2))^(5/2)*(11*b*B*x^2*(35*b^2 - 20*b*c*x^2 + 8*c^2* 
x^4) + 3*A*(105*b^3 - 70*b^2*c*x^2 + 40*b*c^2*x^4 - 16*c^3*x^6)))/(b^4*x^1 
6)
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1940, 1220, 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.

\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{15}} \, dx\)

\(\Big \downarrow \) 1940

\(\displaystyle \frac {1}{2} \int \frac {\left (B x^2+A\right ) \left (c x^4+b x^2\right )^{3/2}}{x^{16}}dx^2\)

\(\Big \downarrow \) 1220

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

\(\Big \downarrow \) 1129

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

\(\Big \downarrow \) 1129

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

\(\Big \downarrow \) 1123

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

Input:

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^15,x]
 

Output:

((-2*A*(b*x^2 + c*x^4)^(5/2))/(11*b*x^16) + ((11*b*B - 6*A*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)))/(11*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 = 0.44 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.65

method	result	size
pseudoelliptic	\(-\frac {\left (c \,x^{2}+b \right )^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (\left (\frac {11 B \,x^{2}}{9}+A \right ) b^{3}-\frac {2 c \,x^{2} \left (\frac {22 B \,x^{2}}{21}+A \right ) b^{2}}{3}+\frac {8 c^{2} \left (\frac {11 B \,x^{2}}{15}+A \right ) x^{4} b}{21}-\frac {16 A \,c^{3} x^{6}}{105}\right )}{11 x^{12} b^{4}}\)	\(87\)
gosper	\(-\frac {\left (c \,x^{2}+b \right ) \left (-48 A \,c^{3} x^{6}+88 B b \,c^{2} x^{6}+120 A b \,c^{2} x^{4}-220 x^{4} B \,b^{2} c -210 A \,b^{2} c \,x^{2}+385 x^{2} B \,b^{3}+315 A \,b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{3465 x^{14} b^{4}}\)	\(94\)
default	\(-\frac {\left (c \,x^{2}+b \right ) \left (-48 A \,c^{3} x^{6}+88 B b \,c^{2} x^{6}+120 A b \,c^{2} x^{4}-220 x^{4} B \,b^{2} c -210 A \,b^{2} c \,x^{2}+385 x^{2} B \,b^{3}+315 A \,b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{3465 x^{14} b^{4}}\)	\(94\)
orering	\(-\frac {\left (c \,x^{2}+b \right ) \left (-48 A \,c^{3} x^{6}+88 B b \,c^{2} x^{6}+120 A b \,c^{2} x^{4}-220 x^{4} B \,b^{2} c -210 A \,b^{2} c \,x^{2}+385 x^{2} B \,b^{3}+315 A \,b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{3465 x^{14} b^{4}}\)	\(94\)
trager	\(-\frac {\left (-48 A \,c^{5} x^{10}+88 B b \,c^{4} x^{10}+24 A b \,c^{4} x^{8}-44 B \,b^{2} c^{3} x^{8}-18 A \,b^{2} c^{3} x^{6}+33 B \,b^{3} c^{2} x^{6}+15 A \,b^{3} c^{2} x^{4}+550 B \,b^{4} c \,x^{4}+420 A \,b^{4} c \,x^{2}+385 B \,b^{5} x^{2}+315 A \,b^{5}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{3465 b^{4} x^{12}}\)	\(135\)
risch	\(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (-48 A \,c^{5} x^{10}+88 B b \,c^{4} x^{10}+24 A b \,c^{4} x^{8}-44 B \,b^{2} c^{3} x^{8}-18 A \,b^{2} c^{3} x^{6}+33 B \,b^{3} c^{2} x^{6}+15 A \,b^{3} c^{2} x^{4}+550 B \,b^{4} c \,x^{4}+420 A \,b^{4} c \,x^{2}+385 B \,b^{5} x^{2}+315 A \,b^{5}\right )}{3465 x^{12} b^{4}}\)	\(135\)

Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^15,x,method=_RETURNVERBOSE)
 

Output:

-1/11*(c*x^2+b)^2*(x^2*(c*x^2+b))^(1/2)*((11/9*B*x^2+A)*b^3-2/3*c*x^2*(22/ 
21*B*x^2+A)*b^2+8/21*c^2*(11/15*B*x^2+A)*x^4*b-16/105*A*c^3*x^6)/x^12/b^4
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.01 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{15}} \, dx=-\frac {{\left (8 \, {\left (11 \, B b c^{4} - 6 \, A c^{5}\right )} x^{10} - 4 \, {\left (11 \, B b^{2} c^{3} - 6 \, A b c^{4}\right )} x^{8} + 3 \, {\left (11 \, B b^{3} c^{2} - 6 \, A b^{2} c^{3}\right )} x^{6} + 315 \, A b^{5} + 5 \, {\left (110 \, B b^{4} c + 3 \, A b^{3} c^{2}\right )} x^{4} + 35 \, {\left (11 \, B b^{5} + 12 \, A b^{4} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{3465 \, b^{4} x^{12}} \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^15,x, algorithm="fricas")
 

Output:

-1/3465*(8*(11*B*b*c^4 - 6*A*c^5)*x^10 - 4*(11*B*b^2*c^3 - 6*A*b*c^4)*x^8 
+ 3*(11*B*b^3*c^2 - 6*A*b^2*c^3)*x^6 + 315*A*b^5 + 5*(110*B*b^4*c + 3*A*b^ 
3*c^2)*x^4 + 35*(11*B*b^5 + 12*A*b^4*c)*x^2)*sqrt(c*x^4 + b*x^2)/(b^4*x^12 
)
 

Sympy [F]

\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{15}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{15}}\, dx \] Input:

integrate((B*x**2+A)*(c*x**4+b*x**2)**(3/2)/x**15,x)
 

Output:

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**15, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 289 vs. \(2 (117) = 234\).

Time = 0.05 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.17 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{15}} \, dx=-\frac {1}{630} \, B {\left (\frac {16 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{b^{3} x^{2}} - \frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{2} x^{4}} + \frac {6 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b x^{6}} - \frac {5 \, \sqrt {c x^{4} + b x^{2}} c}{x^{8}} - \frac {35 \, \sqrt {c x^{4} + b x^{2}} b}{x^{10}} + \frac {105 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{12}}\right )} + \frac {1}{9240} \, A {\left (\frac {128 \, \sqrt {c x^{4} + b x^{2}} c^{5}}{b^{4} x^{2}} - \frac {64 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{b^{3} x^{4}} + \frac {48 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{2} x^{6}} - \frac {40 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b x^{8}} + \frac {35 \, \sqrt {c x^{4} + b x^{2}} c}{x^{10}} + \frac {315 \, \sqrt {c x^{4} + b x^{2}} b}{x^{12}} - \frac {1155 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{14}}\right )} \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^15,x, algorithm="maxima")
 

Output:

-1/630*B*(16*sqrt(c*x^4 + b*x^2)*c^4/(b^3*x^2) - 8*sqrt(c*x^4 + b*x^2)*c^3 
/(b^2*x^4) + 6*sqrt(c*x^4 + b*x^2)*c^2/(b*x^6) - 5*sqrt(c*x^4 + b*x^2)*c/x 
^8 - 35*sqrt(c*x^4 + b*x^2)*b/x^10 + 105*(c*x^4 + b*x^2)^(3/2)/x^12) + 1/9 
240*A*(128*sqrt(c*x^4 + b*x^2)*c^5/(b^4*x^2) - 64*sqrt(c*x^4 + b*x^2)*c^4/ 
(b^3*x^4) + 48*sqrt(c*x^4 + b*x^2)*c^3/(b^2*x^6) - 40*sqrt(c*x^4 + b*x^2)* 
c^2/(b*x^8) + 35*sqrt(c*x^4 + b*x^2)*c/x^10 + 315*sqrt(c*x^4 + b*x^2)*b/x^ 
12 - 1155*(c*x^4 + b*x^2)^(3/2)/x^14)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 490 vs. \(2 (117) = 234\).

Time = 1.65 (sec) , antiderivative size = 490, normalized size of antiderivative = 3.68 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{15}} \, dx=\frac {16 \, {\left (2310 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{16} B c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) - 1155 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{14} B b c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + 6930 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{14} A c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) + 231 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} B b^{2} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + 12474 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} A b c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) - 4851 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} B b^{3} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + 15246 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} A b^{2} c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) + 2475 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} B b^{4} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + 4950 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} A b^{3} c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) + 495 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} B b^{5} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + 990 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} A b^{4} c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) + 605 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b^{6} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) - 330 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A b^{5} c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) - 121 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{7} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + 66 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} A b^{6} c^{\frac {11}{2}} \mathrm {sgn}\left (x\right ) + 11 \, B b^{8} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) - 6 \, A b^{7} c^{\frac {11}{2}} \mathrm {sgn}\left (x\right )\right )}}{3465 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{11}} \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^15,x, algorithm="giac")
 

Output:

16/3465*(2310*(sqrt(c)*x - sqrt(c*x^2 + b))^16*B*c^(9/2)*sgn(x) - 1155*(sq 
rt(c)*x - sqrt(c*x^2 + b))^14*B*b*c^(9/2)*sgn(x) + 6930*(sqrt(c)*x - sqrt( 
c*x^2 + b))^14*A*c^(11/2)*sgn(x) + 231*(sqrt(c)*x - sqrt(c*x^2 + b))^12*B* 
b^2*c^(9/2)*sgn(x) + 12474*(sqrt(c)*x - sqrt(c*x^2 + b))^12*A*b*c^(11/2)*s 
gn(x) - 4851*(sqrt(c)*x - sqrt(c*x^2 + b))^10*B*b^3*c^(9/2)*sgn(x) + 15246 
*(sqrt(c)*x - sqrt(c*x^2 + b))^10*A*b^2*c^(11/2)*sgn(x) + 2475*(sqrt(c)*x 
- sqrt(c*x^2 + b))^8*B*b^4*c^(9/2)*sgn(x) + 4950*(sqrt(c)*x - sqrt(c*x^2 + 
 b))^8*A*b^3*c^(11/2)*sgn(x) + 495*(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*b^5*c 
^(9/2)*sgn(x) + 990*(sqrt(c)*x - sqrt(c*x^2 + b))^6*A*b^4*c^(11/2)*sgn(x) 
+ 605*(sqrt(c)*x - sqrt(c*x^2 + b))^4*B*b^6*c^(9/2)*sgn(x) - 330*(sqrt(c)* 
x - sqrt(c*x^2 + b))^4*A*b^5*c^(11/2)*sgn(x) - 121*(sqrt(c)*x - sqrt(c*x^2 
 + b))^2*B*b^7*c^(9/2)*sgn(x) + 66*(sqrt(c)*x - sqrt(c*x^2 + b))^2*A*b^6*c 
^(11/2)*sgn(x) + 11*B*b^8*c^(9/2)*sgn(x) - 6*A*b^7*c^(11/2)*sgn(x))/((sqrt 
(c)*x - sqrt(c*x^2 + b))^2 - b)^11
 

Mupad [B] (verification not implemented)

Time = 11.10 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.92 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{15}} \, dx=\frac {2\,A\,c^3\,\sqrt {c\,x^4+b\,x^2}}{385\,b^2\,x^6}-\frac {4\,A\,c\,\sqrt {c\,x^4+b\,x^2}}{33\,x^{10}}-\frac {B\,b\,\sqrt {c\,x^4+b\,x^2}}{9\,x^{10}}-\frac {10\,B\,c\,\sqrt {c\,x^4+b\,x^2}}{63\,x^8}-\frac {A\,c^2\,\sqrt {c\,x^4+b\,x^2}}{231\,b\,x^8}-\frac {A\,b\,\sqrt {c\,x^4+b\,x^2}}{11\,x^{12}}-\frac {8\,A\,c^4\,\sqrt {c\,x^4+b\,x^2}}{1155\,b^3\,x^4}+\frac {16\,A\,c^5\,\sqrt {c\,x^4+b\,x^2}}{1155\,b^4\,x^2}-\frac {B\,c^2\,\sqrt {c\,x^4+b\,x^2}}{105\,b\,x^6}+\frac {4\,B\,c^3\,\sqrt {c\,x^4+b\,x^2}}{315\,b^2\,x^4}-\frac {8\,B\,c^4\,\sqrt {c\,x^4+b\,x^2}}{315\,b^3\,x^2} \] Input:

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^15,x)
 

Output:

(2*A*c^3*(b*x^2 + c*x^4)^(1/2))/(385*b^2*x^6) - (4*A*c*(b*x^2 + c*x^4)^(1/ 
2))/(33*x^10) - (B*b*(b*x^2 + c*x^4)^(1/2))/(9*x^10) - (10*B*c*(b*x^2 + c* 
x^4)^(1/2))/(63*x^8) - (A*c^2*(b*x^2 + c*x^4)^(1/2))/(231*b*x^8) - (A*b*(b 
*x^2 + c*x^4)^(1/2))/(11*x^12) - (8*A*c^4*(b*x^2 + c*x^4)^(1/2))/(1155*b^3 
*x^4) + (16*A*c^5*(b*x^2 + c*x^4)^(1/2))/(1155*b^4*x^2) - (B*c^2*(b*x^2 + 
c*x^4)^(1/2))/(105*b*x^6) + (4*B*c^3*(b*x^2 + c*x^4)^(1/2))/(315*b^2*x^4) 
- (8*B*c^4*(b*x^2 + c*x^4)^(1/2))/(315*b^3*x^2)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.73 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{15}} \, dx=\frac {-315 \sqrt {c \,x^{2}+b}\, a \,b^{5}-420 \sqrt {c \,x^{2}+b}\, a \,b^{4} c \,x^{2}-15 \sqrt {c \,x^{2}+b}\, a \,b^{3} c^{2} x^{4}+18 \sqrt {c \,x^{2}+b}\, a \,b^{2} c^{3} x^{6}-24 \sqrt {c \,x^{2}+b}\, a b \,c^{4} x^{8}+48 \sqrt {c \,x^{2}+b}\, a \,c^{5} x^{10}-385 \sqrt {c \,x^{2}+b}\, b^{6} x^{2}-550 \sqrt {c \,x^{2}+b}\, b^{5} c \,x^{4}-33 \sqrt {c \,x^{2}+b}\, b^{4} c^{2} x^{6}+44 \sqrt {c \,x^{2}+b}\, b^{3} c^{3} x^{8}-88 \sqrt {c \,x^{2}+b}\, b^{2} c^{4} x^{10}-48 \sqrt {c}\, a \,c^{5} x^{11}+88 \sqrt {c}\, b^{2} c^{4} x^{11}}{3465 b^{4} x^{11}} \] Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^15,x)
 

Output:

( - 315*sqrt(b + c*x**2)*a*b**5 - 420*sqrt(b + c*x**2)*a*b**4*c*x**2 - 15* 
sqrt(b + c*x**2)*a*b**3*c**2*x**4 + 18*sqrt(b + c*x**2)*a*b**2*c**3*x**6 - 
 24*sqrt(b + c*x**2)*a*b*c**4*x**8 + 48*sqrt(b + c*x**2)*a*c**5*x**10 - 38 
5*sqrt(b + c*x**2)*b**6*x**2 - 550*sqrt(b + c*x**2)*b**5*c*x**4 - 33*sqrt( 
b + c*x**2)*b**4*c**2*x**6 + 44*sqrt(b + c*x**2)*b**3*c**3*x**8 - 88*sqrt( 
b + c*x**2)*b**2*c**4*x**10 - 48*sqrt(c)*a*c**5*x**11 + 88*sqrt(c)*b**2*c* 
*4*x**11)/(3465*b**4*x**11)