\(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^{17}} \, dx\) [182]

Optimal result

Integrand size = 26, antiderivative size = 170 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{17}} \, dx=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{13 b x^{18}}-\frac {(13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{143 b^2 x^{16}}+\frac {2 c (13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{429 b^3 x^{14}}-\frac {8 c^2 (13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{3003 b^4 x^{12}}+\frac {16 c^3 (13 b B-8 A c) \left (b x^2+c x^4\right )^{5/2}}{15015 b^5 x^{10}} \] Output:

-1/13*A*(c*x^4+b*x^2)^(5/2)/b/x^18-1/143*(-8*A*c+13*B*b)*(c*x^4+b*x^2)^(5/ 
2)/b^2/x^16+2/429*c*(-8*A*c+13*B*b)*(c*x^4+b*x^2)^(5/2)/b^3/x^14-8/3003*c^ 
2*(-8*A*c+13*B*b)*(c*x^4+b*x^2)^(5/2)/b^4/x^12+16/15015*c^3*(-8*A*c+13*B*b 
)*(c*x^4+b*x^2)^(5/2)/b^5/x^10
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.65 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{17}} \, dx=\frac {\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (13 b B x^2 \left (-105 b^3+70 b^2 c x^2-40 b c^2 x^4+16 c^3 x^6\right )+A \left (-1155 b^4+840 b^3 c x^2-560 b^2 c^2 x^4+320 b c^3 x^6-128 c^4 x^8\right )\right )}{15015 b^5 x^{18}} \] Input:

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^17,x]
 

Output:

((x^2*(b + c*x^2))^(5/2)*(13*b*B*x^2*(-105*b^3 + 70*b^2*c*x^2 - 40*b*c^2*x 
^4 + 16*c^3*x^6) + A*(-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)))/(15015*b^5*x^18)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1940, 1220, 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.

\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{17}} \, dx\)

\(\Big \downarrow \) 1940

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

\(\Big \downarrow \) 1220

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

\(\Big \downarrow \) 1129

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

\(\Big \downarrow \) 1129

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

\(\Big \downarrow \) 1129

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

\(\Big \downarrow \) 1123

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

Input:

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^17,x]
 

Output:

((-2*A*(b*x^2 + c*x^4)^(5/2))/(13*b*x^18) + ((13*b*B - 8*A*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)))/(11*b)))/(13*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 = 0.46 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.62

method	result	size
pseudoelliptic	\(-\frac {\left (c \,x^{2}+b \right )^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (\left (\frac {13 B \,x^{2}}{11}+A \right ) b^{4}-\frac {8 c \left (\frac {13 B \,x^{2}}{12}+A \right ) x^{2} b^{3}}{11}+\frac {16 \left (\frac {13 B \,x^{2}}{14}+A \right ) c^{2} x^{4} b^{2}}{33}-\frac {64 c^{3} \left (\frac {13 B \,x^{2}}{20}+A \right ) x^{6} b}{231}+\frac {128 A \,c^{4} x^{8}}{1155}\right )}{13 x^{14} b^{5}}\)	\(106\)
gosper	\(-\frac {\left (c \,x^{2}+b \right ) \left (128 A \,c^{4} x^{8}-208 B b \,c^{3} x^{8}-320 A b \,c^{3} x^{6}+520 B \,b^{2} c^{2} x^{6}+560 A \,b^{2} c^{2} x^{4}-910 B \,b^{3} c \,x^{4}-840 A \,b^{3} c \,x^{2}+1365 B \,b^{4} x^{2}+1155 A \,b^{4}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{15015 x^{16} b^{5}}\)	\(118\)
default	\(-\frac {\left (c \,x^{2}+b \right ) \left (128 A \,c^{4} x^{8}-208 B b \,c^{3} x^{8}-320 A b \,c^{3} x^{6}+520 B \,b^{2} c^{2} x^{6}+560 A \,b^{2} c^{2} x^{4}-910 B \,b^{3} c \,x^{4}-840 A \,b^{3} c \,x^{2}+1365 B \,b^{4} x^{2}+1155 A \,b^{4}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{15015 x^{16} b^{5}}\)	\(118\)
orering	\(-\frac {\left (c \,x^{2}+b \right ) \left (128 A \,c^{4} x^{8}-208 B b \,c^{3} x^{8}-320 A b \,c^{3} x^{6}+520 B \,b^{2} c^{2} x^{6}+560 A \,b^{2} c^{2} x^{4}-910 B \,b^{3} c \,x^{4}-840 A \,b^{3} c \,x^{2}+1365 B \,b^{4} x^{2}+1155 A \,b^{4}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{15015 x^{16} b^{5}}\)	\(118\)
trager	\(-\frac {\left (128 A \,c^{6} x^{12}-208 B b \,c^{5} x^{12}-64 A b \,c^{5} x^{10}+104 B \,b^{2} c^{4} x^{10}+48 A \,b^{2} c^{4} x^{8}-78 B \,b^{3} c^{3} x^{8}-40 A \,b^{3} c^{3} x^{6}+65 B \,b^{4} c^{2} x^{6}+35 A \,b^{4} c^{2} x^{4}+1820 B \,b^{5} c \,x^{4}+1470 A \,b^{5} c \,x^{2}+1365 B \,b^{6} x^{2}+1155 A \,b^{6}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{15015 b^{5} x^{14}}\)	\(159\)
risch	\(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (128 A \,c^{6} x^{12}-208 B b \,c^{5} x^{12}-64 A b \,c^{5} x^{10}+104 B \,b^{2} c^{4} x^{10}+48 A \,b^{2} c^{4} x^{8}-78 B \,b^{3} c^{3} x^{8}-40 A \,b^{3} c^{3} x^{6}+65 B \,b^{4} c^{2} x^{6}+35 A \,b^{4} c^{2} x^{4}+1820 B \,b^{5} c \,x^{4}+1470 A \,b^{5} c \,x^{2}+1365 B \,b^{6} x^{2}+1155 A \,b^{6}\right )}{15015 x^{14} b^{5}}\)	\(159\)

Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^17,x,method=_RETURNVERBOSE)
 

Output:

-1/13*(c*x^2+b)^2*(x^2*(c*x^2+b))^(1/2)*((13/11*B*x^2+A)*b^4-8/11*c*(13/12 
*B*x^2+A)*x^2*b^3+16/33*(13/14*B*x^2+A)*c^2*x^4*b^2-64/231*c^3*(13/20*B*x^ 
2+A)*x^6*b+128/1155*A*c^4*x^8)/x^14/b^5
 

Fricas [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.92 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{17}} \, dx=\frac {{\left (16 \, {\left (13 \, B b c^{5} - 8 \, A c^{6}\right )} x^{12} - 8 \, {\left (13 \, B b^{2} c^{4} - 8 \, A b c^{5}\right )} x^{10} + 6 \, {\left (13 \, B b^{3} c^{3} - 8 \, A b^{2} c^{4}\right )} x^{8} - 1155 \, A b^{6} - 5 \, {\left (13 \, B b^{4} c^{2} - 8 \, A b^{3} c^{3}\right )} x^{6} - 35 \, {\left (52 \, B b^{5} c + A b^{4} c^{2}\right )} x^{4} - 105 \, {\left (13 \, B b^{6} + 14 \, A b^{5} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15015 \, b^{5} x^{14}} \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^17,x, algorithm="fricas")
 

Output:

1/15015*(16*(13*B*b*c^5 - 8*A*c^6)*x^12 - 8*(13*B*b^2*c^4 - 8*A*b*c^5)*x^1 
0 + 6*(13*B*b^3*c^3 - 8*A*b^2*c^4)*x^8 - 1155*A*b^6 - 5*(13*B*b^4*c^2 - 8* 
A*b^3*c^3)*x^6 - 35*(52*B*b^5*c + A*b^4*c^2)*x^4 - 105*(13*B*b^6 + 14*A*b^ 
5*c)*x^2)*sqrt(c*x^4 + b*x^2)/(b^5*x^14)
 

Sympy [F]

\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{17}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{17}}\, dx \] Input:

integrate((B*x**2+A)*(c*x**4+b*x**2)**(3/2)/x**17,x)
 

Output:

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**17, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (150) = 300\).

Time = 0.05 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.98 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{17}} \, dx=\frac {1}{9240} \, B {\left (\frac {128 \, \sqrt {c x^{4} + b x^{2}} c^{5}}{b^{4} x^{2}} - \frac {64 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{b^{3} x^{4}} + \frac {48 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{2} x^{6}} - \frac {40 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b x^{8}} + \frac {35 \, \sqrt {c x^{4} + b x^{2}} c}{x^{10}} + \frac {315 \, \sqrt {c x^{4} + b x^{2}} b}{x^{12}} - \frac {1155 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{14}}\right )} - \frac {1}{30030} \, A {\left (\frac {256 \, \sqrt {c x^{4} + b x^{2}} c^{6}}{b^{5} x^{2}} - \frac {128 \, \sqrt {c x^{4} + b x^{2}} c^{5}}{b^{4} x^{4}} + \frac {96 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{b^{3} x^{6}} - \frac {80 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{2} x^{8}} + \frac {70 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b x^{10}} - \frac {63 \, \sqrt {c x^{4} + b x^{2}} c}{x^{12}} - \frac {693 \, \sqrt {c x^{4} + b x^{2}} b}{x^{14}} + \frac {3003 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{16}}\right )} \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^17,x, algorithm="maxima")
 

Output:

1/9240*B*(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) - 1/30030*A*(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)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 550 vs. \(2 (150) = 300\).

Time = 2.06 (sec) , antiderivative size = 550, normalized size of antiderivative = 3.24 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{17}} \, dx =\text {Too large to display} \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^17,x, algorithm="giac")
 

Output:

32/15015*(15015*(sqrt(c)*x - sqrt(c*x^2 + b))^18*B*c^(11/2)*sgn(x) - 3003* 
(sqrt(c)*x - sqrt(c*x^2 + b))^16*B*b*c^(11/2)*sgn(x) + 48048*(sqrt(c)*x - 
sqrt(c*x^2 + b))^16*A*c^(13/2)*sgn(x) - 6006*(sqrt(c)*x - sqrt(c*x^2 + b)) 
^14*B*b^2*c^(11/2)*sgn(x) + 96096*(sqrt(c)*x - sqrt(c*x^2 + b))^14*A*b*c^( 
13/2)*sgn(x) - 28314*(sqrt(c)*x - sqrt(c*x^2 + b))^12*B*b^3*c^(11/2)*sgn(x 
) + 109824*(sqrt(c)*x - sqrt(c*x^2 + b))^12*A*b^2*c^(13/2)*sgn(x) + 13728* 
(sqrt(c)*x - sqrt(c*x^2 + b))^10*B*b^4*c^(11/2)*sgn(x) + 37752*(sqrt(c)*x 
- sqrt(c*x^2 + b))^10*A*b^3*c^(13/2)*sgn(x) + 5720*(sqrt(c)*x - sqrt(c*x^2 
 + b))^8*B*b^5*c^(11/2)*sgn(x) + 5720*(sqrt(c)*x - sqrt(c*x^2 + b))^8*A*b^ 
4*c^(13/2)*sgn(x) + 3718*(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*b^6*c^(11/2)*sg 
n(x) - 2288*(sqrt(c)*x - sqrt(c*x^2 + b))^6*A*b^5*c^(13/2)*sgn(x) - 1014*( 
sqrt(c)*x - sqrt(c*x^2 + b))^4*B*b^7*c^(11/2)*sgn(x) + 624*(sqrt(c)*x - sq 
rt(c*x^2 + b))^4*A*b^6*c^(13/2)*sgn(x) + 169*(sqrt(c)*x - sqrt(c*x^2 + b)) 
^2*B*b^8*c^(11/2)*sgn(x) - 104*(sqrt(c)*x - sqrt(c*x^2 + b))^2*A*b^7*c^(13 
/2)*sgn(x) - 13*B*b^9*c^(11/2)*sgn(x) + 8*A*b^8*c^(13/2)*sgn(x))/((sqrt(c) 
*x - sqrt(c*x^2 + b))^2 - b)^13
 

Mupad [B] (verification not implemented)

Time = 11.83 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.80 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{17}} \, dx=\frac {8\,A\,c^3\,\sqrt {c\,x^4+b\,x^2}}{3003\,b^2\,x^8}-\frac {14\,A\,c\,\sqrt {c\,x^4+b\,x^2}}{143\,x^{12}}-\frac {B\,b\,\sqrt {c\,x^4+b\,x^2}}{11\,x^{12}}-\frac {4\,B\,c\,\sqrt {c\,x^4+b\,x^2}}{33\,x^{10}}-\frac {A\,c^2\,\sqrt {c\,x^4+b\,x^2}}{429\,b\,x^{10}}-\frac {A\,b\,\sqrt {c\,x^4+b\,x^2}}{13\,x^{14}}-\frac {16\,A\,c^4\,\sqrt {c\,x^4+b\,x^2}}{5005\,b^3\,x^6}+\frac {64\,A\,c^5\,\sqrt {c\,x^4+b\,x^2}}{15015\,b^4\,x^4}-\frac {128\,A\,c^6\,\sqrt {c\,x^4+b\,x^2}}{15015\,b^5\,x^2}-\frac {B\,c^2\,\sqrt {c\,x^4+b\,x^2}}{231\,b\,x^8}+\frac {2\,B\,c^3\,\sqrt {c\,x^4+b\,x^2}}{385\,b^2\,x^6}-\frac {8\,B\,c^4\,\sqrt {c\,x^4+b\,x^2}}{1155\,b^3\,x^4}+\frac {16\,B\,c^5\,\sqrt {c\,x^4+b\,x^2}}{1155\,b^4\,x^2} \] Input:

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^17,x)
 

Output:

(8*A*c^3*(b*x^2 + c*x^4)^(1/2))/(3003*b^2*x^8) - (14*A*c*(b*x^2 + c*x^4)^( 
1/2))/(143*x^12) - (B*b*(b*x^2 + c*x^4)^(1/2))/(11*x^12) - (4*B*c*(b*x^2 + 
 c*x^4)^(1/2))/(33*x^10) - (A*c^2*(b*x^2 + c*x^4)^(1/2))/(429*b*x^10) - (A 
*b*(b*x^2 + c*x^4)^(1/2))/(13*x^14) - (16*A*c^4*(b*x^2 + c*x^4)^(1/2))/(50 
05*b^3*x^6) + (64*A*c^5*(b*x^2 + c*x^4)^(1/2))/(15015*b^4*x^4) - (128*A*c^ 
6*(b*x^2 + c*x^4)^(1/2))/(15015*b^5*x^2) - (B*c^2*(b*x^2 + c*x^4)^(1/2))/( 
231*b*x^8) + (2*B*c^3*(b*x^2 + c*x^4)^(1/2))/(385*b^2*x^6) - (8*B*c^4*(b*x 
^2 + c*x^4)^(1/2))/(1155*b^3*x^4) + (16*B*c^5*(b*x^2 + c*x^4)^(1/2))/(1155 
*b^4*x^2)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.58 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{17}} \, dx=\frac {-1155 \sqrt {c \,x^{2}+b}\, a \,b^{6}-1470 \sqrt {c \,x^{2}+b}\, a \,b^{5} c \,x^{2}-35 \sqrt {c \,x^{2}+b}\, a \,b^{4} c^{2} x^{4}+40 \sqrt {c \,x^{2}+b}\, a \,b^{3} c^{3} x^{6}-48 \sqrt {c \,x^{2}+b}\, a \,b^{2} c^{4} x^{8}+64 \sqrt {c \,x^{2}+b}\, a b \,c^{5} x^{10}-128 \sqrt {c \,x^{2}+b}\, a \,c^{6} x^{12}-1365 \sqrt {c \,x^{2}+b}\, b^{7} x^{2}-1820 \sqrt {c \,x^{2}+b}\, b^{6} c \,x^{4}-65 \sqrt {c \,x^{2}+b}\, b^{5} c^{2} x^{6}+78 \sqrt {c \,x^{2}+b}\, b^{4} c^{3} x^{8}-104 \sqrt {c \,x^{2}+b}\, b^{3} c^{4} x^{10}+208 \sqrt {c \,x^{2}+b}\, b^{2} c^{5} x^{12}+128 \sqrt {c}\, a \,c^{6} x^{13}-208 \sqrt {c}\, b^{2} c^{5} x^{13}}{15015 b^{5} x^{13}} \] Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^17,x)
 

Output:

( - 1155*sqrt(b + c*x**2)*a*b**6 - 1470*sqrt(b + c*x**2)*a*b**5*c*x**2 - 3 
5*sqrt(b + c*x**2)*a*b**4*c**2*x**4 + 40*sqrt(b + c*x**2)*a*b**3*c**3*x**6 
 - 48*sqrt(b + c*x**2)*a*b**2*c**4*x**8 + 64*sqrt(b + c*x**2)*a*b*c**5*x** 
10 - 128*sqrt(b + c*x**2)*a*c**6*x**12 - 1365*sqrt(b + c*x**2)*b**7*x**2 - 
 1820*sqrt(b + c*x**2)*b**6*c*x**4 - 65*sqrt(b + c*x**2)*b**5*c**2*x**6 + 
78*sqrt(b + c*x**2)*b**4*c**3*x**8 - 104*sqrt(b + c*x**2)*b**3*c**4*x**10 
+ 208*sqrt(b + c*x**2)*b**2*c**5*x**12 + 128*sqrt(c)*a*c**6*x**13 - 208*sq 
rt(c)*b**2*c**5*x**13)/(15015*b**5*x**13)