\(\int \frac {x^7 (A+B x^2)}{(b x^2+c x^4)^{3/2}} \, dx\) [211]

Optimal result

Integrand size = 26, antiderivative size = 134 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {b (b B-A c) x^2}{c^3 \sqrt {b x^2+c x^4}}-\frac {(7 b B-4 A c) \sqrt {b x^2+c x^4}}{8 c^3}+\frac {B x^2 \sqrt {b x^2+c x^4}}{4 c^2}+\frac {3 b (5 b B-4 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{8 c^{7/2}} \] Output:

-b*(-A*c+B*b)*x^2/c^3/(c*x^4+b*x^2)^(1/2)-1/8*(-4*A*c+7*B*b)*(c*x^4+b*x^2) 
^(1/2)/c^3+1/4*B*x^2*(c*x^4+b*x^2)^(1/2)/c^2+3/8*b*(-4*A*c+5*B*b)*arctanh( 
c^(1/2)*x^2/(c*x^4+b*x^2)^(1/2))/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.99 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {x^3 \left (\sqrt {c} \left (b+c x^2\right ) \left (-15 b^2 B x+b c x \left (12 A-5 B x^2\right )+2 c^2 x^3 \left (2 A+B x^2\right )\right )+6 b (5 b B-4 A c) \left (b+c x^2\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {b}+\sqrt {b+c x^2}}\right )\right )}{8 c^{7/2} \left (x^2 \left (b+c x^2\right )\right )^{3/2}} \] Input:

Integrate[(x^7*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]
 

Output:

(x^3*(Sqrt[c]*(b + c*x^2)*(-15*b^2*B*x + b*c*x*(12*A - 5*B*x^2) + 2*c^2*x^ 
3*(2*A + B*x^2)) + 6*b*(5*b*B - 4*A*c)*(b + c*x^2)^(3/2)*ArcTanh[(Sqrt[c]* 
x)/(-Sqrt[b] + Sqrt[b + c*x^2])]))/(8*c^(7/2)*(x^2*(b + c*x^2))^(3/2))
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {1940, 1211, 2192, 27, 1160, 1091, 219}

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.

\(\displaystyle \int \frac {x^7 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1940

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

\(\Big \downarrow \) 1211

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

\(\Big \downarrow \) 2192

\(\displaystyle \frac {1}{2} \left (\frac {\frac {\int \frac {c \left (4 b (b B-A c)-c (7 b B-4 A c) x^2\right )}{2 \sqrt {c x^4+b x^2}}dx^2}{2 c}+\frac {1}{2} B c x^2 \sqrt {b x^2+c x^4}}{c^3}-\frac {2 b x^2 (b B-A c)}{c^3 \sqrt {b x^2+c x^4}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {\frac {1}{4} \int \frac {4 b (b B-A c)-c (7 b B-4 A c) x^2}{\sqrt {c x^4+b x^2}}dx^2+\frac {1}{2} B c x^2 \sqrt {b x^2+c x^4}}{c^3}-\frac {2 b x^2 (b B-A c)}{c^3 \sqrt {b x^2+c x^4}}\right )\)

\(\Big \downarrow \) 1160

\(\displaystyle \frac {1}{2} \left (\frac {\frac {1}{4} \left (\frac {3}{2} b (5 b B-4 A c) \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2-\sqrt {b x^2+c x^4} (7 b B-4 A c)\right )+\frac {1}{2} B c x^2 \sqrt {b x^2+c x^4}}{c^3}-\frac {2 b x^2 (b B-A c)}{c^3 \sqrt {b x^2+c x^4}}\right )\)

\(\Big \downarrow \) 1091

\(\displaystyle \frac {1}{2} \left (\frac {\frac {1}{4} \left (3 b (5 b B-4 A c) \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}-\sqrt {b x^2+c x^4} (7 b B-4 A c)\right )+\frac {1}{2} B c x^2 \sqrt {b x^2+c x^4}}{c^3}-\frac {2 b x^2 (b B-A c)}{c^3 \sqrt {b x^2+c x^4}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {\frac {1}{4} \left (\frac {3 b (5 b B-4 A c) \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {c}}-\sqrt {b x^2+c x^4} (7 b B-4 A c)\right )+\frac {1}{2} B c x^2 \sqrt {b x^2+c x^4}}{c^3}-\frac {2 b x^2 (b B-A c)}{c^3 \sqrt {b x^2+c x^4}}\right )\)

Input:

Int[(x^7*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]
 

Output:

((-2*b*(b*B - A*c)*x^2)/(c^3*Sqrt[b*x^2 + c*x^4]) + ((B*c*x^2*Sqrt[b*x^2 + 
 c*x^4])/2 + (-((7*b*B - 4*A*c)*Sqrt[b*x^2 + c*x^4]) + (3*b*(5*b*B - 4*A*c 
)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/Sqrt[c])/4)/c^3)/2
 

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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1091

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[Int[1/(1 
 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
 

rule 1160

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

rule 1211

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* 
(x_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[-2*(2*c*d - b*e)^(m - 2)*(c*( 
e*f + d*g) - b*e*g)^n*((d + e*x)/(c^(m + n - 1)*e^(n - 1)*Sqrt[a + b*x + c* 
x^2])), x] + Simp[1/(c^(m + n - 1)*e^(n - 2))   Int[ExpandToSum[((2*c*d - b 
*e)^(m - 1)*(c*(e*f + d*g) - b*e*g)^n - c^(m + n - 1)*e^n*(d + e*x)^(m - 1) 
*(f + g*x)^n)/(c*d - b*e - c*e*x), x]/Sqrt[a + b*x + c*x^2], x], x] /; Free 
Q[{a, b, c, d, e, f, g}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[m, 0] 
&& IGtQ[n, 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]
 

rule 2192

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = 
Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x + 
 c*x^2)^(p + 1)/(c*(q + 2*p + 1))), x] + Simp[1/(c*(q + 2*p + 1))   Int[(a 
+ b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b 
*e*(q + p)*x^(q - 1) - c*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c 
, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.04

method	result	size
default	\(-\frac {x^{3} \left (c \,x^{2}+b \right ) \left (-2 x^{5} B \,c^{\frac {7}{2}}-4 A \,c^{\frac {7}{2}} x^{3}+5 B \,x^{3} b \,c^{\frac {5}{2}}-12 A \,c^{\frac {5}{2}} b x +15 x B \,b^{2} c^{\frac {3}{2}}+12 A \sqrt {c \,x^{2}+b}\, \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b \,c^{2}-15 B \sqrt {c \,x^{2}+b}\, \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{2} c \right )}{8 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{\frac {9}{2}}}\)	\(140\)
risch	\(\frac {x^{2} \left (2 B c \,x^{2}+4 A c -7 B b \right ) \left (c \,x^{2}+b \right )}{8 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {b \left (3 c \left (4 A c -5 B b \right ) \left (-\frac {x}{c \sqrt {c \,x^{2}+b}}+\frac {\ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )}{c^{\frac {3}{2}}}\right )-\frac {7 B b x}{\sqrt {c \,x^{2}+b}}+\frac {4 A c x}{\sqrt {c \,x^{2}+b}}\right ) x \sqrt {c \,x^{2}+b}}{8 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(152\)
pseudoelliptic	\(\frac {4 B \,c^{\frac {5}{2}} x^{6}+8 A \,c^{\frac {5}{2}} x^{4}-10 B b \,c^{\frac {3}{2}} x^{4}+24 A b \,c^{\frac {3}{2}} x^{2}-30 B \,b^{2} x^{2} \sqrt {c}-12 A \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b c \sqrt {x^{2} \left (c \,x^{2}+b \right )}+12 A \ln \left (2\right ) b c \sqrt {x^{2} \left (c \,x^{2}+b \right )}+15 B \ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}-15 B \ln \left (2\right ) b^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{16 c^{\frac {7}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(210\)

Input:

int(x^7*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/8*x^3*(c*x^2+b)*(-2*x^5*B*c^(7/2)-4*A*c^(7/2)*x^3+5*B*x^3*b*c^(5/2)-12* 
A*c^(5/2)*b*x+15*x*B*b^2*c^(3/2)+12*A*(c*x^2+b)^(1/2)*ln(c^(1/2)*x+(c*x^2+ 
b)^(1/2))*b*c^2-15*B*(c*x^2+b)^(1/2)*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*b^2*c)/ 
(c*x^4+b*x^2)^(3/2)/c^(9/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.16 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\left [-\frac {3 \, {\left (5 \, B b^{3} - 4 \, A b^{2} c + {\left (5 \, B b^{2} c - 4 \, A b c^{2}\right )} x^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (2 \, B c^{3} x^{4} - 15 \, B b^{2} c + 12 \, A b c^{2} - {\left (5 \, B b c^{2} - 4 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{16 \, {\left (c^{5} x^{2} + b c^{4}\right )}}, -\frac {3 \, {\left (5 \, B b^{3} - 4 \, A b^{2} c + {\left (5 \, B b^{2} c - 4 \, A b c^{2}\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - {\left (2 \, B c^{3} x^{4} - 15 \, B b^{2} c + 12 \, A b c^{2} - {\left (5 \, B b c^{2} - 4 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{8 \, {\left (c^{5} x^{2} + b c^{4}\right )}}\right ] \] Input:

integrate(x^7*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
 

Output:

[-1/16*(3*(5*B*b^3 - 4*A*b^2*c + (5*B*b^2*c - 4*A*b*c^2)*x^2)*sqrt(c)*log( 
-2*c*x^2 - b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*(2*B*c^3*x^4 - 15*B*b^2* 
c + 12*A*b*c^2 - (5*B*b*c^2 - 4*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/(c^5*x^2 
+ b*c^4), -1/8*(3*(5*B*b^3 - 4*A*b^2*c + (5*B*b^2*c - 4*A*b*c^2)*x^2)*sqrt 
(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) - (2*B*c^3*x^4 - 15* 
B*b^2*c + 12*A*b*c^2 - (5*B*b*c^2 - 4*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/(c^ 
5*x^2 + b*c^4)]
 

Sympy [F]

\[ \int \frac {x^7 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{7} \left (A + B x^{2}\right )}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(x**7*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)
 

Output:

Integral(x**7*(A + B*x**2)/(x**2*(b + c*x**2))**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.40 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {1}{4} \, {\left (\frac {2 \, x^{4}}{\sqrt {c x^{4} + b x^{2}} c} + \frac {6 \, b x^{2}}{\sqrt {c x^{4} + b x^{2}} c^{2}} - \frac {3 \, b \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {5}{2}}}\right )} A + \frac {1}{16} \, {\left (\frac {4 \, x^{6}}{\sqrt {c x^{4} + b x^{2}} c} - \frac {10 \, b x^{4}}{\sqrt {c x^{4} + b x^{2}} c^{2}} - \frac {30 \, b^{2} x^{2}}{\sqrt {c x^{4} + b x^{2}} c^{3}} + \frac {15 \, b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {7}{2}}}\right )} B \] Input:

integrate(x^7*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
 

Output:

1/4*(2*x^4/(sqrt(c*x^4 + b*x^2)*c) + 6*b*x^2/(sqrt(c*x^4 + b*x^2)*c^2) - 3 
*b*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(5/2))*A + 1/16*(4*x 
^6/(sqrt(c*x^4 + b*x^2)*c) - 10*b*x^4/(sqrt(c*x^4 + b*x^2)*c^2) - 30*b^2*x 
^2/(sqrt(c*x^4 + b*x^2)*c^3) + 15*b^2*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x 
^2)*sqrt(c))/c^(7/2))*B
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.08 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {{\left (x^{2} {\left (\frac {2 \, B x^{2}}{c \mathrm {sgn}\left (x\right )} - \frac {5 \, B b c^{3} \mathrm {sgn}\left (x\right ) - 4 \, A c^{4} \mathrm {sgn}\left (x\right )}{c^{5}}\right )} - \frac {3 \, {\left (5 \, B b^{2} c^{2} \mathrm {sgn}\left (x\right ) - 4 \, A b c^{3} \mathrm {sgn}\left (x\right )\right )}}{c^{5}}\right )} x}{8 \, \sqrt {c x^{2} + b}} + \frac {3 \, {\left (5 \, B b^{2} \log \left ({\left | b \right |}\right ) - 4 \, A b c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{16 \, c^{\frac {7}{2}}} - \frac {3 \, {\left (5 \, B b^{2} - 4 \, A b c\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{8 \, c^{\frac {7}{2}} \mathrm {sgn}\left (x\right )} \] Input:

integrate(x^7*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
 

Output:

1/8*(x^2*(2*B*x^2/(c*sgn(x)) - (5*B*b*c^3*sgn(x) - 4*A*c^4*sgn(x))/c^5) - 
3*(5*B*b^2*c^2*sgn(x) - 4*A*b*c^3*sgn(x))/c^5)*x/sqrt(c*x^2 + b) + 3/16*(5 
*B*b^2*log(abs(b)) - 4*A*b*c*log(abs(b)))*sgn(x)/c^(7/2) - 3/8*(5*B*b^2 - 
4*A*b*c)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/(c^(7/2)*sgn(x))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^7\,\left (B\,x^2+A\right )}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \] Input:

int((x^7*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x)
 

Output:

int((x^7*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.89 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {12 \sqrt {c \,x^{2}+b}\, a b \,c^{2} x +4 \sqrt {c \,x^{2}+b}\, a \,c^{3} x^{3}-15 \sqrt {c \,x^{2}+b}\, b^{3} c x -5 \sqrt {c \,x^{2}+b}\, b^{2} c^{2} x^{3}+2 \sqrt {c \,x^{2}+b}\, b \,c^{3} x^{5}-12 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,b^{2} c -12 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a b \,c^{2} x^{2}+15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{4}+15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{3} c \,x^{2}+9 \sqrt {c}\, a \,b^{2} c +9 \sqrt {c}\, a b \,c^{2} x^{2}-10 \sqrt {c}\, b^{4}-10 \sqrt {c}\, b^{3} c \,x^{2}}{8 c^{4} \left (c \,x^{2}+b \right )} \] Input:

int(x^7*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)
 

Output:

(12*sqrt(b + c*x**2)*a*b*c**2*x + 4*sqrt(b + c*x**2)*a*c**3*x**3 - 15*sqrt 
(b + c*x**2)*b**3*c*x - 5*sqrt(b + c*x**2)*b**2*c**2*x**3 + 2*sqrt(b + c*x 
**2)*b*c**3*x**5 - 12*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))* 
a*b**2*c - 12*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*a*b*c**2 
*x**2 + 15*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*b**4 + 15*s 
qrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt(b))*b**3*c*x**2 + 9*sqrt(c) 
*a*b**2*c + 9*sqrt(c)*a*b*c**2*x**2 - 10*sqrt(c)*b**4 - 10*sqrt(c)*b**3*c* 
x**2)/(8*c**4*(b + c*x**2))