\(\int \frac {x^4 (a+b x^2)^2}{(c+d x^2)^{3/2}} \, dx\) [900]

Optimal result

Integrand size = 24, antiderivative size = 177 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\frac {c (b c-a d)^2 x}{d^4 \sqrt {c+d x^2}}+\frac {\left (19 b^2 c^2-28 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 d^4}-\frac {b (11 b c-12 a d) x^3 \sqrt {c+d x^2}}{24 d^3}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d^2}-\frac {c \left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{9/2}} \] Output:

c*(-a*d+b*c)^2*x/d^4/(d*x^2+c)^(1/2)+1/16*(8*a^2*d^2-28*a*b*c*d+19*b^2*c^2 
)*x*(d*x^2+c)^(1/2)/d^4-1/24*b*(-12*a*d+11*b*c)*x^3*(d*x^2+c)^(1/2)/d^3+1/ 
6*b^2*x^5*(d*x^2+c)^(1/2)/d^2-1/16*c*(24*a^2*d^2-60*a*b*c*d+35*b^2*c^2)*ar 
ctanh(d^(1/2)*x/(d*x^2+c)^(1/2))/d^(9/2)
 

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.92 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\frac {x \left (24 a^2 d^2 \left (3 c+d x^2\right )+12 a b d \left (-15 c^2-5 c d x^2+2 d^2 x^4\right )+b^2 \left (105 c^3+35 c^2 d x^2-14 c d^2 x^4+8 d^3 x^6\right )\right )}{48 d^4 \sqrt {c+d x^2}}+\frac {c \left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c}-\sqrt {c+d x^2}}\right )}{8 d^{9/2}} \] Input:

Integrate[(x^4*(a + b*x^2)^2)/(c + d*x^2)^(3/2),x]
 

Output:

(x*(24*a^2*d^2*(3*c + d*x^2) + 12*a*b*d*(-15*c^2 - 5*c*d*x^2 + 2*d^2*x^4) 
+ b^2*(105*c^3 + 35*c^2*d*x^2 - 14*c*d^2*x^4 + 8*d^3*x^6)))/(48*d^4*Sqrt[c 
 + d*x^2]) + (c*(35*b^2*c^2 - 60*a*b*c*d + 24*a^2*d^2)*ArcTanh[(Sqrt[d]*x) 
/(Sqrt[c] - Sqrt[c + d*x^2])])/(8*d^(9/2))
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {366, 25, 363, 262, 262, 224, 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.

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

\(\Big \downarrow \) 366

\(\displaystyle \frac {x^5 (b c-a d)^2}{c d^2 \sqrt {c+d x^2}}-\frac {\int -\frac {x^4 \left (a^2 d^2+b^2 c x^2 d-5 (b c-a d)^2\right )}{\sqrt {d x^2+c}}dx}{c d^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {x^4 \left (a^2 d^2+b^2 c x^2 d-5 (b c-a d)^2\right )}{\sqrt {d x^2+c}}dx}{c d^2}+\frac {x^5 (b c-a d)^2}{c d^2 \sqrt {c+d x^2}}\)

\(\Big \downarrow \) 363

\(\displaystyle \frac {\frac {1}{6} b^2 c x^5 \sqrt {c+d x^2}-\frac {1}{6} \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right ) \int \frac {x^4}{\sqrt {d x^2+c}}dx}{c d^2}+\frac {x^5 (b c-a d)^2}{c d^2 \sqrt {c+d x^2}}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {\frac {1}{6} b^2 c x^5 \sqrt {c+d x^2}-\frac {1}{6} \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right ) \left (\frac {x^3 \sqrt {c+d x^2}}{4 d}-\frac {3 c \int \frac {x^2}{\sqrt {d x^2+c}}dx}{4 d}\right )}{c d^2}+\frac {x^5 (b c-a d)^2}{c d^2 \sqrt {c+d x^2}}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {\frac {1}{6} b^2 c x^5 \sqrt {c+d x^2}-\frac {1}{6} \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right ) \left (\frac {x^3 \sqrt {c+d x^2}}{4 d}-\frac {3 c \left (\frac {x \sqrt {c+d x^2}}{2 d}-\frac {c \int \frac {1}{\sqrt {d x^2+c}}dx}{2 d}\right )}{4 d}\right )}{c d^2}+\frac {x^5 (b c-a d)^2}{c d^2 \sqrt {c+d x^2}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{6} b^2 c x^5 \sqrt {c+d x^2}-\frac {1}{6} \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right ) \left (\frac {x^3 \sqrt {c+d x^2}}{4 d}-\frac {3 c \left (\frac {x \sqrt {c+d x^2}}{2 d}-\frac {c \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}}{2 d}\right )}{4 d}\right )}{c d^2}+\frac {x^5 (b c-a d)^2}{c d^2 \sqrt {c+d x^2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{6} b^2 c x^5 \sqrt {c+d x^2}-\frac {1}{6} \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right ) \left (\frac {x^3 \sqrt {c+d x^2}}{4 d}-\frac {3 c \left (\frac {x \sqrt {c+d x^2}}{2 d}-\frac {c \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 d^{3/2}}\right )}{4 d}\right )}{c d^2}+\frac {x^5 (b c-a d)^2}{c d^2 \sqrt {c+d x^2}}\)

Input:

Int[(x^4*(a + b*x^2)^2)/(c + d*x^2)^(3/2),x]
 

Output:

((b*c - a*d)^2*x^5)/(c*d^2*Sqrt[c + d*x^2]) + ((b^2*c*x^5*Sqrt[c + d*x^2]) 
/6 - ((35*b^2*c^2 - 60*a*b*c*d + 24*a^2*d^2)*((x^3*Sqrt[c + d*x^2])/(4*d) 
- (3*c*((x*Sqrt[c + d*x^2])/(2*d) - (c*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2] 
])/(2*d^(3/2))))/(4*d)))/6)/(c*d^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 224

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

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 363

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), 
 x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3))   Int[(e*x)^ 
m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d 
, 0] && NeQ[m + 2*p + 3, 0]
 

rule 366

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

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.82

method	result	size
pseudoelliptic	\(\frac {-\frac {3 c \sqrt {x^{2} d +c}\, \left (a^{2} d^{2}-\frac {5}{2} a b c d +\frac {35}{24} b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}}{x \sqrt {d}}\right )}{2}+\frac {3 \left (c \left (-\frac {7}{36} b^{2} x^{4}-\frac {5}{6} a b \,x^{2}+a^{2}\right ) d^{\frac {5}{2}}+\frac {\left (\frac {1}{3} b^{2} x^{4}+a b \,x^{2}+a^{2}\right ) x^{2} d^{\frac {7}{2}}}{3}-\frac {5 \left (\left (-\frac {7 b \,x^{2}}{36}+a \right ) d^{\frac {3}{2}}-\frac {7 b c \sqrt {d}}{12}\right ) c^{2} b}{2}\right ) x}{2}}{d^{\frac {9}{2}} \sqrt {x^{2} d +c}}\)	\(146\)
risch	\(\frac {x \left (8 b^{2} d^{2} x^{4}+24 x^{2} a b \,d^{2}-22 x^{2} b^{2} c d +24 a^{2} d^{2}-84 a b c d +57 b^{2} c^{2}\right ) \sqrt {x^{2} d +c}}{48 d^{4}}-\frac {c \left (d \left (24 a^{2} d^{2}-60 a b c d +35 b^{2} c^{2}\right ) \left (-\frac {x}{d \sqrt {x^{2} d +c}}+\frac {\ln \left (\sqrt {d}\, x +\sqrt {x^{2} d +c}\right )}{d^{\frac {3}{2}}}\right )+\frac {19 b^{2} c^{2} x}{\sqrt {x^{2} d +c}}+\frac {8 x \,a^{2} d^{2}}{\sqrt {x^{2} d +c}}-\frac {28 a b c d x}{\sqrt {x^{2} d +c}}\right )}{16 d^{4}}\)	\(191\)
default	\(a^{2} \left (\frac {x^{3}}{2 d \sqrt {x^{2} d +c}}-\frac {3 c \left (-\frac {x}{d \sqrt {x^{2} d +c}}+\frac {\ln \left (\sqrt {d}\, x +\sqrt {x^{2} d +c}\right )}{d^{\frac {3}{2}}}\right )}{2 d}\right )+b^{2} \left (\frac {x^{7}}{6 d \sqrt {x^{2} d +c}}-\frac {7 c \left (\frac {x^{5}}{4 d \sqrt {x^{2} d +c}}-\frac {5 c \left (\frac {x^{3}}{2 d \sqrt {x^{2} d +c}}-\frac {3 c \left (-\frac {x}{d \sqrt {x^{2} d +c}}+\frac {\ln \left (\sqrt {d}\, x +\sqrt {x^{2} d +c}\right )}{d^{\frac {3}{2}}}\right )}{2 d}\right )}{4 d}\right )}{6 d}\right )+2 a b \left (\frac {x^{5}}{4 d \sqrt {x^{2} d +c}}-\frac {5 c \left (\frac {x^{3}}{2 d \sqrt {x^{2} d +c}}-\frac {3 c \left (-\frac {x}{d \sqrt {x^{2} d +c}}+\frac {\ln \left (\sqrt {d}\, x +\sqrt {x^{2} d +c}\right )}{d^{\frac {3}{2}}}\right )}{2 d}\right )}{4 d}\right )\)	\(266\)

Input:

int(x^4*(b*x^2+a)^2/(d*x^2+c)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

3/2/(d*x^2+c)^(1/2)*(-c*(d*x^2+c)^(1/2)*(a^2*d^2-5/2*a*b*c*d+35/24*b^2*c^2 
)*arctanh((d*x^2+c)^(1/2)/x/d^(1/2))+(c*(-7/36*b^2*x^4-5/6*a*b*x^2+a^2)*d^ 
(5/2)+1/3*(1/3*b^2*x^4+a*b*x^2+a^2)*x^2*d^(7/2)-5/2*((-7/36*b*x^2+a)*d^(3/ 
2)-7/12*b*c*d^(1/2))*c^2*b)*x)/d^(9/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 431, normalized size of antiderivative = 2.44 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (35 \, b^{2} c^{4} - 60 \, a b c^{3} d + 24 \, a^{2} c^{2} d^{2} + {\left (35 \, b^{2} c^{3} d - 60 \, a b c^{2} d^{2} + 24 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (8 \, b^{2} d^{4} x^{7} - 2 \, {\left (7 \, b^{2} c d^{3} - 12 \, a b d^{4}\right )} x^{5} + {\left (35 \, b^{2} c^{2} d^{2} - 60 \, a b c d^{3} + 24 \, a^{2} d^{4}\right )} x^{3} + 3 \, {\left (35 \, b^{2} c^{3} d - 60 \, a b c^{2} d^{2} + 24 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{96 \, {\left (d^{6} x^{2} + c d^{5}\right )}}, \frac {3 \, {\left (35 \, b^{2} c^{4} - 60 \, a b c^{3} d + 24 \, a^{2} c^{2} d^{2} + {\left (35 \, b^{2} c^{3} d - 60 \, a b c^{2} d^{2} + 24 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (8 \, b^{2} d^{4} x^{7} - 2 \, {\left (7 \, b^{2} c d^{3} - 12 \, a b d^{4}\right )} x^{5} + {\left (35 \, b^{2} c^{2} d^{2} - 60 \, a b c d^{3} + 24 \, a^{2} d^{4}\right )} x^{3} + 3 \, {\left (35 \, b^{2} c^{3} d - 60 \, a b c^{2} d^{2} + 24 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{48 \, {\left (d^{6} x^{2} + c d^{5}\right )}}\right ] \] Input:

integrate(x^4*(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="fricas")
 

Output:

[1/96*(3*(35*b^2*c^4 - 60*a*b*c^3*d + 24*a^2*c^2*d^2 + (35*b^2*c^3*d - 60* 
a*b*c^2*d^2 + 24*a^2*c*d^3)*x^2)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)* 
sqrt(d)*x - c) + 2*(8*b^2*d^4*x^7 - 2*(7*b^2*c*d^3 - 12*a*b*d^4)*x^5 + (35 
*b^2*c^2*d^2 - 60*a*b*c*d^3 + 24*a^2*d^4)*x^3 + 3*(35*b^2*c^3*d - 60*a*b*c 
^2*d^2 + 24*a^2*c*d^3)*x)*sqrt(d*x^2 + c))/(d^6*x^2 + c*d^5), 1/48*(3*(35* 
b^2*c^4 - 60*a*b*c^3*d + 24*a^2*c^2*d^2 + (35*b^2*c^3*d - 60*a*b*c^2*d^2 + 
 24*a^2*c*d^3)*x^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (8*b^2*d 
^4*x^7 - 2*(7*b^2*c*d^3 - 12*a*b*d^4)*x^5 + (35*b^2*c^2*d^2 - 60*a*b*c*d^3 
 + 24*a^2*d^4)*x^3 + 3*(35*b^2*c^3*d - 60*a*b*c^2*d^2 + 24*a^2*c*d^3)*x)*s 
qrt(d*x^2 + c))/(d^6*x^2 + c*d^5)]
 

Sympy [F]

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

integrate(x**4*(b*x**2+a)**2/(d*x**2+c)**(3/2),x)
 

Output:

Integral(x**4*(a + b*x**2)**2/(c + d*x**2)**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.36 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\frac {b^{2} x^{7}}{6 \, \sqrt {d x^{2} + c} d} - \frac {7 \, b^{2} c x^{5}}{24 \, \sqrt {d x^{2} + c} d^{2}} + \frac {a b x^{5}}{2 \, \sqrt {d x^{2} + c} d} + \frac {35 \, b^{2} c^{2} x^{3}}{48 \, \sqrt {d x^{2} + c} d^{3}} - \frac {5 \, a b c x^{3}}{4 \, \sqrt {d x^{2} + c} d^{2}} + \frac {a^{2} x^{3}}{2 \, \sqrt {d x^{2} + c} d} + \frac {35 \, b^{2} c^{3} x}{16 \, \sqrt {d x^{2} + c} d^{4}} - \frac {15 \, a b c^{2} x}{4 \, \sqrt {d x^{2} + c} d^{3}} + \frac {3 \, a^{2} c x}{2 \, \sqrt {d x^{2} + c} d^{2}} - \frac {35 \, b^{2} c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{16 \, d^{\frac {9}{2}}} + \frac {15 \, a b c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{4 \, d^{\frac {7}{2}}} - \frac {3 \, a^{2} c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{2 \, d^{\frac {5}{2}}} \] Input:

integrate(x^4*(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="maxima")
 

Output:

1/6*b^2*x^7/(sqrt(d*x^2 + c)*d) - 7/24*b^2*c*x^5/(sqrt(d*x^2 + c)*d^2) + 1 
/2*a*b*x^5/(sqrt(d*x^2 + c)*d) + 35/48*b^2*c^2*x^3/(sqrt(d*x^2 + c)*d^3) - 
 5/4*a*b*c*x^3/(sqrt(d*x^2 + c)*d^2) + 1/2*a^2*x^3/(sqrt(d*x^2 + c)*d) + 3 
5/16*b^2*c^3*x/(sqrt(d*x^2 + c)*d^4) - 15/4*a*b*c^2*x/(sqrt(d*x^2 + c)*d^3 
) + 3/2*a^2*c*x/(sqrt(d*x^2 + c)*d^2) - 35/16*b^2*c^3*arcsinh(d*x/sqrt(c*d 
))/d^(9/2) + 15/4*a*b*c^2*arcsinh(d*x/sqrt(c*d))/d^(7/2) - 3/2*a^2*c*arcsi 
nh(d*x/sqrt(c*d))/d^(5/2)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.99 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\frac {{\left ({\left (2 \, {\left (\frac {4 \, b^{2} x^{2}}{d} - \frac {7 \, b^{2} c d^{5} - 12 \, a b d^{6}}{d^{7}}\right )} x^{2} + \frac {35 \, b^{2} c^{2} d^{4} - 60 \, a b c d^{5} + 24 \, a^{2} d^{6}}{d^{7}}\right )} x^{2} + \frac {3 \, {\left (35 \, b^{2} c^{3} d^{3} - 60 \, a b c^{2} d^{4} + 24 \, a^{2} c d^{5}\right )}}{d^{7}}\right )} x}{48 \, \sqrt {d x^{2} + c}} + \frac {{\left (35 \, b^{2} c^{3} - 60 \, a b c^{2} d + 24 \, a^{2} c d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{16 \, d^{\frac {9}{2}}} \] Input:

integrate(x^4*(b*x^2+a)^2/(d*x^2+c)^(3/2),x, algorithm="giac")
 

Output:

1/48*((2*(4*b^2*x^2/d - (7*b^2*c*d^5 - 12*a*b*d^6)/d^7)*x^2 + (35*b^2*c^2* 
d^4 - 60*a*b*c*d^5 + 24*a^2*d^6)/d^7)*x^2 + 3*(35*b^2*c^3*d^3 - 60*a*b*c^2 
*d^4 + 24*a^2*c*d^5)/d^7)*x/sqrt(d*x^2 + c) + 1/16*(35*b^2*c^3 - 60*a*b*c^ 
2*d + 24*a^2*c*d^2)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(9/2)
 

Mupad [F(-1)]

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

int((x^4*(a + b*x^2)^2)/(c + d*x^2)^(3/2),x)
 

Output:

int((x^4*(a + b*x^2)^2)/(c + d*x^2)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 453, normalized size of antiderivative = 2.56 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\frac {576 \sqrt {d \,x^{2}+c}\, a^{2} c \,d^{3} x +192 \sqrt {d \,x^{2}+c}\, a^{2} d^{4} x^{3}-1440 \sqrt {d \,x^{2}+c}\, a b \,c^{2} d^{2} x -480 \sqrt {d \,x^{2}+c}\, a b c \,d^{3} x^{3}+192 \sqrt {d \,x^{2}+c}\, a b \,d^{4} x^{5}+840 \sqrt {d \,x^{2}+c}\, b^{2} c^{3} d x +280 \sqrt {d \,x^{2}+c}\, b^{2} c^{2} d^{2} x^{3}-112 \sqrt {d \,x^{2}+c}\, b^{2} c \,d^{3} x^{5}+64 \sqrt {d \,x^{2}+c}\, b^{2} d^{4} x^{7}-576 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a^{2} c^{2} d^{2}-576 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a^{2} c \,d^{3} x^{2}+1440 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a b \,c^{3} d +1440 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a b \,c^{2} d^{2} x^{2}-840 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {d}\, x}{\sqrt {c}}\right ) b^{2} c^{4}-840 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {d}\, x}{\sqrt {c}}\right ) b^{2} c^{3} d \,x^{2}+432 \sqrt {d}\, a^{2} c^{2} d^{2}+432 \sqrt {d}\, a^{2} c \,d^{3} x^{2}-960 \sqrt {d}\, a b \,c^{3} d -960 \sqrt {d}\, a b \,c^{2} d^{2} x^{2}+525 \sqrt {d}\, b^{2} c^{4}+525 \sqrt {d}\, b^{2} c^{3} d \,x^{2}}{384 d^{5} \left (d \,x^{2}+c \right )} \] Input:

int(x^4*(b*x^2+a)^2/(d*x^2+c)^(3/2),x)
 

Output:

(576*sqrt(c + d*x**2)*a**2*c*d**3*x + 192*sqrt(c + d*x**2)*a**2*d**4*x**3 
- 1440*sqrt(c + d*x**2)*a*b*c**2*d**2*x - 480*sqrt(c + d*x**2)*a*b*c*d**3* 
x**3 + 192*sqrt(c + d*x**2)*a*b*d**4*x**5 + 840*sqrt(c + d*x**2)*b**2*c**3 
*d*x + 280*sqrt(c + d*x**2)*b**2*c**2*d**2*x**3 - 112*sqrt(c + d*x**2)*b** 
2*c*d**3*x**5 + 64*sqrt(c + d*x**2)*b**2*d**4*x**7 - 576*sqrt(d)*log((sqrt 
(c + d*x**2) + sqrt(d)*x)/sqrt(c))*a**2*c**2*d**2 - 576*sqrt(d)*log((sqrt( 
c + d*x**2) + sqrt(d)*x)/sqrt(c))*a**2*c*d**3*x**2 + 1440*sqrt(d)*log((sqr 
t(c + d*x**2) + sqrt(d)*x)/sqrt(c))*a*b*c**3*d + 1440*sqrt(d)*log((sqrt(c 
+ d*x**2) + sqrt(d)*x)/sqrt(c))*a*b*c**2*d**2*x**2 - 840*sqrt(d)*log((sqrt 
(c + d*x**2) + sqrt(d)*x)/sqrt(c))*b**2*c**4 - 840*sqrt(d)*log((sqrt(c + d 
*x**2) + sqrt(d)*x)/sqrt(c))*b**2*c**3*d*x**2 + 432*sqrt(d)*a**2*c**2*d**2 
 + 432*sqrt(d)*a**2*c*d**3*x**2 - 960*sqrt(d)*a*b*c**3*d - 960*sqrt(d)*a*b 
*c**2*d**2*x**2 + 525*sqrt(d)*b**2*c**4 + 525*sqrt(d)*b**2*c**3*d*x**2)/(3 
84*d**5*(c + d*x**2))