\(\int \frac {x^4 (a+b x^2)^2}{(c+d x^2)^{3/2}} \, dx\) [648]

Optimal result

Integrand size = 24, antiderivative size = 197 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\frac {(b c-a d)^2 x^5}{c d^2 \sqrt {c+d x^2}}+\frac {\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 d^4}-\frac {\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x^3 \sqrt {c+d x^2}}{24 c d^3}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d^2}-\frac {c \left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{9/2}} \]

[Out]

-1/16*c*(24*a^2*d^2-60*a*b*c*d+35*b^2*c^2)*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))/d^(9/2)+(-a*d+b*c)^2*x^5/c/d^2/(
d*x^2+c)^(1/2)+1/16*(24*a^2*d^2-60*a*b*c*d+35*b^2*c^2)*x*(d*x^2+c)^(1/2)/d^4-1/24*(24*a^2*d^2-60*a*b*c*d+35*b^
2*c^2)*x^3*(d*x^2+c)^(1/2)/c/d^3+1/6*b^2*x^5*(d*x^2+c)^(1/2)/d^2

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {474, 470, 327, 223, 212} \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=-\frac {c \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{9/2}}+\frac {x \sqrt {c+d x^2} \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right )}{16 d^4}-\frac {x^3 \sqrt {c+d x^2} \left (24 a^2 d^2-60 a b c d+35 b^2 c^2\right )}{24 c d^3}+\frac {x^5 (b c-a d)^2}{c d^2 \sqrt {c+d x^2}}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d^2} \]

[In]

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

[Out]

((b*c - a*d)^2*x^5)/(c*d^2*Sqrt[c + d*x^2]) + ((35*b^2*c^2 - 60*a*b*c*d + 24*a^2*d^2)*x*Sqrt[c + d*x^2])/(16*d
^4) - ((35*b^2*c^2 - 60*a*b*c*d + 24*a^2*d^2)*x^3*Sqrt[c + d*x^2])/(24*c*d^3) + (b^2*x^5*Sqrt[c + d*x^2])/(6*d
^2) - (c*(35*b^2*c^2 - 60*a*b*c*d + 24*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(16*d^(9/2))

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 327

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

Rule 470

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

Rule 474

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

Rubi steps \begin{align*} \text {integral}& = \frac {(b c-a d)^2 x^5}{c d^2 \sqrt {c+d x^2}}-\frac {\int \frac {x^4 \left (-a^2 d^2+5 (b c-a d)^2-b^2 c d x^2\right )}{\sqrt {c+d x^2}} \, dx}{c d^2} \\ & = \frac {(b c-a d)^2 x^5}{c d^2 \sqrt {c+d x^2}}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d^2}-\frac {\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) \int \frac {x^4}{\sqrt {c+d x^2}} \, dx}{6 c d^2} \\ & = \frac {(b c-a d)^2 x^5}{c d^2 \sqrt {c+d x^2}}-\frac {\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x^3 \sqrt {c+d x^2}}{24 c d^3}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d^2}+\frac {\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) \int \frac {x^2}{\sqrt {c+d x^2}} \, dx}{8 d^3} \\ & = \frac {(b c-a d)^2 x^5}{c d^2 \sqrt {c+d x^2}}+\frac {\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 d^4}-\frac {\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x^3 \sqrt {c+d x^2}}{24 c d^3}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d^2}-\frac {\left (c \left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{16 d^4} \\ & = \frac {(b c-a d)^2 x^5}{c d^2 \sqrt {c+d x^2}}+\frac {\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 d^4}-\frac {\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x^3 \sqrt {c+d x^2}}{24 c d^3}+\frac {b^2 x^5 \sqrt {c+d x^2}}{6 d^2}-\frac {\left (c \left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{16 d^4} \\ & = \frac {(b c-a d)^2 x^5}{c d^2 \sqrt {c+d x^2}}+\frac {\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 d^4}-\frac {\left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) x^3 \sqrt {c+d x^2}}{24 c 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 ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 d^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.82 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\frac {x \left (24 a^2 d^2 \left (3 c+d x^2\right )+12 a b d \left (-15 c^2-5 c d x^2+2 d^2 x^4\right )+b^2 \left (105 c^3+35 c^2 d x^2-14 c d^2 x^4+8 d^3 x^6\right )\right )}{48 d^4 \sqrt {c+d x^2}}+\frac {c \left (35 b^2 c^2-60 a b c d+24 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c}-\sqrt {c+d x^2}}\right )}{8 d^{9/2}} \]

[In]

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

[Out]

(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))

Maple [A] (verified)

Time = 3.01 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.74

method	result	size
pseudoelliptic	\(\frac {-\frac {3 \left (a^{2} d^{2}-\frac {5}{2} a b c d +\frac {35}{24} b^{2} c^{2}\right ) c \sqrt {d \,x^{2}+c}\, \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )}{2}+\frac {3 \left (c \left (-\frac {7}{36} b^{2} x^{4}-\frac {5}{6} a b \,x^{2}+a^{2}\right ) d^{\frac {5}{2}}+\frac {x^{2} \left (\frac {1}{3} b^{2} x^{4}+a b \,x^{2}+a^{2}\right ) d^{\frac {7}{2}}}{3}-\frac {5 \left (\left (-\frac {7 b \,x^{2}}{36}+a \right ) d^{\frac {3}{2}}-\frac {7 b \sqrt {d}\, c}{12}\right ) b \,c^{2}}{2}\right ) x}{2}}{d^{\frac {9}{2}} \sqrt {d \,x^{2}+c}}\)	\(146\)
risch	\(\frac {x \left (8 b^{2} d^{2} x^{4}+24 x^{2} a b \,d^{2}-22 x^{2} b^{2} c d +24 a^{2} d^{2}-84 a b c d +57 b^{2} c^{2}\right ) \sqrt {d \,x^{2}+c}}{48 d^{4}}-\frac {c \left (\frac {19 b^{2} c^{2} x}{\sqrt {d \,x^{2}+c}}+\frac {8 a^{2} d^{2} x}{\sqrt {d \,x^{2}+c}}-\frac {28 a b c d x}{\sqrt {d \,x^{2}+c}}+\left (24 a^{2} d^{3}-60 a b c \,d^{2}+35 b^{2} c^{2} d \right ) \left (-\frac {x}{d \sqrt {d \,x^{2}+c}}+\frac {\ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{d^{\frac {3}{2}}}\right )\right )}{16 d^{4}}\)	\(193\)
default	\(b^{2} \left (\frac {x^{7}}{6 d \sqrt {d \,x^{2}+c}}-\frac {7 c \left (\frac {x^{5}}{4 d \sqrt {d \,x^{2}+c}}-\frac {5 c \left (\frac {x^{3}}{2 d \sqrt {d \,x^{2}+c}}-\frac {3 c \left (-\frac {x}{d \sqrt {d \,x^{2}+c}}+\frac {\ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{d^{\frac {3}{2}}}\right )}{2 d}\right )}{4 d}\right )}{6 d}\right )+a^{2} \left (\frac {x^{3}}{2 d \sqrt {d \,x^{2}+c}}-\frac {3 c \left (-\frac {x}{d \sqrt {d \,x^{2}+c}}+\frac {\ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{d^{\frac {3}{2}}}\right )}{2 d}\right )+2 a b \left (\frac {x^{5}}{4 d \sqrt {d \,x^{2}+c}}-\frac {5 c \left (\frac {x^{3}}{2 d \sqrt {d \,x^{2}+c}}-\frac {3 c \left (-\frac {x}{d \sqrt {d \,x^{2}+c}}+\frac {\ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{d^{\frac {3}{2}}}\right )}{2 d}\right )}{4 d}\right )\)	\(266\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 431, normalized size of antiderivative = 2.19 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (35 \, b^{2} c^{4} - 60 \, a b c^{3} d + 24 \, a^{2} c^{2} d^{2} + {\left (35 \, b^{2} c^{3} d - 60 \, a b c^{2} d^{2} + 24 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (8 \, b^{2} d^{4} x^{7} - 2 \, {\left (7 \, b^{2} c d^{3} - 12 \, a b d^{4}\right )} x^{5} + {\left (35 \, b^{2} c^{2} d^{2} - 60 \, a b c d^{3} + 24 \, a^{2} d^{4}\right )} x^{3} + 3 \, {\left (35 \, b^{2} c^{3} d - 60 \, a b c^{2} d^{2} + 24 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{96 \, {\left (d^{6} x^{2} + c d^{5}\right )}}, \frac {3 \, {\left (35 \, b^{2} c^{4} - 60 \, a b c^{3} d + 24 \, a^{2} c^{2} d^{2} + {\left (35 \, b^{2} c^{3} d - 60 \, a b c^{2} d^{2} + 24 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (8 \, b^{2} d^{4} x^{7} - 2 \, {\left (7 \, b^{2} c d^{3} - 12 \, a b d^{4}\right )} x^{5} + {\left (35 \, b^{2} c^{2} d^{2} - 60 \, a b c d^{3} + 24 \, a^{2} d^{4}\right )} x^{3} + 3 \, {\left (35 \, b^{2} c^{3} d - 60 \, a b c^{2} d^{2} + 24 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{48 \, {\left (d^{6} x^{2} + c d^{5}\right )}}\right ] \]

[In]

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

[Out]

[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)*sqr
t(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)*sqrt(d*x^2 + c))/(d^6*x^2 + c*d^5)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.22 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\frac {b^{2} x^{7}}{6 \, \sqrt {d x^{2} + c} d} - \frac {7 \, b^{2} c x^{5}}{24 \, \sqrt {d x^{2} + c} d^{2}} + \frac {a b x^{5}}{2 \, \sqrt {d x^{2} + c} d} + \frac {35 \, b^{2} c^{2} x^{3}}{48 \, \sqrt {d x^{2} + c} d^{3}} - \frac {5 \, a b c x^{3}}{4 \, \sqrt {d x^{2} + c} d^{2}} + \frac {a^{2} x^{3}}{2 \, \sqrt {d x^{2} + c} d} + \frac {35 \, b^{2} c^{3} x}{16 \, \sqrt {d x^{2} + c} d^{4}} - \frac {15 \, a b c^{2} x}{4 \, \sqrt {d x^{2} + c} d^{3}} + \frac {3 \, a^{2} c x}{2 \, \sqrt {d x^{2} + c} d^{2}} - \frac {35 \, b^{2} c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{16 \, d^{\frac {9}{2}}} + \frac {15 \, a b c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{4 \, d^{\frac {7}{2}}} - \frac {3 \, a^{2} c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{2 \, d^{\frac {5}{2}}} \]

[In]

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

[Out]

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) +
35/16*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*a
rcsinh(d*x/sqrt(c*d))/d^(5/2)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.89 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx=\frac {{\left ({\left (2 \, {\left (\frac {4 \, b^{2} x^{2}}{d} - \frac {7 \, b^{2} c d^{5} - 12 \, a b d^{6}}{d^{7}}\right )} x^{2} + \frac {35 \, b^{2} c^{2} d^{4} - 60 \, a b c d^{5} + 24 \, a^{2} d^{6}}{d^{7}}\right )} x^{2} + \frac {3 \, {\left (35 \, b^{2} c^{3} d^{3} - 60 \, a b c^{2} d^{4} + 24 \, a^{2} c d^{5}\right )}}{d^{7}}\right )} x}{48 \, \sqrt {d x^{2} + c}} + \frac {{\left (35 \, b^{2} c^{3} - 60 \, a b c^{2} d + 24 \, a^{2} c d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{16 \, d^{\frac {9}{2}}} \]

[In]

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

[Out]

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

[In]

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

[Out]

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