3.660 \(\int \frac {x^4 (a+b x^2)^2}{(c+d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=202 \[ \frac {\left (8 a^2 d^2-40 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{9/2}}-\frac {x \sqrt {c+d x^2} \left (8 a^2 d^2-40 a b c d+35 b^2 c^2\right )}{8 c d^4}+\frac {x^3 \left (8 a^2 d^2-40 a b c d+35 b^2 c^2\right )}{12 c d^3 \sqrt {c+d x^2}}+\frac {x^5 (b c-a d)^2}{3 c d^2 \left (c+d x^2\right )^{3/2}}+\frac {b^2 x^5}{4 d^2 \sqrt {c+d x^2}} \]

[Out]

1/3*(-a*d+b*c)^2*x^5/c/d^2/(d*x^2+c)^(3/2)+1/8*(8*a^2*d^2-40*a*b*c*d+35*b^2*c^2)*arctanh(x*d^(1/2)/(d*x^2+c)^(
1/2))/d^(9/2)+1/12*(8*a^2*d^2-40*a*b*c*d+35*b^2*c^2)*x^3/c/d^3/(d*x^2+c)^(1/2)+1/4*b^2*x^5/d^2/(d*x^2+c)^(1/2)
-1/8*(8*a^2*d^2-40*a*b*c*d+35*b^2*c^2)*x*(d*x^2+c)^(1/2)/c/d^4

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Rubi [A]  time = 0.15, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {463, 459, 288, 321, 217, 206} \[ \frac {x^3 \left (8 a^2 d^2-40 a b c d+35 b^2 c^2\right )}{12 c d^3 \sqrt {c+d x^2}}-\frac {x \sqrt {c+d x^2} \left (8 a^2 d^2-40 a b c d+35 b^2 c^2\right )}{8 c d^4}+\frac {\left (8 a^2 d^2-40 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{9/2}}+\frac {x^5 (b c-a d)^2}{3 c d^2 \left (c+d x^2\right )^{3/2}}+\frac {b^2 x^5}{4 d^2 \sqrt {c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*x^2)^2)/(c + d*x^2)^(5/2),x]

[Out]

((b*c - a*d)^2*x^5)/(3*c*d^2*(c + d*x^2)^(3/2)) + ((35*b^2*c^2 - 40*a*b*c*d + 8*a^2*d^2)*x^3)/(12*c*d^3*Sqrt[c
+ d*x^2]) + (b^2*x^5)/(4*d^2*Sqrt[c + d*x^2]) - ((35*b^2*c^2 - 40*a*b*c*d + 8*a^2*d^2)*x*Sqrt[c + d*x^2])/(8*
c*d^4) + ((35*b^2*c^2 - 40*a*b*c*d + 8*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(8*d^(9/2))

Rule 206

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

Rule 217

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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

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 459

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 463

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*} \int \frac {x^4 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{5/2}} \, dx &=\frac {(b c-a d)^2 x^5}{3 c d^2 \left (c+d x^2\right )^{3/2}}-\frac {\int \frac {x^4 \left (-3 a^2 d^2+5 (b c-a d)^2-3 b^2 c d x^2\right )}{\left (c+d x^2\right )^{3/2}} \, dx}{3 c d^2}\\ &=\frac {(b c-a d)^2 x^5}{3 c d^2 \left (c+d x^2\right )^{3/2}}+\frac {b^2 x^5}{4 d^2 \sqrt {c+d x^2}}-\frac {\left (35 b^2 c^2-40 a b c d+8 a^2 d^2\right ) \int \frac {x^4}{\left (c+d x^2\right )^{3/2}} \, dx}{12 c d^2}\\ &=\frac {(b c-a d)^2 x^5}{3 c d^2 \left (c+d x^2\right )^{3/2}}+\frac {\left (35 b^2 c^2-40 a b c d+8 a^2 d^2\right ) x^3}{12 c d^3 \sqrt {c+d x^2}}+\frac {b^2 x^5}{4 d^2 \sqrt {c+d x^2}}-\frac {\left (35 b^2 c^2-40 a b c d+8 a^2 d^2\right ) \int \frac {x^2}{\sqrt {c+d x^2}} \, dx}{4 c d^3}\\ &=\frac {(b c-a d)^2 x^5}{3 c d^2 \left (c+d x^2\right )^{3/2}}+\frac {\left (35 b^2 c^2-40 a b c d+8 a^2 d^2\right ) x^3}{12 c d^3 \sqrt {c+d x^2}}+\frac {b^2 x^5}{4 d^2 \sqrt {c+d x^2}}-\frac {\left (35 b^2 c^2-40 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{8 c d^4}+\frac {\left (35 b^2 c^2-40 a b c d+8 a^2 d^2\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{8 d^4}\\ &=\frac {(b c-a d)^2 x^5}{3 c d^2 \left (c+d x^2\right )^{3/2}}+\frac {\left (35 b^2 c^2-40 a b c d+8 a^2 d^2\right ) x^3}{12 c d^3 \sqrt {c+d x^2}}+\frac {b^2 x^5}{4 d^2 \sqrt {c+d x^2}}-\frac {\left (35 b^2 c^2-40 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{8 c d^4}+\frac {\left (35 b^2 c^2-40 a b c d+8 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{8 d^4}\\ &=\frac {(b c-a d)^2 x^5}{3 c d^2 \left (c+d x^2\right )^{3/2}}+\frac {\left (35 b^2 c^2-40 a b c d+8 a^2 d^2\right ) x^3}{12 c d^3 \sqrt {c+d x^2}}+\frac {b^2 x^5}{4 d^2 \sqrt {c+d x^2}}-\frac {\left (35 b^2 c^2-40 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{8 c d^4}+\frac {\left (35 b^2 c^2-40 a b c d+8 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 156, normalized size = 0.77 \[ \frac {\left (8 a^2 d^2-40 a b c d+35 b^2 c^2\right ) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{8 d^{9/2}}+\frac {x \left (-8 a^2 d^2 \left (3 c+4 d x^2\right )+8 a b d \left (15 c^2+20 c d x^2+3 d^2 x^4\right )-\left (b^2 \left (105 c^3+140 c^2 d x^2+21 c d^2 x^4-6 d^3 x^6\right )\right )\right )}{24 d^4 \left (c+d x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*x^2)^2)/(c + d*x^2)^(5/2),x]

[Out]

(x*(-8*a^2*d^2*(3*c + 4*d*x^2) + 8*a*b*d*(15*c^2 + 20*c*d*x^2 + 3*d^2*x^4) - b^2*(105*c^3 + 140*c^2*d*x^2 + 21
*c*d^2*x^4 - 6*d^3*x^6)))/(24*d^4*(c + d*x^2)^(3/2)) + ((35*b^2*c^2 - 40*a*b*c*d + 8*a^2*d^2)*Log[d*x + Sqrt[d
]*Sqrt[c + d*x^2]])/(8*d^(9/2))

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fricas [A]  time = 0.97, size = 522, normalized size = 2.58 \[ \left [\frac {3 \, {\left (35 \, b^{2} c^{4} - 40 \, a b c^{3} d + 8 \, a^{2} c^{2} d^{2} + {\left (35 \, b^{2} c^{2} d^{2} - 40 \, a b c d^{3} + 8 \, a^{2} d^{4}\right )} x^{4} + 2 \, {\left (35 \, b^{2} c^{3} d - 40 \, a b c^{2} d^{2} + 8 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (6 \, b^{2} d^{4} x^{7} - 3 \, {\left (7 \, b^{2} c d^{3} - 8 \, a b d^{4}\right )} x^{5} - 4 \, {\left (35 \, b^{2} c^{2} d^{2} - 40 \, a b c d^{3} + 8 \, a^{2} d^{4}\right )} x^{3} - 3 \, {\left (35 \, b^{2} c^{3} d - 40 \, a b c^{2} d^{2} + 8 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{48 \, {\left (d^{7} x^{4} + 2 \, c d^{6} x^{2} + c^{2} d^{5}\right )}}, -\frac {3 \, {\left (35 \, b^{2} c^{4} - 40 \, a b c^{3} d + 8 \, a^{2} c^{2} d^{2} + {\left (35 \, b^{2} c^{2} d^{2} - 40 \, a b c d^{3} + 8 \, a^{2} d^{4}\right )} x^{4} + 2 \, {\left (35 \, b^{2} c^{3} d - 40 \, a b c^{2} d^{2} + 8 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (6 \, b^{2} d^{4} x^{7} - 3 \, {\left (7 \, b^{2} c d^{3} - 8 \, a b d^{4}\right )} x^{5} - 4 \, {\left (35 \, b^{2} c^{2} d^{2} - 40 \, a b c d^{3} + 8 \, a^{2} d^{4}\right )} x^{3} - 3 \, {\left (35 \, b^{2} c^{3} d - 40 \, a b c^{2} d^{2} + 8 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{24 \, {\left (d^{7} x^{4} + 2 \, c d^{6} x^{2} + c^{2} d^{5}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

[1/48*(3*(35*b^2*c^4 - 40*a*b*c^3*d + 8*a^2*c^2*d^2 + (35*b^2*c^2*d^2 - 40*a*b*c*d^3 + 8*a^2*d^4)*x^4 + 2*(35*
b^2*c^3*d - 40*a*b*c^2*d^2 + 8*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*(6*
b^2*d^4*x^7 - 3*(7*b^2*c*d^3 - 8*a*b*d^4)*x^5 - 4*(35*b^2*c^2*d^2 - 40*a*b*c*d^3 + 8*a^2*d^4)*x^3 - 3*(35*b^2*
c^3*d - 40*a*b*c^2*d^2 + 8*a^2*c*d^3)*x)*sqrt(d*x^2 + c))/(d^7*x^4 + 2*c*d^6*x^2 + c^2*d^5), -1/24*(3*(35*b^2*
c^4 - 40*a*b*c^3*d + 8*a^2*c^2*d^2 + (35*b^2*c^2*d^2 - 40*a*b*c*d^3 + 8*a^2*d^4)*x^4 + 2*(35*b^2*c^3*d - 40*a*
b*c^2*d^2 + 8*a^2*c*d^3)*x^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (6*b^2*d^4*x^7 - 3*(7*b^2*c*d^3 -
8*a*b*d^4)*x^5 - 4*(35*b^2*c^2*d^2 - 40*a*b*c*d^3 + 8*a^2*d^4)*x^3 - 3*(35*b^2*c^3*d - 40*a*b*c^2*d^2 + 8*a^2*
c*d^3)*x)*sqrt(d*x^2 + c))/(d^7*x^4 + 2*c*d^6*x^2 + c^2*d^5)]

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giac [A]  time = 0.51, size = 190, normalized size = 0.94 \[ \frac {{\left ({\left (3 \, {\left (\frac {2 \, b^{2} x^{2}}{d} - \frac {7 \, b^{2} c^{2} d^{5} - 8 \, a b c d^{6}}{c d^{7}}\right )} x^{2} - \frac {4 \, {\left (35 \, b^{2} c^{3} d^{4} - 40 \, a b c^{2} d^{5} + 8 \, a^{2} c d^{6}\right )}}{c d^{7}}\right )} x^{2} - \frac {3 \, {\left (35 \, b^{2} c^{4} d^{3} - 40 \, a b c^{3} d^{4} + 8 \, a^{2} c^{2} d^{5}\right )}}{c d^{7}}\right )} x}{24 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} - \frac {{\left (35 \, b^{2} c^{2} - 40 \, a b c d + 8 \, a^{2} d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{8 \, d^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

1/24*((3*(2*b^2*x^2/d - (7*b^2*c^2*d^5 - 8*a*b*c*d^6)/(c*d^7))*x^2 - 4*(35*b^2*c^3*d^4 - 40*a*b*c^2*d^5 + 8*a^
2*c*d^6)/(c*d^7))*x^2 - 3*(35*b^2*c^4*d^3 - 40*a*b*c^3*d^4 + 8*a^2*c^2*d^5)/(c*d^7))*x/(d*x^2 + c)^(3/2) - 1/8
*(35*b^2*c^2 - 40*a*b*c*d + 8*a^2*d^2)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(9/2)

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maple [A]  time = 0.02, size = 255, normalized size = 1.26 \[ \frac {b^{2} x^{7}}{4 \left (d \,x^{2}+c \right )^{\frac {3}{2}} d}+\frac {a b \,x^{5}}{\left (d \,x^{2}+c \right )^{\frac {3}{2}} d}-\frac {7 b^{2} c \,x^{5}}{8 \left (d \,x^{2}+c \right )^{\frac {3}{2}} d^{2}}-\frac {a^{2} x^{3}}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} d}+\frac {5 a b c \,x^{3}}{3 \left (d \,x^{2}+c \right )^{\frac {3}{2}} d^{2}}-\frac {35 b^{2} c^{2} x^{3}}{24 \left (d \,x^{2}+c \right )^{\frac {3}{2}} d^{3}}-\frac {a^{2} x}{\sqrt {d \,x^{2}+c}\, d^{2}}+\frac {5 a b c x}{\sqrt {d \,x^{2}+c}\, d^{3}}-\frac {35 b^{2} c^{2} x}{8 \sqrt {d \,x^{2}+c}\, d^{4}}+\frac {a^{2} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{d^{\frac {5}{2}}}-\frac {5 a b c \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{d^{\frac {7}{2}}}+\frac {35 b^{2} c^{2} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{8 d^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(b*x^2+a)^2/(d*x^2+c)^(5/2),x)

[Out]

1/4*b^2*x^7/d/(d*x^2+c)^(3/2)-7/8*b^2*c/d^2*x^5/(d*x^2+c)^(3/2)-35/24*b^2*c^2/d^3*x^3/(d*x^2+c)^(3/2)-35/8*b^2
*c^2/d^4*x/(d*x^2+c)^(1/2)+35/8*b^2*c^2/d^(9/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))+a*b*x^5/d/(d*x^2+c)^(3/2)+5/3*a*
b*c/d^2*x^3/(d*x^2+c)^(3/2)+5*a*b*c/d^3*x/(d*x^2+c)^(1/2)-5*a*b*c/d^(7/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))-1/3*a^
2*x^3/d/(d*x^2+c)^(3/2)-a^2/d^2*x/(d*x^2+c)^(1/2)+a^2/d^(5/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))

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maxima [A]  time = 0.89, size = 296, normalized size = 1.47 \[ \frac {b^{2} x^{7}}{4 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d} - \frac {7 \, b^{2} c x^{5}}{8 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{2}} + \frac {a b x^{5}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d} - \frac {1}{3} \, a^{2} x {\left (\frac {3 \, x^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d} + \frac {2 \, c}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{2}}\right )} - \frac {35 \, b^{2} c^{2} x {\left (\frac {3 \, x^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d} + \frac {2 \, c}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{2}}\right )}}{24 \, d^{2}} + \frac {5 \, a b c x {\left (\frac {3 \, x^{2}}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d} + \frac {2 \, c}{{\left (d x^{2} + c\right )}^{\frac {3}{2}} d^{2}}\right )}}{3 \, d} - \frac {35 \, b^{2} c^{2} x}{24 \, \sqrt {d x^{2} + c} d^{4}} + \frac {5 \, a b c x}{3 \, \sqrt {d x^{2} + c} d^{3}} - \frac {a^{2} x}{3 \, \sqrt {d x^{2} + c} d^{2}} + \frac {35 \, b^{2} c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, d^{\frac {9}{2}}} - \frac {5 \, a b c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{d^{\frac {7}{2}}} + \frac {a^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{d^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

1/4*b^2*x^7/((d*x^2 + c)^(3/2)*d) - 7/8*b^2*c*x^5/((d*x^2 + c)^(3/2)*d^2) + a*b*x^5/((d*x^2 + c)^(3/2)*d) - 1/
3*a^2*x*(3*x^2/((d*x^2 + c)^(3/2)*d) + 2*c/((d*x^2 + c)^(3/2)*d^2)) - 35/24*b^2*c^2*x*(3*x^2/((d*x^2 + c)^(3/2
)*d) + 2*c/((d*x^2 + c)^(3/2)*d^2))/d^2 + 5/3*a*b*c*x*(3*x^2/((d*x^2 + c)^(3/2)*d) + 2*c/((d*x^2 + c)^(3/2)*d^
2))/d - 35/24*b^2*c^2*x/(sqrt(d*x^2 + c)*d^4) + 5/3*a*b*c*x/(sqrt(d*x^2 + c)*d^3) - 1/3*a^2*x/(sqrt(d*x^2 + c)
*d^2) + 35/8*b^2*c^2*arcsinh(d*x/sqrt(c*d))/d^(9/2) - 5*a*b*c*arcsinh(d*x/sqrt(c*d))/d^(7/2) + a^2*arcsinh(d*x
/sqrt(c*d))/d^(5/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,{\left (b\,x^2+a\right )}^2}{{\left (d\,x^2+c\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + b*x^2)^2)/(c + d*x^2)^(5/2),x)

[Out]

int((x^4*(a + b*x^2)^2)/(c + d*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} \left (a + b x^{2}\right )^{2}}{\left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(b*x**2+a)**2/(d*x**2+c)**(5/2),x)

[Out]

Integral(x**4*(a + b*x**2)**2/(c + d*x**2)**(5/2), x)

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