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

Optimal. Leaf size=174 \[ \frac{x (11 a d+4 b c)}{6 \sqrt{c+d x^2} (b c-a d)^3}+\frac{x (3 a d+2 b c)}{6 b \left (c+d x^2\right )^{3/2} (b c-a d)^2}+\frac{a x}{2 b \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac{\sqrt{a} (2 a d+3 b c) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 (b c-a d)^{7/2}} \]

[Out]

((2*b*c + 3*a*d)*x)/(6*b*(b*c - a*d)^2*(c + d*x^2)^(3/2)) + (a*x)/(2*b*(b*c - a*d)*(a + b*x^2)*(c + d*x^2)^(3/
2)) + ((4*b*c + 11*a*d)*x)/(6*(b*c - a*d)^3*Sqrt[c + d*x^2]) - (Sqrt[a]*(3*b*c + 2*a*d)*ArcTan[(Sqrt[b*c - a*d
]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*(b*c - a*d)^(7/2))

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Rubi [A]  time = 0.211019, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {470, 527, 12, 377, 205} \[ \frac{x (11 a d+4 b c)}{6 \sqrt{c+d x^2} (b c-a d)^3}+\frac{x (3 a d+2 b c)}{6 b \left (c+d x^2\right )^{3/2} (b c-a d)^2}+\frac{a x}{2 b \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac{\sqrt{a} (2 a d+3 b c) \tan ^{-1}\left (\frac{x \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 (b c-a d)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*b*c + 3*a*d)*x)/(6*b*(b*c - a*d)^2*(c + d*x^2)^(3/2)) + (a*x)/(2*b*(b*c - a*d)*(a + b*x^2)*(c + d*x^2)^(3/
2)) + ((4*b*c + 11*a*d)*x)/(6*(b*c - a*d)^3*Sqrt[c + d*x^2]) - (Sqrt[a]*(3*b*c + 2*a*d)*ArcTan[(Sqrt[b*c - a*d
]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*(b*c - a*d)^(7/2))

Rule 470

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

Rule 527

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 377

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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{a x}{2 b (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac{\int \frac{a c-2 (b c+a d) x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx}{2 b (b c-a d)}\\ &=\frac{(2 b c+3 a d) x}{6 b (b c-a d)^2 \left (c+d x^2\right )^{3/2}}+\frac{a x}{2 b (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}-\frac{\int \frac{5 a b c^2-2 b c (2 b c+3 a d) x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}} \, dx}{6 b c (b c-a d)^2}\\ &=\frac{(2 b c+3 a d) x}{6 b (b c-a d)^2 \left (c+d x^2\right )^{3/2}}+\frac{a x}{2 b (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}+\frac{(4 b c+11 a d) x}{6 (b c-a d)^3 \sqrt{c+d x^2}}-\frac{\int \frac{3 a b c^2 (3 b c+2 a d)}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{6 b c^2 (b c-a d)^3}\\ &=\frac{(2 b c+3 a d) x}{6 b (b c-a d)^2 \left (c+d x^2\right )^{3/2}}+\frac{a x}{2 b (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}+\frac{(4 b c+11 a d) x}{6 (b c-a d)^3 \sqrt{c+d x^2}}-\frac{(a (3 b c+2 a d)) \int \frac{1}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx}{2 (b c-a d)^3}\\ &=\frac{(2 b c+3 a d) x}{6 b (b c-a d)^2 \left (c+d x^2\right )^{3/2}}+\frac{a x}{2 b (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}+\frac{(4 b c+11 a d) x}{6 (b c-a d)^3 \sqrt{c+d x^2}}-\frac{(a (3 b c+2 a d)) \operatorname{Subst}\left (\int \frac{1}{a-(-b c+a d) x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{2 (b c-a d)^3}\\ &=\frac{(2 b c+3 a d) x}{6 b (b c-a d)^2 \left (c+d x^2\right )^{3/2}}+\frac{a x}{2 b (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}+\frac{(4 b c+11 a d) x}{6 (b c-a d)^3 \sqrt{c+d x^2}}-\frac{\sqrt{a} (3 b c+2 a d) \tan ^{-1}\left (\frac{\sqrt{b c-a d} x}{\sqrt{a} \sqrt{c+d x^2}}\right )}{2 (b c-a d)^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.962807, size = 133, normalized size = 0.76 \[ \frac{x^5 \left (\frac{8 x^2 \left (c+d x^2\right ) (b c-a d) \, _2F_1\left (2,3;\frac{11}{2};\frac{(b c-a d) x^2}{c \left (b x^2+a\right )}\right )}{a+b x^2}+9 c \left (7 c+2 d x^2\right ) \, _2F_1\left (1,2;\frac{9}{2};\frac{(b c-a d) x^2}{c \left (b x^2+a\right )}\right )\right )}{315 c^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x^5*(9*c*(7*c + 2*d*x^2)*Hypergeometric2F1[1, 2, 9/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))] + (8*(b*c - a*d)*x^2
*(c + d*x^2)*Hypergeometric2F1[2, 3, 11/2, ((b*c - a*d)*x^2)/(c*(a + b*x^2))])/(a + b*x^2)))/(315*c^3*(a + b*x
^2)^2*(c + d*x^2)^(3/2))

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Maple [B]  time = 0.023, size = 2463, normalized size = 14.2 \begin{align*} \text{result too large to display} \end{align*}

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]

5/12/b^2*a*d*(-a*b)^(1/2)/(a*d-b*c)^2/((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b
*c)/b)^(3/2)-1/4/b*a/(-a*b)^(1/2)/(a*d-b*c)/((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-
(a*d-b*c)/b)^(3/2)-3/4*a/(-a*b)^(1/2)/(a*d-b*c)^2/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(
x+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(
a*d-b*c)/b)^(1/2))/(x+1/b*(-a*b)^(1/2)))-1/4/b^2*a/(a*d-b*c)/(x-1/b*(-a*b)^(1/2))/((x-1/b*(-a*b)^(1/2))^2*d+2*
d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)-1/4/b^2*a/(a*d-b*c)/(x+1/b*(-a*b)^(1/2))/((x+1/b*(-a*
b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)-7/6/b^2*a*d/(a*d-b*c)/c^2/((x+1/b*(-a
*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*x-7/12/b^2*a*d/(a*d-b*c)/c/((x-1/b*(
-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)*x-7/6/b^2*a*d/(a*d-b*c)/c^2/((x-1/
b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*x-5/4/b*a^2*d^2/(a*d-b*c)^3/c/(
(x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*x-5/4/b*a*d*(-a*b)^(1/2)/(
a*d-b*c)^3/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1
/2)*((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x+1/b*(-a*b)^(1/2))
)-7/12/b^2*a*d/(a*d-b*c)/c/((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2
)*x+1/4/b*a/(-a*b)^(1/2)/(a*d-b*c)/((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)
/b)^(3/2)+3/4*a/(-a*b)^(1/2)/(a*d-b*c)^2/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a
*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/
b)^(1/2))/(x-1/b*(-a*b)^(1/2)))+5/4/b*a*d*(-a*b)^(1/2)/(a*d-b*c)^3/((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/
b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)-5/12/b^2*a*d*(-a*b)^(1/2)/(a*d-b*c)^2/((x+1/b*(-a*b)^(1/2))^2*d-2*d*
(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)-5/4/b*a*d*(-a*b)^(1/2)/(a*d-b*c)^3/((x-1/b*(-a*b)^(1/2)
)^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+5/12/b^2*a^2*d^2/(a*d-b*c)^2/c/((x+1/b*(-a*b)
^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)*x+5/12/b^2*a^2*d^2/(a*d-b*c)^2/c/((x-1/
b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)*x+5/6/b^2*a^2*d^2/(a*d-b*c)^2/c
^2/((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*x-5/4/b*a^2*d^2/(a*d-b
*c)^3/c/((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)*x+5/4/b*a*d*(-a*b
)^(1/2)/(a*d-b*c)^3/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))+2*(-(a*d-b
*c)/b)^(1/2)*((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x-1/b*(-a*
b)^(1/2)))+5/6/b^2*a^2*d^2/(a*d-b*c)^2/c^2/((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(
a*d-b*c)/b)^(1/2)*x+3/4/b*a/(a*d-b*c)^2/c/((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a
*d-b*c)/b)^(1/2)*x*d+3/4/b*a/(a*d-b*c)^2/c/((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(
a*d-b*c)/b)^(1/2)*x*d+1/3/b^2*x/c/(d*x^2+c)^(3/2)+2/3/b^2/c^2*x/(d*x^2+c)^(1/2)+3/4*a/(-a*b)^(1/2)/(a*d-b*c)^2
/((x+1/b*(-a*b)^(1/2))^2*d-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)-3/4*a/(-a*b)^(1/2)/(a*d-
b*c)^2/((x-1/b*(-a*b)^(1/2))^2*d+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{{\left (b x^{2} + a\right )}^{2}{\left (d x^{2} + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

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]

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

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Fricas [B]  time = 7.75318, size = 2048, normalized size = 11.77 \begin{align*} \text{result too large to display} \end{align*}

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]

[-1/24*(3*((3*b^2*c*d^2 + 2*a*b*d^3)*x^6 + 3*a*b*c^3 + 2*a^2*c^2*d + (6*b^2*c^2*d + 7*a*b*c*d^2 + 2*a^2*d^3)*x
^4 + (3*b^2*c^3 + 8*a*b*c^2*d + 4*a^2*c*d^2)*x^2)*sqrt(-a/(b*c - a*d))*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*
x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2 + 4*((b^2*c^2 - 3*a*b*c*d + 2*a^2*d^2)*x^3 - (a*b*c^2 - a^2*c*d)
*x)*sqrt(d*x^2 + c)*sqrt(-a/(b*c - a*d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - 4*((4*b^2*c*d + 11*a*b*d^2)*x^5 + 2*(
3*b^2*c^2 + 8*a*b*c*d + 4*a^2*d^2)*x^3 + 3*(3*a*b*c^2 + 2*a^2*c*d)*x)*sqrt(d*x^2 + c))/(a*b^3*c^5 - 3*a^2*b^2*
c^4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*b^2*c*d^4 - a^3*b*d^5)*x^6 + (2
*b^4*c^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^4 + (b^4*c^5 - a*b^3*c^4*d - 3*a^2
*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4)*x^2), 1/12*(3*((3*b^2*c*d^2 + 2*a*b*d^3)*x^6 + 3*a*b*c^3 + 2*a^2
*c^2*d + (6*b^2*c^2*d + 7*a*b*c*d^2 + 2*a^2*d^3)*x^4 + (3*b^2*c^3 + 8*a*b*c^2*d + 4*a^2*c*d^2)*x^2)*sqrt(a/(b*
c - a*d))*arctan(-1/2*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)*sqrt(a/(b*c - a*d))/(a*d*x^3 + a*c*x)) + 2*((4
*b^2*c*d + 11*a*b*d^2)*x^5 + 2*(3*b^2*c^2 + 8*a*b*c*d + 4*a^2*d^2)*x^3 + 3*(3*a*b*c^2 + 2*a^2*c*d)*x)*sqrt(d*x
^2 + c))/(a*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2
*b^2*c*d^4 - a^3*b*d^5)*x^6 + (2*b^4*c^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^4
+ (b^4*c^5 - a*b^3*c^4*d - 3*a^2*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4)*x^2)]

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Giac [B]  time = 6.43751, size = 802, normalized size = 4.61 \begin{align*} \frac{{\left (\frac{2 \,{\left (b^{4} c^{5} d^{2} - a b^{3} c^{4} d^{3} - 3 \, a^{2} b^{2} c^{3} d^{4} + 5 \, a^{3} b c^{2} d^{5} - 2 \, a^{4} c d^{6}\right )} x^{2}}{b^{6} c^{7} d - 6 \, a b^{5} c^{6} d^{2} + 15 \, a^{2} b^{4} c^{5} d^{3} - 20 \, a^{3} b^{3} c^{4} d^{4} + 15 \, a^{4} b^{2} c^{3} d^{5} - 6 \, a^{5} b c^{2} d^{6} + a^{6} c d^{7}} + \frac{3 \,{\left (b^{4} c^{6} d - 2 \, a b^{3} c^{5} d^{2} + 2 \, a^{3} b c^{3} d^{4} - a^{4} c^{2} d^{5}\right )}}{b^{6} c^{7} d - 6 \, a b^{5} c^{6} d^{2} + 15 \, a^{2} b^{4} c^{5} d^{3} - 20 \, a^{3} b^{3} c^{4} d^{4} + 15 \, a^{4} b^{2} c^{3} d^{5} - 6 \, a^{5} b c^{2} d^{6} + a^{6} c d^{7}}\right )} x}{3 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}}} + \frac{{\left (3 \, a b c \sqrt{d} + 2 \, a^{2} d^{\frac{3}{2}}\right )} \arctan \left (\frac{{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt{a b c d - a^{2} d^{2}}}\right )}{2 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{a b c d - a^{2} d^{2}}} - \frac{{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a b c \sqrt{d} - 2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a^{2} d^{\frac{3}{2}} - a b c^{2} \sqrt{d}}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )}{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{4} b - 2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b c + 4 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} a d + b c^{2}\right )}} \end{align*}

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]

1/3*(2*(b^4*c^5*d^2 - a*b^3*c^4*d^3 - 3*a^2*b^2*c^3*d^4 + 5*a^3*b*c^2*d^5 - 2*a^4*c*d^6)*x^2/(b^6*c^7*d - 6*a*
b^5*c^6*d^2 + 15*a^2*b^4*c^5*d^3 - 20*a^3*b^3*c^4*d^4 + 15*a^4*b^2*c^3*d^5 - 6*a^5*b*c^2*d^6 + a^6*c*d^7) + 3*
(b^4*c^6*d - 2*a*b^3*c^5*d^2 + 2*a^3*b*c^3*d^4 - a^4*c^2*d^5)/(b^6*c^7*d - 6*a*b^5*c^6*d^2 + 15*a^2*b^4*c^5*d^
3 - 20*a^3*b^3*c^4*d^4 + 15*a^4*b^2*c^3*d^5 - 6*a^5*b*c^2*d^6 + a^6*c*d^7))*x/(d*x^2 + c)^(3/2) + 1/2*(3*a*b*c
*sqrt(d) + 2*a^2*d^(3/2))*arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a*d)/sqrt(a*b*c*d - a^2*d^2)
)/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(a*b*c*d - a^2*d^2)) - ((sqrt(d)*x - sqrt(d*x^2 + c
))^2*a*b*c*sqrt(d) - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*d^(3/2) - a*b*c^2*sqrt(d))/((b^3*c^3 - 3*a*b^2*c^2*
d + 3*a^2*b*c*d^2 - a^3*d^3)*((sqrt(d)*x - sqrt(d*x^2 + c))^4*b - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c + 4*(s
qrt(d)*x - sqrt(d*x^2 + c))^2*a*d + b*c^2))