3.854 \(\int \frac{(a+b x^2)^2}{(e x)^{3/2} (c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=393 \[ \frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (3 a^2 d^2-2 a b c d+3 b^2 c^2\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{2 c^{7/4} d^{7/4} e^{3/2} \sqrt{c+d x^2}}+\frac{\sqrt{e x} \sqrt{c+d x^2} \left (3 a^2 d^2-2 a b c d+3 b^2 c^2\right )}{c^2 d^{3/2} e^2 \left (\sqrt{c}+\sqrt{d} x\right )}-\frac{(e x)^{3/2} \left (3 a^2 d^2-2 a b c d+b^2 c^2\right )}{c^2 d e^3 \sqrt{c+d x^2}}-\frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (3 a^2 d^2-2 a b c d+3 b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{c^{7/4} d^{7/4} e^{3/2} \sqrt{c+d x^2}}-\frac{2 a^2}{c e \sqrt{e x} \sqrt{c+d x^2}} \]

[Out]

(-2*a^2)/(c*e*Sqrt[e*x]*Sqrt[c + d*x^2]) - ((b^2*c^2 - 2*a*b*c*d + 3*a^2*d^2)*(e*x)^(3/2))/(c^2*d*e^3*Sqrt[c +
d*x^2]) + ((3*b^2*c^2 - 2*a*b*c*d + 3*a^2*d^2)*Sqrt[e*x]*Sqrt[c + d*x^2])/(c^2*d^(3/2)*e^2*(Sqrt[c] + Sqrt[d]
*x)) - ((3*b^2*c^2 - 2*a*b*c*d + 3*a^2*d^2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*El
lipticE[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(c^(7/4)*d^(7/4)*e^(3/2)*Sqrt[c + d*x^2]) + ((3
*b^2*c^2 - 2*a*b*c*d + 3*a^2*d^2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticF[2*
ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(2*c^(7/4)*d^(7/4)*e^(3/2)*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.361189, antiderivative size = 393, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {462, 457, 329, 305, 220, 1196} \[ \frac{\sqrt{e x} \sqrt{c+d x^2} \left (3 a^2 d^2-2 a b c d+3 b^2 c^2\right )}{c^2 d^{3/2} e^2 \left (\sqrt{c}+\sqrt{d} x\right )}-\frac{(e x)^{3/2} \left (3 a^2 d^2-2 a b c d+b^2 c^2\right )}{c^2 d e^3 \sqrt{c+d x^2}}+\frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (3 a^2 d^2-2 a b c d+3 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{2 c^{7/4} d^{7/4} e^{3/2} \sqrt{c+d x^2}}-\frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (3 a^2 d^2-2 a b c d+3 b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{c^{7/4} d^{7/4} e^{3/2} \sqrt{c+d x^2}}-\frac{2 a^2}{c e \sqrt{e x} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2)/(c*e*Sqrt[e*x]*Sqrt[c + d*x^2]) - ((b^2*c^2 - 2*a*b*c*d + 3*a^2*d^2)*(e*x)^(3/2))/(c^2*d*e^3*Sqrt[c +
d*x^2]) + ((3*b^2*c^2 - 2*a*b*c*d + 3*a^2*d^2)*Sqrt[e*x]*Sqrt[c + d*x^2])/(c^2*d^(3/2)*e^2*(Sqrt[c] + Sqrt[d]
*x)) - ((3*b^2*c^2 - 2*a*b*c*d + 3*a^2*d^2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*El
lipticE[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(c^(7/4)*d^(7/4)*e^(3/2)*Sqrt[c + d*x^2]) + ((3
*b^2*c^2 - 2*a*b*c*d + 3*a^2*d^2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticF[2*
ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(2*c^(7/4)*d^(7/4)*e^(3/2)*Sqrt[c + d*x^2])

Rule 462

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

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^2}{(e x)^{3/2} \left (c+d x^2\right )^{3/2}} \, dx &=-\frac{2 a^2}{c e \sqrt{e x} \sqrt{c+d x^2}}+\frac{2 \int \frac{\sqrt{e x} \left (\frac{1}{2} a (2 b c-3 a d)+\frac{1}{2} b^2 c x^2\right )}{\left (c+d x^2\right )^{3/2}} \, dx}{c e^2}\\ &=-\frac{2 a^2}{c e \sqrt{e x} \sqrt{c+d x^2}}-\frac{\left (b^2 c^2-2 a b c d+3 a^2 d^2\right ) (e x)^{3/2}}{c^2 d e^3 \sqrt{c+d x^2}}-\frac{\left (2 a b-\frac{3 b^2 c}{d}-\frac{3 a^2 d}{c}\right ) \int \frac{\sqrt{e x}}{\sqrt{c+d x^2}} \, dx}{2 c e^2}\\ &=-\frac{2 a^2}{c e \sqrt{e x} \sqrt{c+d x^2}}-\frac{\left (b^2 c^2-2 a b c d+3 a^2 d^2\right ) (e x)^{3/2}}{c^2 d e^3 \sqrt{c+d x^2}}-\frac{\left (2 a b-\frac{3 b^2 c}{d}-\frac{3 a^2 d}{c}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{c e^3}\\ &=-\frac{2 a^2}{c e \sqrt{e x} \sqrt{c+d x^2}}-\frac{\left (b^2 c^2-2 a b c d+3 a^2 d^2\right ) (e x)^{3/2}}{c^2 d e^3 \sqrt{c+d x^2}}+\frac{\left (3 b^2 c^2-2 a b c d+3 a^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{c^{3/2} d^{3/2} e^2}-\frac{\left (3 b^2 c^2-2 a b c d+3 a^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{d} x^2}{\sqrt{c} e}}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{c^{3/2} d^{3/2} e^2}\\ &=-\frac{2 a^2}{c e \sqrt{e x} \sqrt{c+d x^2}}-\frac{\left (b^2 c^2-2 a b c d+3 a^2 d^2\right ) (e x)^{3/2}}{c^2 d e^3 \sqrt{c+d x^2}}+\frac{\left (3 b^2 c^2-2 a b c d+3 a^2 d^2\right ) \sqrt{e x} \sqrt{c+d x^2}}{c^2 d^{3/2} e^2 \left (\sqrt{c}+\sqrt{d} x\right )}-\frac{\left (3 b^2 c^2-2 a b c d+3 a^2 d^2\right ) \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{c^{7/4} d^{7/4} e^{3/2} \sqrt{c+d x^2}}+\frac{\left (3 b^2 c^2-2 a b c d+3 a^2 d^2\right ) \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{2 c^{7/4} d^{7/4} e^{3/2} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.107433, size = 126, normalized size = 0.32 \[ \frac{x \left (x^2 \sqrt{\frac{d x^2}{c}+1} \left (3 a^2 d^2-2 a b c d+3 b^2 c^2\right ) \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{d x^2}{c}\right )-3 a^2 d \left (2 c+3 d x^2\right )+6 a b c d x^2-3 b^2 c^2 x^2\right )}{3 c^2 d (e x)^{3/2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-3*b^2*c^2*x^2 + 6*a*b*c*d*x^2 - 3*a^2*d*(2*c + 3*d*x^2) + (3*b^2*c^2 - 2*a*b*c*d + 3*a^2*d^2)*x^2*Sqrt[1
+ (d*x^2)/c]*Hypergeometric2F1[1/2, 3/4, 7/4, -((d*x^2)/c)]))/(3*c^2*d*(e*x)^(3/2)*Sqrt[c + d*x^2])

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Maple [A]  time = 0.025, size = 594, normalized size = 1.5 \begin{align*}{\frac{1}{2\,{d}^{2}e{c}^{2}} \left ( 6\,\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticE} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ){a}^{2}c{d}^{2}-4\,\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticE} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ) ab{c}^{2}d+6\,\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticE} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ){b}^{2}{c}^{3}-3\,\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticF} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ){a}^{2}c{d}^{2}+2\,\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticF} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ) ab{c}^{2}d-3\,\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticF} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ){b}^{2}{c}^{3}-6\,{x}^{2}{a}^{2}{d}^{3}+4\,{x}^{2}abc{d}^{2}-2\,{x}^{2}{b}^{2}{c}^{2}d-4\,{a}^{2}c{d}^{2} \right ){\frac{1}{\sqrt{d{x}^{2}+c}}}{\frac{1}{\sqrt{ex}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(6*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/
2)*d)^(1/2)*EllipticE(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*a^2*c*d^2-4*((d*x+(-c*d)^(1/2))/(-c
*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticE(((d*x+(-
c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*a*b*c^2*d+6*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*
x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticE(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2
),1/2*2^(1/2))*b^2*c^3-3*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1
/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*a^2*c*d^2+2*((d*x
+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*
EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*a*b*c^2*d-3*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(
1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/
(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*b^2*c^3-6*x^2*a^2*d^3+4*x^2*a*b*c*d^2-2*x^2*b^2*c^2*d-4*a^2*c*d^2)/(d*x^2+c)^
(1/2)/d^2/e/(e*x)^(1/2)/c^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{2}}{{\left (d x^{2} + c\right )}^{\frac{3}{2}} \left (e x\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt{d x^{2} + c} \sqrt{e x}}{d^{2} e^{2} x^{6} + 2 \, c d e^{2} x^{4} + c^{2} e^{2} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(d*x^2 + c)*sqrt(e*x)/(d^2*e^2*x^6 + 2*c*d*e^2*x^4 + c^2*e^2*x^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{2}\right )^{2}}{\left (e x\right )^{\frac{3}{2}} \left (c + d x^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{2}}{{\left (d x^{2} + c\right )}^{\frac{3}{2}} \left (e x\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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