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

Optimal. Leaf size=372 \[ -\frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (3 b^2 c^2-5 a d (a d+2 b c)\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{5 c^{3/4} d^{7/4} e^{3/2} \sqrt{c+d x^2}}-\frac{2 a^2 \sqrt{c+d x^2}}{c e \sqrt{e x}}-\frac{2 \sqrt{e x} \sqrt{c+d x^2} \left (3 b^2 c^2-5 a d (a d+2 b c)\right )}{5 c d^{3/2} e^2 \left (\sqrt{c}+\sqrt{d} x\right )}+\frac{2 \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (3 b^2 c^2-5 a d (a d+2 b c)\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{5 c^{3/4} d^{7/4} e^{3/2} \sqrt{c+d x^2}}+\frac{2 b^2 (e x)^{3/2} \sqrt{c+d x^2}}{5 d e^3} \]

[Out]

(-2*a^2*Sqrt[c + d*x^2])/(c*e*Sqrt[e*x]) + (2*b^2*(e*x)^(3/2)*Sqrt[c + d*x^2])/(5*d*e^3) - (2*(3*b^2*c^2 - 5*a
*d*(2*b*c + a*d))*Sqrt[e*x]*Sqrt[c + d*x^2])/(5*c*d^(3/2)*e^2*(Sqrt[c] + Sqrt[d]*x)) + (2*(3*b^2*c^2 - 5*a*d*(
2*b*c + a*d))*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticE[2*ArcTan[(d^(1/4)*Sqrt
[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(5*c^(3/4)*d^(7/4)*e^(3/2)*Sqrt[c + d*x^2]) - ((3*b^2*c^2 - 5*a*d*(2*b*c + a*
d))*(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])/(5*c^(3/4)*d^(7/4)*e^(3/2)*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.340267, antiderivative size = 372, 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, 459, 329, 305, 220, 1196} \[ -\frac{2 a^2 \sqrt{c+d x^2}}{c e \sqrt{e x}}-\frac{2 \sqrt{e x} \sqrt{c+d x^2} \left (3 b^2 c^2-5 a d (a d+2 b c)\right )}{5 c d^{3/2} e^2 \left (\sqrt{c}+\sqrt{d} x\right )}-\frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (3 b^2 c^2-5 a d (a d+2 b c)\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{5 c^{3/4} d^{7/4} e^{3/2} \sqrt{c+d x^2}}+\frac{2 \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (3 b^2 c^2-5 a d (a d+2 b c)\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{5 c^{3/4} d^{7/4} e^{3/2} \sqrt{c+d x^2}}+\frac{2 b^2 (e x)^{3/2} \sqrt{c+d x^2}}{5 d e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*Sqrt[c + d*x^2])/(c*e*Sqrt[e*x]) + (2*b^2*(e*x)^(3/2)*Sqrt[c + d*x^2])/(5*d*e^3) - (2*(3*b^2*c^2 - 5*a
*d*(2*b*c + a*d))*Sqrt[e*x]*Sqrt[c + d*x^2])/(5*c*d^(3/2)*e^2*(Sqrt[c] + Sqrt[d]*x)) + (2*(3*b^2*c^2 - 5*a*d*(
2*b*c + a*d))*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticE[2*ArcTan[(d^(1/4)*Sqrt
[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(5*c^(3/4)*d^(7/4)*e^(3/2)*Sqrt[c + d*x^2]) - ((3*b^2*c^2 - 5*a*d*(2*b*c + a*
d))*(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])/(5*c^(3/4)*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 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 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} \sqrt{c+d x^2}} \, dx &=-\frac{2 a^2 \sqrt{c+d x^2}}{c e \sqrt{e x}}+\frac{2 \int \frac{\sqrt{e x} \left (\frac{1}{2} a (2 b c+a d)+\frac{1}{2} b^2 c x^2\right )}{\sqrt{c+d x^2}} \, dx}{c e^2}\\ &=-\frac{2 a^2 \sqrt{c+d x^2}}{c e \sqrt{e x}}+\frac{2 b^2 (e x)^{3/2} \sqrt{c+d x^2}}{5 d e^3}-\frac{\left (4 \left (\frac{3 b^2 c^2}{4}-\frac{5}{4} a d (2 b c+a d)\right )\right ) \int \frac{\sqrt{e x}}{\sqrt{c+d x^2}} \, dx}{5 c d e^2}\\ &=-\frac{2 a^2 \sqrt{c+d x^2}}{c e \sqrt{e x}}+\frac{2 b^2 (e x)^{3/2} \sqrt{c+d x^2}}{5 d e^3}-\frac{\left (8 \left (\frac{3 b^2 c^2}{4}-\frac{5}{4} a d (2 b c+a d)\right )\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{5 c d e^3}\\ &=-\frac{2 a^2 \sqrt{c+d x^2}}{c e \sqrt{e x}}+\frac{2 b^2 (e x)^{3/2} \sqrt{c+d x^2}}{5 d e^3}-\frac{\left (2 \left (3 b^2 c^2-5 a d (2 b c+a d)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{5 \sqrt{c} d^{3/2} e^2}+\frac{\left (2 \left (3 b^2 c^2-5 a d (2 b c+a d)\right )\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 )}{5 \sqrt{c} d^{3/2} e^2}\\ &=-\frac{2 a^2 \sqrt{c+d x^2}}{c e \sqrt{e x}}+\frac{2 b^2 (e x)^{3/2} \sqrt{c+d x^2}}{5 d e^3}-\frac{2 \left (3 b^2 c^2-5 a d (2 b c+a d)\right ) \sqrt{e x} \sqrt{c+d x^2}}{5 c d^{3/2} e^2 \left (\sqrt{c}+\sqrt{d} x\right )}+\frac{2 \left (3 b^2 c^2-5 a d (2 b c+a d)\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 )}{5 c^{3/4} d^{7/4} e^{3/2} \sqrt{c+d x^2}}-\frac{\left (3 b^2 c^2-5 a d (2 b c+a d)\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 )}{5 c^{3/4} d^{7/4} e^{3/2} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.111314, size = 115, normalized size = 0.31 \[ \frac{x \left (2 x^2 \sqrt{\frac{c}{d x^2}+1} \left (5 a^2 d^2+10 a b c d-3 b^2 c^2\right ) \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-\frac{c}{d x^2}\right )+2 \left (c+d x^2\right ) \left (b^2 c x^2-5 a^2 d\right )\right )}{5 c d (e x)^{3/2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(2*(-5*a^2*d + b^2*c*x^2)*(c + d*x^2) + 2*(-3*b^2*c^2 + 10*a*b*c*d + 5*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x^2*Hyp
ergeometric2F1[-1/4, 1/2, 3/4, -(c/(d*x^2))]))/(5*c*d*(e*x)^(3/2)*Sqrt[c + d*x^2])

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Maple [A]  time = 0.02, size = 595, normalized size = 1.6 \begin{align*}{\frac{1}{5\,e{d}^{2}c} \left ( 10\,\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}+20\,\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}-5\,\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}-10\,\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}+2\,{x}^{4}{b}^{2}c{d}^{2}-10\,{x}^{2}{a}^{2}{d}^{3}+2\,{x}^{2}{b}^{2}{c}^{2}d-10\,{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)^(1/2),x)

[Out]

1/5*(10*((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+20*((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^3-5*((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^2-10*((
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+2*x^4*b^2*c*d^2-10*x^2*a^2*d^3+2*x^2*b^2*c^2*d-10*a^2*c*d^2)/(d*x^
2+c)^(1/2)/d^2/e/(e*x)^(1/2)/c

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{2}}{\sqrt{d x^{2} + c} \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)^(1/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2/(sqrt(d*x^2 + c)*(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 e^{2} x^{4} + c 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)^(1/2),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*e^2*x^4 + c*e^2*x^2), x)

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Sympy [C]  time = 6.53332, size = 148, normalized size = 0.4 \begin{align*} \frac{a^{2} \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{4}, \frac{1}{2} \\ \frac{3}{4} \end{matrix}\middle |{\frac{d x^{2} e^{i \pi }}{c}} \right )}}{2 \sqrt{c} e^{\frac{3}{2}} \sqrt{x} \Gamma \left (\frac{3}{4}\right )} + \frac{a b x^{\frac{3}{2}} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{d x^{2} e^{i \pi }}{c}} \right )}}{\sqrt{c} e^{\frac{3}{2}} \Gamma \left (\frac{7}{4}\right )} + \frac{b^{2} x^{\frac{7}{2}} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{d x^{2} e^{i \pi }}{c}} \right )}}{2 \sqrt{c} e^{\frac{3}{2}} \Gamma \left (\frac{11}{4}\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)**(1/2),x)

[Out]

a**2*gamma(-1/4)*hyper((-1/4, 1/2), (3/4,), d*x**2*exp_polar(I*pi)/c)/(2*sqrt(c)*e**(3/2)*sqrt(x)*gamma(3/4))
+ a*b*x**(3/2)*gamma(3/4)*hyper((1/2, 3/4), (7/4,), d*x**2*exp_polar(I*pi)/c)/(sqrt(c)*e**(3/2)*gamma(7/4)) +
b**2*x**(7/2)*gamma(7/4)*hyper((1/2, 7/4), (11/4,), d*x**2*exp_polar(I*pi)/c)/(2*sqrt(c)*e**(3/2)*gamma(11/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{2}}{\sqrt{d x^{2} + c} \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)^(1/2),x, algorithm="giac")

[Out]

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