3.837 \(\int \frac{(a+b x^2)^2 (c+d x^2)^{3/2}}{(e x)^{5/2}} \, dx\)

Optimal. Leaf size=288 \[ -\frac{4 c^{3/4} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (3 b^2 c^2-11 a d (7 a d+6 b c)\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{231 d^{5/4} e^{5/2} \sqrt{c+d x^2}}-\frac{2 a^2 \left (c+d x^2\right )^{5/2}}{3 c e (e x)^{3/2}}-\frac{2 \sqrt{e x} \left (c+d x^2\right )^{3/2} \left (3 b^2 c^2-11 a d (7 a d+6 b c)\right )}{231 c d e^3}-\frac{4 \sqrt{e x} \sqrt{c+d x^2} \left (3 b^2 c^2-11 a d (7 a d+6 b c)\right )}{231 d e^3}+\frac{2 b^2 \sqrt{e x} \left (c+d x^2\right )^{5/2}}{11 d e^3} \]

[Out]

(-4*(3*b^2*c^2 - 11*a*d*(6*b*c + 7*a*d))*Sqrt[e*x]*Sqrt[c + d*x^2])/(231*d*e^3) - (2*(3*b^2*c^2 - 11*a*d*(6*b*
c + 7*a*d))*Sqrt[e*x]*(c + d*x^2)^(3/2))/(231*c*d*e^3) - (2*a^2*(c + d*x^2)^(5/2))/(3*c*e*(e*x)^(3/2)) + (2*b^
2*Sqrt[e*x]*(c + d*x^2)^(5/2))/(11*d*e^3) - (4*c^(3/4)*(3*b^2*c^2 - 11*a*d*(6*b*c + 7*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])
/(231*d^(5/4)*e^(5/2)*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.241653, antiderivative size = 288, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {462, 459, 279, 329, 220} \[ -\frac{2 a^2 \left (c+d x^2\right )^{5/2}}{3 c e (e x)^{3/2}}-\frac{4 c^{3/4} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (3 b^2 c^2-11 a d (7 a d+6 b c)\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{231 d^{5/4} e^{5/2} \sqrt{c+d x^2}}-\frac{2 \sqrt{e x} \left (c+d x^2\right )^{3/2} \left (3 b^2 c^2-11 a d (7 a d+6 b c)\right )}{231 c d e^3}-\frac{4 \sqrt{e x} \sqrt{c+d x^2} \left (3 b^2 c^2-11 a d (7 a d+6 b c)\right )}{231 d e^3}+\frac{2 b^2 \sqrt{e x} \left (c+d x^2\right )^{5/2}}{11 d e^3} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*(c + d*x^2)^(3/2))/(e*x)^(5/2),x]

[Out]

(-4*(3*b^2*c^2 - 11*a*d*(6*b*c + 7*a*d))*Sqrt[e*x]*Sqrt[c + d*x^2])/(231*d*e^3) - (2*(3*b^2*c^2 - 11*a*d*(6*b*
c + 7*a*d))*Sqrt[e*x]*(c + d*x^2)^(3/2))/(231*c*d*e^3) - (2*a^2*(c + d*x^2)^(5/2))/(3*c*e*(e*x)^(3/2)) + (2*b^
2*Sqrt[e*x]*(c + d*x^2)^(5/2))/(11*d*e^3) - (4*c^(3/4)*(3*b^2*c^2 - 11*a*d*(6*b*c + 7*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])
/(231*d^(5/4)*e^(5/2)*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 279

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

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 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]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2}}{(e x)^{5/2}} \, dx &=-\frac{2 a^2 \left (c+d x^2\right )^{5/2}}{3 c e (e x)^{3/2}}+\frac{2 \int \frac{\left (\frac{1}{2} a (6 b c+7 a d)+\frac{3}{2} b^2 c x^2\right ) \left (c+d x^2\right )^{3/2}}{\sqrt{e x}} \, dx}{3 c e^2}\\ &=-\frac{2 a^2 \left (c+d x^2\right )^{5/2}}{3 c e (e x)^{3/2}}+\frac{2 b^2 \sqrt{e x} \left (c+d x^2\right )^{5/2}}{11 d e^3}-\frac{\left (3 b^2 c^2-11 a d (6 b c+7 a d)\right ) \int \frac{\left (c+d x^2\right )^{3/2}}{\sqrt{e x}} \, dx}{33 c d e^2}\\ &=-\frac{2 \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right ) \sqrt{e x} \left (c+d x^2\right )^{3/2}}{231 c d e^3}-\frac{2 a^2 \left (c+d x^2\right )^{5/2}}{3 c e (e x)^{3/2}}+\frac{2 b^2 \sqrt{e x} \left (c+d x^2\right )^{5/2}}{11 d e^3}-\frac{\left (2 \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right )\right ) \int \frac{\sqrt{c+d x^2}}{\sqrt{e x}} \, dx}{77 d e^2}\\ &=-\frac{4 \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right ) \sqrt{e x} \sqrt{c+d x^2}}{231 d e^3}-\frac{2 \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right ) \sqrt{e x} \left (c+d x^2\right )^{3/2}}{231 c d e^3}-\frac{2 a^2 \left (c+d x^2\right )^{5/2}}{3 c e (e x)^{3/2}}+\frac{2 b^2 \sqrt{e x} \left (c+d x^2\right )^{5/2}}{11 d e^3}-\frac{\left (4 c \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right )\right ) \int \frac{1}{\sqrt{e x} \sqrt{c+d x^2}} \, dx}{231 d e^2}\\ &=-\frac{4 \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right ) \sqrt{e x} \sqrt{c+d x^2}}{231 d e^3}-\frac{2 \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right ) \sqrt{e x} \left (c+d x^2\right )^{3/2}}{231 c d e^3}-\frac{2 a^2 \left (c+d x^2\right )^{5/2}}{3 c e (e x)^{3/2}}+\frac{2 b^2 \sqrt{e x} \left (c+d x^2\right )^{5/2}}{11 d e^3}-\frac{\left (8 c \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{231 d e^3}\\ &=-\frac{4 \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right ) \sqrt{e x} \sqrt{c+d x^2}}{231 d e^3}-\frac{2 \left (3 b^2 c^2-11 a d (6 b c+7 a d)\right ) \sqrt{e x} \left (c+d x^2\right )^{3/2}}{231 c d e^3}-\frac{2 a^2 \left (c+d x^2\right )^{5/2}}{3 c e (e x)^{3/2}}+\frac{2 b^2 \sqrt{e x} \left (c+d x^2\right )^{5/2}}{11 d e^3}-\frac{4 c^{3/4} \left (3 b^2 c^2-11 a d (6 b c+7 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 )}{231 d^{5/4} e^{5/2} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.228935, size = 202, normalized size = 0.7 \[ \frac{x^{5/2} \left (\frac{2 \left (c+d x^2\right ) \left (77 a^2 d \left (d x^2-c\right )+66 a b d x^2 \left (3 c+d x^2\right )+3 b^2 x^2 \left (4 c^2+13 c d x^2+7 d^2 x^4\right )\right )}{d x^{3/2}}+\frac{8 i c x \sqrt{\frac{c}{d x^2}+1} \left (77 a^2 d^2+66 a b c d-3 b^2 c^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}{\sqrt{x}}\right ),-1\right )}{d \sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}\right )}{231 (e x)^{5/2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*(c + d*x^2)^(3/2))/(e*x)^(5/2),x]

[Out]

(x^(5/2)*((2*(c + d*x^2)*(77*a^2*d*(-c + d*x^2) + 66*a*b*d*x^2*(3*c + d*x^2) + 3*b^2*x^2*(4*c^2 + 13*c*d*x^2 +
7*d^2*x^4)))/(d*x^(3/2)) + ((8*I)*c*(-3*b^2*c^2 + 66*a*b*c*d + 77*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x*EllipticF[I*
ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[d]]/Sqrt[x]], -1])/(Sqrt[(I*Sqrt[c])/Sqrt[d]]*d)))/(231*(e*x)^(5/2)*Sqrt[c + d*x
^2])

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Maple [A]  time = 0.017, size = 415, normalized size = 1.4 \begin{align*}{\frac{2}{231\,x{d}^{2}{e}^{2}} \left ( 21\,{x}^{8}{b}^{2}{d}^{4}+154\,\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 ) \sqrt{-cd}x{a}^{2}c{d}^{2}+132\,\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 ) \sqrt{-cd}xab{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 EllipticF} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ) \sqrt{-cd}x{b}^{2}{c}^{3}+66\,{x}^{6}ab{d}^{4}+60\,{x}^{6}{b}^{2}c{d}^{3}+77\,{x}^{4}{a}^{2}{d}^{4}+264\,{x}^{4}abc{d}^{3}+51\,{x}^{4}{b}^{2}{c}^{2}{d}^{2}+198\,{x}^{2}ab{c}^{2}{d}^{2}+12\,{x}^{2}{b}^{2}{c}^{3}d-77\,{a}^{2}{c}^{2}{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*(d*x^2+c)^(3/2)/(e*x)^(5/2),x)

[Out]

2/231/(d*x^2+c)^(1/2)/x*(21*x^8*b^2*d^4+154*((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)
)*(-c*d)^(1/2)*x*a^2*c*d^2+132*((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))*(-c*d)^(1/2
)*x*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)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*(-c*d)^(1/2)*x*b^2*c^3+66*
x^6*a*b*d^4+60*x^6*b^2*c*d^3+77*x^4*a^2*d^4+264*x^4*a*b*c*d^3+51*x^4*b^2*c^2*d^2+198*x^2*a*b*c^2*d^2+12*x^2*b^
2*c^3*d-77*a^2*c^2*d^2)/d^2/e^2/(e*x)^(1/2)

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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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(3/2)/(e*x)^(5/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2*(d*x^2 + c)^(3/2)/(e*x)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{2} d x^{6} +{\left (b^{2} c + 2 \, a b d\right )} x^{4} + a^{2} c +{\left (2 \, a b c + a^{2} d\right )} x^{2}\right )} \sqrt{d x^{2} + c} \sqrt{e x}}{e^{3} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(3/2)/(e*x)^(5/2),x, algorithm="fricas")

[Out]

integral((b^2*d*x^6 + (b^2*c + 2*a*b*d)*x^4 + a^2*c + (2*a*b*c + a^2*d)*x^2)*sqrt(d*x^2 + c)*sqrt(e*x)/(e^3*x^
3), x)

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Sympy [C]  time = 55.4485, size = 309, normalized size = 1.07 \begin{align*} \frac{a^{2} c^{\frac{3}{2}} \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, - \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{d x^{2} e^{i \pi }}{c}} \right )}}{2 e^{\frac{5}{2}} x^{\frac{3}{2}} \Gamma \left (\frac{1}{4}\right )} + \frac{a^{2} \sqrt{c} d \sqrt{x} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{d x^{2} e^{i \pi }}{c}} \right )}}{2 e^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right )} + \frac{a b c^{\frac{3}{2}} \sqrt{x} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{d x^{2} e^{i \pi }}{c}} \right )}}{e^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right )} + \frac{a b \sqrt{c} d x^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{d x^{2} e^{i \pi }}{c}} \right )}}{e^{\frac{5}{2}} \Gamma \left (\frac{9}{4}\right )} + \frac{b^{2} c^{\frac{3}{2}} x^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{d x^{2} e^{i \pi }}{c}} \right )}}{2 e^{\frac{5}{2}} \Gamma \left (\frac{9}{4}\right )} + \frac{b^{2} \sqrt{c} d x^{\frac{9}{2}} \Gamma \left (\frac{9}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{9}{4} \\ \frac{13}{4} \end{matrix}\middle |{\frac{d x^{2} e^{i \pi }}{c}} \right )}}{2 e^{\frac{5}{2}} \Gamma \left (\frac{13}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(3/2)/(e*x)**(5/2),x)

[Out]

a**2*c**(3/2)*gamma(-3/4)*hyper((-3/4, -1/2), (1/4,), d*x**2*exp_polar(I*pi)/c)/(2*e**(5/2)*x**(3/2)*gamma(1/4
)) + a**2*sqrt(c)*d*sqrt(x)*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), d*x**2*exp_polar(I*pi)/c)/(2*e**(5/2)*gamma(
5/4)) + a*b*c**(3/2)*sqrt(x)*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), d*x**2*exp_polar(I*pi)/c)/(e**(5/2)*gamma(5
/4)) + a*b*sqrt(c)*d*x**(5/2)*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), d*x**2*exp_polar(I*pi)/c)/(e**(5/2)*gamma(
9/4)) + b**2*c**(3/2)*x**(5/2)*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), d*x**2*exp_polar(I*pi)/c)/(2*e**(5/2)*gam
ma(9/4)) + b**2*sqrt(c)*d*x**(9/2)*gamma(9/4)*hyper((-1/2, 9/4), (13/4,), d*x**2*exp_polar(I*pi)/c)/(2*e**(5/2
)*gamma(13/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{2}{\left (d x^{2} + c\right )}^{\frac{3}{2}}}{\left (e x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(3/2)/(e*x)^(5/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^2*(d*x^2 + c)^(3/2)/(e*x)^(5/2), x)