3.838 \(\int \frac{(a+b x^2)^2 (c+d x^2)^{3/2}}{(e x)^{7/2}} \, dx\)

Optimal. Leaf size=468 \[ \frac{4 \sqrt [4]{c} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (9 a d (a d+2 b c)+b^2 c^2\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{15 d^{3/4} e^{7/2} \sqrt{c+d x^2}}-\frac{2 a^2 \left (c+d x^2\right )^{5/2}}{5 c e (e x)^{5/2}}-\frac{8 \sqrt [4]{c} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (9 a d (a d+2 b c)+b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 d^{3/4} e^{7/2} \sqrt{c+d x^2}}+\frac{2 (e x)^{3/2} \left (c+d x^2\right )^{3/2} \left (9 a d (a d+2 b c)+b^2 c^2\right )}{9 c^2 e^5}+\frac{4 (e x)^{3/2} \sqrt{c+d x^2} \left (9 a d (a d+2 b c)+b^2 c^2\right )}{15 c e^5}+\frac{8 \sqrt{e x} \sqrt{c+d x^2} \left (9 a d (a d+2 b c)+b^2 c^2\right )}{15 \sqrt{d} e^4 \left (\sqrt{c}+\sqrt{d} x\right )}-\frac{2 a \left (c+d x^2\right )^{5/2} (a d+2 b c)}{c^2 e^3 \sqrt{e x}} \]

[Out]

(4*(b^2*c^2 + 9*a*d*(2*b*c + a*d))*(e*x)^(3/2)*Sqrt[c + d*x^2])/(15*c*e^5) + (8*(b^2*c^2 + 9*a*d*(2*b*c + a*d)
)*Sqrt[e*x]*Sqrt[c + d*x^2])/(15*Sqrt[d]*e^4*(Sqrt[c] + Sqrt[d]*x)) + (2*(b^2*c^2 + 9*a*d*(2*b*c + a*d))*(e*x)
^(3/2)*(c + d*x^2)^(3/2))/(9*c^2*e^5) - (2*a^2*(c + d*x^2)^(5/2))/(5*c*e*(e*x)^(5/2)) - (2*a*(2*b*c + a*d)*(c
+ d*x^2)^(5/2))/(c^2*e^3*Sqrt[e*x]) - (8*c^(1/4)*(b^2*c^2 + 9*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])/(15*d^(3/4
)*e^(7/2)*Sqrt[c + d*x^2]) + (4*c^(1/4)*(b^2*c^2 + 9*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])/(15*d^(3/4)*e^(7/2)
*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.444668, antiderivative size = 468, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {462, 453, 279, 329, 305, 220, 1196} \[ -\frac{2 a^2 \left (c+d x^2\right )^{5/2}}{5 c e (e x)^{5/2}}+\frac{4 \sqrt [4]{c} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (9 a d (a d+2 b c)+b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 d^{3/4} e^{7/2} \sqrt{c+d x^2}}-\frac{8 \sqrt [4]{c} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (9 a d (a d+2 b c)+b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 d^{3/4} e^{7/2} \sqrt{c+d x^2}}+\frac{2 (e x)^{3/2} \left (c+d x^2\right )^{3/2} \left (9 a d (a d+2 b c)+b^2 c^2\right )}{9 c^2 e^5}+\frac{4 (e x)^{3/2} \sqrt{c+d x^2} \left (9 a d (a d+2 b c)+b^2 c^2\right )}{15 c e^5}+\frac{8 \sqrt{e x} \sqrt{c+d x^2} \left (9 a d (a d+2 b c)+b^2 c^2\right )}{15 \sqrt{d} e^4 \left (\sqrt{c}+\sqrt{d} x\right )}-\frac{2 a \left (c+d x^2\right )^{5/2} (a d+2 b c)}{c^2 e^3 \sqrt{e x}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*(c + d*x^2)^(3/2))/(e*x)^(7/2),x]

[Out]

(4*(b^2*c^2 + 9*a*d*(2*b*c + a*d))*(e*x)^(3/2)*Sqrt[c + d*x^2])/(15*c*e^5) + (8*(b^2*c^2 + 9*a*d*(2*b*c + a*d)
)*Sqrt[e*x]*Sqrt[c + d*x^2])/(15*Sqrt[d]*e^4*(Sqrt[c] + Sqrt[d]*x)) + (2*(b^2*c^2 + 9*a*d*(2*b*c + a*d))*(e*x)
^(3/2)*(c + d*x^2)^(3/2))/(9*c^2*e^5) - (2*a^2*(c + d*x^2)^(5/2))/(5*c*e*(e*x)^(5/2)) - (2*a*(2*b*c + a*d)*(c
+ d*x^2)^(5/2))/(c^2*e^3*Sqrt[e*x]) - (8*c^(1/4)*(b^2*c^2 + 9*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])/(15*d^(3/4
)*e^(7/2)*Sqrt[c + d*x^2]) + (4*c^(1/4)*(b^2*c^2 + 9*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])/(15*d^(3/4)*e^(7/2)
*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 453

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

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 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 \left (c+d x^2\right )^{3/2}}{(e x)^{7/2}} \, dx &=-\frac{2 a^2 \left (c+d x^2\right )^{5/2}}{5 c e (e x)^{5/2}}+\frac{2 \int \frac{\left (\frac{5}{2} a (2 b c+a d)+\frac{5}{2} b^2 c x^2\right ) \left (c+d x^2\right )^{3/2}}{(e x)^{3/2}} \, dx}{5 c e^2}\\ &=-\frac{2 a^2 \left (c+d x^2\right )^{5/2}}{5 c e (e x)^{5/2}}-\frac{2 a (2 b c+a d) \left (c+d x^2\right )^{5/2}}{c^2 e^3 \sqrt{e x}}+\frac{\left (b^2 c^2+9 a d (2 b c+a d)\right ) \int \sqrt{e x} \left (c+d x^2\right )^{3/2} \, dx}{c^2 e^4}\\ &=\frac{2 \left (b^2 c^2+9 a d (2 b c+a d)\right ) (e x)^{3/2} \left (c+d x^2\right )^{3/2}}{9 c^2 e^5}-\frac{2 a^2 \left (c+d x^2\right )^{5/2}}{5 c e (e x)^{5/2}}-\frac{2 a (2 b c+a d) \left (c+d x^2\right )^{5/2}}{c^2 e^3 \sqrt{e x}}+\frac{\left (2 \left (b^2 c^2+9 a d (2 b c+a d)\right )\right ) \int \sqrt{e x} \sqrt{c+d x^2} \, dx}{3 c e^4}\\ &=\frac{4 \left (b^2 c^2+9 a d (2 b c+a d)\right ) (e x)^{3/2} \sqrt{c+d x^2}}{15 c e^5}+\frac{2 \left (b^2 c^2+9 a d (2 b c+a d)\right ) (e x)^{3/2} \left (c+d x^2\right )^{3/2}}{9 c^2 e^5}-\frac{2 a^2 \left (c+d x^2\right )^{5/2}}{5 c e (e x)^{5/2}}-\frac{2 a (2 b c+a d) \left (c+d x^2\right )^{5/2}}{c^2 e^3 \sqrt{e x}}+\frac{\left (4 \left (b^2 c^2+9 a d (2 b c+a d)\right )\right ) \int \frac{\sqrt{e x}}{\sqrt{c+d x^2}} \, dx}{15 e^4}\\ &=\frac{4 \left (b^2 c^2+9 a d (2 b c+a d)\right ) (e x)^{3/2} \sqrt{c+d x^2}}{15 c e^5}+\frac{2 \left (b^2 c^2+9 a d (2 b c+a d)\right ) (e x)^{3/2} \left (c+d x^2\right )^{3/2}}{9 c^2 e^5}-\frac{2 a^2 \left (c+d x^2\right )^{5/2}}{5 c e (e x)^{5/2}}-\frac{2 a (2 b c+a d) \left (c+d x^2\right )^{5/2}}{c^2 e^3 \sqrt{e x}}+\frac{\left (8 \left (b^2 c^2+9 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 )}{15 e^5}\\ &=\frac{4 \left (b^2 c^2+9 a d (2 b c+a d)\right ) (e x)^{3/2} \sqrt{c+d x^2}}{15 c e^5}+\frac{2 \left (b^2 c^2+9 a d (2 b c+a d)\right ) (e x)^{3/2} \left (c+d x^2\right )^{3/2}}{9 c^2 e^5}-\frac{2 a^2 \left (c+d x^2\right )^{5/2}}{5 c e (e x)^{5/2}}-\frac{2 a (2 b c+a d) \left (c+d x^2\right )^{5/2}}{c^2 e^3 \sqrt{e x}}+\frac{\left (8 \sqrt{c} \left (b^2 c^2+9 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 )}{15 \sqrt{d} e^4}-\frac{\left (8 \sqrt{c} \left (b^2 c^2+9 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 )}{15 \sqrt{d} e^4}\\ &=\frac{4 \left (b^2 c^2+9 a d (2 b c+a d)\right ) (e x)^{3/2} \sqrt{c+d x^2}}{15 c e^5}+\frac{8 \left (b^2 c^2+9 a d (2 b c+a d)\right ) \sqrt{e x} \sqrt{c+d x^2}}{15 \sqrt{d} e^4 \left (\sqrt{c}+\sqrt{d} x\right )}+\frac{2 \left (b^2 c^2+9 a d (2 b c+a d)\right ) (e x)^{3/2} \left (c+d x^2\right )^{3/2}}{9 c^2 e^5}-\frac{2 a^2 \left (c+d x^2\right )^{5/2}}{5 c e (e x)^{5/2}}-\frac{2 a (2 b c+a d) \left (c+d x^2\right )^{5/2}}{c^2 e^3 \sqrt{e x}}-\frac{8 \sqrt [4]{c} \left (b^2 c^2+9 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 )}{15 d^{3/4} e^{7/2} \sqrt{c+d x^2}}+\frac{4 \sqrt [4]{c} \left (b^2 c^2+9 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 )}{15 d^{3/4} e^{7/2} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.162997, size = 141, normalized size = 0.3 \[ \frac{x \left (24 x^4 \sqrt{\frac{c}{d x^2}+1} \left (9 a^2 d^2+18 a b c d+b^2 c^2\right ) \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-\frac{c}{d x^2}\right )-2 \left (c+d x^2\right ) \left (9 a^2 \left (c+7 d x^2\right )-18 a b x^2 \left (d x^2-5 c\right )-b^2 x^4 \left (11 c+5 d x^2\right )\right )\right )}{45 (e x)^{7/2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*(c + d*x^2)^(3/2))/(e*x)^(7/2),x]

[Out]

(x*(-2*(c + d*x^2)*(-18*a*b*x^2*(-5*c + d*x^2) - b^2*x^4*(11*c + 5*d*x^2) + 9*a^2*(c + 7*d*x^2)) + 24*(b^2*c^2
+ 18*a*b*c*d + 9*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x^4*Hypergeometric2F1[-1/4, 1/2, 3/4, -(c/(d*x^2))]))/(45*(e*x)
^(7/2)*Sqrt[c + d*x^2])

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Maple [A]  time = 0.022, size = 668, normalized size = 1.4 \begin{align*}{\frac{2}{45\,d{x}^{2}{e}^{3}} \left ( 5\,{b}^{2}{d}^{3}{x}^{8}+108\,\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 ){x}^{2}{a}^{2}c{d}^{2}+216\,\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 ){x}^{2}ab{c}^{2}d+12\,\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 ){x}^{2}{b}^{2}{c}^{3}-54\,\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 ){x}^{2}{a}^{2}c{d}^{2}-108\,\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 ){x}^{2}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 EllipticF} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ){x}^{2}{b}^{2}{c}^{3}+18\,{x}^{6}ab{d}^{3}+16\,{x}^{6}{b}^{2}c{d}^{2}-63\,{x}^{4}{a}^{2}{d}^{3}-72\,{x}^{4}abc{d}^{2}+11\,{x}^{4}{b}^{2}{c}^{2}d-72\,{x}^{2}{a}^{2}c{d}^{2}-90\,{x}^{2}ab{c}^{2}d-9\,{a}^{2}{c}^{2}d \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)^(7/2),x)

[Out]

2/45/x^2*(5*b^2*d^3*x^8+108*((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))*x^2*a^2*c*d^2+
216*((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))*x^2*a*b*c^2*d+12*((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))*x^2*b^2*c^3-54*((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))*x^2*a^2*c*d^2-108*((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))*x^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)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*x^2*b^2*c^3+18*x^6*a*b*d^3+16*
x^6*b^2*c*d^2-63*x^4*a^2*d^3-72*x^4*a*b*c*d^2+11*x^4*b^2*c^2*d-72*x^2*a^2*c*d^2-90*x^2*a*b*c^2*d-9*a^2*c^2*d)/
(d*x^2+c)^(1/2)/d/e^3/(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{7}{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)^(7/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2*(d*x^2 + c)^(3/2)/(e*x)^(7/2), 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^{4} x^{4}}, 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)^(7/2),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^4*x^
4), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**(7/2),x)

[Out]

Timed out

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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{7}{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)^(7/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^2*(d*x^2 + c)^(3/2)/(e*x)^(7/2), x)