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

Optimal. Leaf size=430 \[ -\frac{c^{5/4} e^{5/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (117 a^2 d^2+7 b c (11 b c-26 a d)\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{195 d^{15/4} \sqrt{c+d x^2}}+\frac{2 c^{5/4} e^{5/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (117 a^2 d^2+7 b c (11 b c-26 a d)\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{195 d^{15/4} \sqrt{c+d x^2}}-\frac{2 c e^2 \sqrt{e x} \sqrt{c+d x^2} \left (117 a^2 d^2+7 b c (11 b c-26 a d)\right )}{195 d^{7/2} \left (\sqrt{c}+\sqrt{d} x\right )}+\frac{2 e (e x)^{3/2} \sqrt{c+d x^2} \left (117 a^2 d^2+7 b c (11 b c-26 a d)\right )}{585 d^3}-\frac{2 b (e x)^{7/2} \sqrt{c+d x^2} (11 b c-26 a d)}{117 d^2 e}+\frac{2 b^2 (e x)^{11/2} \sqrt{c+d x^2}}{13 d e^3} \]

[Out]

(2*(117*a^2*d^2 + 7*b*c*(11*b*c - 26*a*d))*e*(e*x)^(3/2)*Sqrt[c + d*x^2])/(585*d^3) - (2*b*(11*b*c - 26*a*d)*(
e*x)^(7/2)*Sqrt[c + d*x^2])/(117*d^2*e) + (2*b^2*(e*x)^(11/2)*Sqrt[c + d*x^2])/(13*d*e^3) - (2*c*(117*a^2*d^2
+ 7*b*c*(11*b*c - 26*a*d))*e^2*Sqrt[e*x]*Sqrt[c + d*x^2])/(195*d^(7/2)*(Sqrt[c] + Sqrt[d]*x)) + (2*c^(5/4)*(11
7*a^2*d^2 + 7*b*c*(11*b*c - 26*a*d))*e^(5/2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*E
llipticE[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(195*d^(15/4)*Sqrt[c + d*x^2]) - (c^(5/4)*(117
*a^2*d^2 + 7*b*c*(11*b*c - 26*a*d))*e^(5/2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*El
lipticF[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(195*d^(15/4)*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.405569, antiderivative size = 430, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {464, 459, 321, 329, 305, 220, 1196} \[ -\frac{c^{5/4} e^{5/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (117 a^2 d^2+7 b c (11 b c-26 a d)\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{195 d^{15/4} \sqrt{c+d x^2}}+\frac{2 c^{5/4} e^{5/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (117 a^2 d^2+7 b c (11 b c-26 a d)\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{195 d^{15/4} \sqrt{c+d x^2}}-\frac{2 c e^2 \sqrt{e x} \sqrt{c+d x^2} \left (117 a^2 d^2+7 b c (11 b c-26 a d)\right )}{195 d^{7/2} \left (\sqrt{c}+\sqrt{d} x\right )}+\frac{2 e (e x)^{3/2} \sqrt{c+d x^2} \left (117 a^2 d^2+7 b c (11 b c-26 a d)\right )}{585 d^3}-\frac{2 b (e x)^{7/2} \sqrt{c+d x^2} (11 b c-26 a d)}{117 d^2 e}+\frac{2 b^2 (e x)^{11/2} \sqrt{c+d x^2}}{13 d e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(117*a^2*d^2 + 7*b*c*(11*b*c - 26*a*d))*e*(e*x)^(3/2)*Sqrt[c + d*x^2])/(585*d^3) - (2*b*(11*b*c - 26*a*d)*(
e*x)^(7/2)*Sqrt[c + d*x^2])/(117*d^2*e) + (2*b^2*(e*x)^(11/2)*Sqrt[c + d*x^2])/(13*d*e^3) - (2*c*(117*a^2*d^2
+ 7*b*c*(11*b*c - 26*a*d))*e^2*Sqrt[e*x]*Sqrt[c + d*x^2])/(195*d^(7/2)*(Sqrt[c] + Sqrt[d]*x)) + (2*c^(5/4)*(11
7*a^2*d^2 + 7*b*c*(11*b*c - 26*a*d))*e^(5/2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*E
llipticE[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(195*d^(15/4)*Sqrt[c + d*x^2]) - (c^(5/4)*(117
*a^2*d^2 + 7*b*c*(11*b*c - 26*a*d))*e^(5/2)*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*El
lipticF[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(195*d^(15/4)*Sqrt[c + d*x^2])

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(d^2*(e*x)^
(m + n + 1)*(a + b*x^n)^(p + 1))/(b*e^(n + 1)*(m + n*(p + 2) + 1)), x] + Dist[1/(b*(m + n*(p + 2) + 1)), Int[(
e*x)^m*(a + b*x^n)^p*Simp[b*c^2*(m + n*(p + 2) + 1) + d*((2*b*c - a*d)*(m + n + 1) + 2*b*c*n*(p + 1))*x^n, x],
x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && NeQ[m + n*(p + 2) + 1, 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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && 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{(e x)^{5/2} \left (a+b x^2\right )^2}{\sqrt{c+d x^2}} \, dx &=\frac{2 b^2 (e x)^{11/2} \sqrt{c+d x^2}}{13 d e^3}+\frac{2 \int \frac{(e x)^{5/2} \left (\frac{13 a^2 d}{2}-\frac{1}{2} b (11 b c-26 a d) x^2\right )}{\sqrt{c+d x^2}} \, dx}{13 d}\\ &=-\frac{2 b (11 b c-26 a d) (e x)^{7/2} \sqrt{c+d x^2}}{117 d^2 e}+\frac{2 b^2 (e x)^{11/2} \sqrt{c+d x^2}}{13 d e^3}-\frac{1}{117} \left (-117 a^2-\frac{7 b c (11 b c-26 a d)}{d^2}\right ) \int \frac{(e x)^{5/2}}{\sqrt{c+d x^2}} \, dx\\ &=\frac{2 \left (117 a^2+\frac{7 b c (11 b c-26 a d)}{d^2}\right ) e (e x)^{3/2} \sqrt{c+d x^2}}{585 d}-\frac{2 b (11 b c-26 a d) (e x)^{7/2} \sqrt{c+d x^2}}{117 d^2 e}+\frac{2 b^2 (e x)^{11/2} \sqrt{c+d x^2}}{13 d e^3}-\frac{\left (c \left (117 a^2+\frac{7 b c (11 b c-26 a d)}{d^2}\right ) e^2\right ) \int \frac{\sqrt{e x}}{\sqrt{c+d x^2}} \, dx}{195 d}\\ &=\frac{2 \left (117 a^2+\frac{7 b c (11 b c-26 a d)}{d^2}\right ) e (e x)^{3/2} \sqrt{c+d x^2}}{585 d}-\frac{2 b (11 b c-26 a d) (e x)^{7/2} \sqrt{c+d x^2}}{117 d^2 e}+\frac{2 b^2 (e x)^{11/2} \sqrt{c+d x^2}}{13 d e^3}-\frac{\left (2 c \left (117 a^2+\frac{7 b c (11 b c-26 a d)}{d^2}\right ) e\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{195 d}\\ &=\frac{2 \left (117 a^2+\frac{7 b c (11 b c-26 a d)}{d^2}\right ) e (e x)^{3/2} \sqrt{c+d x^2}}{585 d}-\frac{2 b (11 b c-26 a d) (e x)^{7/2} \sqrt{c+d x^2}}{117 d^2 e}+\frac{2 b^2 (e x)^{11/2} \sqrt{c+d x^2}}{13 d e^3}-\frac{\left (2 c^{3/2} \left (117 a^2+\frac{7 b c (11 b c-26 a d)}{d^2}\right ) e^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{195 d^{3/2}}+\frac{\left (2 c^{3/2} \left (117 a^2+\frac{7 b c (11 b c-26 a d)}{d^2}\right ) e^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 )}{195 d^{3/2}}\\ &=\frac{2 \left (117 a^2+\frac{7 b c (11 b c-26 a d)}{d^2}\right ) e (e x)^{3/2} \sqrt{c+d x^2}}{585 d}-\frac{2 b (11 b c-26 a d) (e x)^{7/2} \sqrt{c+d x^2}}{117 d^2 e}+\frac{2 b^2 (e x)^{11/2} \sqrt{c+d x^2}}{13 d e^3}-\frac{2 c \left (117 a^2+\frac{7 b c (11 b c-26 a d)}{d^2}\right ) e^2 \sqrt{e x} \sqrt{c+d x^2}}{195 d^{3/2} \left (\sqrt{c}+\sqrt{d} x\right )}+\frac{2 c^{5/4} \left (117 a^2+\frac{7 b c (11 b c-26 a d)}{d^2}\right ) e^{5/2} \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 )}{195 d^{7/4} \sqrt{c+d x^2}}-\frac{c^{5/4} \left (117 a^2+\frac{7 b c (11 b c-26 a d)}{d^2}\right ) e^{5/2} \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 )}{195 d^{7/4} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.137503, size = 143, normalized size = 0.33 \[ \frac{2 e (e x)^{3/2} \left (\left (c+d x^2\right ) \left (117 a^2 d^2+26 a b d \left (5 d x^2-7 c\right )+b^2 \left (77 c^2-55 c d x^2+45 d^2 x^4\right )\right )-3 c \sqrt{\frac{c}{d x^2}+1} \left (117 a^2 d^2-182 a b c d+77 b^2 c^2\right ) \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-\frac{c}{d x^2}\right )\right )}{585 d^3 \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*(e*x)^(3/2)*((c + d*x^2)*(117*a^2*d^2 + 26*a*b*d*(-7*c + 5*d*x^2) + b^2*(77*c^2 - 55*c*d*x^2 + 45*d^2*x^4
)) - 3*c*(77*b^2*c^2 - 182*a*b*c*d + 117*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*Hypergeometric2F1[-1/4, 1/2, 3/4, -(c/(d
*x^2))]))/(585*d^3*Sqrt[c + d*x^2])

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Maple [A]  time = 0.028, size = 661, normalized size = 1.5 \begin{align*} -{\frac{{e}^{2}}{585\,x{d}^{4}}\sqrt{ex} \left ( -90\,{x}^{8}{b}^{2}{d}^{4}-260\,{x}^{6}ab{d}^{4}+20\,{x}^{6}{b}^{2}c{d}^{3}+702\,\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}^{2}{d}^{2}-1092\,\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}^{3}d+462\,\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}^{4}-351\,\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}^{2}{d}^{2}+546\,\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}^{3}d-231\,\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}^{4}-234\,{x}^{4}{a}^{2}{d}^{4}+104\,{x}^{4}abc{d}^{3}-44\,{x}^{4}{b}^{2}{c}^{2}{d}^{2}-234\,{x}^{2}{a}^{2}c{d}^{3}+364\,{x}^{2}ab{c}^{2}{d}^{2}-154\,{x}^{2}{b}^{2}{c}^{3}d \right ){\frac{1}{\sqrt{d{x}^{2}+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(5/2)*(b*x^2+a)^2/(d*x^2+c)^(1/2),x)

[Out]

-1/585/x*e^2*(e*x)^(1/2)/(d*x^2+c)^(1/2)/d^4*(-90*x^8*b^2*d^4-260*x^6*a*b*d^4+20*x^6*b^2*c*d^3+702*((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)*Ellipt
icE(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*a^2*c^2*d^2-1092*((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^3*d+462*((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^4-351*((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^2*d^2+546*((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)*Ellip
ticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*a*b*c^3*d-231*((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^4-234*x^4*a^2*d^4+104*x^4*a*b*c*d^3-44*x^4*b^2*c^2*d^2-234*x^2*a^2*c*d^3+3
64*x^2*a*b*c^2*d^2-154*x^2*b^2*c^3*d)

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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 (e x\right )^{\frac{5}{2}}}{\sqrt{d x^{2} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

integral((b^2*e^2*x^6 + 2*a*b*e^2*x^4 + a^2*e^2*x^2)*sqrt(e*x)/sqrt(d*x^2 + c), 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((e*x)**(5/2)*(b*x**2+a)**2/(d*x**2+c)**(1/2),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 (e x\right )^{\frac{5}{2}}}{\sqrt{d x^{2} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(b*x^2+a)^2/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

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