3.823 \(\int \sqrt{e x} (a+b x^2)^2 \sqrt{c+d x^2} \, dx\)

Optimal. Leaf size=425 \[ \frac{2 c^{5/4} \sqrt{e} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (39 a^2 d^2+b c (7 b c-26 a d)\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{195 d^{11/4} \sqrt{c+d x^2}}-\frac{4 c^{5/4} \sqrt{e} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (39 a^2 d^2+b c (7 b c-26 a d)\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{195 d^{11/4} \sqrt{c+d x^2}}+\frac{2 (e x)^{3/2} \sqrt{c+d x^2} \left (39 a^2 d^2+b c (7 b c-26 a d)\right )}{195 d^2 e}+\frac{4 c \sqrt{e x} \sqrt{c+d x^2} \left (39 a^2 d^2+b c (7 b c-26 a d)\right )}{195 d^{5/2} \left (\sqrt{c}+\sqrt{d} x\right )}-\frac{2 b (e x)^{3/2} \left (c+d x^2\right )^{3/2} (7 b c-26 a d)}{117 d^2 e}+\frac{2 b^2 (e x)^{7/2} \left (c+d x^2\right )^{3/2}}{13 d e^3} \]

[Out]

(2*(39*a^2*d^2 + b*c*(7*b*c - 26*a*d))*(e*x)^(3/2)*Sqrt[c + d*x^2])/(195*d^2*e) + (4*c*(39*a^2*d^2 + b*c*(7*b*
c - 26*a*d))*Sqrt[e*x]*Sqrt[c + d*x^2])/(195*d^(5/2)*(Sqrt[c] + Sqrt[d]*x)) - (2*b*(7*b*c - 26*a*d)*(e*x)^(3/2
)*(c + d*x^2)^(3/2))/(117*d^2*e) + (2*b^2*(e*x)^(7/2)*(c + d*x^2)^(3/2))/(13*d*e^3) - (4*c^(5/4)*(39*a^2*d^2 +
 b*c*(7*b*c - 26*a*d))*Sqrt[e]*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticE[2*Arc
Tan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(195*d^(11/4)*Sqrt[c + d*x^2]) + (2*c^(5/4)*(39*a^2*d^2 + b*
c*(7*b*c - 26*a*d))*Sqrt[e]*(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])/(195*d^(11/4)*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.414861, antiderivative size = 425, 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, 279, 329, 305, 220, 1196} \[ \frac{2 c^{5/4} \sqrt{e} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (39 a^2 d^2+b c (7 b c-26 a d)\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{195 d^{11/4} \sqrt{c+d x^2}}-\frac{4 c^{5/4} \sqrt{e} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (39 a^2 d^2+b c (7 b c-26 a d)\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{195 d^{11/4} \sqrt{c+d x^2}}+\frac{2 (e x)^{3/2} \sqrt{c+d x^2} \left (39 a^2 d^2+b c (7 b c-26 a d)\right )}{195 d^2 e}+\frac{4 c \sqrt{e x} \sqrt{c+d x^2} \left (39 a^2 d^2+b c (7 b c-26 a d)\right )}{195 d^{5/2} \left (\sqrt{c}+\sqrt{d} x\right )}-\frac{2 b (e x)^{3/2} \left (c+d x^2\right )^{3/2} (7 b c-26 a d)}{117 d^2 e}+\frac{2 b^2 (e x)^{7/2} \left (c+d x^2\right )^{3/2}}{13 d e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(39*a^2*d^2 + b*c*(7*b*c - 26*a*d))*(e*x)^(3/2)*Sqrt[c + d*x^2])/(195*d^2*e) + (4*c*(39*a^2*d^2 + b*c*(7*b*
c - 26*a*d))*Sqrt[e*x]*Sqrt[c + d*x^2])/(195*d^(5/2)*(Sqrt[c] + Sqrt[d]*x)) - (2*b*(7*b*c - 26*a*d)*(e*x)^(3/2
)*(c + d*x^2)^(3/2))/(117*d^2*e) + (2*b^2*(e*x)^(7/2)*(c + d*x^2)^(3/2))/(13*d*e^3) - (4*c^(5/4)*(39*a^2*d^2 +
 b*c*(7*b*c - 26*a*d))*Sqrt[e]*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticE[2*Arc
Tan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(195*d^(11/4)*Sqrt[c + d*x^2]) + (2*c^(5/4)*(39*a^2*d^2 + b*
c*(7*b*c - 26*a*d))*Sqrt[e]*(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])/(195*d^(11/4)*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 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 \sqrt{e x} \left (a+b x^2\right )^2 \sqrt{c+d x^2} \, dx &=\frac{2 b^2 (e x)^{7/2} \left (c+d x^2\right )^{3/2}}{13 d e^3}+\frac{2 \int \sqrt{e x} \sqrt{c+d x^2} \left (\frac{13 a^2 d}{2}-\frac{1}{2} b (7 b c-26 a d) x^2\right ) \, dx}{13 d}\\ &=-\frac{2 b (7 b c-26 a d) (e x)^{3/2} \left (c+d x^2\right )^{3/2}}{117 d^2 e}+\frac{2 b^2 (e x)^{7/2} \left (c+d x^2\right )^{3/2}}{13 d e^3}+\frac{1}{39} \left (39 a^2+\frac{b c (7 b c-26 a d)}{d^2}\right ) \int \sqrt{e x} \sqrt{c+d x^2} \, dx\\ &=\frac{2 \left (39 a^2+\frac{b c (7 b c-26 a d)}{d^2}\right ) (e x)^{3/2} \sqrt{c+d x^2}}{195 e}-\frac{2 b (7 b c-26 a d) (e x)^{3/2} \left (c+d x^2\right )^{3/2}}{117 d^2 e}+\frac{2 b^2 (e x)^{7/2} \left (c+d x^2\right )^{3/2}}{13 d e^3}+\frac{1}{195} \left (2 c \left (39 a^2+\frac{b c (7 b c-26 a d)}{d^2}\right )\right ) \int \frac{\sqrt{e x}}{\sqrt{c+d x^2}} \, dx\\ &=\frac{2 \left (39 a^2+\frac{b c (7 b c-26 a d)}{d^2}\right ) (e x)^{3/2} \sqrt{c+d x^2}}{195 e}-\frac{2 b (7 b c-26 a d) (e x)^{3/2} \left (c+d x^2\right )^{3/2}}{117 d^2 e}+\frac{2 b^2 (e x)^{7/2} \left (c+d x^2\right )^{3/2}}{13 d e^3}+\frac{\left (4 c \left (39 a^2+\frac{b c (7 b c-26 a d)}{d^2}\right )\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{195 e}\\ &=\frac{2 \left (39 a^2+\frac{b c (7 b c-26 a d)}{d^2}\right ) (e x)^{3/2} \sqrt{c+d x^2}}{195 e}-\frac{2 b (7 b c-26 a d) (e x)^{3/2} \left (c+d x^2\right )^{3/2}}{117 d^2 e}+\frac{2 b^2 (e x)^{7/2} \left (c+d x^2\right )^{3/2}}{13 d e^3}+\frac{\left (4 c^{3/2} \left (39 a^2+\frac{b c (7 b c-26 a d)}{d^2}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{195 \sqrt{d}}-\frac{\left (4 c^{3/2} \left (39 a^2+\frac{b c (7 b c-26 a d)}{d^2}\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 )}{195 \sqrt{d}}\\ &=\frac{2 \left (39 a^2+\frac{b c (7 b c-26 a d)}{d^2}\right ) (e x)^{3/2} \sqrt{c+d x^2}}{195 e}+\frac{4 c \left (39 a^2+\frac{b c (7 b c-26 a d)}{d^2}\right ) \sqrt{e x} \sqrt{c+d x^2}}{195 \sqrt{d} \left (\sqrt{c}+\sqrt{d} x\right )}-\frac{2 b (7 b c-26 a d) (e x)^{3/2} \left (c+d x^2\right )^{3/2}}{117 d^2 e}+\frac{2 b^2 (e x)^{7/2} \left (c+d x^2\right )^{3/2}}{13 d e^3}-\frac{4 c^{5/4} \left (39 a^2+\frac{b c (7 b c-26 a d)}{d^2}\right ) \sqrt{e} \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^{3/4} \sqrt{c+d x^2}}+\frac{2 c^{5/4} \left (39 a^2+\frac{b c (7 b c-26 a d)}{d^2}\right ) \sqrt{e} \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^{3/4} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.132363, size = 145, normalized size = 0.34 \[ \frac{2 \sqrt{e x} \left (6 c x \sqrt{\frac{c}{d x^2}+1} \left (39 a^2 d^2-26 a b c d+7 b^2 c^2\right ) \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-\frac{c}{d x^2}\right )-x \left (c+d x^2\right ) \left (-117 a^2 d^2-26 a b d \left (2 c+5 d x^2\right )+b^2 \left (14 c^2-10 c d x^2-45 d^2 x^4\right )\right )\right )}{585 d^2 \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[e*x]*(-(x*(c + d*x^2)*(-117*a^2*d^2 - 26*a*b*d*(2*c + 5*d*x^2) + b^2*(14*c^2 - 10*c*d*x^2 - 45*d^2*x^4
))) + 6*c*(7*b^2*c^2 - 26*a*b*c*d + 39*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x*Hypergeometric2F1[-1/4, 1/2, 3/4, -(c/(d
*x^2))]))/(585*d^2*Sqrt[c + d*x^2])

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Maple [A]  time = 0.04, size = 658, normalized size = 1.6 \begin{align*}{\frac{2}{585\,{d}^{3}x}\sqrt{ex} \left ( 45\,{x}^{8}{b}^{2}{d}^{4}+130\,{x}^{6}ab{d}^{4}+55\,{x}^{6}{b}^{2}c{d}^{3}+234\,\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}-156\,\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+42\,\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}-117\,\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}+78\,\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-21\,\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}+117\,{x}^{4}{a}^{2}{d}^{4}+182\,{x}^{4}abc{d}^{3}-4\,{x}^{4}{b}^{2}{c}^{2}{d}^{2}+117\,{x}^{2}{a}^{2}c{d}^{3}+52\,{x}^{2}ab{c}^{2}{d}^{2}-14\,{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((b*x^2+a)^2*(e*x)^(1/2)*(d*x^2+c)^(1/2),x)

[Out]

2/585*(e*x)^(1/2)/(d*x^2+c)^(1/2)/d^3*(45*x^8*b^2*d^4+130*x^6*a*b*d^4+55*x^6*b^2*c*d^3+234*((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^2*d^2-156*((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+42*((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^
4-117*((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+78*((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^3*d-21*((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+117*x^4*a^2*d^4+182*x^4*a*b*c*d^3-4*x^4*b^2*c^2*d^2+117*x^2*a^2*c*d^3+52*x^2*a*b*c^2
*d^2-14*x^2*b^2*c^3*d)/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(e*x)^(1/2)*(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), x)

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Sympy [C]  time = 6.68144, size = 148, normalized size = 0.35 \begin{align*} \frac{a^{2} \sqrt{c} \left (e x\right )^{\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 )}}{2 e \Gamma \left (\frac{7}{4}\right )} + \frac{a b \sqrt{c} \left (e x\right )^{\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 )}}{e^{3} \Gamma \left (\frac{11}{4}\right )} + \frac{b^{2} \sqrt{c} \left (e x\right )^{\frac{11}{2}} \Gamma \left (\frac{11}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{11}{4} \\ \frac{15}{4} \end{matrix}\middle |{\frac{d x^{2} e^{i \pi }}{c}} \right )}}{2 e^{5} \Gamma \left (\frac{15}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(e*x)**(1/2)*(d*x**2+c)**(1/2),x)

[Out]

a**2*sqrt(c)*(e*x)**(3/2)*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), d*x**2*exp_polar(I*pi)/c)/(2*e*gamma(7/4)) + a
*b*sqrt(c)*(e*x)**(7/2)*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), d*x**2*exp_polar(I*pi)/c)/(e**3*gamma(11/4)) +
b**2*sqrt(c)*(e*x)**(11/2)*gamma(11/4)*hyper((-1/2, 11/4), (15/4,), d*x**2*exp_polar(I*pi)/c)/(2*e**5*gamma(15
/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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