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

Optimal. Leaf size=375 \[ \frac{\sqrt [4]{c} \sqrt{e} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (15 a^2 d^2+b c (7 b c-18 a d)\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{15 d^{11/4} \sqrt{c+d x^2}}+\frac{2 \sqrt{e x} \sqrt{c+d x^2} \left (15 a^2 d^2+b c (7 b c-18 a d)\right )}{15 d^{5/2} \left (\sqrt{c}+\sqrt{d} x\right )}-\frac{2 \sqrt [4]{c} \sqrt{e} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (15 a^2 d^2+b c (7 b c-18 a d)\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 d^{11/4} \sqrt{c+d x^2}}-\frac{2 b (e x)^{3/2} \sqrt{c+d x^2} (7 b c-18 a d)}{45 d^2 e}+\frac{2 b^2 (e x)^{7/2} \sqrt{c+d x^2}}{9 d e^3} \]

[Out]

(-2*b*(7*b*c - 18*a*d)*(e*x)^(3/2)*Sqrt[c + d*x^2])/(45*d^2*e) + (2*b^2*(e*x)^(7/2)*Sqrt[c + d*x^2])/(9*d*e^3)
+ (2*(15*a^2*d^2 + b*c*(7*b*c - 18*a*d))*Sqrt[e*x]*Sqrt[c + d*x^2])/(15*d^(5/2)*(Sqrt[c] + Sqrt[d]*x)) - (2*c
^(1/4)*(15*a^2*d^2 + b*c*(7*b*c - 18*a*d))*Sqrt[e]*(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^(11/4)*Sqrt[c + d*x^2]) + (c^(1/4)
*(15*a^2*d^2 + b*c*(7*b*c - 18*a*d))*Sqrt[e]*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*E
llipticF[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(15*d^(11/4)*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.355625, antiderivative size = 375, 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 = {464, 459, 329, 305, 220, 1196} \[ \frac{2 \sqrt{e x} \sqrt{c+d x^2} \left (15 a^2 d^2+b c (7 b c-18 a d)\right )}{15 d^{5/2} \left (\sqrt{c}+\sqrt{d} x\right )}+\frac{\sqrt [4]{c} \sqrt{e} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (15 a^2 d^2+b c (7 b c-18 a d)\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 d^{11/4} \sqrt{c+d x^2}}-\frac{2 \sqrt [4]{c} \sqrt{e} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (15 a^2 d^2+b c (7 b c-18 a d)\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 d^{11/4} \sqrt{c+d x^2}}-\frac{2 b (e x)^{3/2} \sqrt{c+d x^2} (7 b c-18 a d)}{45 d^2 e}+\frac{2 b^2 (e x)^{7/2} \sqrt{c+d x^2}}{9 d e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*(7*b*c - 18*a*d)*(e*x)^(3/2)*Sqrt[c + d*x^2])/(45*d^2*e) + (2*b^2*(e*x)^(7/2)*Sqrt[c + d*x^2])/(9*d*e^3)
+ (2*(15*a^2*d^2 + b*c*(7*b*c - 18*a*d))*Sqrt[e*x]*Sqrt[c + d*x^2])/(15*d^(5/2)*(Sqrt[c] + Sqrt[d]*x)) - (2*c
^(1/4)*(15*a^2*d^2 + b*c*(7*b*c - 18*a*d))*Sqrt[e]*(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^(11/4)*Sqrt[c + d*x^2]) + (c^(1/4)
*(15*a^2*d^2 + b*c*(7*b*c - 18*a*d))*Sqrt[e]*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*E
llipticF[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(15*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 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{\sqrt{e x} \left (a+b x^2\right )^2}{\sqrt{c+d x^2}} \, dx &=\frac{2 b^2 (e x)^{7/2} \sqrt{c+d x^2}}{9 d e^3}+\frac{2 \int \frac{\sqrt{e x} \left (\frac{9 a^2 d}{2}-\frac{1}{2} b (7 b c-18 a d) x^2\right )}{\sqrt{c+d x^2}} \, dx}{9 d}\\ &=-\frac{2 b (7 b c-18 a d) (e x)^{3/2} \sqrt{c+d x^2}}{45 d^2 e}+\frac{2 b^2 (e x)^{7/2} \sqrt{c+d x^2}}{9 d e^3}+\frac{1}{15} \left (15 a^2+\frac{b c (7 b c-18 a d)}{d^2}\right ) \int \frac{\sqrt{e x}}{\sqrt{c+d x^2}} \, dx\\ &=-\frac{2 b (7 b c-18 a d) (e x)^{3/2} \sqrt{c+d x^2}}{45 d^2 e}+\frac{2 b^2 (e x)^{7/2} \sqrt{c+d x^2}}{9 d e^3}+\frac{\left (2 \left (15 a^2+\frac{b c (7 b c-18 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 )}{15 e}\\ &=-\frac{2 b (7 b c-18 a d) (e x)^{3/2} \sqrt{c+d x^2}}{45 d^2 e}+\frac{2 b^2 (e x)^{7/2} \sqrt{c+d x^2}}{9 d e^3}+\frac{\left (2 \sqrt{c} \left (15 a^2+\frac{b c (7 b c-18 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 )}{15 \sqrt{d}}-\frac{\left (2 \sqrt{c} \left (15 a^2+\frac{b c (7 b c-18 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 )}{15 \sqrt{d}}\\ &=-\frac{2 b (7 b c-18 a d) (e x)^{3/2} \sqrt{c+d x^2}}{45 d^2 e}+\frac{2 b^2 (e x)^{7/2} \sqrt{c+d x^2}}{9 d e^3}+\frac{2 \left (15 a^2+\frac{b c (7 b c-18 a d)}{d^2}\right ) \sqrt{e x} \sqrt{c+d x^2}}{15 \sqrt{d} \left (\sqrt{c}+\sqrt{d} x\right )}-\frac{2 \sqrt [4]{c} \left (15 a^2+\frac{b c (7 b c-18 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 )}{15 d^{3/4} \sqrt{c+d x^2}}+\frac{\sqrt [4]{c} \left (15 a^2+\frac{b c (7 b c-18 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 )}{15 d^{3/4} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.110666, size = 111, normalized size = 0.3 \[ \frac{2 \sqrt{e x} \left (3 x \sqrt{\frac{c}{d x^2}+1} \left (15 a^2 d^2-18 a b c d+7 b^2 c^2\right ) \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-\frac{c}{d x^2}\right )+b x \left (c+d x^2\right ) \left (18 a d-7 b c+5 b d x^2\right )\right )}{45 d^2 \sqrt{c+d x^2}} \]

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]*(b*x*(c + d*x^2)*(-7*b*c + 18*a*d + 5*b*d*x^2) + 3*(7*b^2*c^2 - 18*a*b*c*d + 15*a^2*d^2)*Sqrt[1 +
c/(d*x^2)]*x*Hypergeometric2F1[-1/4, 1/2, 3/4, -(c/(d*x^2))]))/(45*d^2*Sqrt[c + d*x^2])

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Maple [A]  time = 0.017, size = 604, normalized size = 1.6 \begin{align*}{\frac{1}{45\,{d}^{3}x}\sqrt{ex} \left ( 10\,{x}^{6}{b}^{2}{d}^{3}+90\,\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}-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 ) ab{c}^{2}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}^{3}-45\,\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}+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 ) ab{c}^{2}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}^{3}+36\,{x}^{4}ab{d}^{3}-4\,{x}^{4}{b}^{2}c{d}^{2}+36\,{x}^{2}abc{d}^{2}-14\,{x}^{2}{b}^{2}{c}^{2}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]

1/45*(e*x)^(1/2)/(d*x^2+c)^(1/2)/d^3*(10*x^6*b^2*d^3+90*((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^2-108*((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))*a*b*c^2*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^3-45*((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+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))*a*b*c^2*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^3+36*x^4*a*b*d^3
-4*x^4*b^2*c*d^2+36*x^2*a*b*c*d^2-14*x^2*b^2*c^2*d)/x

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

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Sympy [C]  time = 5.29669, size = 143, normalized size = 0.38 \begin{align*} \frac{a^{2} \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 \sqrt{c} e \Gamma \left (\frac{7}{4}\right )} + \frac{a b \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 )}}{\sqrt{c} e^{3} \Gamma \left (\frac{11}{4}\right )} + \frac{b^{2} \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 \sqrt{c} 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*(e*x)**(3/2)*gamma(3/4)*hyper((1/2, 3/4), (7/4,), d*x**2*exp_polar(I*pi)/c)/(2*sqrt(c)*e*gamma(7/4)) + a*
b*(e*x)**(7/2)*gamma(7/4)*hyper((1/2, 7/4), (11/4,), d*x**2*exp_polar(I*pi)/c)/(sqrt(c)*e**3*gamma(11/4)) + b*
*2*(e*x)**(11/2)*gamma(11/4)*hyper((1/2, 11/4), (15/4,), d*x**2*exp_polar(I*pi)/c)/(2*sqrt(c)*e**5*gamma(15/4)
)

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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{e x}}{\sqrt{d x^{2} + c}}\,{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(e*x)/sqrt(d*x^2 + c), x)