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

Optimal. Leaf size=193 \[ -\frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (-3 a^2 d^2-6 a b c d+5 b^2 c^2\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{6 c^{5/4} d^{9/4} \sqrt{e} \sqrt{c+d x^2}}+\frac{\sqrt{e x} (b c-a d)^2}{c d^2 e \sqrt{c+d x^2}}+\frac{2 b^2 \sqrt{e x} \sqrt{c+d x^2}}{3 d^2 e} \]

[Out]

((b*c - a*d)^2*Sqrt[e*x])/(c*d^2*e*Sqrt[c + d*x^2]) + (2*b^2*Sqrt[e*x]*Sqrt[c + d*x^2])/(3*d^2*e) - ((5*b^2*c^
2 - 6*a*b*c*d - 3*a^2*d^2)*(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])/(6*c^(5/4)*d^(9/4)*Sqrt[e]*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.155666, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {463, 459, 329, 220} \[ -\frac{\left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (-3 a^2 d^2-6 a b c d+5 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{6 c^{5/4} d^{9/4} \sqrt{e} \sqrt{c+d x^2}}+\frac{\sqrt{e x} (b c-a d)^2}{c d^2 e \sqrt{c+d x^2}}+\frac{2 b^2 \sqrt{e x} \sqrt{c+d x^2}}{3 d^2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^2*Sqrt[e*x])/(c*d^2*e*Sqrt[c + d*x^2]) + (2*b^2*Sqrt[e*x]*Sqrt[c + d*x^2])/(3*d^2*e) - ((5*b^2*c^
2 - 6*a*b*c*d - 3*a^2*d^2)*(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])/(6*c^(5/4)*d^(9/4)*Sqrt[e]*Sqrt[c + d*x^2])

Rule 463

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

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 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}{\sqrt{e x} \left (c+d x^2\right )^{3/2}} \, dx &=\frac{(b c-a d)^2 \sqrt{e x}}{c d^2 e \sqrt{c+d x^2}}-\frac{\int \frac{\frac{1}{2} \left (-2 a^2 d^2+(b c-a d)^2\right )-b^2 c d x^2}{\sqrt{e x} \sqrt{c+d x^2}} \, dx}{c d^2}\\ &=\frac{(b c-a d)^2 \sqrt{e x}}{c d^2 e \sqrt{c+d x^2}}+\frac{2 b^2 \sqrt{e x} \sqrt{c+d x^2}}{3 d^2 e}-\frac{\left (5 b^2 c^2-6 a b c d-3 a^2 d^2\right ) \int \frac{1}{\sqrt{e x} \sqrt{c+d x^2}} \, dx}{6 c d^2}\\ &=\frac{(b c-a d)^2 \sqrt{e x}}{c d^2 e \sqrt{c+d x^2}}+\frac{2 b^2 \sqrt{e x} \sqrt{c+d x^2}}{3 d^2 e}-\frac{\left (5 b^2 c^2-6 a b c d-3 a^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{3 c d^2 e}\\ &=\frac{(b c-a d)^2 \sqrt{e x}}{c d^2 e \sqrt{c+d x^2}}+\frac{2 b^2 \sqrt{e x} \sqrt{c+d x^2}}{3 d^2 e}-\frac{\left (5 b^2 c^2-6 a b c d-3 a^2 d^2\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 )}{6 c^{5/4} d^{9/4} \sqrt{e} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.15125, size = 174, normalized size = 0.9 \[ \frac{i x^{3/2} \sqrt{\frac{c}{d x^2}+1} \left (3 a^2 d^2+6 a b c d-5 b^2 c^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}{\sqrt{x}}\right ),-1\right )+x \sqrt{\frac{i \sqrt{c}}{\sqrt{d}}} \left (3 a^2 d^2-6 a b c d+b^2 c \left (5 c+2 d x^2\right )\right )}{3 c d^2 \sqrt{\frac{i \sqrt{c}}{\sqrt{d}}} \sqrt{e x} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(I*Sqrt[c])/Sqrt[d]]*x*(-6*a*b*c*d + 3*a^2*d^2 + b^2*c*(5*c + 2*d*x^2)) + I*(-5*b^2*c^2 + 6*a*b*c*d + 3*
a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[c])/Sqrt[d]]/Sqrt[x]], -1])/(3*c*Sqrt[(I
*Sqrt[c])/Sqrt[d]]*d^2*Sqrt[e*x]*Sqrt[c + d*x^2])

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Maple [A]  time = 0.022, size = 341, normalized size = 1.8 \begin{align*}{\frac{1}{6\,c{d}^{3}} \left ( 3\,\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{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-cd}{a}^{2}{d}^{2}+6\,\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{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-cd}abcd-5\,\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{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-cd}{b}^{2}{c}^{2}+4\,{x}^{3}{b}^{2}c{d}^{2}+6\,x{a}^{2}{d}^{3}-12\,xabc{d}^{2}+10\,x{b}^{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)^(1/2),x)

[Out]

1/6*(3*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))*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-c*d)^(1/2)*a^2*d^2+6*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))*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-c*d)^(1/2)*a*b*c*d-5*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))*((d
*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-c*d)^(1/2)*b^2*c^2+4*x^3*b^2*c*d^2+6*x*a^2*d^3-12*x*a*b*c*d^2+10*x*b^2*
c^2*d)/(d*x^2+c)^(1/2)/c/(e*x)^(1/2)/d^3

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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}} \sqrt{e x}}\,{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)^(1/2),x, algorithm="maxima")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{2}\right )^{2}}{\sqrt{e x} \left (c + d x^{2}\right )^{\frac{3}{2}}}\, dx \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)**(1/2),x)

[Out]

Integral((a + b*x**2)**2/(sqrt(e*x)*(c + d*x**2)**(3/2)), x)

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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}} \sqrt{e x}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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