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

Optimal. Leaf size=384 \[ -\frac{\sqrt{e} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (5 a^2 d^2-30 a b c d+21 b^2 c^2\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{10 c^{3/4} d^{11/4} \sqrt{c+d x^2}}-\frac{\sqrt{e x} \sqrt{c+d x^2} \left (5 a^2 d^2-30 a b c d+21 b^2 c^2\right )}{5 c d^{5/2} \left (\sqrt{c}+\sqrt{d} x\right )}+\frac{\sqrt{e} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (5 a^2 d^2-30 a b c d+21 b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{5 c^{3/4} d^{11/4} \sqrt{c+d x^2}}+\frac{(e x)^{3/2} (b c-a d)^2}{c d^2 e \sqrt{c+d x^2}}+\frac{2 b^2 (e x)^{3/2} \sqrt{c+d x^2}}{5 d^2 e} \]

[Out]

((b*c - a*d)^2*(e*x)^(3/2))/(c*d^2*e*Sqrt[c + d*x^2]) + (2*b^2*(e*x)^(3/2)*Sqrt[c + d*x^2])/(5*d^2*e) - ((21*b
^2*c^2 - 30*a*b*c*d + 5*a^2*d^2)*Sqrt[e*x]*Sqrt[c + d*x^2])/(5*c*d^(5/2)*(Sqrt[c] + Sqrt[d]*x)) + ((21*b^2*c^2
- 30*a*b*c*d + 5*a^2*d^2)*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])/(5*c^(3/4)*d^(11/4)*Sqrt[c + d*x^2]) - ((21*b^2*c^2 - 30
*a*b*c*d + 5*a^2*d^2)*Sqrt[e]*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticF[2*ArcT
an[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(10*c^(3/4)*d^(11/4)*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.326508, antiderivative size = 384, 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 = {463, 459, 329, 305, 220, 1196} \[ -\frac{\sqrt{e x} \sqrt{c+d x^2} \left (5 a^2 d^2-30 a b c d+21 b^2 c^2\right )}{5 c d^{5/2} \left (\sqrt{c}+\sqrt{d} x\right )}-\frac{\sqrt{e} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (5 a^2 d^2-30 a b c d+21 b^2 c^2\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{10 c^{3/4} d^{11/4} \sqrt{c+d x^2}}+\frac{\sqrt{e} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (5 a^2 d^2-30 a b c d+21 b^2 c^2\right ) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{5 c^{3/4} d^{11/4} \sqrt{c+d x^2}}+\frac{(e x)^{3/2} (b c-a d)^2}{c d^2 e \sqrt{c+d x^2}}+\frac{2 b^2 (e x)^{3/2} \sqrt{c+d x^2}}{5 d^2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^2*(e*x)^(3/2))/(c*d^2*e*Sqrt[c + d*x^2]) + (2*b^2*(e*x)^(3/2)*Sqrt[c + d*x^2])/(5*d^2*e) - ((21*b
^2*c^2 - 30*a*b*c*d + 5*a^2*d^2)*Sqrt[e*x]*Sqrt[c + d*x^2])/(5*c*d^(5/2)*(Sqrt[c] + Sqrt[d]*x)) + ((21*b^2*c^2
- 30*a*b*c*d + 5*a^2*d^2)*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])/(5*c^(3/4)*d^(11/4)*Sqrt[c + d*x^2]) - ((21*b^2*c^2 - 30
*a*b*c*d + 5*a^2*d^2)*Sqrt[e]*(Sqrt[c] + Sqrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticF[2*ArcT
an[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])], 1/2])/(10*c^(3/4)*d^(11/4)*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 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}{\left (c+d x^2\right )^{3/2}} \, dx &=\frac{(b c-a d)^2 (e x)^{3/2}}{c d^2 e \sqrt{c+d x^2}}-\frac{\int \frac{\sqrt{e x} \left (\frac{1}{2} \left (-2 a^2 d^2+3 (b c-a d)^2\right )-b^2 c d x^2\right )}{\sqrt{c+d x^2}} \, dx}{c d^2}\\ &=\frac{(b c-a d)^2 (e x)^{3/2}}{c d^2 e \sqrt{c+d x^2}}+\frac{2 b^2 (e x)^{3/2} \sqrt{c+d x^2}}{5 d^2 e}-\frac{\left (21 b^2 c^2-30 a b c d+5 a^2 d^2\right ) \int \frac{\sqrt{e x}}{\sqrt{c+d x^2}} \, dx}{10 c d^2}\\ &=\frac{(b c-a d)^2 (e x)^{3/2}}{c d^2 e \sqrt{c+d x^2}}+\frac{2 b^2 (e x)^{3/2} \sqrt{c+d x^2}}{5 d^2 e}-\frac{\left (21 b^2 c^2-30 a b c d+5 a^2 d^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{5 c d^2 e}\\ &=\frac{(b c-a d)^2 (e x)^{3/2}}{c d^2 e \sqrt{c+d x^2}}+\frac{2 b^2 (e x)^{3/2} \sqrt{c+d x^2}}{5 d^2 e}-\frac{\left (21 b^2 c^2-30 a b c d+5 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 )}{5 \sqrt{c} d^{5/2}}+\frac{\left (21 b^2 c^2-30 a b c d+5 a^2 d^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 )}{5 \sqrt{c} d^{5/2}}\\ &=\frac{(b c-a d)^2 (e x)^{3/2}}{c d^2 e \sqrt{c+d x^2}}+\frac{2 b^2 (e x)^{3/2} \sqrt{c+d x^2}}{5 d^2 e}-\frac{\left (21 b^2 c^2-30 a b c d+5 a^2 d^2\right ) \sqrt{e x} \sqrt{c+d x^2}}{5 c d^{5/2} \left (\sqrt{c}+\sqrt{d} x\right )}+\frac{\left (21 b^2 c^2-30 a b c d+5 a^2 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 )}{5 c^{3/4} d^{11/4} \sqrt{c+d x^2}}-\frac{\left (21 b^2 c^2-30 a b c d+5 a^2 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 )}{10 c^{3/4} d^{11/4} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.121057, size = 119, normalized size = 0.31 \[ \frac{\sqrt{e x} \left (x \sqrt{\frac{c}{d x^2}+1} \left (-5 a^2 d^2+30 a b c d-21 b^2 c^2\right ) \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-\frac{c}{d x^2}\right )+x \left (5 a^2 d^2-10 a b c d+b^2 c \left (7 c+2 d x^2\right )\right )\right )}{5 c d^2 \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[e*x]*(x*(-10*a*b*c*d + 5*a^2*d^2 + b^2*c*(7*c + 2*d*x^2)) + (-21*b^2*c^2 + 30*a*b*c*d - 5*a^2*d^2)*Sqrt[
1 + c/(d*x^2)]*x*Hypergeometric2F1[-1/4, 1/2, 3/4, -(c/(d*x^2))]))/(5*c*d^2*Sqrt[c + d*x^2])

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Maple [A]  time = 0.021, size = 597, normalized size = 1.6 \begin{align*} -{\frac{1}{10\,{d}^{3}xc}\sqrt{ex} \left ( 10\,\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}-60\,\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}-5\,\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}+30\,\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}-4\,{x}^{4}{b}^{2}c{d}^{2}-10\,{x}^{2}{a}^{2}{d}^{3}+20\,{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)^(3/2),x)

[Out]

-1/10*(e*x)^(1/2)*(10*((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^2-60*((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)*El
lipticE(((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^3-5*((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+30*((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^3-4*x^4*b^2*c*d^2-10*x^2*a^2*d^3+20*x^2*a*b*c*d^2-14
*x^2*b^2*c^2*d)/(d*x^2+c)^(1/2)/d^3/x/c

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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}}{{\left (d x^{2} + c\right )}^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="maxima")

[Out]

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

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

[Out]

Integral(sqrt(e*x)*(a + b*x**2)**2/(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} \sqrt{e x}}{{\left (d x^{2} + c\right )}^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="giac")

[Out]

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