3.600 \(\int \sqrt{c x} (a+b x^2)^{3/2} \, dx\)

Optimal. Leaf size=297 \[ \frac{4 a^{9/4} \sqrt{c} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right ),\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+b x^2}}-\frac{8 a^{9/4} \sqrt{c} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+b x^2}}+\frac{8 a^2 \sqrt{c x} \sqrt{a+b x^2}}{15 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{4 a (c x)^{3/2} \sqrt{a+b x^2}}{15 c}+\frac{2 (c x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 c} \]

[Out]

(4*a*(c*x)^(3/2)*Sqrt[a + b*x^2])/(15*c) + (8*a^2*Sqrt[c*x]*Sqrt[a + b*x^2])/(15*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x)
) + (2*(c*x)^(3/2)*(a + b*x^2)^(3/2))/(9*c) - (8*a^(9/4)*Sqrt[c]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[
a] + Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(15*b^(3/4)*Sqrt[a + b*x^2
]) + (4*a^(9/4)*Sqrt[c]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^
(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(15*b^(3/4)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.223943, antiderivative size = 297, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {279, 329, 305, 220, 1196} \[ \frac{4 a^{9/4} \sqrt{c} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+b x^2}}-\frac{8 a^{9/4} \sqrt{c} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+b x^2}}+\frac{8 a^2 \sqrt{c x} \sqrt{a+b x^2}}{15 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{4 a (c x)^{3/2} \sqrt{a+b x^2}}{15 c}+\frac{2 (c x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 c} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*x]*(a + b*x^2)^(3/2),x]

[Out]

(4*a*(c*x)^(3/2)*Sqrt[a + b*x^2])/(15*c) + (8*a^2*Sqrt[c*x]*Sqrt[a + b*x^2])/(15*Sqrt[b]*(Sqrt[a] + Sqrt[b]*x)
) + (2*(c*x)^(3/2)*(a + b*x^2)^(3/2))/(9*c) - (8*a^(9/4)*Sqrt[c]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[
a] + Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(15*b^(3/4)*Sqrt[a + b*x^2
]) + (4*a^(9/4)*Sqrt[c]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^
(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(15*b^(3/4)*Sqrt[a + b*x^2])

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{c x} \left (a+b x^2\right )^{3/2} \, dx &=\frac{2 (c x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 c}+\frac{1}{3} (2 a) \int \sqrt{c x} \sqrt{a+b x^2} \, dx\\ &=\frac{4 a (c x)^{3/2} \sqrt{a+b x^2}}{15 c}+\frac{2 (c x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 c}+\frac{1}{15} \left (4 a^2\right ) \int \frac{\sqrt{c x}}{\sqrt{a+b x^2}} \, dx\\ &=\frac{4 a (c x)^{3/2} \sqrt{a+b x^2}}{15 c}+\frac{2 (c x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 c}+\frac{\left (8 a^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+\frac{b x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{15 c}\\ &=\frac{4 a (c x)^{3/2} \sqrt{a+b x^2}}{15 c}+\frac{2 (c x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 c}+\frac{\left (8 a^{5/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{15 \sqrt{b}}-\frac{\left (8 a^{5/2}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{b} x^2}{\sqrt{a} c}}{\sqrt{a+\frac{b x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{15 \sqrt{b}}\\ &=\frac{4 a (c x)^{3/2} \sqrt{a+b x^2}}{15 c}+\frac{8 a^2 \sqrt{c x} \sqrt{a+b x^2}}{15 \sqrt{b} \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{2 (c x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 c}-\frac{8 a^{9/4} \sqrt{c} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+b x^2}}+\frac{4 a^{9/4} \sqrt{c} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0118379, size = 57, normalized size = 0.19 \[ \frac{2 a x \sqrt{c x} \sqrt{a+b x^2} \, _2F_1\left (-\frac{3}{2},\frac{3}{4};\frac{7}{4};-\frac{b x^2}{a}\right )}{3 \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 218, normalized size = 0.7 \begin{align*}{\frac{2}{45\,bx}\sqrt{cx} \left ( 5\,{b}^{3}{x}^{6}+12\,\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticE} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ){a}^{3}-6\,\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ){a}^{3}+16\,a{b}^{2}{x}^{4}+11\,{a}^{2}b{x}^{2} \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(1/2)*(b*x^2+a)^(3/2),x)

[Out]

2/45*(c*x)^(1/2)/(b*x^2+a)^(1/2)/b*(5*b^3*x^6+12*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)
^(1/2))/(-a*b)^(1/2))^(1/2)*(-x*b/(-a*b)^(1/2))^(1/2)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^
(1/2))*a^3-6*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-x*b/(-
a*b)^(1/2))^(1/2)*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*a^3+16*a*b^2*x^4+11*a^2*b*x^2
)/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(1/2)*(b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b x^{2} + a\right )}^{\frac{3}{2}} \sqrt{c x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(1/2)*(b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(3/2)*sqrt(c*x), x)

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Sympy [C]  time = 4.2133, size = 46, normalized size = 0.15 \begin{align*} \frac{a^{\frac{3}{2}} \sqrt{c} x^{\frac{3}{2}} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(1/2)*(b*x**2+a)**(3/2),x)

[Out]

a**(3/2)*sqrt(c)*x**(3/2)*gamma(3/4)*hyper((-3/2, 3/4), (7/4,), b*x**2*exp_polar(I*pi)/a)/(2*gamma(7/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(1/2)*(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

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