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

Optimal. Leaf size=186 \[ \frac{4 a^{15/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{231 b^{9/4} \sqrt{a x+b x^3}}-\frac{8 a^3 \sqrt{a x+b x^3}}{231 b^2}+\frac{8 a^2 x^2 \sqrt{a x+b x^3}}{385 b}+\frac{4}{55} a x^4 \sqrt{a x+b x^3}+\frac{2}{15} x^3 \left (a x+b x^3\right )^{3/2} \]

[Out]

(-8*a^3*Sqrt[a*x + b*x^3])/(231*b^2) + (8*a^2*x^2*Sqrt[a*x + b*x^3])/(385*b) + (4*a*x^4*Sqrt[a*x + b*x^3])/55
+ (2*x^3*(a*x + b*x^3)^(3/2))/15 + (4*a^(15/4)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[
b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(231*b^(9/4)*Sqrt[a*x + b*x^3])

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Rubi [A]  time = 0.229545, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {2021, 2024, 2011, 329, 220} \[ -\frac{8 a^3 \sqrt{a x+b x^3}}{231 b^2}+\frac{4 a^{15/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{231 b^{9/4} \sqrt{a x+b x^3}}+\frac{8 a^2 x^2 \sqrt{a x+b x^3}}{385 b}+\frac{4}{55} a x^4 \sqrt{a x+b x^3}+\frac{2}{15} x^3 \left (a x+b x^3\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a*x + b*x^3)^(3/2),x]

[Out]

(-8*a^3*Sqrt[a*x + b*x^3])/(231*b^2) + (8*a^2*x^2*Sqrt[a*x + b*x^3])/(385*b) + (4*a*x^4*Sqrt[a*x + b*x^3])/55
+ (2*x^3*(a*x + b*x^3)^(3/2))/15 + (4*a^(15/4)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[
b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(231*b^(9/4)*Sqrt[a*x + b*x^3])

Rule 2021

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b
*x^n)^p)/(c*(m + n*p + 1)), x] + Dist[(a*(n - j)*p)/(c^j*(m + n*p + 1)), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rule 2024

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^(n - j)*(m + j*p - n + j + 1))/(b*(m + n*p + 1)
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]

Rule 2011

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

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 x^2 \left (a x+b x^3\right )^{3/2} \, dx &=\frac{2}{15} x^3 \left (a x+b x^3\right )^{3/2}+\frac{1}{5} (2 a) \int x^3 \sqrt{a x+b x^3} \, dx\\ &=\frac{4}{55} a x^4 \sqrt{a x+b x^3}+\frac{2}{15} x^3 \left (a x+b x^3\right )^{3/2}+\frac{1}{55} \left (4 a^2\right ) \int \frac{x^4}{\sqrt{a x+b x^3}} \, dx\\ &=\frac{8 a^2 x^2 \sqrt{a x+b x^3}}{385 b}+\frac{4}{55} a x^4 \sqrt{a x+b x^3}+\frac{2}{15} x^3 \left (a x+b x^3\right )^{3/2}-\frac{\left (4 a^3\right ) \int \frac{x^2}{\sqrt{a x+b x^3}} \, dx}{77 b}\\ &=-\frac{8 a^3 \sqrt{a x+b x^3}}{231 b^2}+\frac{8 a^2 x^2 \sqrt{a x+b x^3}}{385 b}+\frac{4}{55} a x^4 \sqrt{a x+b x^3}+\frac{2}{15} x^3 \left (a x+b x^3\right )^{3/2}+\frac{\left (4 a^4\right ) \int \frac{1}{\sqrt{a x+b x^3}} \, dx}{231 b^2}\\ &=-\frac{8 a^3 \sqrt{a x+b x^3}}{231 b^2}+\frac{8 a^2 x^2 \sqrt{a x+b x^3}}{385 b}+\frac{4}{55} a x^4 \sqrt{a x+b x^3}+\frac{2}{15} x^3 \left (a x+b x^3\right )^{3/2}+\frac{\left (4 a^4 \sqrt{x} \sqrt{a+b x^2}\right ) \int \frac{1}{\sqrt{x} \sqrt{a+b x^2}} \, dx}{231 b^2 \sqrt{a x+b x^3}}\\ &=-\frac{8 a^3 \sqrt{a x+b x^3}}{231 b^2}+\frac{8 a^2 x^2 \sqrt{a x+b x^3}}{385 b}+\frac{4}{55} a x^4 \sqrt{a x+b x^3}+\frac{2}{15} x^3 \left (a x+b x^3\right )^{3/2}+\frac{\left (8 a^4 \sqrt{x} \sqrt{a+b x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^4}} \, dx,x,\sqrt{x}\right )}{231 b^2 \sqrt{a x+b x^3}}\\ &=-\frac{8 a^3 \sqrt{a x+b x^3}}{231 b^2}+\frac{8 a^2 x^2 \sqrt{a x+b x^3}}{385 b}+\frac{4}{55} a x^4 \sqrt{a x+b x^3}+\frac{2}{15} x^3 \left (a x+b x^3\right )^{3/2}+\frac{4 a^{15/4} \sqrt{x} \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{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{231 b^{9/4} \sqrt{a x+b x^3}}\\ \end{align*}

Mathematica [C]  time = 0.0545894, size = 94, normalized size = 0.51 \[ \frac{2 \sqrt{x \left (a+b x^2\right )} \left (5 a^3 \, _2F_1\left (-\frac{3}{2},\frac{1}{4};\frac{5}{4};-\frac{b x^2}{a}\right )-\left (5 a-11 b x^2\right ) \left (a+b x^2\right )^2 \sqrt{\frac{b x^2}{a}+1}\right )}{165 b^2 \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a*x + b*x^3)^(3/2),x]

[Out]

(2*Sqrt[x*(a + b*x^2)]*(-((5*a - 11*b*x^2)*(a + b*x^2)^2*Sqrt[1 + (b*x^2)/a]) + 5*a^3*Hypergeometric2F1[-3/2,
1/4, 5/4, -((b*x^2)/a)]))/(165*b^2*Sqrt[1 + (b*x^2)/a])

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Maple [A]  time = 0.014, size = 188, normalized size = 1. \begin{align*}{\frac{2\,b{x}^{6}}{15}\sqrt{b{x}^{3}+ax}}+{\frac{34\,a{x}^{4}}{165}\sqrt{b{x}^{3}+ax}}+{\frac{8\,{a}^{2}{x}^{2}}{385\,b}\sqrt{b{x}^{3}+ax}}-{\frac{8\,{a}^{3}}{231\,{b}^{2}}\sqrt{b{x}^{3}+ax}}+{\frac{4\,{a}^{4}}{231\,{b}^{3}}\sqrt{-ab}\sqrt{{b \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{-2\,{\frac{b}{\sqrt{-ab}} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }}\sqrt{-{bx{\frac{1}{\sqrt{-ab}}}}}{\it EllipticF} \left ( \sqrt{{b \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ){\frac{1}{\sqrt{-ab}}}}},{\frac{\sqrt{2}}{2}} \right ){\frac{1}{\sqrt{b{x}^{3}+ax}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*b*x^6*(b*x^3+a*x)^(1/2)+34/165*a*x^4*(b*x^3+a*x)^(1/2)+8/385*a^2*x^2*(b*x^3+a*x)^(1/2)/b-8/231*a^3*(b*x^3
+a*x)^(1/2)/b^2+4/231*a^4/b^3*(-a*b)^(1/2)*((x+1/b*(-a*b)^(1/2))*b/(-a*b)^(1/2))^(1/2)*(-2*(x-1/b*(-a*b)^(1/2)
)*b/(-a*b)^(1/2))^(1/2)*(-x*b/(-a*b)^(1/2))^(1/2)/(b*x^3+a*x)^(1/2)*EllipticF(((x+1/b*(-a*b)^(1/2))*b/(-a*b)^(
1/2))^(1/2),1/2*2^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a*x)^(3/2)*x^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b x^{5} + a x^{3}\right )} \sqrt{b x^{3} + a x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^5 + a*x^3)*sqrt(b*x^3 + a*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*(x*(a + b*x**2))**(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a*x)^(3/2)*x^2, x)