3.146 \(\int \frac{(b \sqrt [3]{x}+a x)^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=213 \[ \frac{4 a^{15/4} \sqrt [6]{x} \left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right ) \sqrt{\frac{a x^{2/3}+b}{\left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{77 b^{9/4} \sqrt{a x+b \sqrt [3]{x}}}+\frac{8 a^3 \sqrt{a x+b \sqrt [3]{x}}}{77 b^2 x^{2/3}}-\frac{24 a^2 \sqrt{a x+b \sqrt [3]{x}}}{385 b x^{4/3}}-\frac{12 a \sqrt{a x+b \sqrt [3]{x}}}{55 x^2}-\frac{2 \left (a x+b \sqrt [3]{x}\right )^{3/2}}{5 x^3} \]

[Out]

(-12*a*Sqrt[b*x^(1/3) + a*x])/(55*x^2) - (24*a^2*Sqrt[b*x^(1/3) + a*x])/(385*b*x^(4/3)) + (8*a^3*Sqrt[b*x^(1/3
) + a*x])/(77*b^2*x^(2/3)) - (2*(b*x^(1/3) + a*x)^(3/2))/(5*x^3) + (4*a^(15/4)*(Sqrt[b] + Sqrt[a]*x^(1/3))*Sqr
t[(b + a*x^(2/3))/(Sqrt[b] + Sqrt[a]*x^(1/3))^2]*x^(1/6)*EllipticF[2*ArcTan[(a^(1/4)*x^(1/6))/b^(1/4)], 1/2])/
(77*b^(9/4)*Sqrt[b*x^(1/3) + a*x])

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Rubi [A]  time = 0.308573, antiderivative size = 213, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {2018, 2020, 2025, 2011, 329, 220} \[ \frac{8 a^3 \sqrt{a x+b \sqrt [3]{x}}}{77 b^2 x^{2/3}}+\frac{4 a^{15/4} \sqrt [6]{x} \left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right ) \sqrt{\frac{a x^{2/3}+b}{\left (\sqrt{a} \sqrt [3]{x}+\sqrt{b}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{77 b^{9/4} \sqrt{a x+b \sqrt [3]{x}}}-\frac{24 a^2 \sqrt{a x+b \sqrt [3]{x}}}{385 b x^{4/3}}-\frac{12 a \sqrt{a x+b \sqrt [3]{x}}}{55 x^2}-\frac{2 \left (a x+b \sqrt [3]{x}\right )^{3/2}}{5 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^(1/3) + a*x)^(3/2)/x^4,x]

[Out]

(-12*a*Sqrt[b*x^(1/3) + a*x])/(55*x^2) - (24*a^2*Sqrt[b*x^(1/3) + a*x])/(385*b*x^(4/3)) + (8*a^3*Sqrt[b*x^(1/3
) + a*x])/(77*b^2*x^(2/3)) - (2*(b*x^(1/3) + a*x)^(3/2))/(5*x^3) + (4*a^(15/4)*(Sqrt[b] + Sqrt[a]*x^(1/3))*Sqr
t[(b + a*x^(2/3))/(Sqrt[b] + Sqrt[a]*x^(1/3))^2]*x^(1/6)*EllipticF[2*ArcTan[(a^(1/4)*x^(1/6))/b^(1/4)], 1/2])/
(77*b^(9/4)*Sqrt[b*x^(1/3) + a*x])

Rule 2018

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rule 2020

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 + j*p + 1)), x] - Dist[(b*p*(n - j))/(c^n*(m + j*p + 1)), Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p -
1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*p + 1, 0]

Rule 2025

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*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]) && LtQ[m + j*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 \frac{\left (b \sqrt [3]{x}+a x\right )^{3/2}}{x^4} \, dx &=3 \operatorname{Subst}\left (\int \frac{\left (b x+a x^3\right )^{3/2}}{x^{10}} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac{2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{5 x^3}+\frac{1}{5} (6 a) \operatorname{Subst}\left (\int \frac{\sqrt{b x+a x^3}}{x^7} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac{12 a \sqrt{b \sqrt [3]{x}+a x}}{55 x^2}-\frac{2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{5 x^3}+\frac{1}{55} \left (12 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{x^4 \sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac{12 a \sqrt{b \sqrt [3]{x}+a x}}{55 x^2}-\frac{24 a^2 \sqrt{b \sqrt [3]{x}+a x}}{385 b x^{4/3}}-\frac{2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{5 x^3}-\frac{\left (12 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{77 b}\\ &=-\frac{12 a \sqrt{b \sqrt [3]{x}+a x}}{55 x^2}-\frac{24 a^2 \sqrt{b \sqrt [3]{x}+a x}}{385 b x^{4/3}}+\frac{8 a^3 \sqrt{b \sqrt [3]{x}+a x}}{77 b^2 x^{2/3}}-\frac{2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{5 x^3}+\frac{\left (4 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b x+a x^3}} \, dx,x,\sqrt [3]{x}\right )}{77 b^2}\\ &=-\frac{12 a \sqrt{b \sqrt [3]{x}+a x}}{55 x^2}-\frac{24 a^2 \sqrt{b \sqrt [3]{x}+a x}}{385 b x^{4/3}}+\frac{8 a^3 \sqrt{b \sqrt [3]{x}+a x}}{77 b^2 x^{2/3}}-\frac{2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{5 x^3}+\frac{\left (4 a^4 \sqrt{b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{b+a x^2}} \, dx,x,\sqrt [3]{x}\right )}{77 b^2 \sqrt{b \sqrt [3]{x}+a x}}\\ &=-\frac{12 a \sqrt{b \sqrt [3]{x}+a x}}{55 x^2}-\frac{24 a^2 \sqrt{b \sqrt [3]{x}+a x}}{385 b x^{4/3}}+\frac{8 a^3 \sqrt{b \sqrt [3]{x}+a x}}{77 b^2 x^{2/3}}-\frac{2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{5 x^3}+\frac{\left (8 a^4 \sqrt{b+a x^{2/3}} \sqrt [6]{x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b+a x^4}} \, dx,x,\sqrt [6]{x}\right )}{77 b^2 \sqrt{b \sqrt [3]{x}+a x}}\\ &=-\frac{12 a \sqrt{b \sqrt [3]{x}+a x}}{55 x^2}-\frac{24 a^2 \sqrt{b \sqrt [3]{x}+a x}}{385 b x^{4/3}}+\frac{8 a^3 \sqrt{b \sqrt [3]{x}+a x}}{77 b^2 x^{2/3}}-\frac{2 \left (b \sqrt [3]{x}+a x\right )^{3/2}}{5 x^3}+\frac{4 a^{15/4} \left (\sqrt{b}+\sqrt{a} \sqrt [3]{x}\right ) \sqrt{\frac{b+a x^{2/3}}{\left (\sqrt{b}+\sqrt{a} \sqrt [3]{x}\right )^2}} \sqrt [6]{x} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt [6]{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{77 b^{9/4} \sqrt{b \sqrt [3]{x}+a x}}\\ \end{align*}

Mathematica [C]  time = 0.0661396, size = 62, normalized size = 0.29 \[ -\frac{2 b \sqrt{a x+b \sqrt [3]{x}} \, _2F_1\left (-\frac{15}{4},-\frac{3}{2};-\frac{11}{4};-\frac{a x^{2/3}}{b}\right )}{5 x^{8/3} \sqrt{\frac{a x^{2/3}}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^(1/3) + a*x)^(3/2)/x^4,x]

[Out]

(-2*b*Sqrt[b*x^(1/3) + a*x]*Hypergeometric2F1[-15/4, -3/2, -11/4, -((a*x^(2/3))/b)])/(5*Sqrt[1 + (a*x^(2/3))/b
]*x^(8/3))

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Maple [A]  time = 0.023, size = 168, normalized size = 0.8 \begin{align*}{\frac{2}{385\,{b}^{2}} \left ( 10\,{a}^{3}\sqrt{-ab}\sqrt{{\frac{a\sqrt [3]{x}+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-2\,{\frac{a\sqrt [3]{x}-\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{a\sqrt [3]{x}}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{a\sqrt [3]{x}+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ){x}^{14/3}-131\,{x}^{11/3}{a}^{2}{b}^{2}+8\,{x}^{13/3}{a}^{3}b-196\,a{b}^{3}{x}^{3}+20\,{x}^{5}{a}^{4}-77\,{x}^{7/3}{b}^{4} \right ){\frac{1}{\sqrt{\sqrt [3]{x} \left ( b+a{x}^{{\frac{2}{3}}} \right ) }}}{x}^{-{\frac{14}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^(1/3)+a*x)^(3/2)/x^4,x)

[Out]

2/385*(10*a^3*(-a*b)^(1/2)*((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-2*(a*x^(1/3)-(-a*b)^(1/2))/(-a*b)^(
1/2))^(1/2)*(-x^(1/3)*a/(-a*b)^(1/2))^(1/2)*EllipticF(((a*x^(1/3)+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2
))*x^(14/3)-131*x^(11/3)*a^2*b^2+8*x^(13/3)*a^3*b-196*a*b^3*x^3+20*x^5*a^4-77*x^(7/3)*b^4)/b^2/(x^(1/3)*(b+a*x
^(2/3)))^(1/2)/x^(14/3)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + b x^{\frac{1}{3}}\right )}^{\frac{3}{2}}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^(1/3)+a*x)^(3/2)/x^4,x, algorithm="maxima")

[Out]

integrate((a*x + b*x^(1/3))^(3/2)/x^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a x + b x^{\frac{1}{3}}\right )}^{\frac{3}{2}}}{x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^(1/3)+a*x)^(3/2)/x^4,x, algorithm="fricas")

[Out]

integral((a*x + b*x^(1/3))^(3/2)/x^4, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**(1/3)+a*x)**(3/2)/x**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + b x^{\frac{1}{3}}\right )}^{\frac{3}{2}}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^(1/3)+a*x)^(3/2)/x^4,x, algorithm="giac")

[Out]

integrate((a*x + b*x^(1/3))^(3/2)/x^4, x)