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

Optimal. Leaf size=113 \[ \frac{3 a^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [3]{x}}{\sqrt{a x+b x^{2/3}}}\right )}{8 b^{3/2}}-\frac{3 a^2 \sqrt{a x+b x^{2/3}}}{8 b x^{2/3}}-\frac{3 a \sqrt{a x+b x^{2/3}}}{4 x}-\frac{\left (a x+b x^{2/3}\right )^{3/2}}{x^2} \]

[Out]

(-3*a*Sqrt[b*x^(2/3) + a*x])/(4*x) - (3*a^2*Sqrt[b*x^(2/3) + a*x])/(8*b*x^(2/3)) - (b*x^(2/3) + a*x)^(3/2)/x^2
+ (3*a^3*ArcTanh[(Sqrt[b]*x^(1/3))/Sqrt[b*x^(2/3) + a*x]])/(8*b^(3/2))

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Rubi [A]  time = 0.183797, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2020, 2025, 2029, 206} \[ \frac{3 a^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [3]{x}}{\sqrt{a x+b x^{2/3}}}\right )}{8 b^{3/2}}-\frac{3 a^2 \sqrt{a x+b x^{2/3}}}{8 b x^{2/3}}-\frac{3 a \sqrt{a x+b x^{2/3}}}{4 x}-\frac{\left (a x+b x^{2/3}\right )^{3/2}}{x^2} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^(2/3) + a*x)^(3/2)/x^3,x]

[Out]

(-3*a*Sqrt[b*x^(2/3) + a*x])/(4*x) - (3*a^2*Sqrt[b*x^(2/3) + a*x])/(8*b*x^(2/3)) - (b*x^(2/3) + a*x)^(3/2)/x^2
+ (3*a^3*ArcTanh[(Sqrt[b]*x^(1/3))/Sqrt[b*x^(2/3) + a*x]])/(8*b^(3/2))

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 2029

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\left (b x^{2/3}+a x\right )^{3/2}}{x^3} \, dx &=-\frac{\left (b x^{2/3}+a x\right )^{3/2}}{x^2}+\frac{1}{2} a \int \frac{\sqrt{b x^{2/3}+a x}}{x^2} \, dx\\ &=-\frac{3 a \sqrt{b x^{2/3}+a x}}{4 x}-\frac{\left (b x^{2/3}+a x\right )^{3/2}}{x^2}+\frac{1}{8} a^2 \int \frac{1}{x \sqrt{b x^{2/3}+a x}} \, dx\\ &=-\frac{3 a \sqrt{b x^{2/3}+a x}}{4 x}-\frac{3 a^2 \sqrt{b x^{2/3}+a x}}{8 b x^{2/3}}-\frac{\left (b x^{2/3}+a x\right )^{3/2}}{x^2}-\frac{a^3 \int \frac{1}{x^{2/3} \sqrt{b x^{2/3}+a x}} \, dx}{16 b}\\ &=-\frac{3 a \sqrt{b x^{2/3}+a x}}{4 x}-\frac{3 a^2 \sqrt{b x^{2/3}+a x}}{8 b x^{2/3}}-\frac{\left (b x^{2/3}+a x\right )^{3/2}}{x^2}+\frac{\left (3 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sqrt [3]{x}}{\sqrt{b x^{2/3}+a x}}\right )}{8 b}\\ &=-\frac{3 a \sqrt{b x^{2/3}+a x}}{4 x}-\frac{3 a^2 \sqrt{b x^{2/3}+a x}}{8 b x^{2/3}}-\frac{\left (b x^{2/3}+a x\right )^{3/2}}{x^2}+\frac{3 a^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [3]{x}}{\sqrt{b x^{2/3}+a x}}\right )}{8 b^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0525169, size = 61, normalized size = 0.54 \[ \frac{6 a^3 \left (a \sqrt [3]{x}+b\right )^2 \sqrt{a x+b x^{2/3}} \, _2F_1\left (\frac{5}{2},4;\frac{7}{2};\frac{\sqrt [3]{x} a}{b}+1\right )}{5 b^4 \sqrt [3]{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a^3*(b + a*x^(1/3))^2*Sqrt[b*x^(2/3) + a*x]*Hypergeometric2F1[5/2, 4, 7/2, 1 + (a*x^(1/3))/b])/(5*b^4*x^(1/
3))

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Maple [A]  time = 0.01, size = 93, normalized size = 0.8 \begin{align*}{\frac{1}{8\,{x}^{2}} \left ( b{x}^{{\frac{2}{3}}}+ax \right ) ^{{\frac{3}{2}}} \left ( 3\,{b}^{7/2}\sqrt{b+a\sqrt [3]{x}}-8\,{b}^{5/2} \left ( b+a\sqrt [3]{x} \right ) ^{3/2}-3\,{b}^{3/2} \left ( b+a\sqrt [3]{x} \right ) ^{5/2}+3\,{\it Artanh} \left ({\frac{\sqrt{b+a\sqrt [3]{x}}}{\sqrt{b}}} \right ) b{a}^{3}x \right ){b}^{-{\frac{5}{2}}} \left ( b+a\sqrt [3]{x} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*x^(2/3)+a*x)^(3/2)*(3*b^(7/2)*(b+a*x^(1/3))^(1/2)-8*b^(5/2)*(b+a*x^(1/3))^(3/2)-3*b^(3/2)*(b+a*x^(1/3))
^(5/2)+3*arctanh((b+a*x^(1/3))^(1/2)/b^(1/2))*b*a^3*x)/x^2/(b+a*x^(1/3))^(3/2)/b^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [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^(2/3)+a*x)^(3/2)/x^3,x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.24507, size = 124, normalized size = 1.1 \begin{align*} -\frac{\frac{3 \, a^{4} \arctan \left (\frac{\sqrt{a x^{\frac{1}{3}} + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b} + \frac{3 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{5}{2}} a^{4} + 8 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{3}{2}} a^{4} b - 3 \, \sqrt{a x^{\frac{1}{3}} + b} a^{4} b^{2}}{a^{3} b x}}{8 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(3*a^4*arctan(sqrt(a*x^(1/3) + b)/sqrt(-b))/(sqrt(-b)*b) + (3*(a*x^(1/3) + b)^(5/2)*a^4 + 8*(a*x^(1/3) +
b)^(3/2)*a^4*b - 3*sqrt(a*x^(1/3) + b)*a^4*b^2)/(a^3*b*x))/a