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

Optimal. Leaf size=178 \[ \frac{21 a^4 \sqrt{a x+b x^{2/3}}}{128 b^4 x^{2/3}}-\frac{7 a^3 \sqrt{a x+b x^{2/3}}}{64 b^3 x}+\frac{7 a^2 \sqrt{a x+b x^{2/3}}}{80 b^2 x^{4/3}}-\frac{21 a^5 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [3]{x}}{\sqrt{a x+b x^{2/3}}}\right )}{128 b^{9/2}}-\frac{3 a \sqrt{a x+b x^{2/3}}}{40 b x^{5/3}}-\frac{3 \sqrt{a x+b x^{2/3}}}{5 x^2} \]

[Out]

(-3*Sqrt[b*x^(2/3) + a*x])/(5*x^2) - (3*a*Sqrt[b*x^(2/3) + a*x])/(40*b*x^(5/3)) + (7*a^2*Sqrt[b*x^(2/3) + a*x]
)/(80*b^2*x^(4/3)) - (7*a^3*Sqrt[b*x^(2/3) + a*x])/(64*b^3*x) + (21*a^4*Sqrt[b*x^(2/3) + a*x])/(128*b^4*x^(2/3
)) - (21*a^5*ArcTanh[(Sqrt[b]*x^(1/3))/Sqrt[b*x^(2/3) + a*x]])/(128*b^(9/2))

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Rubi [A]  time = 0.295909, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2020, 2025, 2029, 206} \[ \frac{21 a^4 \sqrt{a x+b x^{2/3}}}{128 b^4 x^{2/3}}-\frac{7 a^3 \sqrt{a x+b x^{2/3}}}{64 b^3 x}+\frac{7 a^2 \sqrt{a x+b x^{2/3}}}{80 b^2 x^{4/3}}-\frac{21 a^5 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [3]{x}}{\sqrt{a x+b x^{2/3}}}\right )}{128 b^{9/2}}-\frac{3 a \sqrt{a x+b x^{2/3}}}{40 b x^{5/3}}-\frac{3 \sqrt{a x+b x^{2/3}}}{5 x^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x^(2/3) + a*x]/x^3,x]

[Out]

(-3*Sqrt[b*x^(2/3) + a*x])/(5*x^2) - (3*a*Sqrt[b*x^(2/3) + a*x])/(40*b*x^(5/3)) + (7*a^2*Sqrt[b*x^(2/3) + a*x]
)/(80*b^2*x^(4/3)) - (7*a^3*Sqrt[b*x^(2/3) + a*x])/(64*b^3*x) + (21*a^4*Sqrt[b*x^(2/3) + a*x])/(128*b^4*x^(2/3
)) - (21*a^5*ArcTanh[(Sqrt[b]*x^(1/3))/Sqrt[b*x^(2/3) + a*x]])/(128*b^(9/2))

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{\sqrt{b x^{2/3}+a x}}{x^3} \, dx &=-\frac{3 \sqrt{b x^{2/3}+a x}}{5 x^2}+\frac{1}{10} a \int \frac{1}{x^2 \sqrt{b x^{2/3}+a x}} \, dx\\ &=-\frac{3 \sqrt{b x^{2/3}+a x}}{5 x^2}-\frac{3 a \sqrt{b x^{2/3}+a x}}{40 b x^{5/3}}-\frac{\left (7 a^2\right ) \int \frac{1}{x^{5/3} \sqrt{b x^{2/3}+a x}} \, dx}{80 b}\\ &=-\frac{3 \sqrt{b x^{2/3}+a x}}{5 x^2}-\frac{3 a \sqrt{b x^{2/3}+a x}}{40 b x^{5/3}}+\frac{7 a^2 \sqrt{b x^{2/3}+a x}}{80 b^2 x^{4/3}}+\frac{\left (7 a^3\right ) \int \frac{1}{x^{4/3} \sqrt{b x^{2/3}+a x}} \, dx}{96 b^2}\\ &=-\frac{3 \sqrt{b x^{2/3}+a x}}{5 x^2}-\frac{3 a \sqrt{b x^{2/3}+a x}}{40 b x^{5/3}}+\frac{7 a^2 \sqrt{b x^{2/3}+a x}}{80 b^2 x^{4/3}}-\frac{7 a^3 \sqrt{b x^{2/3}+a x}}{64 b^3 x}-\frac{\left (7 a^4\right ) \int \frac{1}{x \sqrt{b x^{2/3}+a x}} \, dx}{128 b^3}\\ &=-\frac{3 \sqrt{b x^{2/3}+a x}}{5 x^2}-\frac{3 a \sqrt{b x^{2/3}+a x}}{40 b x^{5/3}}+\frac{7 a^2 \sqrt{b x^{2/3}+a x}}{80 b^2 x^{4/3}}-\frac{7 a^3 \sqrt{b x^{2/3}+a x}}{64 b^3 x}+\frac{21 a^4 \sqrt{b x^{2/3}+a x}}{128 b^4 x^{2/3}}+\frac{\left (7 a^5\right ) \int \frac{1}{x^{2/3} \sqrt{b x^{2/3}+a x}} \, dx}{256 b^4}\\ &=-\frac{3 \sqrt{b x^{2/3}+a x}}{5 x^2}-\frac{3 a \sqrt{b x^{2/3}+a x}}{40 b x^{5/3}}+\frac{7 a^2 \sqrt{b x^{2/3}+a x}}{80 b^2 x^{4/3}}-\frac{7 a^3 \sqrt{b x^{2/3}+a x}}{64 b^3 x}+\frac{21 a^4 \sqrt{b x^{2/3}+a x}}{128 b^4 x^{2/3}}-\frac{\left (21 a^5\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 )}{128 b^4}\\ &=-\frac{3 \sqrt{b x^{2/3}+a x}}{5 x^2}-\frac{3 a \sqrt{b x^{2/3}+a x}}{40 b x^{5/3}}+\frac{7 a^2 \sqrt{b x^{2/3}+a x}}{80 b^2 x^{4/3}}-\frac{7 a^3 \sqrt{b x^{2/3}+a x}}{64 b^3 x}+\frac{21 a^4 \sqrt{b x^{2/3}+a x}}{128 b^4 x^{2/3}}-\frac{21 a^5 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [3]{x}}{\sqrt{b x^{2/3}+a x}}\right )}{128 b^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0408602, size = 57, normalized size = 0.32 \[ \frac{2 a^5 \left (a \sqrt [3]{x}+b\right ) \sqrt{a x+b x^{2/3}} \, _2F_1\left (\frac{3}{2},6;\frac{5}{2};\frac{\sqrt [3]{x} a}{b}+1\right )}{b^6 \sqrt [3]{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.011, size = 125, normalized size = 0.7 \begin{align*} -{\frac{1}{640\,{x}^{2}}\sqrt{b{x}^{{\frac{2}{3}}}+ax} \left ( 105\,{b}^{17/2}\sqrt{b+a\sqrt [3]{x}}+790\,{b}^{15/2} \left ( b+a\sqrt [3]{x} \right ) ^{3/2}-896\,{b}^{13/2} \left ( b+a\sqrt [3]{x} \right ) ^{5/2}+490\,{b}^{11/2} \left ( b+a\sqrt [3]{x} \right ) ^{7/2}-105\,{b}^{9/2} \left ( b+a\sqrt [3]{x} \right ) ^{9/2}+105\,{\it Artanh} \left ({\frac{\sqrt{b+a\sqrt [3]{x}}}{\sqrt{b}}} \right ){b}^{4}{a}^{5}{x}^{5/3} \right ){b}^{-{\frac{17}{2}}}{\frac{1}{\sqrt{b+a\sqrt [3]{x}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/640*(b*x^(2/3)+a*x)^(1/2)*(105*b^(17/2)*(b+a*x^(1/3))^(1/2)+790*b^(15/2)*(b+a*x^(1/3))^(3/2)-896*b^(13/2)*(
b+a*x^(1/3))^(5/2)+490*b^(11/2)*(b+a*x^(1/3))^(7/2)-105*b^(9/2)*(b+a*x^(1/3))^(9/2)+105*arctanh((b+a*x^(1/3))^
(1/2)/b^(1/2))*b^4*a^5*x^(5/3))/x^2/(b+a*x^(1/3))^(1/2)/b^(17/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a*x + b*x^(2/3))/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)^(1/2)/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{\sqrt{a x + b x^{\frac{2}{3}}}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20498, size = 170, normalized size = 0.96 \begin{align*} \frac{\frac{105 \, a^{6} \arctan \left (\frac{\sqrt{a x^{\frac{1}{3}} + b}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{4}} + \frac{105 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{9}{2}} a^{6} - 490 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{7}{2}} a^{6} b + 896 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{5}{2}} a^{6} b^{2} - 790 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{3}{2}} a^{6} b^{3} - 105 \, \sqrt{a x^{\frac{1}{3}} + b} a^{6} b^{4}}{a^{5} b^{4} x^{\frac{5}{3}}}}{640 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/640*(105*a^6*arctan(sqrt(a*x^(1/3) + b)/sqrt(-b))/(sqrt(-b)*b^4) + (105*(a*x^(1/3) + b)^(9/2)*a^6 - 490*(a*x
^(1/3) + b)^(7/2)*a^6*b + 896*(a*x^(1/3) + b)^(5/2)*a^6*b^2 - 790*(a*x^(1/3) + b)^(3/2)*a^6*b^3 - 105*sqrt(a*x
^(1/3) + b)*a^6*b^4)/(a^5*b^4*x^(5/3)))/a