3.674 \(\int \frac{x^{5/3}}{a+b x} \, dx\)

Optimal. Leaf size=125 \[ -\frac{3 a^{5/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 b^{8/3}}+\frac{a^{5/3} \log (a+b x)}{2 b^{8/3}}-\frac{\sqrt{3} a^{5/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt{3} \sqrt [3]{a}}\right )}{b^{8/3}}-\frac{3 a x^{2/3}}{2 b^2}+\frac{3 x^{5/3}}{5 b} \]

[Out]

(-3*a*x^(2/3))/(2*b^2) + (3*x^(5/3))/(5*b) - (Sqrt[3]*a^(5/3)*ArcTan[(a^(1/3) -
2*b^(1/3)*x^(1/3))/(Sqrt[3]*a^(1/3))])/b^(8/3) - (3*a^(5/3)*Log[a^(1/3) + b^(1/3
)*x^(1/3)])/(2*b^(8/3)) + (a^(5/3)*Log[a + b*x])/(2*b^(8/3))

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Rubi [A]  time = 0.142331, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385 \[ -\frac{3 a^{5/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 b^{8/3}}+\frac{a^{5/3} \log (a+b x)}{2 b^{8/3}}-\frac{\sqrt{3} a^{5/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt{3} \sqrt [3]{a}}\right )}{b^{8/3}}-\frac{3 a x^{2/3}}{2 b^2}+\frac{3 x^{5/3}}{5 b} \]

Antiderivative was successfully verified.

[In]  Int[x^(5/3)/(a + b*x),x]

[Out]

(-3*a*x^(2/3))/(2*b^2) + (3*x^(5/3))/(5*b) - (Sqrt[3]*a^(5/3)*ArcTan[(a^(1/3) -
2*b^(1/3)*x^(1/3))/(Sqrt[3]*a^(1/3))])/b^(8/3) - (3*a^(5/3)*Log[a^(1/3) + b^(1/3
)*x^(1/3)])/(2*b^(8/3)) + (a^(5/3)*Log[a + b*x])/(2*b^(8/3))

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Rubi in Sympy [A]  time = 15.3539, size = 119, normalized size = 0.95 \[ - \frac{3 a^{\frac{5}{3}} \log{\left (\sqrt [3]{a} + \sqrt [3]{b} \sqrt [3]{x} \right )}}{2 b^{\frac{8}{3}}} + \frac{a^{\frac{5}{3}} \log{\left (a + b x \right )}}{2 b^{\frac{8}{3}}} - \frac{\sqrt{3} a^{\frac{5}{3}} \operatorname{atan}{\left (\frac{\sqrt{3} \left (\frac{\sqrt [3]{a}}{3} - \frac{2 \sqrt [3]{b} \sqrt [3]{x}}{3}\right )}{\sqrt [3]{a}} \right )}}{b^{\frac{8}{3}}} - \frac{3 a x^{\frac{2}{3}}}{2 b^{2}} + \frac{3 x^{\frac{5}{3}}}{5 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x**(5/3)/(b*x+a),x)

[Out]

-3*a**(5/3)*log(a**(1/3) + b**(1/3)*x**(1/3))/(2*b**(8/3)) + a**(5/3)*log(a + b*
x)/(2*b**(8/3)) - sqrt(3)*a**(5/3)*atan(sqrt(3)*(a**(1/3)/3 - 2*b**(1/3)*x**(1/3
)/3)/a**(1/3))/b**(8/3) - 3*a*x**(2/3)/(2*b**2) + 3*x**(5/3)/(5*b)

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Mathematica [A]  time = 0.078127, size = 140, normalized size = 1.12 \[ \frac{5 a^{5/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sqrt [3]{x}+b^{2/3} x^{2/3}\right )-10 a^{5/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )-10 \sqrt{3} a^{5/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a}}}{\sqrt{3}}\right )-15 a b^{2/3} x^{2/3}+6 b^{5/3} x^{5/3}}{10 b^{8/3}} \]

Antiderivative was successfully verified.

[In]  Integrate[x^(5/3)/(a + b*x),x]

[Out]

(-15*a*b^(2/3)*x^(2/3) + 6*b^(5/3)*x^(5/3) - 10*Sqrt[3]*a^(5/3)*ArcTan[(1 - (2*b
^(1/3)*x^(1/3))/a^(1/3))/Sqrt[3]] - 10*a^(5/3)*Log[a^(1/3) + b^(1/3)*x^(1/3)] +
5*a^(5/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x^(1/3) + b^(2/3)*x^(2/3)])/(10*b^(8/3))

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Maple [A]  time = 0.01, size = 122, normalized size = 1. \[{\frac{3}{5\,b}{x}^{{\frac{5}{3}}}}-{\frac{3\,a}{2\,{b}^{2}}{x}^{{\frac{2}{3}}}}-{\frac{{a}^{2}}{{b}^{3}}\ln \left ( \sqrt [3]{x}+\sqrt [3]{{\frac{a}{b}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}+{\frac{{a}^{2}}{2\,{b}^{3}}\ln \left ({x}^{{\frac{2}{3}}}-\sqrt [3]{x}\sqrt [3]{{\frac{a}{b}}}+ \left ({\frac{a}{b}} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}+{\frac{{a}^{2}\sqrt{3}}{{b}^{3}}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\sqrt [3]{x}{\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-1 \right ) } \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(x^(5/3)/(b*x+a),x)

[Out]

3/5*x^(5/3)/b-3/2*a*x^(2/3)/b^2-a^2/b^3/(a/b)^(1/3)*ln(x^(1/3)+(a/b)^(1/3))+1/2*
a^2/b^3/(a/b)^(1/3)*ln(x^(2/3)-x^(1/3)*(a/b)^(1/3)+(a/b)^(2/3))+a^2/b^3*3^(1/2)/
(a/b)^(1/3)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x^(1/3)-1))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^(5/3)/(b*x + a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.220763, size = 208, normalized size = 1.66 \[ -\frac{10 \, \sqrt{3} a \left (-\frac{a^{2}}{b^{2}}\right )^{\frac{1}{3}} \arctan \left (-\frac{\sqrt{3}{\left (b \left (-\frac{a^{2}}{b^{2}}\right )^{\frac{2}{3}} - 2 \, a x^{\frac{1}{3}}\right )}}{3 \, b \left (-\frac{a^{2}}{b^{2}}\right )^{\frac{2}{3}}}\right ) + 5 \, a \left (-\frac{a^{2}}{b^{2}}\right )^{\frac{1}{3}} \log \left (-b x^{\frac{1}{3}} \left (-\frac{a^{2}}{b^{2}}\right )^{\frac{2}{3}} + a x^{\frac{2}{3}} - a \left (-\frac{a^{2}}{b^{2}}\right )^{\frac{1}{3}}\right ) - 10 \, a \left (-\frac{a^{2}}{b^{2}}\right )^{\frac{1}{3}} \log \left (b \left (-\frac{a^{2}}{b^{2}}\right )^{\frac{2}{3}} + a x^{\frac{1}{3}}\right ) - 3 \,{\left (2 \, b x - 5 \, a\right )} x^{\frac{2}{3}}}{10 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^(5/3)/(b*x + a),x, algorithm="fricas")

[Out]

-1/10*(10*sqrt(3)*a*(-a^2/b^2)^(1/3)*arctan(-1/3*sqrt(3)*(b*(-a^2/b^2)^(2/3) - 2
*a*x^(1/3))/(b*(-a^2/b^2)^(2/3))) + 5*a*(-a^2/b^2)^(1/3)*log(-b*x^(1/3)*(-a^2/b^
2)^(2/3) + a*x^(2/3) - a*(-a^2/b^2)^(1/3)) - 10*a*(-a^2/b^2)^(1/3)*log(b*(-a^2/b
^2)^(2/3) + a*x^(1/3)) - 3*(2*b*x - 5*a)*x^(2/3))/b^2

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Sympy [A]  time = 6.29798, size = 206, normalized size = 1.65 \[ - \frac{8 a^{\frac{5}{3}} e^{\frac{10 i \pi }{3}} \log{\left (1 - \frac{\sqrt [3]{b} \sqrt [3]{x} e^{\frac{i \pi }{3}}}{\sqrt [3]{a}} \right )} \Gamma \left (\frac{8}{3}\right )}{3 b^{\frac{8}{3}} \Gamma \left (\frac{11}{3}\right )} - \frac{8 a^{\frac{5}{3}} \log{\left (1 - \frac{\sqrt [3]{b} \sqrt [3]{x} e^{i \pi }}{\sqrt [3]{a}} \right )} \Gamma \left (\frac{8}{3}\right )}{3 b^{\frac{8}{3}} \Gamma \left (\frac{11}{3}\right )} - \frac{8 a^{\frac{5}{3}} e^{\frac{2 i \pi }{3}} \log{\left (1 - \frac{\sqrt [3]{b} \sqrt [3]{x} e^{\frac{5 i \pi }{3}}}{\sqrt [3]{a}} \right )} \Gamma \left (\frac{8}{3}\right )}{3 b^{\frac{8}{3}} \Gamma \left (\frac{11}{3}\right )} - \frac{4 a x^{\frac{2}{3}} \Gamma \left (\frac{8}{3}\right )}{b^{2} \Gamma \left (\frac{11}{3}\right )} + \frac{8 x^{\frac{5}{3}} \Gamma \left (\frac{8}{3}\right )}{5 b \Gamma \left (\frac{11}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x**(5/3)/(b*x+a),x)

[Out]

-8*a**(5/3)*exp(10*I*pi/3)*log(1 - b**(1/3)*x**(1/3)*exp_polar(I*pi/3)/a**(1/3))
*gamma(8/3)/(3*b**(8/3)*gamma(11/3)) - 8*a**(5/3)*log(1 - b**(1/3)*x**(1/3)*exp_
polar(I*pi)/a**(1/3))*gamma(8/3)/(3*b**(8/3)*gamma(11/3)) - 8*a**(5/3)*exp(2*I*p
i/3)*log(1 - b**(1/3)*x**(1/3)*exp_polar(5*I*pi/3)/a**(1/3))*gamma(8/3)/(3*b**(8
/3)*gamma(11/3)) - 4*a*x**(2/3)*gamma(8/3)/(b**2*gamma(11/3)) + 8*x**(5/3)*gamma
(8/3)/(5*b*gamma(11/3))

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GIAC/XCAS [A]  time = 0.217947, size = 186, normalized size = 1.49 \[ -\frac{a \left (-\frac{a}{b}\right )^{\frac{2}{3}}{\rm ln}\left ({\left | x^{\frac{1}{3}} - \left (-\frac{a}{b}\right )^{\frac{1}{3}} \right |}\right )}{b^{2}} - \frac{\sqrt{3} \left (-a b^{2}\right )^{\frac{2}{3}} a \arctan \left (\frac{\sqrt{3}{\left (2 \, x^{\frac{1}{3}} + \left (-\frac{a}{b}\right )^{\frac{1}{3}}\right )}}{3 \, \left (-\frac{a}{b}\right )^{\frac{1}{3}}}\right )}{b^{4}} + \frac{\left (-a b^{2}\right )^{\frac{2}{3}} a{\rm ln}\left (x^{\frac{2}{3}} + x^{\frac{1}{3}} \left (-\frac{a}{b}\right )^{\frac{1}{3}} + \left (-\frac{a}{b}\right )^{\frac{2}{3}}\right )}{2 \, b^{4}} + \frac{3 \,{\left (2 \, b^{4} x^{\frac{5}{3}} - 5 \, a b^{3} x^{\frac{2}{3}}\right )}}{10 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^(5/3)/(b*x + a),x, algorithm="giac")

[Out]

-a*(-a/b)^(2/3)*ln(abs(x^(1/3) - (-a/b)^(1/3)))/b^2 - sqrt(3)*(-a*b^2)^(2/3)*a*a
rctan(1/3*sqrt(3)*(2*x^(1/3) + (-a/b)^(1/3))/(-a/b)^(1/3))/b^4 + 1/2*(-a*b^2)^(2
/3)*a*ln(x^(2/3) + x^(1/3)*(-a/b)^(1/3) + (-a/b)^(2/3))/b^4 + 3/10*(2*b^4*x^(5/3
) - 5*a*b^3*x^(2/3))/b^5