3.682 \(\int \frac {x^{5/3}}{(a+b x)^2} \, dx\)

Optimal. Leaf size=129 \[ \frac {5 a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 b^{8/3}}-\frac {5 a^{2/3} \log (a+b x)}{6 b^{8/3}}+\frac {5 a^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{8/3}}-\frac {x^{5/3}}{b (a+b x)}+\frac {5 x^{2/3}}{2 b^2} \]

[Out]

5/2*x^(2/3)/b^2-x^(5/3)/b/(b*x+a)+5/2*a^(2/3)*ln(a^(1/3)+b^(1/3)*x^(1/3))/b^(8/3)-5/6*a^(2/3)*ln(b*x+a)/b^(8/3
)+5/3*a^(2/3)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x^(1/3))/a^(1/3)*3^(1/2))/b^(8/3)*3^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {47, 50, 56, 617, 204, 31} \[ \frac {5 a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 b^{8/3}}-\frac {5 a^{2/3} \log (a+b x)}{6 b^{8/3}}+\frac {5 a^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{8/3}}-\frac {x^{5/3}}{b (a+b x)}+\frac {5 x^{2/3}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/3)/(a + b*x)^2,x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 56

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, Simp
[Log[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(
1/3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && Ne
gQ[(b*c - a*d)/b]

Rule 204

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

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {x^{5/3}}{(a+b x)^2} \, dx &=-\frac {x^{5/3}}{b (a+b x)}+\frac {5 \int \frac {x^{2/3}}{a+b x} \, dx}{3 b}\\ &=\frac {5 x^{2/3}}{2 b^2}-\frac {x^{5/3}}{b (a+b x)}-\frac {(5 a) \int \frac {1}{\sqrt [3]{x} (a+b x)} \, dx}{3 b^2}\\ &=\frac {5 x^{2/3}}{2 b^2}-\frac {x^{5/3}}{b (a+b x)}-\frac {5 a^{2/3} \log (a+b x)}{6 b^{8/3}}-\frac {(5 a) \operatorname {Subst}\left (\int \frac {1}{\frac {a^{2/3}}{b^{2/3}}-\frac {\sqrt [3]{a} x}{\sqrt [3]{b}}+x^2} \, dx,x,\sqrt [3]{x}\right )}{2 b^3}+\frac {\left (5 a^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt [3]{a}}{\sqrt [3]{b}}+x} \, dx,x,\sqrt [3]{x}\right )}{2 b^{8/3}}\\ &=\frac {5 x^{2/3}}{2 b^2}-\frac {x^{5/3}}{b (a+b x)}+\frac {5 a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 b^{8/3}}-\frac {5 a^{2/3} \log (a+b x)}{6 b^{8/3}}-\frac {\left (5 a^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a}}\right )}{b^{8/3}}\\ &=\frac {5 x^{2/3}}{2 b^2}-\frac {x^{5/3}}{b (a+b x)}+\frac {5 a^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt {3} b^{8/3}}+\frac {5 a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 b^{8/3}}-\frac {5 a^{2/3} \log (a+b x)}{6 b^{8/3}}\\ \end {align*}

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Mathematica [C]  time = 0.00, size = 27, normalized size = 0.21 \[ \frac {3 x^{8/3} \, _2F_1\left (2,\frac {8}{3};\frac {11}{3};-\frac {b x}{a}\right )}{8 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/3)/(a + b*x)^2,x]

[Out]

(3*x^(8/3)*Hypergeometric2F1[2, 8/3, 11/3, -((b*x)/a)])/(8*a^2)

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fricas [A]  time = 0.47, size = 162, normalized size = 1.26 \[ -\frac {10 \, \sqrt {3} {\left (b x + a\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x^{\frac {1}{3}} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} - \sqrt {3} a}{3 \, a}\right ) + 5 \, {\left (b x + a\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (-b x^{\frac {1}{3}} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}} + a x^{\frac {2}{3}} + a \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}}\right ) - 10 \, {\left (b x + a\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (b \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}} + a x^{\frac {1}{3}}\right ) - 3 \, {\left (3 \, b x + 5 \, a\right )} x^{\frac {2}{3}}}{6 \, {\left (b^{3} x + a b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/3)/(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/6*(10*sqrt(3)*(b*x + a)*(a^2/b^2)^(1/3)*arctan(1/3*(2*sqrt(3)*b*x^(1/3)*(a^2/b^2)^(1/3) - sqrt(3)*a)/a) + 5
*(b*x + 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*(b*x + a)*(a^2
/b^2)^(1/3)*log(b*(a^2/b^2)^(2/3) + a*x^(1/3)) - 3*(3*b*x + 5*a)*x^(2/3))/(b^3*x + a*b^2)

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giac [A]  time = 1.05, size = 135, normalized size = 1.05 \[ \frac {5 \, \left (-\frac {a}{b}\right )^{\frac {2}{3}} \log \left ({\left | x^{\frac {1}{3}} - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, b^{2}} + \frac {a x^{\frac {2}{3}}}{{\left (b x + a\right )} b^{2}} + \frac {3 \, x^{\frac {2}{3}}}{2 \, b^{2}} + \frac {5 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, b^{4}} - \frac {5 \, \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (x^{\frac {2}{3}} + x^{\frac {1}{3}} \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/3)/(b*x+a)^2,x, algorithm="giac")

[Out]

5/3*(-a/b)^(2/3)*log(abs(x^(1/3) - (-a/b)^(1/3)))/b^2 + a*x^(2/3)/((b*x + a)*b^2) + 3/2*x^(2/3)/b^2 + 5/3*sqrt
(3)*(-a*b^2)^(2/3)*arctan(1/3*sqrt(3)*(2*x^(1/3) + (-a/b)^(1/3))/(-a/b)^(1/3))/b^4 - 5/6*(-a*b^2)^(2/3)*log(x^
(2/3) + x^(1/3)*(-a/b)^(1/3) + (-a/b)^(2/3))/b^4

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maple [A]  time = 0.01, size = 123, normalized size = 0.95 \[ \frac {a \,x^{\frac {2}{3}}}{\left (b x +a \right ) b^{2}}-\frac {5 \sqrt {3}\, a \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x^{\frac {1}{3}}}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{3}}+\frac {5 a \ln \left (x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{3}}-\frac {5 a \ln \left (x^{\frac {2}{3}}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{3}}+\frac {3 x^{\frac {2}{3}}}{2 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/3)/(b*x+a)^2,x)

[Out]

3/2*x^(2/3)/b^2+a/b^2*x^(2/3)/(b*x+a)+5/3*a/b^3/(a/b)^(1/3)*ln(x^(1/3)+(a/b)^(1/3))-5/6*a/b^3/(a/b)^(1/3)*ln(x
^(2/3)-(a/b)^(1/3)*x^(1/3)+(a/b)^(2/3))-5/3*a/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 [A]  time = 2.96, size = 133, normalized size = 1.03 \[ \frac {a x^{\frac {2}{3}}}{b^{3} x + a b^{2}} - \frac {5 \, \sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (2 \, x^{\frac {1}{3}} - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {3 \, x^{\frac {2}{3}}}{2 \, b^{2}} - \frac {5 \, a \log \left (x^{\frac {2}{3}} - x^{\frac {1}{3}} \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {5 \, a \log \left (x^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/3)/(b*x+a)^2,x, algorithm="maxima")

[Out]

a*x^(2/3)/(b^3*x + a*b^2) - 5/3*sqrt(3)*a*arctan(1/3*sqrt(3)*(2*x^(1/3) - (a/b)^(1/3))/(a/b)^(1/3))/(b^3*(a/b)
^(1/3)) + 3/2*x^(2/3)/b^2 - 5/6*a*log(x^(2/3) - x^(1/3)*(a/b)^(1/3) + (a/b)^(2/3))/(b^3*(a/b)^(1/3)) + 5/3*a*l
og(x^(1/3) + (a/b)^(1/3))/(b^3*(a/b)^(1/3))

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mupad [B]  time = 0.26, size = 150, normalized size = 1.16 \[ \frac {3\,x^{2/3}}{2\,b^2}+\frac {5\,a^{2/3}\,\ln \left (\frac {25\,a^{7/3}}{b^{10/3}}+\frac {25\,a^2\,x^{1/3}}{b^3}\right )}{3\,b^{8/3}}+\frac {a\,x^{2/3}}{x\,b^3+a\,b^2}+\frac {5\,a^{2/3}\,\ln \left (\frac {25\,a^{7/3}\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{b^{10/3}}+\frac {25\,a^2\,x^{1/3}}{b^3}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,b^{8/3}}-\frac {5\,a^{2/3}\,\ln \left (\frac {25\,a^{7/3}\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{b^{10/3}}+\frac {25\,a^2\,x^{1/3}}{b^3}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,b^{8/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/3)/(a + b*x)^2,x)

[Out]

(3*x^(2/3))/(2*b^2) + (5*a^(2/3)*log((25*a^(7/3))/b^(10/3) + (25*a^2*x^(1/3))/b^3))/(3*b^(8/3)) + (a*x^(2/3))/
(a*b^2 + b^3*x) + (5*a^(2/3)*log((25*a^(7/3)*((3^(1/2)*1i)/2 - 1/2)^2)/b^(10/3) + (25*a^2*x^(1/3))/b^3)*((3^(1
/2)*1i)/2 - 1/2))/(3*b^(8/3)) - (5*a^(2/3)*log((25*a^(7/3)*((3^(1/2)*1i)/2 + 1/2)^2)/b^(10/3) + (25*a^2*x^(1/3
))/b^3)*((3^(1/2)*1i)/2 + 1/2))/(3*b^(8/3))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/3)/(b*x+a)**2,x)

[Out]

Timed out

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