\(\int \frac {x^{5/3}}{(a+b x)^2} \, dx\) [682]

Optimal result

Integrand size = 13, antiderivative size = 129 \[ \int \frac {x^{5/3}}{(a+b x)^2} \, dx=\frac {5 x^{2/3}}{2 b^2}-\frac {x^{5/3}}{b (a+b x)}+\frac {5 a^{2/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{8/3}}+\frac {5 a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 b^{8/3}}-\frac {5 a^{2/3} \log (a+b x)}{6 b^{8/3}} \]

[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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {43, 52, 58, 631, 210, 31} \[ \int \frac {x^{5/3}}{(a+b x)^2} \, dx=\frac {5 a^{2/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{8/3}}+\frac {5 a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )}{2 b^{8/3}}-\frac {5 a^{2/3} \log (a+b x)}{6 b^{8/3}}-\frac {x^{5/3}}{b (a+b x)}+\frac {5 x^{2/3}}{2 b^2} \]

[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 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

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 58

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[-(b*c - a*d)/b, 3]}, Simp[L
og[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] && NegQ
[(b*c - a*d)/b]

Rule 210

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

Rule 631

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*} \text {integral}& = -\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) \text {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 ) \text {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 ) \text {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*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.10 \[ \int \frac {x^{5/3}}{(a+b x)^2} \, dx=\frac {\frac {3 b^{2/3} x^{2/3} (5 a+3 b x)}{a+b x}+10 \sqrt {3} a^{2/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} \sqrt [3]{x}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+10 a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt [3]{x}\right )-5 a^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sqrt [3]{x}+b^{2/3} x^{2/3}\right )}{6 b^{8/3}} \]

[In]

Integrate[x^(5/3)/(a + b*x)^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {3 x^{\frac {2}{3}}}{2 b^{2}}-\frac {3 a \left (-\frac {x^{\frac {2}{3}}}{3 \left (b x +a \right )}-\frac {5 \ln \left (x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {5 \ln \left (x^{\frac {2}{3}}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {5 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x^{\frac {1}{3}}}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{b^{2}}\)	\(124\)
default	\(\frac {3 x^{\frac {2}{3}}}{2 b^{2}}-\frac {3 a \left (-\frac {x^{\frac {2}{3}}}{3 \left (b x +a \right )}-\frac {5 \ln \left (x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {5 \ln \left (x^{\frac {2}{3}}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {5 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x^{\frac {1}{3}}}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{b^{2}}\)	\(124\)
risch	\(\frac {3 x^{\frac {2}{3}}}{2 b^{2}}-\frac {a \left (-\frac {x^{\frac {2}{3}}}{b x +a}-\frac {5 \ln \left (x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {5 \ln \left (x^{\frac {2}{3}}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x^{\frac {1}{3}}+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {5 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x^{\frac {1}{3}}}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{b^{2}}\)	\(124\)

[In]

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

[Out]

3/2*x^(2/3)/b^2-3*a/b^2*(-1/3*x^(2/3)/(b*x+a)-5/9/b/(a/b)^(1/3)*ln(x^(1/3)+(a/b)^(1/3))+5/18/b/(a/b)^(1/3)*ln(
x^(2/3)-(a/b)^(1/3)*x^(1/3)+(a/b)^(2/3))+5/9*3^(1/2)/b/(a/b)^(1/3)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x^(1/3)-1
)))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.26 \[ \int \frac {x^{5/3}}{(a+b x)^2} \, dx=-\frac {10 \, \sqrt {3} {\left (b x + a\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x^{\frac {1}{3}} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} - \sqrt {3} a}{3 \, a}\right ) + 5 \, {\left (b x + a\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (-b x^{\frac {1}{3}} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}} + a x^{\frac {2}{3}} + a \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}}\right ) - 10 \, {\left (b x + a\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (b \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}} + a x^{\frac {1}{3}}\right ) - 3 \, {\left (3 \, b x + 5 \, a\right )} x^{\frac {2}{3}}}{6 \, {\left (b^{3} x + a b^{2}\right )}} \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{5/3}}{(a+b x)^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.03 \[ \int \frac {x^{5/3}}{(a+b x)^2} \, dx=\frac {a x^{\frac {2}{3}}}{b^{3} x + a b^{2}} - \frac {5 \, \sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (2 \, x^{\frac {1}{3}} - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {3 \, x^{\frac {2}{3}}}{2 \, b^{2}} - \frac {5 \, a \log \left (x^{\frac {2}{3}} - x^{\frac {1}{3}} \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {5 \, a \log \left (x^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.05 \[ \int \frac {x^{5/3}}{(a+b x)^2} \, dx=\frac {5 \, \left (-\frac {a}{b}\right )^{\frac {2}{3}} \log \left ({\left | x^{\frac {1}{3}} - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, b^{2}} + \frac {a x^{\frac {2}{3}}}{{\left (b x + a\right )} b^{2}} + \frac {3 \, x^{\frac {2}{3}}}{2 \, b^{2}} + \frac {5 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, b^{4}} - \frac {5 \, \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (x^{\frac {2}{3}} + x^{\frac {1}{3}} \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b^{4}} \]

[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

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.16 \[ \int \frac {x^{5/3}}{(a+b x)^2} \, dx=\frac {3\,x^{2/3}}{2\,b^2}+\frac {5\,a^{2/3}\,\ln \left (\frac {25\,a^{7/3}}{b^{10/3}}+\frac {25\,a^2\,x^{1/3}}{b^3}\right )}{3\,b^{8/3}}+\frac {a\,x^{2/3}}{x\,b^3+a\,b^2}+\frac {5\,a^{2/3}\,\ln \left (\frac {25\,a^{7/3}\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{b^{10/3}}+\frac {25\,a^2\,x^{1/3}}{b^3}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,b^{8/3}}-\frac {5\,a^{2/3}\,\ln \left (\frac {25\,a^{7/3}\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{b^{10/3}}+\frac {25\,a^2\,x^{1/3}}{b^3}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,b^{8/3}} \]

[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))