Integrand size = 13, antiderivative size = 130 \[ \int \frac {1}{x^3 (a+b x)^{2/3}} \, dx=-\frac {\sqrt [3]{a+b x}}{2 a x^2}+\frac {5 b \sqrt [3]{a+b x}}{6 a^2 x}-\frac {5 b^2 \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{8/3}}-\frac {5 b^2 \log (x)}{18 a^{8/3}}+\frac {5 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{6 a^{8/3}} \] Output:
-1/2*(b*x+a)^(1/3)/a/x^2+5/6*b*(b*x+a)^(1/3)/a^2/x-5/9*b^2*arctan(1/3*(a^( 1/3)+2*(b*x+a)^(1/3))*3^(1/2)/a^(1/3))*3^(1/2)/a^(8/3)-5/18*b^2*ln(x)/a^(8 /3)+5/6*b^2*ln(a^(1/3)-(b*x+a)^(1/3))/a^(8/3)
Time = 0.12 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.15 \[ \int \frac {1}{x^3 (a+b x)^{2/3}} \, dx=-\frac {\sqrt [3]{a+b x} (8 a-5 (a+b x))}{6 a^2 x^2}-\frac {5 b^2 \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{8/3}}+\frac {5 b^2 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{9 a^{8/3}}-\frac {5 b^2 \log \left (a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x}+(a+b x)^{2/3}\right )}{18 a^{8/3}} \] Input:
Integrate[1/(x^3*(a + b*x)^(2/3)),x]
Output:
-1/6*((a + b*x)^(1/3)*(8*a - 5*(a + b*x)))/(a^2*x^2) - (5*b^2*ArcTan[1/Sqr t[3] + (2*(a + b*x)^(1/3))/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(8/3)) + (5*b^ 2*Log[a^(1/3) - (a + b*x)^(1/3)])/(9*a^(8/3)) - (5*b^2*Log[a^(2/3) + a^(1/ 3)*(a + b*x)^(1/3) + (a + b*x)^(2/3)])/(18*a^(8/3))
Time = 0.20 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {52, 52, 69, 16, 1082, 217}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {1}{x^3 (a+b x)^{2/3}} \, dx\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {5 b \int \frac {1}{x^2 (a+b x)^{2/3}}dx}{6 a}-\frac {\sqrt [3]{a+b x}}{2 a x^2}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {5 b \left (-\frac {2 b \int \frac {1}{x (a+b x)^{2/3}}dx}{3 a}-\frac {\sqrt [3]{a+b x}}{a x}\right )}{6 a}-\frac {\sqrt [3]{a+b x}}{2 a x^2}\) |
| \(\Big \downarrow \) 69 |
| \(\displaystyle -\frac {5 b \left (-\frac {2 b \left (-\frac {3 \int \frac {1}{\sqrt [3]{a}-\sqrt [3]{a+b x}}d\sqrt [3]{a+b x}}{2 a^{2/3}}-\frac {3 \int \frac {1}{a^{2/3}+\sqrt [3]{a+b x} \sqrt [3]{a}+(a+b x)^{2/3}}d\sqrt [3]{a+b x}}{2 \sqrt [3]{a}}-\frac {\log (x)}{2 a^{2/3}}\right )}{3 a}-\frac {\sqrt [3]{a+b x}}{a x}\right )}{6 a}-\frac {\sqrt [3]{a+b x}}{2 a x^2}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle -\frac {5 b \left (-\frac {2 b \left (-\frac {3 \int \frac {1}{a^{2/3}+\sqrt [3]{a+b x} \sqrt [3]{a}+(a+b x)^{2/3}}d\sqrt [3]{a+b x}}{2 \sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{2 a^{2/3}}-\frac {\log (x)}{2 a^{2/3}}\right )}{3 a}-\frac {\sqrt [3]{a+b x}}{a x}\right )}{6 a}-\frac {\sqrt [3]{a+b x}}{2 a x^2}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle -\frac {5 b \left (-\frac {2 b \left (\frac {3 \int \frac {1}{-(a+b x)^{2/3}-3}d\left (\frac {2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}+1\right )}{a^{2/3}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{2 a^{2/3}}-\frac {\log (x)}{2 a^{2/3}}\right )}{3 a}-\frac {\sqrt [3]{a+b x}}{a x}\right )}{6 a}-\frac {\sqrt [3]{a+b x}}{2 a x^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {5 b \left (-\frac {2 b \left (-\frac {\sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{a+b x}}{\sqrt [3]{a}}+1}{\sqrt {3}}\right )}{a^{2/3}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+b x}\right )}{2 a^{2/3}}-\frac {\log (x)}{2 a^{2/3}}\right )}{3 a}-\frac {\sqrt [3]{a+b x}}{a x}\right )}{6 a}-\frac {\sqrt [3]{a+b x}}{2 a x^2}\) |
Input:
Int[1/(x^3*(a + b*x)^(2/3)),x]
Output:
-1/2*(a + b*x)^(1/3)/(a*x^2) - (5*b*(-((a + b*x)^(1/3)/(a*x)) - (2*b*(-((S qrt[3]*ArcTan[(1 + (2*(a + b*x)^(1/3))/a^(1/3))/Sqrt[3]])/a^(2/3)) - Log[x ]/(2*a^(2/3)) + (3*Log[a^(1/3) - (a + b*x)^(1/3)])/(2*a^(2/3))))/(3*a)))/( 6*a)
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ {q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Simp[3/(2*b*q) Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 /3)], x] - Simp[3/(2*b*q^2) Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
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])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[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])] /; Fre eQ[{a, b, c}, x]
Time = 0.23 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.94
| method | result | size |
| pseudoelliptic | \(\frac {-10 b^{2} \sqrt {3}\, \arctan \left (\frac {\left (a^{\frac {1}{3}}+2 \left (b x +a \right )^{\frac {1}{3}}\right ) \sqrt {3}}{3 a^{\frac {1}{3}}}\right ) x^{2}+10 b^{2} \ln \left (\left (b x +a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right ) x^{2}-5 b^{2} \ln \left (\left (b x +a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b x +a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right ) x^{2}+15 b x \left (b x +a \right )^{\frac {1}{3}} a^{\frac {2}{3}}-9 \left (b x +a \right )^{\frac {1}{3}} a^{\frac {5}{3}}}{18 a^{\frac {8}{3}} x^{2}}\) | \(122\) |
| derivativedivides | \(3 b^{2} \left (-\frac {\left (b x +a \right )^{\frac {1}{3}}}{6 a \,b^{2} x^{2}}-\frac {5 \left (-\frac {\left (b x +a \right )^{\frac {1}{3}}}{3 a b x}+\frac {-\frac {2 \ln \left (\left (b x +a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )}{9 a^{\frac {2}{3}}}+\frac {\ln \left (\left (b x +a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b x +a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )}{9 a^{\frac {2}{3}}}+\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b x +a \right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{9 a^{\frac {2}{3}}}}{a}\right )}{6 a}\right )\) | \(130\) |
| default | \(3 b^{2} \left (-\frac {\left (b x +a \right )^{\frac {1}{3}}}{6 a \,b^{2} x^{2}}-\frac {5 \left (-\frac {\left (b x +a \right )^{\frac {1}{3}}}{3 a b x}+\frac {-\frac {2 \ln \left (\left (b x +a \right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )}{9 a^{\frac {2}{3}}}+\frac {\ln \left (\left (b x +a \right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (b x +a \right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )}{9 a^{\frac {2}{3}}}+\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b x +a \right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{9 a^{\frac {2}{3}}}}{a}\right )}{6 a}\right )\) | \(130\) |
Input:
int(1/x^3/(b*x+a)^(2/3),x,method=_RETURNVERBOSE)
Output:
1/18*(-10*b^2*3^(1/2)*arctan(1/3*(a^(1/3)+2*(b*x+a)^(1/3))*3^(1/2)/a^(1/3) )*x^2+10*b^2*ln((b*x+a)^(1/3)-a^(1/3))*x^2-5*b^2*ln((b*x+a)^(2/3)+a^(1/3)* (b*x+a)^(1/3)+a^(2/3))*x^2+15*b*x*(b*x+a)^(1/3)*a^(2/3)-9*(b*x+a)^(1/3)*a^ (5/3))/a^(8/3)/x^2
Time = 0.08 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.22 \[ \int \frac {1}{x^3 (a+b x)^{2/3}} \, dx=-\frac {30 \, \sqrt {\frac {1}{3}} {\left (a^{2}\right )}^{\frac {1}{6}} a b^{2} x^{2} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (a^{2}\right )}^{\frac {1}{6}} {\left ({\left (a^{2}\right )}^{\frac {1}{3}} a + 2 \, {\left (a^{2}\right )}^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {1}{3}}\right )}}{a^{2}}\right ) + 5 \, {\left (a^{2}\right )}^{\frac {2}{3}} b^{2} x^{2} \log \left ({\left (b x + a\right )}^{\frac {2}{3}} a + {\left (a^{2}\right )}^{\frac {1}{3}} a + {\left (a^{2}\right )}^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {1}{3}}\right ) - 10 \, {\left (a^{2}\right )}^{\frac {2}{3}} b^{2} x^{2} \log \left ({\left (b x + a\right )}^{\frac {1}{3}} a - {\left (a^{2}\right )}^{\frac {2}{3}}\right ) - 3 \, {\left (5 \, a^{2} b x - 3 \, a^{3}\right )} {\left (b x + a\right )}^{\frac {1}{3}}}{18 \, a^{4} x^{2}} \] Input:
integrate(1/x^3/(b*x+a)^(2/3),x, algorithm="fricas")
Output:
-1/18*(30*sqrt(1/3)*(a^2)^(1/6)*a*b^2*x^2*arctan(sqrt(1/3)*(a^2)^(1/6)*((a ^2)^(1/3)*a + 2*(a^2)^(2/3)*(b*x + a)^(1/3))/a^2) + 5*(a^2)^(2/3)*b^2*x^2* log((b*x + a)^(2/3)*a + (a^2)^(1/3)*a + (a^2)^(2/3)*(b*x + a)^(1/3)) - 10* (a^2)^(2/3)*b^2*x^2*log((b*x + a)^(1/3)*a - (a^2)^(2/3)) - 3*(5*a^2*b*x - 3*a^3)*(b*x + a)^(1/3))/(a^4*x^2)
Result contains complex when optimal does not.
Time = 1.94 (sec) , antiderivative size = 2728, normalized size of antiderivative = 20.98 \[ \int \frac {1}{x^3 (a+b x)^{2/3}} \, dx=\text {Too large to display} \] Input:
integrate(1/x**3/(b*x+a)**(2/3),x)
Output:
10*a**(13/3)*b**(8/3)*(a/b + x)**(2/3)*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)/a**(1/3))*gamma(1/3)/(54*a**7*b**(2/3)*(a/b + x)**(2/3)*exp(2 *I*pi/3)*gamma(4/3) - 162*a**6*b**(5/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*gam ma(4/3) + 162*a**5*b**(8/3)*(a/b + x)**(8/3)*exp(2*I*pi/3)*gamma(4/3) - 54 *a**4*b**(11/3)*(a/b + x)**(11/3)*exp(2*I*pi/3)*gamma(4/3)) + 10*a**(13/3) *b**(8/3)*(a/b + x)**(2/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_polar(2*I *pi/3)/a**(1/3))*gamma(1/3)/(54*a**7*b**(2/3)*(a/b + x)**(2/3)*exp(2*I*pi/ 3)*gamma(4/3) - 162*a**6*b**(5/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*gamma(4/3 ) + 162*a**5*b**(8/3)*(a/b + x)**(8/3)*exp(2*I*pi/3)*gamma(4/3) - 54*a**4* b**(11/3)*(a/b + x)**(11/3)*exp(2*I*pi/3)*gamma(4/3)) + 10*a**(13/3)*b**(8 /3)*(a/b + x)**(2/3)*exp(-2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**(1/3)*exp_ polar(4*I*pi/3)/a**(1/3))*gamma(1/3)/(54*a**7*b**(2/3)*(a/b + x)**(2/3)*ex p(2*I*pi/3)*gamma(4/3) - 162*a**6*b**(5/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)* gamma(4/3) + 162*a**5*b**(8/3)*(a/b + x)**(8/3)*exp(2*I*pi/3)*gamma(4/3) - 54*a**4*b**(11/3)*(a/b + x)**(11/3)*exp(2*I*pi/3)*gamma(4/3)) - 30*a**(10 /3)*b**(11/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*log(1 - b**(1/3)*(a/b + x)**( 1/3)/a**(1/3))*gamma(1/3)/(54*a**7*b**(2/3)*(a/b + x)**(2/3)*exp(2*I*pi/3) *gamma(4/3) - 162*a**6*b**(5/3)*(a/b + x)**(5/3)*exp(2*I*pi/3)*gamma(4/3) + 162*a**5*b**(8/3)*(a/b + x)**(8/3)*exp(2*I*pi/3)*gamma(4/3) - 54*a**4*b* *(11/3)*(a/b + x)**(11/3)*exp(2*I*pi/3)*gamma(4/3)) - 30*a**(10/3)*b**(...
Time = 0.11 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x^3 (a+b x)^{2/3}} \, dx=-\frac {5 \, \sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{9 \, a^{\frac {8}{3}}} - \frac {5 \, b^{2} \log \left ({\left (b x + a\right )}^{\frac {2}{3}} + {\left (b x + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{18 \, a^{\frac {8}{3}}} + \frac {5 \, b^{2} \log \left ({\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{9 \, a^{\frac {8}{3}}} + \frac {5 \, {\left (b x + a\right )}^{\frac {4}{3}} b^{2} - 8 \, {\left (b x + a\right )}^{\frac {1}{3}} a b^{2}}{6 \, {\left ({\left (b x + a\right )}^{2} a^{2} - 2 \, {\left (b x + a\right )} a^{3} + a^{4}\right )}} \] Input:
integrate(1/x^3/(b*x+a)^(2/3),x, algorithm="maxima")
Output:
-5/9*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*(b*x + a)^(1/3) + a^(1/3))/a^(1/3)) /a^(8/3) - 5/18*b^2*log((b*x + a)^(2/3) + (b*x + a)^(1/3)*a^(1/3) + a^(2/3 ))/a^(8/3) + 5/9*b^2*log((b*x + a)^(1/3) - a^(1/3))/a^(8/3) + 1/6*(5*(b*x + a)^(4/3)*b^2 - 8*(b*x + a)^(1/3)*a*b^2)/((b*x + a)^2*a^2 - 2*(b*x + a)*a ^3 + a^4)
Time = 0.41 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^3 (a+b x)^{2/3}} \, dx=-\frac {\frac {10 \, \sqrt {3} b^{3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right )}{a^{\frac {8}{3}}} + \frac {5 \, b^{3} \log \left ({\left (b x + a\right )}^{\frac {2}{3}} + {\left (b x + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right )}{a^{\frac {8}{3}}} - \frac {10 \, b^{3} \log \left ({\left | {\left (b x + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}} \right |}\right )}{a^{\frac {8}{3}}} - \frac {3 \, {\left (5 \, {\left (b x + a\right )}^{\frac {4}{3}} b^{3} - 8 \, {\left (b x + a\right )}^{\frac {1}{3}} a b^{3}\right )}}{a^{2} b^{2} x^{2}}}{18 \, b} \] Input:
integrate(1/x^3/(b*x+a)^(2/3),x, algorithm="giac")
Output:
-1/18*(10*sqrt(3)*b^3*arctan(1/3*sqrt(3)*(2*(b*x + a)^(1/3) + a^(1/3))/a^( 1/3))/a^(8/3) + 5*b^3*log((b*x + a)^(2/3) + (b*x + a)^(1/3)*a^(1/3) + a^(2 /3))/a^(8/3) - 10*b^3*log(abs((b*x + a)^(1/3) - a^(1/3)))/a^(8/3) - 3*(5*( b*x + a)^(4/3)*b^3 - 8*(b*x + a)^(1/3)*a*b^3)/(a^2*b^2*x^2))/b
Time = 0.08 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.35 \[ \int \frac {1}{x^3 (a+b x)^{2/3}} \, dx=\frac {5\,b^2\,\ln \left ({\left (a+b\,x\right )}^{1/3}-a^{1/3}\right )}{9\,a^{8/3}}-\frac {\frac {4\,b^2\,{\left (a+b\,x\right )}^{1/3}}{3\,a}-\frac {5\,b^2\,{\left (a+b\,x\right )}^{4/3}}{6\,a^2}}{{\left (a+b\,x\right )}^2-2\,a\,\left (a+b\,x\right )+a^2}+\frac {5\,b^2\,\ln \left (\frac {5\,b^2\,{\left (a+b\,x\right )}^{1/3}}{a^2}-\frac {5\,b^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{5/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{8/3}}-\frac {5\,b^2\,\ln \left (\frac {5\,b^2\,{\left (a+b\,x\right )}^{1/3}}{a^2}+\frac {5\,b^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{5/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,a^{8/3}} \] Input:
int(1/(x^3*(a + b*x)^(2/3)),x)
Output:
(5*b^2*log((a + b*x)^(1/3) - a^(1/3)))/(9*a^(8/3)) - ((4*b^2*(a + b*x)^(1/ 3))/(3*a) - (5*b^2*(a + b*x)^(4/3))/(6*a^2))/((a + b*x)^2 - 2*a*(a + b*x) + a^2) + (5*b^2*log((5*b^2*(a + b*x)^(1/3))/a^2 - (5*b^2*((3^(1/2)*1i)/2 - 1/2))/a^(5/3))*((3^(1/2)*1i)/2 - 1/2))/(9*a^(8/3)) - (5*b^2*log((5*b^2*(a + b*x)^(1/3))/a^2 + (5*b^2*((3^(1/2)*1i)/2 + 1/2))/a^(5/3))*((3^(1/2)*1i) /2 + 1/2))/(9*a^(8/3))
Time = 0.17 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.58 \[ \int \frac {1}{x^3 (a+b x)^{2/3}} \, dx=\frac {10 \sqrt {3}\, \mathit {atan} \left (\frac {2 \left (b x +a \right )^{\frac {1}{6}}+a^{\frac {1}{6}}}{a^{\frac {1}{6}} \sqrt {3}}\right ) b^{2} x^{2}-10 \sqrt {3}\, \mathit {atan} \left (\frac {2 \left (b x +a \right )^{\frac {1}{6}}-a^{\frac {1}{6}}}{a^{\frac {1}{6}} \sqrt {3}}\right ) b^{2} x^{2}-9 a^{\frac {5}{3}} \left (b x +a \right )^{\frac {1}{3}}+15 a^{\frac {2}{3}} \left (b x +a \right )^{\frac {1}{3}} b x +10 \,\mathrm {log}\left (\left (b x +a \right )^{\frac {1}{6}}+a^{\frac {1}{6}}\right ) b^{2} x^{2}+10 \,\mathrm {log}\left (\left (b x +a \right )^{\frac {1}{6}}-a^{\frac {1}{6}}\right ) b^{2} x^{2}-5 \,\mathrm {log}\left (-a^{\frac {1}{6}} \left (b x +a \right )^{\frac {1}{6}}+\left (b x +a \right )^{\frac {1}{3}}+a^{\frac {1}{3}}\right ) b^{2} x^{2}-5 \,\mathrm {log}\left (a^{\frac {1}{6}} \left (b x +a \right )^{\frac {1}{6}}+\left (b x +a \right )^{\frac {1}{3}}+a^{\frac {1}{3}}\right ) b^{2} x^{2}}{18 a^{\frac {8}{3}} x^{2}} \] Input:
int(1/x^3/(b*x+a)^(2/3),x)
Output:
(10*sqrt(3)*atan((2*(a + b*x)**(1/6) + a**(1/6))/(a**(1/6)*sqrt(3)))*b**2* x**2 - 10*sqrt(3)*atan((2*(a + b*x)**(1/6) - a**(1/6))/(a**(1/6)*sqrt(3))) *b**2*x**2 - 9*a**(2/3)*(a + b*x)**(1/3)*a + 15*a**(2/3)*(a + b*x)**(1/3)* b*x + 10*log((a + b*x)**(1/6) + a**(1/6))*b**2*x**2 + 10*log((a + b*x)**(1 /6) - a**(1/6))*b**2*x**2 - 5*log( - a**(1/6)*(a + b*x)**(1/6) + (a + b*x) **(1/3) + a**(1/3))*b**2*x**2 - 5*log(a**(1/6)*(a + b*x)**(1/6) + (a + b*x )**(1/3) + a**(1/3))*b**2*x**2)/(18*a**(2/3)*a**2*x**2)