Integrand size = 25, antiderivative size = 332 \[ \int \frac {1}{(b+a x) \sqrt [3]{-b^3+a^3 x^3}} \, dx=\frac {(-1)^{5/6} \sqrt {3} \text {arctanh}\left (\frac {\frac {i b+\sqrt {3} b}{2^{2/3} \sqrt {3}}+\frac {\left (-i a-\sqrt {3} a\right ) x}{2^{2/3} \sqrt {3}}+\frac {i \sqrt [3]{-b^3+a^3 x^3}}{\sqrt {3}}}{\sqrt [3]{-b^3+a^3 x^3}}\right )}{2 \sqrt [3]{2} a b}+\frac {\sqrt [3]{-\frac {1}{2}} \log \left (-(-1)^{2/3} \sqrt {a} b^{3/2}+(-1)^{2/3} a^{3/2} \sqrt {b} x-2^{2/3} \sqrt {a} \sqrt {b} \sqrt [3]{-b^3+a^3 x^3}\right )}{2 a b}-\frac {\sqrt [3]{-\frac {1}{2}} \log \left (\sqrt [3]{-1} a b^3-2 \sqrt [3]{-1} a^2 b^2 x+\sqrt [3]{-1} a^3 b x^2+\left ((-2)^{2/3} a b^2-(-2)^{2/3} a^2 b x\right ) \sqrt [3]{-b^3+a^3 x^3}-2 \sqrt [3]{2} a b \left (-b^3+a^3 x^3\right )^{2/3}\right )}{4 a b} \] Output:
1/4*(-1)^(5/6)*3^(1/2)*arctanh((1/6*(I*b+3^(1/2)*b)*2^(1/3)*3^(1/2)+1/6*(- I*a-3^(1/2)*a)*x*2^(1/3)*3^(1/2)+1/3*I*(a^3*x^3-b^3)^(1/3)*3^(1/2))/(a^3*x ^3-b^3)^(1/3))*2^(2/3)/a/b+1/4*(-1)^(1/3)*2^(2/3)*ln(-(-1)^(2/3)*a^(1/2)*b ^(3/2)+(-1)^(2/3)*a^(3/2)*b^(1/2)*x-2^(2/3)*a^(1/2)*b^(1/2)*(a^3*x^3-b^3)^ (1/3))/a/b-1/8*(-1)^(1/3)*2^(2/3)*ln((-1)^(1/3)*a*b^3-2*(-1)^(1/3)*a^2*b^2 *x+(-1)^(1/3)*a^3*b*x^2+((-2)^(2/3)*a*b^2-(-2)^(2/3)*a^2*b*x)*(a^3*x^3-b^3 )^(1/3)-2*2^(1/3)*a*b*(a^3*x^3-b^3)^(2/3))/a/b
Time = 2.30 (sec) , antiderivative size = 310, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(b+a x) \sqrt [3]{-b^3+a^3 x^3}} \, dx=\frac {\sqrt [3]{-\frac {1}{2}} \left (2 i \sqrt {3} \text {arctanh}\left (\frac {\sqrt [3]{2} \left (3+i \sqrt {3}\right ) b+\sqrt [3]{2} \left (-3-i \sqrt {3}\right ) a x+2 i \sqrt {3} \sqrt [3]{-b^3+a^3 x^3}}{6 \sqrt [3]{-b^3+a^3 x^3}}\right )+2 \log \left (\sqrt {a} \sqrt {b} \left (-b+i \sqrt {3} b+a x-i \sqrt {3} a x+2\ 2^{2/3} \sqrt [3]{-b^3+a^3 x^3}\right )\right )-\log \left (a b \left (\left (-1-i \sqrt {3}\right ) b^2+\left (-1-i \sqrt {3}\right ) a^2 x^2+2 a b \left (x+i \sqrt {3} x\right )-2 (-2)^{2/3} b \sqrt [3]{-b^3+a^3 x^3}+2 (-2)^{2/3} a x \sqrt [3]{-b^3+a^3 x^3}+4 \sqrt [3]{2} \left (-b^3+a^3 x^3\right )^{2/3}\right )\right )\right )}{4 a b} \] Input:
Integrate[1/((b + a*x)*(-b^3 + a^3*x^3)^(1/3)),x]
Output:
((-1/2)^(1/3)*((2*I)*Sqrt[3]*ArcTanh[(2^(1/3)*(3 + I*Sqrt[3])*b + 2^(1/3)* (-3 - I*Sqrt[3])*a*x + (2*I)*Sqrt[3]*(-b^3 + a^3*x^3)^(1/3))/(6*(-b^3 + a^ 3*x^3)^(1/3))] + 2*Log[Sqrt[a]*Sqrt[b]*(-b + I*Sqrt[3]*b + a*x - I*Sqrt[3] *a*x + 2*2^(2/3)*(-b^3 + a^3*x^3)^(1/3))] - Log[a*b*((-1 - I*Sqrt[3])*b^2 + (-1 - I*Sqrt[3])*a^2*x^2 + 2*a*b*(x + I*Sqrt[3]*x) - 2*(-2)^(2/3)*b*(-b^ 3 + a^3*x^3)^(1/3) + 2*(-2)^(2/3)*a*x*(-b^3 + a^3*x^3)^(1/3) + 4*2^(1/3)*( -b^3 + a^3*x^3)^(2/3))]))/(4*a*b)
Time = 0.27 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.42, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2574}
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.
| \(\displaystyle \int \frac {1}{(a x+b) \sqrt [3]{a^3 x^3-b^3}} \, dx\) |
| \(\Big \downarrow \) 2574 |
| \(\displaystyle \frac {\sqrt {3} \arctan \left (\frac {1-\frac {\sqrt [3]{2} (b-a x)}{\sqrt [3]{a^3 x^3-b^3}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} a b}-\frac {3 \log \left (2^{2/3} a \sqrt [3]{a^3 x^3-b^3}+a (b-a x)\right )}{4 \sqrt [3]{2} a b}+\frac {\log \left ((b-a x) (a x+b)^2\right )}{4 \sqrt [3]{2} a b}\) |
Input:
Int[1/((b + a*x)*(-b^3 + a^3*x^3)^(1/3)),x]
Output:
(Sqrt[3]*ArcTan[(1 - (2^(1/3)*(b - a*x))/(-b^3 + a^3*x^3)^(1/3))/Sqrt[3]]) /(2*2^(1/3)*a*b) + Log[(b - a*x)*(b + a*x)^2]/(4*2^(1/3)*a*b) - (3*Log[a*( b - a*x) + 2^(2/3)*a*(-b^3 + a^3*x^3)^(1/3)])/(4*2^(1/3)*a*b)
Int[1/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^3)^(1/3)), x_Symbol] :> Simp[ Sqrt[3]*(ArcTan[(1 - 2^(1/3)*Rt[b, 3]*((c - d*x)/(d*(a + b*x^3)^(1/3))))/Sq rt[3]]/(2^(4/3)*Rt[b, 3]*c)), x] + (Simp[Log[(c + d*x)^2*(c - d*x)]/(2^(7/3 )*Rt[b, 3]*c), x] - Simp[(3*Log[Rt[b, 3]*(c - d*x) + 2^(2/3)*d*(a + b*x^3)^ (1/3)])/(2^(7/3)*Rt[b, 3]*c), x]) /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^3 + a*d^3, 0]
\[\int \frac {1}{\left (a x +b \right ) \left (a^{3} x^{3}-b^{3}\right )^{\frac {1}{3}}}d x\]
Input:
int(1/(a*x+b)/(a^3*x^3-b^3)^(1/3),x)
Output:
int(1/(a*x+b)/(a^3*x^3-b^3)^(1/3),x)
Timed out. \[ \int \frac {1}{(b+a x) \sqrt [3]{-b^3+a^3 x^3}} \, dx=\text {Timed out} \] Input:
integrate(1/(a*x+b)/(a^3*x^3-b^3)^(1/3),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{(b+a x) \sqrt [3]{-b^3+a^3 x^3}} \, dx=\int \frac {1}{\sqrt [3]{\left (a x - b\right ) \left (a^{2} x^{2} + a b x + b^{2}\right )} \left (a x + b\right )}\, dx \] Input:
integrate(1/(a*x+b)/(a**3*x**3-b**3)**(1/3),x)
Output:
Integral(1/(((a*x - b)*(a**2*x**2 + a*b*x + b**2))**(1/3)*(a*x + b)), x)
\[ \int \frac {1}{(b+a x) \sqrt [3]{-b^3+a^3 x^3}} \, dx=\int { \frac {1}{{\left (a^{3} x^{3} - b^{3}\right )}^{\frac {1}{3}} {\left (a x + b\right )}} \,d x } \] Input:
integrate(1/(a*x+b)/(a^3*x^3-b^3)^(1/3),x, algorithm="maxima")
Output:
integrate(1/((a^3*x^3 - b^3)^(1/3)*(a*x + b)), x)
\[ \int \frac {1}{(b+a x) \sqrt [3]{-b^3+a^3 x^3}} \, dx=\int { \frac {1}{{\left (a^{3} x^{3} - b^{3}\right )}^{\frac {1}{3}} {\left (a x + b\right )}} \,d x } \] Input:
integrate(1/(a*x+b)/(a^3*x^3-b^3)^(1/3),x, algorithm="giac")
Output:
integrate(1/((a^3*x^3 - b^3)^(1/3)*(a*x + b)), x)
Timed out. \[ \int \frac {1}{(b+a x) \sqrt [3]{-b^3+a^3 x^3}} \, dx=\int \frac {1}{{\left (a^3\,x^3-b^3\right )}^{1/3}\,\left (b+a\,x\right )} \,d x \] Input:
int(1/((a^3*x^3 - b^3)^(1/3)*(b + a*x)),x)
Output:
int(1/((a^3*x^3 - b^3)^(1/3)*(b + a*x)), x)
\[ \int \frac {1}{(b+a x) \sqrt [3]{-b^3+a^3 x^3}} \, dx=\int \frac {1}{\left (a^{3} x^{3}-b^{3}\right )^{\frac {1}{3}} a x +\left (a^{3} x^{3}-b^{3}\right )^{\frac {1}{3}} b}d x \] Input:
int(1/(a*x+b)/(a^3*x^3-b^3)^(1/3),x)
Output:
int(1/((a**3*x**3 - b**3)**(1/3)*a*x + (a**3*x**3 - b**3)**(1/3)*b),x)