Integrand size = 53, antiderivative size = 564 \[ \int \frac {1}{(d+e x) \sqrt [3]{-c^2 d^2+b c d e+2 b^2 e^2+9 b c e^2 x+9 c^2 e^2 x^2}} \, dx=-\frac {\sqrt {3} \sqrt [3]{-c e (c d-2 b e)+3 c^2 e^2 x} \sqrt [3]{c e (c d+b e)+3 c^2 e^2 x} \arctan \left (\frac {1}{\sqrt {3}}-\frac {\sqrt [3]{2} \left (-c e (c d-2 b e)+3 c^2 e^2 x\right )^{2/3}}{\sqrt {3} \sqrt [3]{c} \sqrt [3]{e} \sqrt [3]{2 c d-b e} \sqrt [3]{c e (c d+b e)+3 c^2 e^2 x}}\right )}{2 \sqrt [3]{2} c^{2/3} e^{5/3} (2 c d-b e)^{2/3} \sqrt [3]{-((c d-2 b e) (c d+b e))+9 b c e^2 x+9 c^2 e^2 x^2}}-\frac {\sqrt [3]{-c e (c d-2 b e)+3 c^2 e^2 x} \sqrt [3]{c e (c d+b e)+3 c^2 e^2 x} \log (d+e x)}{2 \sqrt [3]{2} c^{2/3} e^{5/3} (2 c d-b e)^{2/3} \sqrt [3]{-((c d-2 b e) (c d+b e))+9 b c e^2 x+9 c^2 e^2 x^2}}+\frac {3 \sqrt [3]{-c e (c d-2 b e)+3 c^2 e^2 x} \sqrt [3]{c e (c d+b e)+3 c^2 e^2 x} \log \left (-\frac {\sqrt [3]{\frac {3}{2}} \left (-c e (c d-2 b e)+3 c^2 e^2 x\right )^{2/3}}{\sqrt [3]{c} \sqrt [3]{e} \sqrt [3]{2 c d-b e}}-\sqrt [3]{6} \sqrt [3]{c e (c d+b e)+3 c^2 e^2 x}\right )}{4 \sqrt [3]{2} c^{2/3} e^{5/3} (2 c d-b e)^{2/3} \sqrt [3]{-((c d-2 b e) (c d+b e))+9 b c e^2 x+9 c^2 e^2 x^2}} \]
-1/4*(-c*e*(-2*b*e+c*d)+3*c^2*e^2*x)^(1/3)*(c*e*(b*e+c*d)+3*c^2*e^2*x)^(1/ 3)*ln(e*x+d)*2^(2/3)/c^(2/3)/e^(5/3)/(-b*e+2*c*d)^(2/3)/(-(-2*b*e+c*d)*(b* e+c*d)+9*b*c*e^2*x+9*c^2*e^2*x^2)^(1/3)+3/8*(-c*e*(-2*b*e+c*d)+3*c^2*e^2*x )^(1/3)*(c*e*(b*e+c*d)+3*c^2*e^2*x)^(1/3)*ln(-1/2*3^(1/3)*2^(2/3)*(-c*e*(- 2*b*e+c*d)+3*c^2*e^2*x)^(2/3)/c^(1/3)/e^(1/3)/(-b*e+2*c*d)^(1/3)-6^(1/3)*( c*e*(b*e+c*d)+3*c^2*e^2*x)^(1/3))*2^(2/3)/c^(2/3)/e^(5/3)/(-b*e+2*c*d)^(2/ 3)/(-(-2*b*e+c*d)*(b*e+c*d)+9*b*c*e^2*x+9*c^2*e^2*x^2)^(1/3)+1/4*(-c*e*(-2 *b*e+c*d)+3*c^2*e^2*x)^(1/3)*(c*e*(b*e+c*d)+3*c^2*e^2*x)^(1/3)*arctan(-1/3 *3^(1/2)+1/3*2^(1/3)*(-c*e*(-2*b*e+c*d)+3*c^2*e^2*x)^(2/3)/c^(1/3)/e^(1/3) /(-b*e+2*c*d)^(1/3)/(c*e*(b*e+c*d)+3*c^2*e^2*x)^(1/3)*3^(1/2))*3^(1/2)*2^( 2/3)/c^(2/3)/e^(5/3)/(-b*e+2*c*d)^(2/3)/(-(-2*b*e+c*d)*(b*e+c*d)+9*b*c*e^2 *x+9*c^2*e^2*x^2)^(1/3)
Time = 1.55 (sec) , antiderivative size = 473, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(d+e x) \sqrt [3]{-c^2 d^2+b c d e+2 b^2 e^2+9 b c e^2 x+9 c^2 e^2 x^2}} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{2 c d-b e} \sqrt [3]{2 b^2 e^2+b c e (d+9 e x)-c^2 \left (d^2-9 e^2 x^2\right )}}{-2 \sqrt [3]{2} b e+\sqrt [3]{2} c (d-3 e x)+\sqrt [3]{2 c d-b e} \sqrt [3]{2 b^2 e^2+b c e (d+9 e x)-c^2 \left (d^2-9 e^2 x^2\right )}}\right )+2 \log \left (\sqrt {e} \left (-\sqrt [3]{2} c d+\sqrt [3]{2} e (2 b+3 c x)+2 \sqrt [3]{2 c d-b e} \sqrt [3]{2 b^2 e^2+b c e (d+9 e x)-c^2 \left (d^2-9 e^2 x^2\right )}\right )\right )-\log \left (2^{2/3} c^2 d^2 e-4\ 2^{2/3} b c d e^2+4\ 2^{2/3} b^2 e^3-6\ 2^{2/3} c^2 d e^2 x+12\ 2^{2/3} b c e^3 x+9\ 2^{2/3} c^2 e^3 x^2+2 e \sqrt [3]{4 c d-2 b e} (c d-2 b e-3 c e x) \sqrt [3]{2 b^2 e^2+b c e (d+9 e x)-c^2 \left (d^2-9 e^2 x^2\right )}+4 e (2 c d-b e)^{2/3} \left (2 b^2 e^2+b c e (d+9 e x)-c^2 \left (d^2-9 e^2 x^2\right )\right )^{2/3}\right )}{4 \sqrt [3]{2} e (2 c d-b e)^{2/3}} \]
Integrate[1/((d + e*x)*(-(c^2*d^2) + b*c*d*e + 2*b^2*e^2 + 9*b*c*e^2*x + 9 *c^2*e^2*x^2)^(1/3)),x]
(2*Sqrt[3]*ArcTan[(Sqrt[3]*(2*c*d - b*e)^(1/3)*(2*b^2*e^2 + b*c*e*(d + 9*e *x) - c^2*(d^2 - 9*e^2*x^2))^(1/3))/(-2*2^(1/3)*b*e + 2^(1/3)*c*(d - 3*e*x ) + (2*c*d - b*e)^(1/3)*(2*b^2*e^2 + b*c*e*(d + 9*e*x) - c^2*(d^2 - 9*e^2* x^2))^(1/3))] + 2*Log[Sqrt[e]*(-(2^(1/3)*c*d) + 2^(1/3)*e*(2*b + 3*c*x) + 2*(2*c*d - b*e)^(1/3)*(2*b^2*e^2 + b*c*e*(d + 9*e*x) - c^2*(d^2 - 9*e^2*x^ 2))^(1/3))] - Log[2^(2/3)*c^2*d^2*e - 4*2^(2/3)*b*c*d*e^2 + 4*2^(2/3)*b^2* e^3 - 6*2^(2/3)*c^2*d*e^2*x + 12*2^(2/3)*b*c*e^3*x + 9*2^(2/3)*c^2*e^3*x^2 + 2*e*(4*c*d - 2*b*e)^(1/3)*(c*d - 2*b*e - 3*c*e*x)*(2*b^2*e^2 + b*c*e*(d + 9*e*x) - c^2*(d^2 - 9*e^2*x^2))^(1/3) + 4*e*(2*c*d - b*e)^(2/3)*(2*b^2* e^2 + b*c*e*(d + 9*e*x) - c^2*(d^2 - 9*e^2*x^2))^(2/3)])/(4*2^(1/3)*e*(2*c *d - b*e)^(2/3))
Time = 0.52 (sec) , antiderivative size = 376, normalized size of antiderivative = 0.67, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {1177, 27, 133}
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.
\(\displaystyle \int \frac {1}{(d+e x) \sqrt [3]{2 b^2 e^2+b c d e+9 b c e^2 x-c^2 d^2+9 c^2 e^2 x^2}} \, dx\) |
\(\Big \downarrow \) 1177 |
\(\displaystyle \frac {6^{2/3} \sqrt [3]{3 c^2 e^2 x-c e (c d-2 b e)} \sqrt [3]{c e (b e+c d)+3 c^2 e^2 x} \int \frac {1}{6^{2/3} (d+e x) \sqrt [3]{3 c^2 e^2 x-c e (c d-2 b e)} \sqrt [3]{3 c^2 x e^2+c (c d+b e) e}}dx}{\sqrt [3]{-(c d-2 b e) (b e+c d)+9 b c e^2 x+9 c^2 e^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt [3]{3 c^2 e^2 x-c e (c d-2 b e)} \sqrt [3]{c e (b e+c d)+3 c^2 e^2 x} \int \frac {1}{(d+e x) \sqrt [3]{3 c^2 e^2 x-c e (c d-2 b e)} \sqrt [3]{3 c^2 x e^2+c (c d+b e) e}}dx}{\sqrt [3]{-(c d-2 b e) (b e+c d)+9 b c e^2 x+9 c^2 e^2 x^2}}\) |
\(\Big \downarrow \) 133 |
\(\displaystyle \frac {\sqrt [3]{3 c^2 e^2 x-c e (c d-2 b e)} \sqrt [3]{c e (b e+c d)+3 c^2 e^2 x} \left (-\frac {\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {\sqrt [3]{2} \left (3 c^2 e^2 x-c e (c d-2 b e)\right )^{2/3}}{\sqrt {3} \sqrt [3]{c} \sqrt [3]{e} \sqrt [3]{2 c d-b e} \sqrt [3]{c e (b e+c d)+3 c^2 e^2 x}}\right )}{2 \sqrt [3]{2} c^{2/3} e^{5/3} (2 c d-b e)^{2/3}}-\frac {\log (d+e x)}{2 \sqrt [3]{2} c^{2/3} e^{5/3} (2 c d-b e)^{2/3}}+\frac {3 \log \left (-\frac {\left (3 c^2 e^2 x-c e (c d-2 b e)\right )^{2/3}}{2^{2/3} \sqrt [3]{c} \sqrt [3]{e} \sqrt [3]{2 c d-b e}}-\sqrt [3]{c e (b e+c d)+3 c^2 e^2 x}\right )}{4 \sqrt [3]{2} c^{2/3} e^{5/3} (2 c d-b e)^{2/3}}\right )}{\sqrt [3]{-(c d-2 b e) (b e+c d)+9 b c e^2 x+9 c^2 e^2 x^2}}\) |
((-(c*e*(c*d - 2*b*e)) + 3*c^2*e^2*x)^(1/3)*(c*e*(c*d + b*e) + 3*c^2*e^2*x )^(1/3)*(-1/2*(Sqrt[3]*ArcTan[1/Sqrt[3] - (2^(1/3)*(-(c*e*(c*d - 2*b*e)) + 3*c^2*e^2*x)^(2/3))/(Sqrt[3]*c^(1/3)*e^(1/3)*(2*c*d - b*e)^(1/3)*(c*e*(c* d + b*e) + 3*c^2*e^2*x)^(1/3))])/(2^(1/3)*c^(2/3)*e^(5/3)*(2*c*d - b*e)^(2 /3)) - Log[d + e*x]/(2*2^(1/3)*c^(2/3)*e^(5/3)*(2*c*d - b*e)^(2/3)) + (3*L og[-((-(c*e*(c*d - 2*b*e)) + 3*c^2*e^2*x)^(2/3)/(2^(2/3)*c^(1/3)*e^(1/3)*( 2*c*d - b*e)^(1/3))) - (c*e*(c*d + b*e) + 3*c^2*e^2*x)^(1/3)])/(4*2^(1/3)* c^(2/3)*e^(5/3)*(2*c*d - b*e)^(2/3))))/(-((c*d - 2*b*e)*(c*d + b*e)) + 9*b *c*e^2*x + 9*c^2*e^2*x^2)^(1/3)
3.26.9.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)*((e_.) + (f_.)*(x_)) ^(1/3)), x_] :> With[{q = Rt[b*((b*e - a*f)/(b*c - a*d)^2), 3]}, Simp[-Log[ a + b*x]/(2*q*(b*c - a*d)), x] + (-Simp[Sqrt[3]*(ArcTan[1/Sqrt[3] + 2*q*((c + d*x)^(2/3)/(Sqrt[3]*(e + f*x)^(1/3)))]/(2*q*(b*c - a*d))), x] + Simp[3*( Log[q*(c + d*x)^(2/3) - (e + f*x)^(1/3)]/(4*q*(b*c - a*d))), x])] /; FreeQ[ {a, b, c, d, e, f}, x] && EqQ[2*b*d*e - b*c*f - a*d*f, 0]
Int[1/(((d_.) + (e_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(1/3)), x_Sy mbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(b + q + 2*c*x)^(1/3)*((b - q + 2*c*x)^(1/3)/(a + b*x + c*x^2)^(1/3)) Int[1/((d + e*x)*(b + q + 2*c*x)^ (1/3)*(b - q + 2*c*x)^(1/3)), x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c ^2*d^2 - b*c*d*e - 2*b^2*e^2 + 9*a*c*e^2, 0]
\[\int \frac {1}{\left (e x +d \right ) \left (9 c^{2} e^{2} x^{2}+9 b c \,e^{2} x +2 b^{2} e^{2}+b c d e -c^{2} d^{2}\right )^{\frac {1}{3}}}d x\]
Timed out. \[ \int \frac {1}{(d+e x) \sqrt [3]{-c^2 d^2+b c d e+2 b^2 e^2+9 b c e^2 x+9 c^2 e^2 x^2}} \, dx=\text {Timed out} \]
integrate(1/(e*x+d)/(9*c^2*e^2*x^2+9*b*c*e^2*x+2*b^2*e^2+b*c*d*e-c^2*d^2)^ (1/3),x, algorithm="fricas")
\[ \int \frac {1}{(d+e x) \sqrt [3]{-c^2 d^2+b c d e+2 b^2 e^2+9 b c e^2 x+9 c^2 e^2 x^2}} \, dx=\int \frac {1}{\sqrt [3]{\left (b e + c d + 3 c e x\right ) \left (2 b e - c d + 3 c e x\right )} \left (d + e x\right )}\, dx \]
\[ \int \frac {1}{(d+e x) \sqrt [3]{-c^2 d^2+b c d e+2 b^2 e^2+9 b c e^2 x+9 c^2 e^2 x^2}} \, dx=\int { \frac {1}{{\left (9 \, c^{2} e^{2} x^{2} + 9 \, b c e^{2} x - c^{2} d^{2} + b c d e + 2 \, b^{2} e^{2}\right )}^{\frac {1}{3}} {\left (e x + d\right )}} \,d x } \]
integrate(1/(e*x+d)/(9*c^2*e^2*x^2+9*b*c*e^2*x+2*b^2*e^2+b*c*d*e-c^2*d^2)^ (1/3),x, algorithm="maxima")
integrate(1/((9*c^2*e^2*x^2 + 9*b*c*e^2*x - c^2*d^2 + b*c*d*e + 2*b^2*e^2) ^(1/3)*(e*x + d)), x)
\[ \int \frac {1}{(d+e x) \sqrt [3]{-c^2 d^2+b c d e+2 b^2 e^2+9 b c e^2 x+9 c^2 e^2 x^2}} \, dx=\int { \frac {1}{{\left (9 \, c^{2} e^{2} x^{2} + 9 \, b c e^{2} x - c^{2} d^{2} + b c d e + 2 \, b^{2} e^{2}\right )}^{\frac {1}{3}} {\left (e x + d\right )}} \,d x } \]
integrate(1/(e*x+d)/(9*c^2*e^2*x^2+9*b*c*e^2*x+2*b^2*e^2+b*c*d*e-c^2*d^2)^ (1/3),x, algorithm="giac")
integrate(1/((9*c^2*e^2*x^2 + 9*b*c*e^2*x - c^2*d^2 + b*c*d*e + 2*b^2*e^2) ^(1/3)*(e*x + d)), x)
Timed out. \[ \int \frac {1}{(d+e x) \sqrt [3]{-c^2 d^2+b c d e+2 b^2 e^2+9 b c e^2 x+9 c^2 e^2 x^2}} \, dx=\int \frac {1}{\left (d+e\,x\right )\,{\left (2\,b^2\,e^2+b\,c\,d\,e+9\,b\,c\,e^2\,x-c^2\,d^2+9\,c^2\,e^2\,x^2\right )}^{1/3}} \,d x \]