Integrand size = 14, antiderivative size = 415 \[ \int \sqrt [3]{c+d \tan (e+f x)} \, dx=-\frac {1}{4} \sqrt [3]{c-\sqrt {-d^2}} x-\frac {1}{4} \sqrt [3]{c+\sqrt {-d^2}} x+\frac {\sqrt {3} d \sqrt [3]{c-\sqrt {-d^2}} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-\sqrt {-d^2}}}}{\sqrt {3}}\right )}{2 \sqrt {-d^2} f}-\frac {\sqrt {3} d \sqrt [3]{c+\sqrt {-d^2}} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+\sqrt {-d^2}}}}{\sqrt {3}}\right )}{2 \sqrt {-d^2} f}-\frac {d \sqrt [3]{c-\sqrt {-d^2}} \log (\cos (e+f x))}{4 \sqrt {-d^2} f}+\frac {d \sqrt [3]{c+\sqrt {-d^2}} \log (\cos (e+f x))}{4 \sqrt {-d^2} f}-\frac {3 d \sqrt [3]{c-\sqrt {-d^2}} \log \left (\sqrt [3]{c-\sqrt {-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt {-d^2} f}+\frac {3 d \sqrt [3]{c+\sqrt {-d^2}} \log \left (\sqrt [3]{c+\sqrt {-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt {-d^2} f} \]
-1/4*x*(c-(-d^2)^(1/2))^(1/3)-1/4*d*ln(cos(f*x+e))*(c-(-d^2)^(1/2))^(1/3)/ f/(-d^2)^(1/2)-3/4*d*ln((c-(-d^2)^(1/2))^(1/3)-(c+d*tan(f*x+e))^(1/3))*(c- (-d^2)^(1/2))^(1/3)/f/(-d^2)^(1/2)+1/2*d*arctan(1/3*(1+2*(c+d*tan(f*x+e))^ (1/3)/(c-(-d^2)^(1/2))^(1/3))*3^(1/2))*3^(1/2)*(c-(-d^2)^(1/2))^(1/3)/f/(- d^2)^(1/2)-1/4*x*(c+(-d^2)^(1/2))^(1/3)+1/4*d*ln(cos(f*x+e))*(c+(-d^2)^(1/ 2))^(1/3)/f/(-d^2)^(1/2)+3/4*d*ln((c+(-d^2)^(1/2))^(1/3)-(c+d*tan(f*x+e))^ (1/3))*(c+(-d^2)^(1/2))^(1/3)/f/(-d^2)^(1/2)-1/2*d*arctan(1/3*(1+2*(c+d*ta n(f*x+e))^(1/3)/(c+(-d^2)^(1/2))^(1/3))*3^(1/2))*3^(1/2)*(c+(-d^2)^(1/2))^ (1/3)/f/(-d^2)^(1/2)
Result contains complex when optimal does not.
Time = 0.23 (sec) , antiderivative size = 294, normalized size of antiderivative = 0.71 \[ \int \sqrt [3]{c+d \tan (e+f x)} \, dx=\frac {-i \sqrt [3]{c-i d} \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-i d}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{c-i d}-\sqrt [3]{c+d \tan (e+f x)}\right )+\log \left ((c-i d)^{2/3}+\sqrt [3]{c-i d} \sqrt [3]{c+d \tan (e+f x)}+(c+d \tan (e+f x))^{2/3}\right )\right )+i \sqrt [3]{c+i d} \left (2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+i d}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{c+i d}-\sqrt [3]{c+d \tan (e+f x)}\right )+\log \left ((c+i d)^{2/3}+\sqrt [3]{c+i d} \sqrt [3]{c+d \tan (e+f x)}+(c+d \tan (e+f x))^{2/3}\right )\right )}{4 f} \]
((-I)*(c - I*d)^(1/3)*(2*Sqrt[3]*ArcTan[(1 + (2*(c + d*Tan[e + f*x])^(1/3) )/(c - I*d)^(1/3))/Sqrt[3]] - 2*Log[(c - I*d)^(1/3) - (c + d*Tan[e + f*x]) ^(1/3)] + Log[(c - I*d)^(2/3) + (c - I*d)^(1/3)*(c + d*Tan[e + f*x])^(1/3) + (c + d*Tan[e + f*x])^(2/3)]) + I*(c + I*d)^(1/3)*(2*Sqrt[3]*ArcTan[(1 + (2*(c + d*Tan[e + f*x])^(1/3))/(c + I*d)^(1/3))/Sqrt[3]] - 2*Log[(c + I*d )^(1/3) - (c + d*Tan[e + f*x])^(1/3)] + Log[(c + I*d)^(2/3) + (c + I*d)^(1 /3)*(c + d*Tan[e + f*x])^(1/3) + (c + d*Tan[e + f*x])^(2/3)]))/(4*f)
Time = 0.56 (sec) , antiderivative size = 379, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3966, 485, 2009}
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.
\(\displaystyle \int \sqrt [3]{c+d \tan (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt [3]{c+d \tan (e+f x)}dx\) |
\(\Big \downarrow \) 3966 |
\(\displaystyle \frac {d \int \frac {\sqrt [3]{c+d \tan (e+f x)}}{\tan ^2(e+f x) d^2+d^2}d(d \tan (e+f x))}{f}\) |
\(\Big \downarrow \) 485 |
\(\displaystyle \frac {d \int \left (\frac {\sqrt [3]{c+d \tan (e+f x)} \sqrt {-d^2}}{2 d^2 \left (\sqrt {-d^2}-d \tan (e+f x)\right )}+\frac {\sqrt [3]{c+d \tan (e+f x)} \sqrt {-d^2}}{2 d^2 \left (d \tan (e+f x)+\sqrt {-d^2}\right )}\right )d(d \tan (e+f x))}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {d \left (\frac {\sqrt {3} \sqrt [3]{c-\sqrt {-d^2}} \arctan \left (\frac {\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c-\sqrt {-d^2}}}+1}{\sqrt {3}}\right )}{2 \sqrt {-d^2}}-\frac {\sqrt {3} \sqrt [3]{c+\sqrt {-d^2}} \arctan \left (\frac {\frac {2 \sqrt [3]{c+d \tan (e+f x)}}{\sqrt [3]{c+\sqrt {-d^2}}}+1}{\sqrt {3}}\right )}{2 \sqrt {-d^2}}-\frac {\sqrt [3]{c+\sqrt {-d^2}} \log \left (\sqrt {-d^2}-d \tan (e+f x)\right )}{4 \sqrt {-d^2}}+\frac {\sqrt [3]{c-\sqrt {-d^2}} \log \left (\sqrt {-d^2}+d \tan (e+f x)\right )}{4 \sqrt {-d^2}}-\frac {3 \sqrt [3]{c-\sqrt {-d^2}} \log \left (\sqrt [3]{c-\sqrt {-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt {-d^2}}+\frac {3 \sqrt [3]{c+\sqrt {-d^2}} \log \left (\sqrt [3]{c+\sqrt {-d^2}}-\sqrt [3]{c+d \tan (e+f x)}\right )}{4 \sqrt {-d^2}}\right )}{f}\) |
(d*((Sqrt[3]*(c - Sqrt[-d^2])^(1/3)*ArcTan[(1 + (2*(c + d*Tan[e + f*x])^(1 /3))/(c - Sqrt[-d^2])^(1/3))/Sqrt[3]])/(2*Sqrt[-d^2]) - (Sqrt[3]*(c + Sqrt [-d^2])^(1/3)*ArcTan[(1 + (2*(c + d*Tan[e + f*x])^(1/3))/(c + Sqrt[-d^2])^ (1/3))/Sqrt[3]])/(2*Sqrt[-d^2]) - ((c + Sqrt[-d^2])^(1/3)*Log[Sqrt[-d^2] - d*Tan[e + f*x]])/(4*Sqrt[-d^2]) + ((c - Sqrt[-d^2])^(1/3)*Log[Sqrt[-d^2] + d*Tan[e + f*x]])/(4*Sqrt[-d^2]) - (3*(c - Sqrt[-d^2])^(1/3)*Log[(c - Sqr t[-d^2])^(1/3) - (c + d*Tan[e + f*x])^(1/3)])/(4*Sqrt[-d^2]) + (3*(c + Sqr t[-d^2])^(1/3)*Log[(c + Sqrt[-d^2])^(1/3) - (c + d*Tan[e + f*x])^(1/3)])/( 4*Sqrt[-d^2])))/f
3.7.85.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[Expand Integrand[(c + d*x)^n, 1/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d, n}, x] & & !IntegerQ[2*n]
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d Su bst[Int[(a + x)^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c , d, n}, x] && NeQ[a^2 + b^2, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.97 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.14
method | result | size |
derivativedivides | \(\frac {d \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 c \,\textit {\_Z}^{3}+c^{2}+d^{2}\right )}{\sum }\frac {\textit {\_R}^{3} \ln \left (\left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} c}\right )}{2 f}\) | \(60\) |
default | \(\frac {d \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 c \,\textit {\_Z}^{3}+c^{2}+d^{2}\right )}{\sum }\frac {\textit {\_R}^{3} \ln \left (\left (c +d \tan \left (f x +e \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} c}\right )}{2 f}\) | \(60\) |
1/2/f*d*sum(_R^3/(_R^5-_R^2*c)*ln((c+d*tan(f*x+e))^(1/3)-_R),_R=RootOf(_Z^ 6-2*_Z^3*c+c^2+d^2))
Time = 0.29 (sec) , antiderivative size = 536, normalized size of antiderivative = 1.29 \[ \int \sqrt [3]{c+d \tan (e+f x)} \, dx=-\frac {1}{4} \, {\left (\sqrt {-3} + 1\right )} \left (-\frac {f^{3} \sqrt {-\frac {c^{2}}{f^{6}}} + d}{f^{3}}\right )^{\frac {1}{3}} \log \left (-\frac {1}{2} \, {\left (\sqrt {-3} f^{4} + f^{4}\right )} \left (-\frac {f^{3} \sqrt {-\frac {c^{2}}{f^{6}}} + d}{f^{3}}\right )^{\frac {1}{3}} \sqrt {-\frac {c^{2}}{f^{6}}} + {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {1}{3}} c\right ) + \frac {1}{4} \, {\left (\sqrt {-3} - 1\right )} \left (-\frac {f^{3} \sqrt {-\frac {c^{2}}{f^{6}}} + d}{f^{3}}\right )^{\frac {1}{3}} \log \left (\frac {1}{2} \, {\left (\sqrt {-3} f^{4} - f^{4}\right )} \left (-\frac {f^{3} \sqrt {-\frac {c^{2}}{f^{6}}} + d}{f^{3}}\right )^{\frac {1}{3}} \sqrt {-\frac {c^{2}}{f^{6}}} + {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {1}{3}} c\right ) - \frac {1}{4} \, {\left (\sqrt {-3} + 1\right )} \left (\frac {f^{3} \sqrt {-\frac {c^{2}}{f^{6}}} - d}{f^{3}}\right )^{\frac {1}{3}} \log \left (\frac {1}{2} \, {\left (\sqrt {-3} f^{4} + f^{4}\right )} \left (\frac {f^{3} \sqrt {-\frac {c^{2}}{f^{6}}} - d}{f^{3}}\right )^{\frac {1}{3}} \sqrt {-\frac {c^{2}}{f^{6}}} + {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {1}{3}} c\right ) + \frac {1}{4} \, {\left (\sqrt {-3} - 1\right )} \left (\frac {f^{3} \sqrt {-\frac {c^{2}}{f^{6}}} - d}{f^{3}}\right )^{\frac {1}{3}} \log \left (-\frac {1}{2} \, {\left (\sqrt {-3} f^{4} - f^{4}\right )} \left (\frac {f^{3} \sqrt {-\frac {c^{2}}{f^{6}}} - d}{f^{3}}\right )^{\frac {1}{3}} \sqrt {-\frac {c^{2}}{f^{6}}} + {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {1}{3}} c\right ) + \frac {1}{2} \, \left (-\frac {f^{3} \sqrt {-\frac {c^{2}}{f^{6}}} + d}{f^{3}}\right )^{\frac {1}{3}} \log \left (f^{4} \left (-\frac {f^{3} \sqrt {-\frac {c^{2}}{f^{6}}} + d}{f^{3}}\right )^{\frac {1}{3}} \sqrt {-\frac {c^{2}}{f^{6}}} + {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {1}{3}} c\right ) + \frac {1}{2} \, \left (\frac {f^{3} \sqrt {-\frac {c^{2}}{f^{6}}} - d}{f^{3}}\right )^{\frac {1}{3}} \log \left (-f^{4} \left (\frac {f^{3} \sqrt {-\frac {c^{2}}{f^{6}}} - d}{f^{3}}\right )^{\frac {1}{3}} \sqrt {-\frac {c^{2}}{f^{6}}} + {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {1}{3}} c\right ) \]
-1/4*(sqrt(-3) + 1)*(-(f^3*sqrt(-c^2/f^6) + d)/f^3)^(1/3)*log(-1/2*(sqrt(- 3)*f^4 + f^4)*(-(f^3*sqrt(-c^2/f^6) + d)/f^3)^(1/3)*sqrt(-c^2/f^6) + (d*ta n(f*x + e) + c)^(1/3)*c) + 1/4*(sqrt(-3) - 1)*(-(f^3*sqrt(-c^2/f^6) + d)/f ^3)^(1/3)*log(1/2*(sqrt(-3)*f^4 - f^4)*(-(f^3*sqrt(-c^2/f^6) + d)/f^3)^(1/ 3)*sqrt(-c^2/f^6) + (d*tan(f*x + e) + c)^(1/3)*c) - 1/4*(sqrt(-3) + 1)*((f ^3*sqrt(-c^2/f^6) - d)/f^3)^(1/3)*log(1/2*(sqrt(-3)*f^4 + f^4)*((f^3*sqrt( -c^2/f^6) - d)/f^3)^(1/3)*sqrt(-c^2/f^6) + (d*tan(f*x + e) + c)^(1/3)*c) + 1/4*(sqrt(-3) - 1)*((f^3*sqrt(-c^2/f^6) - d)/f^3)^(1/3)*log(-1/2*(sqrt(-3 )*f^4 - f^4)*((f^3*sqrt(-c^2/f^6) - d)/f^3)^(1/3)*sqrt(-c^2/f^6) + (d*tan( f*x + e) + c)^(1/3)*c) + 1/2*(-(f^3*sqrt(-c^2/f^6) + d)/f^3)^(1/3)*log(f^4 *(-(f^3*sqrt(-c^2/f^6) + d)/f^3)^(1/3)*sqrt(-c^2/f^6) + (d*tan(f*x + e) + c)^(1/3)*c) + 1/2*((f^3*sqrt(-c^2/f^6) - d)/f^3)^(1/3)*log(-f^4*((f^3*sqrt (-c^2/f^6) - d)/f^3)^(1/3)*sqrt(-c^2/f^6) + (d*tan(f*x + e) + c)^(1/3)*c)
\[ \int \sqrt [3]{c+d \tan (e+f x)} \, dx=\int \sqrt [3]{c + d \tan {\left (e + f x \right )}}\, dx \]
\[ \int \sqrt [3]{c+d \tan (e+f x)} \, dx=\int { {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {1}{3}} \,d x } \]
\[ \int \sqrt [3]{c+d \tan (e+f x)} \, dx=\int { {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {1}{3}} \,d x } \]
Time = 8.60 (sec) , antiderivative size = 863, normalized size of antiderivative = 2.08 \[ \int \sqrt [3]{c+d \tan (e+f x)} \, dx=\text {Too large to display} \]
log((c + d*tan(e + f*x))^(1/3) + f*(-(c*1i + d)/f^3)^(1/3)*1i)*(-(c*1i + d )/(8*f^3))^(1/3) + log(d*(c + d*tan(e + f*x))^(1/3)*1i - c*(c + d*tan(e + f*x))^(1/3) + f^4*((c*1i - d)/f^3)^(4/3) + 2*d*f*((c*1i - d)/f^3)^(1/3))*( (c*1i - d)/(8*f^3))^(1/3) - log((((c*1i - d)/f^3)^(1/3)*((3^(1/2)*1i)/2 + 1/2)*((((c*1i - d)/f^3)^(2/3)*((3^(1/2)*1i)/2 - 1/2)*((3888*d^5*(c^2 + d^2 )*(c + d*tan(e + f*x))^(1/3))/f - 3888*c*d^4*((c*1i - d)/f^3)^(1/3)*((3^(1 /2)*1i)/2 + 1/2)*(c^2 + d^2)))/4 + (1944*c*d^5*(c^2 + d^2))/f^3))/2 - (486 *(d^8 - c^4*d^4)*(c + d*tan(e + f*x))^(1/3))/f^4)*((3^(1/2)*1i)/2 + 1/2)*( (c*1i - d)/(8*f^3))^(1/3) + log((486*(d^8 - c^4*d^4)*(c + d*tan(e + f*x))^ (1/3))/f^4 - (((c*1i - d)/f^3)^(1/3)*((3^(1/2)*1i)/2 - 1/2)*((((c*1i - d)/ f^3)^(2/3)*((3^(1/2)*1i)/2 + 1/2)*((3888*d^5*(c^2 + d^2)*(c + d*tan(e + f* x))^(1/3))/f + 3888*c*d^4*((c*1i - d)/f^3)^(1/3)*((3^(1/2)*1i)/2 - 1/2)*(c ^2 + d^2)))/4 - (1944*c*d^5*(c^2 + d^2))/f^3))/2)*((3^(1/2)*1i)/2 - 1/2)*( (c*1i - d)/(8*f^3))^(1/3) - log((((3^(1/2)*1i)/2 + 1/2)*((((3888*d^5*(c^2 + d^2)*(c + d*tan(e + f*x))^(1/3))/f - 3888*c*d^4*((3^(1/2)*1i)/2 + 1/2)*( -(c*1i + d)/f^3)^(1/3)*(c^2 + d^2))*((3^(1/2)*1i)/2 - 1/2)*(-(c*1i + d)/f^ 3)^(2/3))/4 + (1944*c*d^5*(c^2 + d^2))/f^3)*(-(c*1i + d)/f^3)^(1/3))/2 - ( 486*(d^8 - c^4*d^4)*(c + d*tan(e + f*x))^(1/3))/f^4)*((3^(1/2)*1i)/2 + 1/2 )*(-(c*1i + d)/(8*f^3))^(1/3) + log((486*(d^8 - c^4*d^4)*(c + d*tan(e + f* x))^(1/3))/f^4 - (((3^(1/2)*1i)/2 - 1/2)*((((3888*d^5*(c^2 + d^2)*(c + ...