Integrand size = 30, antiderivative size = 340 \[ \int \frac {(d \sec (e+f x))^{2/3}}{\sqrt [3]{a+i a \tan (e+f x)}} \, dx=-\frac {\sqrt [3]{a} x (d \sec (e+f x))^{2/3}}{2\ 2^{2/3} \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}+\frac {i \sqrt {3} \sqrt [3]{a} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a-i a \tan (e+f x)}}{\sqrt {3} \sqrt [3]{a}}\right ) (d \sec (e+f x))^{2/3}}{2^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {i \sqrt [3]{a} \log (\cos (e+f x)) (d \sec (e+f x))^{2/3}}{2\ 2^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {3 i \sqrt [3]{a} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a-i a \tan (e+f x)}\right ) (d \sec (e+f x))^{2/3}}{2\ 2^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}} \]
-1/4*a^(1/3)*x*(d*sec(f*x+e))^(2/3)*2^(1/3)/(a-I*a*tan(f*x+e))^(1/3)/(a+I* a*tan(f*x+e))^(1/3)-1/4*I*a^(1/3)*ln(cos(f*x+e))*(d*sec(f*x+e))^(2/3)*2^(1 /3)/f/(a-I*a*tan(f*x+e))^(1/3)/(a+I*a*tan(f*x+e))^(1/3)-3/4*I*a^(1/3)*ln(2 ^(1/3)*a^(1/3)-(a-I*a*tan(f*x+e))^(1/3))*(d*sec(f*x+e))^(2/3)*2^(1/3)/f/(a -I*a*tan(f*x+e))^(1/3)/(a+I*a*tan(f*x+e))^(1/3)+1/2*I*a^(1/3)*arctan(1/3*( a^(1/3)+2^(2/3)*(a-I*a*tan(f*x+e))^(1/3))/a^(1/3)*3^(1/2))*(d*sec(f*x+e))^ (2/3)*3^(1/2)*2^(1/3)/f/(a-I*a*tan(f*x+e))^(1/3)/(a+I*a*tan(f*x+e))^(1/3)
Time = 1.16 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.47 \[ \int \frac {(d \sec (e+f x))^{2/3}}{\sqrt [3]{a+i a \tan (e+f x)}} \, dx=-\frac {\left (\frac {d e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{2/3} \sqrt [3]{1+e^{2 i (e+f x)}} \left (2 f x+2 i \sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{1+e^{2 i (e+f x)}}}{\sqrt {3}}\right )+3 i \log \left (1-\sqrt [3]{1+e^{2 i (e+f x)}}\right )\right )}{2\ 2^{2/3} \sqrt [3]{\frac {a e^{2 i (e+f x)}}{1+e^{2 i (e+f x)}}} f} \]
-1/2*(((d*E^(I*(e + f*x)))/(1 + E^((2*I)*(e + f*x))))^(2/3)*(1 + E^((2*I)* (e + f*x)))^(1/3)*(2*f*x + (2*I)*Sqrt[3]*ArcTan[(1 + 2*(1 + E^((2*I)*(e + f*x)))^(1/3))/Sqrt[3]] + (3*I)*Log[1 - (1 + E^((2*I)*(e + f*x)))^(1/3)]))/ (2^(2/3)*((a*E^((2*I)*(e + f*x)))/(1 + E^((2*I)*(e + f*x))))^(1/3)*f)
Time = 0.45 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.45, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3042, 3973, 3042, 3962, 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.
| \(\displaystyle \int \frac {(d \sec (e+f x))^{2/3}}{\sqrt [3]{a+i a \tan (e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(d \sec (e+f x))^{2/3}}{\sqrt [3]{a+i a \tan (e+f x)}}dx\) |
| \(\Big \downarrow \) 3973 |
| \(\displaystyle \frac {(d \sec (e+f x))^{2/3} \int \sqrt [3]{a-i a \tan (e+f x)}dx}{\sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(d \sec (e+f x))^{2/3} \int \sqrt [3]{a-i a \tan (e+f x)}dx}{\sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3962 |
| \(\displaystyle \frac {i a (d \sec (e+f x))^{2/3} \int \frac {1}{(a-i a \tan (e+f x))^{2/3} (i \tan (e+f x) a+a)}d(-i a \tan (e+f x))}{f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 69 |
| \(\displaystyle \frac {i a (d \sec (e+f x))^{2/3} \left (\frac {3 \int \frac {1}{i a \tan (e+f x)+\sqrt [3]{2} \sqrt [3]{a}}d\sqrt [3]{a-i a \tan (e+f x)}}{2\ 2^{2/3} a^{2/3}}+\frac {3 \int \frac {1}{-a^2 \tan ^2(e+f x)-i \sqrt [3]{2} a^{4/3} \tan (e+f x)+2^{2/3} a^{2/3}}d\sqrt [3]{a-i a \tan (e+f x)}}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a+i a \tan (e+f x))}{2\ 2^{2/3} a^{2/3}}\right )}{f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {i a (d \sec (e+f x))^{2/3} \left (\frac {3 \int \frac {1}{-a^2 \tan ^2(e+f x)-i \sqrt [3]{2} a^{4/3} \tan (e+f x)+2^{2/3} a^{2/3}}d\sqrt [3]{a-i a \tan (e+f x)}}{2 \sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}+i a \tan (e+f x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a+i a \tan (e+f x))}{2\ 2^{2/3} a^{2/3}}\right )}{f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {i a (d \sec (e+f x))^{2/3} \left (-\frac {3 \int \frac {1}{a^2 \tan ^2(e+f x)-3}d\left (1-i 2^{2/3} a^{2/3} \tan (e+f x)\right )}{2^{2/3} a^{2/3}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}+i a \tan (e+f x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a+i a \tan (e+f x))}{2\ 2^{2/3} a^{2/3}}\right )}{f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {i a (d \sec (e+f x))^{2/3} \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (e+f x)}{\sqrt {3}}\right )}{2^{2/3} a^{2/3}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}+i a \tan (e+f x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a+i a \tan (e+f x))}{2\ 2^{2/3} a^{2/3}}\right )}{f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\) |
(I*a*(((-I)*Sqrt[3]*ArcTanh[(a*Tan[e + f*x])/Sqrt[3]])/(2^(2/3)*a^(2/3)) - (3*Log[2^(1/3)*a^(1/3) + I*a*Tan[e + f*x]])/(2*2^(2/3)*a^(2/3)) + Log[a + I*a*Tan[e + f*x]]/(2*2^(2/3)*a^(2/3)))*(d*Sec[e + f*x])^(2/3))/(f*(a - I* a*Tan[e + f*x])^(1/3)*(a + I*a*Tan[e + f*x])^(1/3))
3.5.45.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
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]
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[-b/d S ubst[Int[(a + x)^(n - 1)/(a - x), x], x, b*Tan[c + d*x]], x] /; FreeQ[{a, b , c, d, n}, x] && EqQ[a^2 + b^2, 0]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x _)])^(n_), x_Symbol] :> Simp[(a/d)^(2*IntPart[n])*(a + b*Tan[e + f*x])^Frac Part[n]*((a - b*Tan[e + f*x])^FracPart[n]/(d*Sec[e + f*x])^(2*FracPart[n])) Int[1/(a - b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] && EqQ[Simplify[m/2 + n], 0]
\[\int \frac {\left (d \sec \left (f x +e \right )\right )^{\frac {2}{3}}}{\left (a +i a \tan \left (f x +e \right )\right )^{\frac {1}{3}}}d x\]
Time = 0.26 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.08 \[ \int \frac {(d \sec (e+f x))^{2/3}}{\sqrt [3]{a+i a \tan (e+f x)}} \, dx=\frac {1}{2} \, {\left (i \, \sqrt {3} - 1\right )} \left (\frac {i \, d^{2}}{4 \, a f^{3}}\right )^{\frac {1}{3}} \log \left (2 \, {\left (2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (\sqrt {3} a f + i \, a f\right )} \left (\frac {i \, d^{2}}{4 \, a f^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) + \frac {1}{2} \, {\left (-i \, \sqrt {3} - 1\right )} \left (\frac {i \, d^{2}}{4 \, a f^{3}}\right )^{\frac {1}{3}} \log \left (2 \, {\left (2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - {\left (\sqrt {3} a f - i \, a f\right )} \left (\frac {i \, d^{2}}{4 \, a f^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) + \left (\frac {i \, d^{2}}{4 \, a f^{3}}\right )^{\frac {1}{3}} \log \left (2 \, {\left (2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, a f \left (\frac {i \, d^{2}}{4 \, a f^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) \]
1/2*(I*sqrt(3) - 1)*(1/4*I*d^2/(a*f^3))^(1/3)*log(2*(2^(1/3)*(a/(e^(2*I*f* x + 2*I*e) + 1))^(2/3)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*(e^(2*I*f*x + 2 *I*e) + 1)*e^(2*I*f*x + 2*I*e) + (sqrt(3)*a*f + I*a*f)*(1/4*I*d^2/(a*f^3)) ^(1/3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) + 1/2*(-I*sqrt(3) - 1)*( 1/4*I*d^2/(a*f^3))^(1/3)*log(2*(2^(1/3)*(a/(e^(2*I*f*x + 2*I*e) + 1))^(2/3 )*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*(e^(2*I*f*x + 2*I*e) + 1)*e^(2*I*f*x + 2*I*e) - (sqrt(3)*a*f - I*a*f)*(1/4*I*d^2/(a*f^3))^(1/3)*e^(2*I*f*x + 2 *I*e))*e^(-2*I*f*x - 2*I*e)) + (1/4*I*d^2/(a*f^3))^(1/3)*log(2*(2^(1/3)*(a /(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*(e^( 2*I*f*x + 2*I*e) + 1)*e^(2*I*f*x + 2*I*e) - 2*I*a*f*(1/4*I*d^2/(a*f^3))^(1 /3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e))
\[ \int \frac {(d \sec (e+f x))^{2/3}}{\sqrt [3]{a+i a \tan (e+f x)}} \, dx=\int \frac {\left (d \sec {\left (e + f x \right )}\right )^{\frac {2}{3}}}{\sqrt [3]{i a \left (\tan {\left (e + f x \right )} - i\right )}}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1753 vs. \(2 (251) = 502\).
Time = 0.47 (sec) , antiderivative size = 1753, normalized size of antiderivative = 5.16 \[ \int \frac {(d \sec (e+f x))^{2/3}}{\sqrt [3]{a+i a \tan (e+f x)}} \, dx=\text {Too large to display} \]
1/8*(-2*I*sqrt(3)*2^(1/3)*arctan2(2/3*sqrt(3)*(cos(2*f*x + 2*e)^2 + sin(2* f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/6)*cos(1/3*arctan2(sin(2*f*x + 2 *e), cos(2*f*x + 2*e) + 1)) + 1/3*sqrt(3), 1/3*sqrt(3)*(2*(cos(2*f*x + 2*e )^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/6)*sin(1/3*arctan2(s in(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + sqrt(3))) - 2*I*sqrt(3)*2^(1/3)* arctan2(2/3*sqrt(3)*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/6)*cos(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) ) + 1/3*sqrt(3), -1/3*sqrt(3)*(2*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/6)*sin(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f* x + 2*e) + 1)) - sqrt(3))) + sqrt(3)*2^(1/3)*log(4/3*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/3)*(cos(1/3*arctan2(sin(2 *f*x + 2*e), cos(2*f*x + 2*e) + 1))^2 + sin(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))^2) + 4/3*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/6)*(sqrt(3)*sin(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + cos(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e ) + 1))) + 4/3) - sqrt(3)*2^(1/3)*log(4/3*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/3)*(cos(1/3*arctan2(sin(2*f*x + 2*e) , cos(2*f*x + 2*e) + 1))^2 + sin(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))^2) - 4/3*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f* x + 2*e) + 1)^(1/6)*(sqrt(3)*sin(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*...
\[ \int \frac {(d \sec (e+f x))^{2/3}}{\sqrt [3]{a+i a \tan (e+f x)}} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {2}{3}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
Timed out. \[ \int \frac {(d \sec (e+f x))^{2/3}}{\sqrt [3]{a+i a \tan (e+f x)}} \, dx=\int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{2/3}}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{1/3}} \,d x \]