Integrand size = 34, antiderivative size = 289 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=-\frac {a^{2/3} (i A+B) x}{2 \sqrt [3]{2}}+\frac {\sqrt {3} a^{2/3} A \arctan \left (\frac {\sqrt [3]{a}+2 \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{d}-\frac {\sqrt {3} a^{2/3} (A-i B) \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{2} d}-\frac {a^{2/3} (A-i B) \log (\cos (c+d x))}{2 \sqrt [3]{2} d}-\frac {a^{2/3} A \log (\tan (c+d x))}{2 d}+\frac {3 a^{2/3} A \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 d}-\frac {3 a^{2/3} (A-i B) \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{2} d} \]
-1/4*a^(2/3)*(I*A+B)*x*2^(2/3)-1/4*a^(2/3)*(A-I*B)*ln(cos(d*x+c))*2^(2/3)/ d-1/2*a^(2/3)*A*ln(tan(d*x+c))/d+3/2*a^(2/3)*A*ln(a^(1/3)-(a+I*a*tan(d*x+c ))^(1/3))/d-3/4*a^(2/3)*(A-I*B)*ln(2^(1/3)*a^(1/3)-(a+I*a*tan(d*x+c))^(1/3 ))*2^(2/3)/d+a^(2/3)*A*arctan(1/3*(a^(1/3)+2*(a+I*a*tan(d*x+c))^(1/3))/a^( 1/3)*3^(1/2))*3^(1/2)/d-1/2*a^(2/3)*(A-I*B)*arctan(1/3*(a^(1/3)+2^(2/3)*(a +I*a*tan(d*x+c))^(1/3))/a^(1/3)*3^(1/2))*3^(1/2)*2^(2/3)/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 4.29 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.44 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\frac {3 \left (\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{2/3} \left ((A-i B) \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},\frac {e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )-2 A \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},\frac {2 e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )\right )}{2 \sqrt [3]{2} d} \]
(3*((a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x))))^(2/3)*((A - I*B)*Hy pergeometric2F1[2/3, 1, 5/3, E^((2*I)*(c + d*x))/(1 + E^((2*I)*(c + d*x))) ] - 2*A*Hypergeometric2F1[2/3, 1, 5/3, (2*E^((2*I)*(c + d*x)))/(1 + E^((2* I)*(c + d*x)))]))/(2*2^(1/3)*d)
Time = 0.76 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.75, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.382, Rules used = {3042, 4083, 3042, 3962, 67, 16, 1082, 217, 4082, 67, 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.
\(\displaystyle \int \cot (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x))}{\tan (c+d x)}dx\) |
\(\Big \downarrow \) 4083 |
\(\displaystyle (B+i A) \int (i \tan (c+d x) a+a)^{2/3}dx+\frac {A \int \cot (c+d x) (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{2/3}dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (B+i A) \int (i \tan (c+d x) a+a)^{2/3}dx+\frac {A \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{2/3}}{\tan (c+d x)}dx}{a}\) |
\(\Big \downarrow \) 3962 |
\(\displaystyle \frac {A \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{2/3}}{\tan (c+d x)}dx}{a}-\frac {i a (B+i A) \int \frac {1}{(a-i a \tan (c+d x)) \sqrt [3]{i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 67 |
\(\displaystyle \frac {A \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{2/3}}{\tan (c+d x)}dx}{a}-\frac {i a (B+i A) \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}+\frac {3 \int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)}d\sqrt [3]{i \tan (c+d x) a+a}}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {A \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{2/3}}{\tan (c+d x)}dx}{a}-\frac {i a (B+i A) \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {A \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{2/3}}{\tan (c+d x)}dx}{a}-\frac {i a (B+i A) \left (\frac {3 \int \frac {1}{a^2 \tan ^2(c+d x)-3}d\left (i 2^{2/3} a^{2/3} \tan (c+d x)+1\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {A \int \frac {(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{2/3}}{\tan (c+d x)}dx}{a}-\frac {i a (B+i A) \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (c+d x)}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}\) |
\(\Big \downarrow \) 4082 |
\(\displaystyle \frac {a A \int \frac {\cot (c+d x)}{\sqrt [3]{i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {i a (B+i A) \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (c+d x)}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}\) |
\(\Big \downarrow \) 67 |
\(\displaystyle \frac {a A \left (\frac {3}{2} \int \frac {1}{a^{2/3}+\sqrt [3]{i \tan (c+d x) a+a} \sqrt [3]{a}+(i \tan (c+d x) a+a)^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}-\frac {3 \int \frac {1}{\sqrt [3]{a}-\sqrt [3]{i \tan (c+d x) a+a}}d\sqrt [3]{i \tan (c+d x) a+a}}{2 \sqrt [3]{a}}-\frac {\log (\tan (c+d x))}{2 \sqrt [3]{a}}\right )}{d}-\frac {i a (B+i A) \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (c+d x)}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {a A \left (\frac {3}{2} \int \frac {1}{a^{2/3}+\sqrt [3]{i \tan (c+d x) a+a} \sqrt [3]{a}+(i \tan (c+d x) a+a)^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}-\frac {\log (\tan (c+d x))}{2 \sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{a}}\right )}{d}-\frac {i a (B+i A) \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (c+d x)}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {a A \left (-\frac {3 \int \frac {1}{-(i \tan (c+d x) a+a)^{2/3}-3}d\left (\frac {2 \sqrt [3]{i \tan (c+d x) a+a}}{\sqrt [3]{a}}+1\right )}{\sqrt [3]{a}}-\frac {\log (\tan (c+d x))}{2 \sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{a}}\right )}{d}-\frac {i a (B+i A) \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (c+d x)}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {a A \left (\frac {\sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{a}}-\frac {\log (\tan (c+d x))}{2 \sqrt [3]{a}}+\frac {3 \log \left (\sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2 \sqrt [3]{a}}\right )}{d}-\frac {i a (B+i A) \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (c+d x)}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}\) |
((-I)*a*(I*A + B)*(((-I)*Sqrt[3]*ArcTanh[(a*Tan[c + d*x])/Sqrt[3]])/(2^(1/ 3)*a^(1/3)) - (3*Log[2^(1/3)*a^(1/3) - I*a*Tan[c + d*x]])/(2*2^(1/3)*a^(1/ 3)) + Log[a - I*a*Tan[c + d*x]]/(2*2^(1/3)*a^(1/3))))/d + (a*A*((Sqrt[3]*A rcTan[(1 + (2*(a + I*a*Tan[c + d*x])^(1/3))/a^(1/3))/Sqrt[3]])/a^(1/3) - L og[Tan[c + d*x]]/(2*a^(1/3)) + (3*Log[a^(1/3) - (a + I*a*Tan[c + d*x])^(1/ 3)])/(2*a^(1/3))))/d
3.2.100.3.1 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_))^(1/3)), x_Symbol] :> With[ {q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q), x ] + (Simp[3/(2*b) Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/3)], x] - Simp[3/(2*b*q) 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[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(B/f) Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[ a^2 + b^2, 0] && EqQ[A*b + a*B, 0]
Int[(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[( A*b + a*B)/(b*c + a*d) Int[(a + b*Tan[e + f*x])^m, x], x] - Simp[(B*c - A *d)/(b*c + a*d) Int[(a + b*Tan[e + f*x])^m*((a - b*Tan[e + f*x])/(c + d*T an[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
Time = 0.22 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.86
method | result | size |
derivativedivides | \(\frac {3 a \left (\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) \left (i B -A \right )+\left (\frac {\ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )}{3 a^{\frac {1}{3}}}-\frac {\ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )}{6 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{3 a^{\frac {1}{3}}}\right ) A \right )}{d}\) | \(248\) |
default | \(\frac {3 a \left (\left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right ) \left (i B -A \right )+\left (\frac {\ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-a^{\frac {1}{3}}\right )}{3 a^{\frac {1}{3}}}-\frac {\ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+a^{\frac {2}{3}}\right )}{6 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{3 a^{\frac {1}{3}}}\right ) A \right )}{d}\) | \(248\) |
3/d*a*((1/6*2^(2/3)/a^(1/3)*ln((a+I*a*tan(d*x+c))^(1/3)-2^(1/3)*a^(1/3))-1 /12*2^(2/3)/a^(1/3)*ln((a+I*a*tan(d*x+c))^(2/3)+2^(1/3)*a^(1/3)*(a+I*a*tan (d*x+c))^(1/3)+2^(2/3)*a^(2/3))+1/6*3^(1/2)*2^(2/3)/a^(1/3)*arctan(1/3*3^( 1/2)*(2^(2/3)/a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+1)))*(-A+I*B)+(1/3/a^(1/3)* ln((a+I*a*tan(d*x+c))^(1/3)-a^(1/3))-1/6/a^(1/3)*ln((a+I*a*tan(d*x+c))^(2/ 3)+a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+a^(2/3))+1/3*3^(1/2)/a^(1/3)*arctan(1/ 3*3^(1/2)*(2/a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+1)))*A)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 711 vs. \(2 (211) = 422\).
Time = 0.27 (sec) , antiderivative size = 711, normalized size of antiderivative = 2.46 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
1/2*(1/2)^(1/3)*(-I*sqrt(3) - 1)*(-(A^3 - 3*I*A^2*B - 3*A*B^2 + I*B^3)*a^2 /d^3)^(1/3)*log((2^(1/3)*(A^2 - 2*I*A*B - B^2)*a*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) - (1/2)^(2/3)*(I*sqrt(3)*d^2 - d^2)*(-( A^3 - 3*I*A^2*B - 3*A*B^2 + I*B^3)*a^2/d^3)^(2/3))/((A^2 - 2*I*A*B - B^2)* a)) + 1/2*(1/2)^(1/3)*(I*sqrt(3) - 1)*(-(A^3 - 3*I*A^2*B - 3*A*B^2 + I*B^3 )*a^2/d^3)^(1/3)*log((2^(1/3)*(A^2 - 2*I*A*B - B^2)*a*(a/(e^(2*I*d*x + 2*I *c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) - (1/2)^(2/3)*(-I*sqrt(3)*d^2 - d^ 2)*(-(A^3 - 3*I*A^2*B - 3*A*B^2 + I*B^3)*a^2/d^3)^(2/3))/((A^2 - 2*I*A*B - B^2)*a)) + 1/2*(A^3*a^2/d^3)^(1/3)*(I*sqrt(3) - 1)*log(1/2*(2*2^(1/3)*A^2 *a*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) + (I*sqrt(3 )*d^2 + d^2)*(A^3*a^2/d^3)^(2/3))/(A^2*a)) + 1/2*(A^3*a^2/d^3)^(1/3)*(-I*s qrt(3) - 1)*log(1/2*(2*2^(1/3)*A^2*a*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e ^(2/3*I*d*x + 2/3*I*c) + (-I*sqrt(3)*d^2 + d^2)*(A^3*a^2/d^3)^(2/3))/(A^2* a)) + (1/2)^(1/3)*(-(A^3 - 3*I*A^2*B - 3*A*B^2 + I*B^3)*a^2/d^3)^(1/3)*log ((2^(1/3)*(A^2 - 2*I*A*B - B^2)*a*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2 /3*I*d*x + 2/3*I*c) - 2*(1/2)^(2/3)*d^2*(-(A^3 - 3*I*A^2*B - 3*A*B^2 + I*B ^3)*a^2/d^3)^(2/3))/((A^2 - 2*I*A*B - B^2)*a)) + (A^3*a^2/d^3)^(1/3)*log(( 2^(1/3)*A^2*a*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c) - (A^3*a^2/d^3)^(2/3)*d^2)/(A^2*a))
\[ \int \cot (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {2}{3}} \left (A + B \tan {\left (c + d x \right )}\right ) \cot {\left (c + d x \right )}\, dx \]
Time = 0.41 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.87 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=-\frac {2 \, \sqrt {3} 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 2 \cdot 2^{\frac {2}{3}} {\left (A - i \, B\right )} a^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - 4 \, \sqrt {3} A a^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} + a^{\frac {1}{3}}\right )}}{3 \, a^{\frac {1}{3}}}\right ) + 2 \, A a^{\frac {2}{3}} \log \left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + a^{\frac {2}{3}}\right ) - 4 \, A a^{\frac {2}{3}} \log \left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} - a^{\frac {1}{3}}\right )}{4 \, d} \]
-1/4*(2*sqrt(3)*2^(2/3)*(A - I*B)*a^(2/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1 /3)*a^(1/3) + 2*(I*a*tan(d*x + c) + a)^(1/3))/a^(1/3)) - 2^(2/3)*(A - I*B) *a^(2/3)*log(2^(2/3)*a^(2/3) + 2^(1/3)*(I*a*tan(d*x + c) + a)^(1/3)*a^(1/3 ) + (I*a*tan(d*x + c) + a)^(2/3)) + 2*2^(2/3)*(A - I*B)*a^(2/3)*log(-2^(1/ 3)*a^(1/3) + (I*a*tan(d*x + c) + a)^(1/3)) - 4*sqrt(3)*A*a^(2/3)*arctan(1/ 3*sqrt(3)*(2*(I*a*tan(d*x + c) + a)^(1/3) + a^(1/3))/a^(1/3)) + 2*A*a^(2/3 )*log((I*a*tan(d*x + c) + a)^(2/3) + (I*a*tan(d*x + c) + a)^(1/3)*a^(1/3) + a^(2/3)) - 4*A*a^(2/3)*log((I*a*tan(d*x + c) + a)^(1/3) - a^(1/3)))/d
\[ \int \cot (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} \cot \left (d x + c\right ) \,d x } \]
Time = 8.45 (sec) , antiderivative size = 1761, normalized size of antiderivative = 6.09 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{2/3} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
log(- (486*d^3*(3*A*B^2*a^9 - B^3*a^9*1i + A^2*B*a^9*3i) - (1458*a^7*d^6*( (A^3*a^2)/d^3)^(2/3) + 243*d*(a + a*tan(c + d*x)*1i)^(1/3)*(B^2*a^8*d^3 - 5*A^2*a^8*d^3 + A*B*a^8*d^3*2i))*((A^3*a^2)/d^3)^(1/3))*((A^3*a^2)/d^3)^(2 /3) - 243*d*(a + a*tan(c + d*x)*1i)^(1/3)*(A^5*a^10 - A^4*B*a^10*4i + A^2* B^3*a^10*2i - 5*A^3*B^2*a^10))*((A^3*a^2)/d^3)^(1/3) + log(- (486*d^3*(3*A *B^2*a^9 - B^3*a^9*1i + A^2*B*a^9*3i) - (1458*a^7*d^6*(-(A^3*a^2 + B^3*a^2 *1i - 3*A*B^2*a^2 - A^2*B*a^2*3i)/(2*d^3))^(2/3) + 243*d*(a + a*tan(c + d* x)*1i)^(1/3)*(B^2*a^8*d^3 - 5*A^2*a^8*d^3 + A*B*a^8*d^3*2i))*(-(A^3*a^2 + B^3*a^2*1i - 3*A*B^2*a^2 - A^2*B*a^2*3i)/(2*d^3))^(1/3))*(-(A^3*a^2 + B^3* a^2*1i - 3*A*B^2*a^2 - A^2*B*a^2*3i)/(2*d^3))^(2/3) - 243*d*(a + a*tan(c + d*x)*1i)^(1/3)*(A^5*a^10 - A^4*B*a^10*4i + A^2*B^3*a^10*2i - 5*A^3*B^2*a^ 10))*(-(A^3*a^2 + B^3*a^2*1i - 3*A*B^2*a^2 - A^2*B*a^2*3i)/(2*d^3))^(1/3) + (log(- ((3^(1/2)*1i - 1)^2*(486*d^3*(3*A*B^2*a^9 - B^3*a^9*1i + A^2*B*a^ 9*3i) - ((3^(1/2)*1i - 1)*(243*d*(a + a*tan(c + d*x)*1i)^(1/3)*(B^2*a^8*d^ 3 - 5*A^2*a^8*d^3 + A*B*a^8*d^3*2i) + (729*a^7*d^6*(3^(1/2)*1i - 1)^2*((A^ 3*a^2)/d^3)^(2/3))/2)*((A^3*a^2)/d^3)^(1/3))/2)*((A^3*a^2)/d^3)^(2/3))/4 - 243*d*(a + a*tan(c + d*x)*1i)^(1/3)*(A^5*a^10 - A^4*B*a^10*4i + A^2*B^3*a ^10*2i - 5*A^3*B^2*a^10))*(3^(1/2)*1i - 1)*((A^3*a^2)/d^3)^(1/3))/2 - (log (- ((3^(1/2)*1i + 1)^2*(486*d^3*(3*A*B^2*a^9 - B^3*a^9*1i + A^2*B*a^9*3i) + ((3^(1/2)*1i + 1)*(243*d*(a + a*tan(c + d*x)*1i)^(1/3)*(B^2*a^8*d^3 -...