Optimal. Leaf size=303 \[ -\frac {\tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {\tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{12 a d}+\frac {\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {i \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 \sqrt {3} a d}-\frac {i \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{6 a d}+\frac {\log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}-\frac {\log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}+\frac {i \log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{12 a d} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.46, antiderivative size = 303, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 13, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3549, 3538, 3476, 329, 275, 200, 31, 634, 618, 204, 628, 295, 203} \[ -\frac {\tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {\tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{12 a d}+\frac {i \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 \sqrt {3} a d}+\frac {\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {i \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{6 a d}+\frac {\log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}-\frac {\log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}+\frac {i \log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{12 a d} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 31
Rule 200
Rule 203
Rule 204
Rule 275
Rule 295
Rule 329
Rule 618
Rule 628
Rule 634
Rule 3476
Rule 3538
Rule 3549
Rubi steps
\begin {align*} \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx &=\frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\int \frac {\frac {2 i a}{3}-\frac {1}{3} a \tan (c+d x)}{\sqrt [3]{\tan (c+d x)}} \, dx}{2 a^2}\\ &=\frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {i \int \frac {1}{\sqrt [3]{\tan (c+d x)}} \, dx}{3 a}+\frac {\int \tan ^{\frac {2}{3}}(c+d x) \, dx}{6 a}\\ &=\frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {i \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{3 a d}+\frac {\operatorname {Subst}\left (\int \frac {x^{2/3}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{6 a d}\\ &=\frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {i \operatorname {Subst}\left (\int \frac {x}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{a d}+\frac {\operatorname {Subst}\left (\int \frac {x^4}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{2 a d}\\ &=\frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {i \operatorname {Subst}\left (\int \frac {1}{1+x^3} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{2 a d}+\frac {\operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {1}{2}+\frac {\sqrt {3} x}{2}}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {1}{2}-\frac {\sqrt {3} x}{2}}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{6 a d}\\ &=\frac {\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {i \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a d}-\frac {i \operatorname {Subst}\left (\int \frac {2-x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a d}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{24 a d}+\frac {\operatorname {Subst}\left (\int \frac {1}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{24 a d}+\frac {\operatorname {Subst}\left (\int \frac {-\sqrt {3}+2 x}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{8 \sqrt {3} a d}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {3}+2 x}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{8 \sqrt {3} a d}\\ &=\frac {\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a d}+\frac {\log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {\log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}+\frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {i \operatorname {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{12 a d}-\frac {i \operatorname {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{4 a d}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,-\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}\\ &=-\frac {\tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {\tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a d}+\frac {\log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {\log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}+\frac {i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{12 a d}+\frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {i \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 \tan ^{\frac {2}{3}}(c+d x)\right )}{2 a d}\\ &=-\frac {\tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {\tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {i \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 \sqrt {3} a d}+\frac {\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a d}+\frac {\log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {\log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}+\frac {i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{12 a d}+\frac {i \tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\\ \end {align*}
________________________________________________________________________________________
Mathematica [C] time = 0.98, size = 163, normalized size = 0.54 \[ \frac {i e^{-2 i (c+d x)} \left (-3 \sqrt [3]{2} e^{2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {2}{3};\frac {5}{3};\frac {1}{2} \left (1-e^{2 i (c+d x)}\right )\right )-2 e^{2 i (c+d x)} \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};-\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )+4 e^{2 i (c+d x)}+4\right ) \tan ^{\frac {2}{3}}(c+d x)}{16 a d} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [B] time = 0.61, size = 493, normalized size = 1.63 \[ -\frac {{\left (3 \, {\left (\sqrt {3} a d \sqrt {\frac {1}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (\frac {1}{2} \, \sqrt {3} a d \sqrt {\frac {1}{a^{2} d^{2}}} + \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} + \frac {1}{2} i\right ) - 3 \, {\left (\sqrt {3} a d \sqrt {\frac {1}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (-\frac {1}{2} \, \sqrt {3} a d \sqrt {\frac {1}{a^{2} d^{2}}} + \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} + \frac {1}{2} i\right ) - {\left (3 \, \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (\frac {3}{2} \, \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} d^{2}}} + \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} - \frac {1}{2} i\right ) + {\left (3 \, \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (-\frac {3}{2} \, \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} d^{2}}} + \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} - \frac {1}{2} i\right ) + 2 i \, e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} + i\right ) + 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} - i\right ) - \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} {\left (6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 6 i\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{24 \, a d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (d x + c\right )^{\frac {2}{3}}}{i \, a \tan \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.26, size = 258, normalized size = 0.85 \[ -\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{12 d a}+\frac {1}{6 d a \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}+\frac {i \ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{8 d a}-\frac {\sqrt {3}\, \arctanh \left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{4 d a}+\frac {\tan ^{\frac {1}{3}}\left (d x +c \right )}{3 d a \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}-\frac {i}{6 d a \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}+\frac {i \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{24 d a}+\frac {\sqrt {3}\, \arctanh \left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{12 d a}-\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{4 d a} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 5.29, size = 599, normalized size = 1.98 \[ \ln \left (\left (a^3\,d^3\,3744{}\mathrm {i}-a^4\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}\,17280{}\mathrm {i}\right )\,{\left (\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{2/3}-36\,a\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\right )\,{\left (\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}+\ln \left (\left (a^3\,d^3\,3744{}\mathrm {i}-a^4\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (\frac {1{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}\,17280{}\mathrm {i}\right )\,{\left (\frac {1{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{2/3}-36\,a\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\right )\,{\left (\frac {1{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}+\frac {\ln \left (\frac {{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (a^3\,d^3\,3744{}\mathrm {i}-a^4\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}\,8640{}\mathrm {i}\right )\,{\left (\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{2/3}}{4}-36\,a\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}}{2}-\frac {\ln \left (\frac {{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (a^3\,d^3\,3744{}\mathrm {i}+a^4\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}\,8640{}\mathrm {i}\right )\,{\left (\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{2/3}}{4}-36\,a\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1{}\mathrm {i}}{64\,a^3\,d^3}\right )}^{1/3}}{2}+\frac {\ln \left (\frac {{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (a^3\,d^3\,3744{}\mathrm {i}-a^4\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}\,8640{}\mathrm {i}\right )\,{\left (\frac {1{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{2/3}}{4}-36\,a\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}}{2}-\frac {\ln \left (\frac {{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (a^3\,d^3\,3744{}\mathrm {i}+a^4\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}\,8640{}\mathrm {i}\right )\,{\left (\frac {1{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{2/3}}{4}-36\,a\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1{}\mathrm {i}}{1728\,a^3\,d^3}\right )}^{1/3}}{2}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^{2/3}\,1{}\mathrm {i}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {i \int \frac {\tan ^{\frac {2}{3}}{\left (c + d x \right )}}{\tan {\left (c + d x \right )} - i}\, dx}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________