Integrand size = 26, antiderivative size = 87 \[ \int \frac {\tan ^m(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {3 i \operatorname {AppellF1}\left (-\frac {1}{3},-m,1,\frac {2}{3},1+i \tan (c+d x),\frac {1}{2} (1+i \tan (c+d x))\right ) (-i \tan (c+d x))^{-m} \tan ^m(c+d x)}{2 d \sqrt [3]{a+i a \tan (c+d x)}} \] Output:
3/2*I*AppellF1(-1/3,-m,1,2/3,1+I*tan(d*x+c),1/2+1/2*I*tan(d*x+c))*tan(d*x+ c)^m/d/((-I*tan(d*x+c))^m)/(a+I*a*tan(d*x+c))^(1/3)
\[ \int \frac {\tan ^m(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\int \frac {\tan ^m(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx \] Input:
Integrate[Tan[c + d*x]^m/(a + I*a*Tan[c + d*x])^(1/3),x]
Output:
Integrate[Tan[c + d*x]^m/(a + I*a*Tan[c + d*x])^(1/3), x]
Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3042, 4047, 25, 27, 152, 150}
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.
\(\displaystyle \int \frac {\tan ^m(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^m}{\sqrt [3]{a+i a \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4047 |
\(\displaystyle \frac {i a^2 \int -\frac {\tan ^m(c+d x)}{a (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{4/3}}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {i a^2 \int \frac {\tan ^m(c+d x)}{a (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{4/3}}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {i a \int \frac {\tan ^m(c+d x)}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{4/3}}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 152 |
\(\displaystyle -\frac {i \sqrt [3]{1+i \tan (c+d x)} \int \frac {\tan ^m(c+d x)}{(i \tan (c+d x)+1)^{4/3} (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d \sqrt [3]{a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {\sqrt [3]{1+i \tan (c+d x)} \tan ^{m+1}(c+d x) \operatorname {AppellF1}\left (m+1,\frac {4}{3},1,m+2,-i \tan (c+d x),i \tan (c+d x)\right )}{d (m+1) \sqrt [3]{a+i a \tan (c+d x)}}\) |
Input:
Int[Tan[c + d*x]^m/(a + I*a*Tan[c + d*x])^(1/3),x]
Output:
(AppellF1[1 + m, 4/3, 1, 2 + m, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*(1 + I* Tan[c + d*x])^(1/3)*Tan[c + d*x]^(1 + m))/(d*(1 + m)*(a + I*a*Tan[c + d*x] )^(1/3))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 , (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !GtQ[c, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(b/f) Subst[Int[(a + x)^(m - 1)*(( c + (d/b)*x)^n/(b^2 + a*x)), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d ^2, 0]
\[\int \frac {\tan \left (d x +c \right )^{m}}{\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]
Input:
int(tan(d*x+c)^m/(a+I*a*tan(d*x+c))^(1/3),x)
Output:
int(tan(d*x+c)^m/(a+I*a*tan(d*x+c))^(1/3),x)
\[ \int \frac {\tan ^m(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\int { \frac {\tan \left (d x + c\right )^{m}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate(tan(d*x+c)^m/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="fricas")
Output:
integral(1/2*2^(2/3)*((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^m*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(e^(2*I*d*x + 2*I*c) + 1)*e^(-2/ 3*I*d*x - 2/3*I*c)/a, x)
\[ \int \frac {\tan ^m(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\int \frac {\tan ^{m}{\left (c + d x \right )}}{\sqrt [3]{i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \] Input:
integrate(tan(d*x+c)**m/(a+I*a*tan(d*x+c))**(1/3),x)
Output:
Integral(tan(c + d*x)**m/(I*a*(tan(c + d*x) - I))**(1/3), x)
\[ \int \frac {\tan ^m(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\int { \frac {\tan \left (d x + c\right )^{m}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate(tan(d*x+c)^m/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="maxima")
Output:
integrate(tan(d*x + c)^m/(I*a*tan(d*x + c) + a)^(1/3), x)
\[ \int \frac {\tan ^m(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\int { \frac {\tan \left (d x + c\right )^{m}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate(tan(d*x+c)^m/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="giac")
Output:
integrate(tan(d*x + c)^m/(I*a*tan(d*x + c) + a)^(1/3), x)
Timed out. \[ \int \frac {\tan ^m(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^m}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}} \,d x \] Input:
int(tan(c + d*x)^m/(a + a*tan(c + d*x)*1i)^(1/3),x)
Output:
int(tan(c + d*x)^m/(a + a*tan(c + d*x)*1i)^(1/3), x)
\[ \int \frac {\tan ^m(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {\int \frac {\tan \left (d x +c \right )^{m}}{\left (\tan \left (d x +c \right ) i +1\right )^{\frac {1}{3}}}d x}{a^{\frac {1}{3}}} \] Input:
int(tan(d*x+c)^m/(a+I*a*tan(d*x+c))^(1/3),x)
Output:
int(tan(c + d*x)**m/(tan(c + d*x)*i + 1)**(1/3),x)/a**(1/3)