Integrand size = 23, antiderivative size = 98 \[ \int \frac {(c+d x)^m}{a+i a \tan (e+f x)} \, dx=\frac {(c+d x)^{1+m}}{2 a d (1+m)}+\frac {i 2^{-2-m} e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )}{a f} \] Output:
1/2*(d*x+c)^(1+m)/a/d/(1+m)+I*2^(-2-m)*(d*x+c)^m*GAMMA(1+m,2*I*f*(d*x+c)/d )/a/exp(2*I*(e-c*f/d))/f/((I*f*(d*x+c)/d)^m)
Time = 1.59 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.36 \[ \int \frac {(c+d x)^m}{a+i a \tan (e+f x)} \, dx=\frac {e^{-i e} (c+d x)^m \left (\frac {2 e^{2 i e} f (c+d x)}{d (1+m)}+i 2^{-m} e^{\frac {2 i c f}{d}} \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )\right ) \sec (e+f x) (\cos (f x)+i \sin (f x))}{4 f (a+i a \tan (e+f x))} \] Input:
Integrate[(c + d*x)^m/(a + I*a*Tan[e + f*x]),x]
Output:
((c + d*x)^m*((2*E^((2*I)*e)*f*(c + d*x))/(d*(1 + m)) + (I*E^(((2*I)*c*f)/ d)*Gamma[1 + m, ((2*I)*f*(c + d*x))/d])/(2^m*((I*f*(c + d*x))/d)^m))*Sec[e + f*x]*(Cos[f*x] + I*Sin[f*x]))/(4*E^(I*e)*f*(a + I*a*Tan[e + f*x]))
Time = 0.33 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 4210, 2612}
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.
\(\displaystyle \int \frac {(c+d x)^m}{a+i a \tan (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c+d x)^m}{a+i a \tan (e+f x)}dx\) |
\(\Big \downarrow \) 4210 |
\(\displaystyle \frac {(c+d x)^{m+1}}{2 a d (m+1)}+\frac {\int e^{-2 i (e+f x)} (c+d x)^mdx}{2 a}\) |
\(\Big \downarrow \) 2612 |
\(\displaystyle \frac {(c+d x)^{m+1}}{2 a d (m+1)}+\frac {i 2^{-m-2} e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 i f (c+d x)}{d}\right )}{a f}\) |
Input:
Int[(c + d*x)^m/(a + I*a*Tan[e + f*x]),x]
Output:
(c + d*x)^(1 + m)/(2*a*d*(1 + m)) + (I*2^(-2 - m)*(c + d*x)^m*Gamma[1 + m, ((2*I)*f*(c + d*x))/d])/(a*E^((2*I)*(e - (c*f)/d))*f*((I*f*(c + d*x))/d)^ m)
Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c + d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d) )^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m + 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] && !IntegerQ[m]
Int[((c_.) + (d_.)*(x_))^(m_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Sym bol] :> Simp[(c + d*x)^(m + 1)/(2*a*d*(m + 1)), x] + Simp[1/(2*a) Int[(c + d*x)^m*E^(2*(a/b)*(e + f*x)), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] & & EqQ[a^2 + b^2, 0] && !IntegerQ[m]
\[\int \frac {\left (d x +c \right )^{m}}{a +i a \tan \left (f x +e \right )}d x\]
Input:
int((d*x+c)^m/(a+I*a*tan(f*x+e)),x)
Output:
int((d*x+c)^m/(a+I*a*tan(f*x+e)),x)
Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.87 \[ \int \frac {(c+d x)^m}{a+i a \tan (e+f x)} \, dx=\frac {{\left (i \, d m + i \, d\right )} e^{\left (-\frac {d m \log \left (\frac {2 i \, f}{d}\right ) + 2 i \, d e - 2 i \, c f}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (-i \, d f x - i \, c f\right )}}{d}\right ) + 2 \, {\left (d f x + c f\right )} {\left (d x + c\right )}^{m}}{4 \, {\left (a d f m + a d f\right )}} \] Input:
integrate((d*x+c)^m/(a+I*a*tan(f*x+e)),x, algorithm="fricas")
Output:
1/4*((I*d*m + I*d)*e^(-(d*m*log(2*I*f/d) + 2*I*d*e - 2*I*c*f)/d)*gamma(m + 1, -2*(-I*d*f*x - I*c*f)/d) + 2*(d*f*x + c*f)*(d*x + c)^m)/(a*d*f*m + a*d *f)
\[ \int \frac {(c+d x)^m}{a+i a \tan (e+f x)} \, dx=- \frac {i \int \frac {\left (c + d x\right )^{m}}{\tan {\left (e + f x \right )} - i}\, dx}{a} \] Input:
integrate((d*x+c)**m/(a+I*a*tan(f*x+e)),x)
Output:
-I*Integral((c + d*x)**m/(tan(e + f*x) - I), x)/a
\[ \int \frac {(c+d x)^m}{a+i a \tan (e+f x)} \, dx=\int { \frac {{\left (d x + c\right )}^{m}}{i \, a \tan \left (f x + e\right ) + a} \,d x } \] Input:
integrate((d*x+c)^m/(a+I*a*tan(f*x+e)),x, algorithm="maxima")
Output:
1/2*((d*m + d)*integrate((d*x + c)^m*cos(2*f*x + 2*e), x) - (I*d*m + I*d)* integrate((d*x + c)^m*sin(2*f*x + 2*e), x) + e^(m*log(d*x + c) + log(d*x + c)))/(a*d*m + a*d)
\[ \int \frac {(c+d x)^m}{a+i a \tan (e+f x)} \, dx=\int { \frac {{\left (d x + c\right )}^{m}}{i \, a \tan \left (f x + e\right ) + a} \,d x } \] Input:
integrate((d*x+c)^m/(a+I*a*tan(f*x+e)),x, algorithm="giac")
Output:
integrate((d*x + c)^m/(I*a*tan(f*x + e) + a), x)
Timed out. \[ \int \frac {(c+d x)^m}{a+i a \tan (e+f x)} \, dx=\int \frac {{\left (c+d\,x\right )}^m}{a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}} \,d x \] Input:
int((c + d*x)^m/(a + a*tan(e + f*x)*1i),x)
Output:
int((c + d*x)^m/(a + a*tan(e + f*x)*1i), x)
\[ \int \frac {(c+d x)^m}{a+i a \tan (e+f x)} \, dx=\frac {\int \frac {\left (d x +c \right )^{m}}{\tan \left (f x +e \right ) i +1}d x}{a} \] Input:
int((d*x+c)^m/(a+I*a*tan(f*x+e)),x)
Output:
int((c + d*x)**m/(tan(e + f*x)*i + 1),x)/a