\(\int \frac {(a+i a \tan (e+f x))^m}{c-i c \tan (e+f x)} \, dx\) [1055]

Optimal result

Integrand size = 31, antiderivative size = 52 \[ \int \frac {(a+i a \tan (e+f x))^m}{c-i c \tan (e+f x)} \, dx=-\frac {i \operatorname {Hypergeometric2F1}\left (2,m,1+m,\frac {1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m}{4 c f m} \]

[Out]

-1/4*I*hypergeom([2, m],[1+m],1/2+1/2*I*tan(f*x+e))*(a+I*a*tan(f*x+e))^m/c/f/m

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3603, 3568, 70} \[ \int \frac {(a+i a \tan (e+f x))^m}{c-i c \tan (e+f x)} \, dx=-\frac {i (a+i a \tan (e+f x))^m \operatorname {Hypergeometric2F1}\left (2,m,m+1,\frac {1}{2} (i \tan (e+f x)+1)\right )}{4 c f m} \]

[In]

Int[(a + I*a*Tan[e + f*x])^m/(c - I*c*Tan[e + f*x]),x]

[Out]

((-1/4*I)*Hypergeometric2F1[2, m, 1 + m, (1 + I*Tan[e + f*x])/2]*(a + I*a*Tan[e + f*x])^m)/(c*f*m)

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 3603

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rubi steps \begin{align*} \text {integral}& = \frac {\int \cos ^2(e+f x) (a+i a \tan (e+f x))^{1+m} \, dx}{a c} \\ & = -\frac {\left (i a^2\right ) \text {Subst}\left (\int \frac {(a+x)^{-1+m}}{(a-x)^2} \, dx,x,i a \tan (e+f x)\right )}{c f} \\ & = -\frac {i \operatorname {Hypergeometric2F1}\left (2,m,1+m,\frac {1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m}{4 c f m} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.19 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int \frac {(a+i a \tan (e+f x))^m}{c-i c \tan (e+f x)} \, dx=-\frac {i \operatorname {Hypergeometric2F1}\left (2,m,1+m,\frac {1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m}{4 c f m} \]

[In]

Integrate[(a + I*a*Tan[e + f*x])^m/(c - I*c*Tan[e + f*x]),x]

[Out]

((-1/4*I)*Hypergeometric2F1[2, m, 1 + m, (1 + I*Tan[e + f*x])/2]*(a + I*a*Tan[e + f*x])^m)/(c*f*m)

Maple [F]

\[\int \frac {\left (a +i a \tan \left (f x +e \right )\right )^{m}}{c -i c \tan \left (f x +e \right )}d x\]

[In]

int((a+I*a*tan(f*x+e))^m/(c-I*c*tan(f*x+e)),x)

[Out]

int((a+I*a*tan(f*x+e))^m/(c-I*c*tan(f*x+e)),x)

Fricas [F]

\[ \int \frac {(a+i a \tan (e+f x))^m}{c-i c \tan (e+f x)} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m}}{-i \, c \tan \left (f x + e\right ) + c} \,d x } \]

[In]

integrate((a+I*a*tan(f*x+e))^m/(c-I*c*tan(f*x+e)),x, algorithm="fricas")

[Out]

integral(1/2*(2*a*e^(2*I*f*x + 2*I*e)/(e^(2*I*f*x + 2*I*e) + 1))^m*(e^(2*I*f*x + 2*I*e) + 1)/c, x)

Sympy [F]

\[ \int \frac {(a+i a \tan (e+f x))^m}{c-i c \tan (e+f x)} \, dx=\frac {i \int \frac {\left (i a \tan {\left (e + f x \right )} + a\right )^{m}}{\tan {\left (e + f x \right )} + i}\, dx}{c} \]

[In]

integrate((a+I*a*tan(f*x+e))**m/(c-I*c*tan(f*x+e)),x)

[Out]

I*Integral((I*a*tan(e + f*x) + a)**m/(tan(e + f*x) + I), x)/c

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^m}{c-i c \tan (e+f x)} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((a+I*a*tan(f*x+e))^m/(c-I*c*tan(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F]

\[ \int \frac {(a+i a \tan (e+f x))^m}{c-i c \tan (e+f x)} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m}}{-i \, c \tan \left (f x + e\right ) + c} \,d x } \]

[In]

integrate((a+I*a*tan(f*x+e))^m/(c-I*c*tan(f*x+e)),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^m/(-I*c*tan(f*x + e) + c), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (e+f x))^m}{c-i c \tan (e+f x)} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^m}{c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}} \,d x \]

[In]

int((a + a*tan(e + f*x)*1i)^m/(c - c*tan(e + f*x)*1i),x)

[Out]

int((a + a*tan(e + f*x)*1i)^m/(c - c*tan(e + f*x)*1i), x)