\(\int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^n \, dx\) [1176]

Optimal result

Integrand size = 26, antiderivative size = 61 \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^n \, dx=\frac {a \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {c+d \tan (e+f x)}{c-i d}\right ) (c+d \tan (e+f x))^{1+n}}{(i c+d) f (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3618, 70} \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^n \, dx=\frac {a (c+d \tan (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {c+d \tan (e+f x)}{c-i d}\right )}{f (n+1) (d+i c)} \]

[In]

Int[(a + I*a*Tan[e + f*x])*(c + d*Tan[e + f*x])^n,x]

[Out]

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

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 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(
d/f), Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]

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

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.05 \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^n \, dx=-\frac {i a \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {c+d \tan (e+f x)}{c-i d}\right ) (c+d \tan (e+f x))^{1+n}}{(c-i d) f (1+n)} \]

[In]

Integrate[(a + I*a*Tan[e + f*x])*(c + d*Tan[e + f*x])^n,x]

[Out]

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

Maple [F]

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

[In]

int((a+I*a*tan(f*x+e))*(c+d*tan(f*x+e))^n,x)

[Out]

int((a+I*a*tan(f*x+e))*(c+d*tan(f*x+e))^n,x)

Fricas [F]

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

[In]

integrate((a+I*a*tan(f*x+e))*(c+d*tan(f*x+e))^n,x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

integrate((a+I*a*tan(f*x+e))*(c+d*tan(f*x+e))**n,x)

[Out]

I*a*(Integral(-I*(c + d*tan(e + f*x))**n, x) + Integral((c + d*tan(e + f*x))**n*tan(e + f*x), x))

Maxima [F]

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

[In]

integrate((a+I*a*tan(f*x+e))*(c+d*tan(f*x+e))^n,x, algorithm="maxima")

[Out]

integrate((I*a*tan(f*x + e) + a)*(d*tan(f*x + e) + c)^n, x)

Giac [F]

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

[In]

integrate((a+I*a*tan(f*x+e))*(c+d*tan(f*x+e))^n,x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)*(d*tan(f*x + e) + c)^n, x)

Mupad [F(-1)]

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

[In]

int((a + a*tan(e + f*x)*1i)*(c + d*tan(e + f*x))^n,x)

[Out]

int((a + a*tan(e + f*x)*1i)*(c + d*tan(e + f*x))^n, x)