\(\int \tan (c+d x) (a+i a \tan (c+d x))^m \, dx\) [328]

Optimal result

Integrand size = 22, antiderivative size = 70 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^m \, dx=\frac {(a+i a \tan (c+d x))^m}{d m}-\frac {\operatorname {Hypergeometric2F1}\left (1,m,1+m,\frac {1}{2} (1+i \tan (c+d x))\right ) (a+i a \tan (c+d x))^m}{2 d m} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3608, 3562, 70} \[ \int \tan (c+d x) (a+i a \tan (c+d x))^m \, dx=\frac {(a+i a \tan (c+d x))^m}{d m}-\frac {(a+i a \tan (c+d x))^m \operatorname {Hypergeometric2F1}\left (1,m,m+1,\frac {1}{2} (i \tan (c+d x)+1)\right )}{2 d m} \]

[In]

Int[Tan[c + d*x]*(a + I*a*Tan[c + d*x])^m,x]

[Out]

(a + I*a*Tan[c + d*x])^m/(d*m) - (Hypergeometric2F1[1, m, 1 + m, (1 + I*Tan[c + d*x])/2]*(a + I*a*Tan[c + d*x]
)^m)/(2*d*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 3562

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[-b/d, Subst[Int[(a + x)^(n - 1)/(a - x), x]
, x, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 + b^2, 0]

Rule 3608

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

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.70 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^m \, dx=-\frac {\left (-2+\operatorname {Hypergeometric2F1}\left (1,m,1+m,\frac {1}{2} (1+i \tan (c+d x))\right )\right ) (a+i a \tan (c+d x))^m}{2 d m} \]

[In]

Integrate[Tan[c + d*x]*(a + I*a*Tan[c + d*x])^m,x]

[Out]

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

Maple [F]

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

[In]

int(tan(d*x+c)*(a+I*a*tan(d*x+c))^m,x)

[Out]

int(tan(d*x+c)*(a+I*a*tan(d*x+c))^m,x)

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))**m,x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**m*tan(c + d*x), x)

Maxima [F]

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

[In]

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

[Out]

integrate((I*a*tan(d*x + c) + a)^m*tan(d*x + c), x)

Giac [F]

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

[In]

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

[Out]

integrate((I*a*tan(d*x + c) + a)^m*tan(d*x + c), x)

Mupad [F(-1)]

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

[In]

int(tan(c + d*x)*(a + a*tan(c + d*x)*1i)^m,x)

[Out]

int(tan(c + d*x)*(a + a*tan(c + d*x)*1i)^m, x)