3.328 \(\int \tan (c+d x) (a+i a \tan (c+d x))^m \, dx\)

Optimal. Leaf size=70 \[ \frac{(a+i a \tan (c+d x))^m}{d m}-\frac{(a+i a \tan (c+d x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (c+d x)+1)\right )}{2 d m} \]

[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)

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Rubi [A]  time = 0.0536973, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {3527, 3481, 68} \[ \frac{(a+i a \tan (c+d x))^m}{d m}-\frac{(a+i a \tan (c+d x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (c+d x)+1)\right )}{2 d m} \]

Antiderivative was successfully verified.

[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 3527

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]

Rule 3481

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 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \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}-i \int (a+i a \tan (c+d x))^m \, dx\\ &=\frac{(a+i a \tan (c+d x))^m}{d m}-\frac{a \operatorname{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{\, _2F_1\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]  time = 4.66536, size = 134, normalized size = 1.91 \[ \frac{2^{m-1} \left (e^{i d x}\right )^m \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^m \left (m \left (-e^{2 i (c+d x)}\right ) \, _2F_1\left (1,1;m+2;-e^{2 i (c+d x)}\right )+m+1\right ) \sec ^{-m}(c+d x) (\cos (d x)+i \sin (d x))^{-m} (a+i a \tan (c+d x))^m}{d m (m+1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2^(-1 + m)*(E^(I*d*x))^m*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^m*(1 + m - E^((2*I)*(c + d*x))*m*Hyperge
ometric2F1[1, 1, 2 + m, -E^((2*I)*(c + d*x))])*(a + I*a*Tan[c + d*x])^m)/(d*m*(1 + m)*Sec[c + d*x]^m*(Cos[d*x]
+ I*Sin[d*x])^m)

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Maple [F]  time = 0.664, size = 0, normalized size = 0. \begin{align*} \int \tan \left ( dx+c \right ) \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{m} \tan \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (\frac{2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{m}{\left (-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (i \tan{\left (c + d x \right )} + 1\right )\right )^{m} \tan{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(I*tan(c + d*x) + 1))**m*tan(c + d*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{m} \tan \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)