3.1183 \(\int \frac{(a+i a \tan (e+f x))^m}{c+d \tan (e+f x)} \, dx\)

Optimal. Leaf size=122 \[ \frac{(a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (e+f x)+1)\right )}{2 f m (d+i c)}-\frac{d (a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;-\frac{d (i \tan (e+f x)+1)}{i c-d}\right )}{f m \left (c^2+d^2\right )} \]

[Out]

(Hypergeometric2F1[1, m, 1 + m, (1 + I*Tan[e + f*x])/2]*(a + I*a*Tan[e + f*x])^m)/(2*(I*c + d)*f*m) - (d*Hyper
geometric2F1[1, m, 1 + m, -((d*(1 + I*Tan[e + f*x]))/(I*c - d))]*(a + I*a*Tan[e + f*x])^m)/((c^2 + d^2)*f*m)

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Rubi [A]  time = 0.259604, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3562, 3481, 68, 3599} \[ \frac{(a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (e+f x)+1)\right )}{2 f m (d+i c)}-\frac{d (a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;-\frac{d (i \tan (e+f x)+1)}{i c-d}\right )}{f m \left (c^2+d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Hypergeometric2F1[1, m, 1 + m, (1 + I*Tan[e + f*x])/2]*(a + I*a*Tan[e + f*x])^m)/(2*(I*c + d)*f*m) - (d*Hyper
geometric2F1[1, m, 1 + m, -((d*(1 + I*Tan[e + f*x]))/(I*c - d))]*(a + I*a*Tan[e + f*x])^m)/((c^2 + d^2)*f*m)

Rule 3562

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

Rule 3599

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*B)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x
]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,
0]

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (e+f x))^m}{c+d \tan (e+f x)} \, dx &=\frac{\int (a+i a \tan (e+f x))^m \, dx}{c-i d}-\frac{d \int \frac{(a+i a \tan (e+f x))^m (i a+a \tan (e+f x))}{c+d \tan (e+f x)} \, dx}{a (c-i d)}\\ &=\frac{a \operatorname{Subst}\left (\int \frac{(a+x)^{-1+m}}{a-x} \, dx,x,i a \tan (e+f x)\right )}{(i c+d) f}+\frac{(a d) \operatorname{Subst}\left (\int \frac{(a+i a x)^{-1+m}}{c+d x} \, dx,x,\tan (e+f x)\right )}{(i c+d) f}\\ &=\frac{\, _2F_1\left (1,m;1+m;\frac{1}{2} (1+i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m}{2 (i c+d) f m}-\frac{d \, _2F_1\left (1,m;1+m;-\frac{d (1+i \tan (e+f x))}{i c-d}\right ) (a+i a \tan (e+f x))^m}{\left (c^2+d^2\right ) f m}\\ \end{align*}

Mathematica [F]  time = 13.8814, size = 0, normalized size = 0. \[ \int \frac{(a+i a \tan (e+f x))^m}{c+d \tan (e+f x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.348, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{m}}{c+d\tan \left ( fx+e \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((I*a*tan(f*x + e) + a)^m/(d*tan(f*x + e) + 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 \, f x + 2 i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{m}{\left (i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )}}{{\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c - d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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