3.1184 \(\int \frac{(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=180 \[ -\frac{d (c (2-m)+i d m) (a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;-\frac{d (i \tan (e+f x)+1)}{i c-d}\right )}{f m \left (c^2+d^2\right )^2}-\frac{d (a+i a \tan (e+f x))^m}{f \left (c^2+d^2\right ) (c+d \tan (e+f x))}-\frac{i (a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (e+f x)+1)\right )}{2 f m (c-i d)^2} \]

[Out]

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

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Rubi [A]  time = 0.471646, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3561, 3600, 3481, 68, 3599} \[ -\frac{d (c (2-m)+i d m) (a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;-\frac{d (i \tan (e+f x)+1)}{i c-d}\right )}{f m \left (c^2+d^2\right )^2}-\frac{d (a+i a \tan (e+f x))^m}{f \left (c^2+d^2\right ) (c+d \tan (e+f x))}-\frac{i (a+i a \tan (e+f x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (e+f x)+1)\right )}{2 f m (c-i d)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^m/(c + d*Tan[e + f*x])^2,x]

[Out]

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

Rule 3561

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(d*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(c^2 + d^2)*
(n + 1)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*d*m - a*c*(n + 1) + a*d*(m + n + 1)*T
an[e + f*x], 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] && LtQ[n, -1] && (IntegerQ[n] || IntegersQ[2*m, 2*n])

Rule 3600

Int[(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(A*b + a*B)/(b*c + a*d), Int[(a + b*Tan[e + f*x])^m, x], x] - Dist[(B*c
- A*d)/(b*c + a*d), Int[((a + b*Tan[e + f*x])^m*(a - b*Tan[e + f*x]))/(c + d*Tan[e + f*x]), x], x] /; FreeQ[{
a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 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))^2} \, dx &=-\frac{d (a+i a \tan (e+f x))^m}{\left (c^2+d^2\right ) f (c+d \tan (e+f x))}+\frac{\int \frac{(a+i a \tan (e+f x))^m (a (c+i d m)-a d (1-m) \tan (e+f x))}{c+d \tan (e+f x)} \, dx}{a \left (c^2+d^2\right )}\\ &=-\frac{d (a+i a \tan (e+f x))^m}{\left (c^2+d^2\right ) f (c+d \tan (e+f x))}+\frac{\int (a+i a \tan (e+f x))^m \, dx}{(c-i d)^2}-\frac{(d (i c (2-m)-d m)) \int \frac{(a-i a \tan (e+f x)) (a+i a \tan (e+f x))^m}{c+d \tan (e+f x)} \, dx}{a (c-i d)^2 (c+i d)}\\ &=-\frac{d (a+i a \tan (e+f x))^m}{\left (c^2+d^2\right ) f (c+d \tan (e+f x))}-\frac{(i a) \operatorname{Subst}\left (\int \frac{(a+x)^{-1+m}}{a-x} \, dx,x,i a \tan (e+f x)\right )}{(c-i d)^2 f}-\frac{(a d (i c (2-m)-d m)) \operatorname{Subst}\left (\int \frac{(a+i a x)^{-1+m}}{c+d x} \, dx,x,\tan (e+f x)\right )}{(c-i d)^2 (c+i d) f}\\ &=-\frac{i \, _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 (c-i d)^2 f m}-\frac{d (c (2-m)+i d m) \, _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 )^2 f m}-\frac{d (a+i a \tan (e+f x))^m}{\left (c^2+d^2\right ) f (c+d \tan (e+f x))}\\ \end{align*}

Mathematica [F]  time = 26.3085, size = 0, normalized size = 0. \[ \int \frac{(a+i a \tan (e+f x))^m}{(c+d \tan (e+f x))^2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + I*a*Tan[e + f*x])^m/(c + d*Tan[e + f*x])^2,x]

[Out]

Integrate[(a + I*a*Tan[e + f*x])^m/(c + d*Tan[e + f*x])^2, x]

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Maple [F]  time = 0.54, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{m}}{ \left ( c+d\tan \left ( fx+e \right ) \right ) ^{2}}}\, 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))^2,x)

[Out]

int((a+I*a*tan(f*x+e))^m/(c+d*tan(f*x+e))^2,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}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{2}}\,{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))^2,x, algorithm="maxima")

[Out]

integrate((I*a*tan(f*x + e) + a)^m/(d*tan(f*x + e) + c)^2, 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 (e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}}{c^{2} + 2 i \, c d - d^{2} +{\left (c^{2} - 2 i \, c d - d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \,{\left (c^{2} + d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}, 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))^2,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*(e^(4*I*f*x + 4*I*e) + 2*e^(2*I*f*x + 2*I*e) +
1)/(c^2 + 2*I*c*d - d^2 + (c^2 - 2*I*c*d - d^2)*e^(4*I*f*x + 4*I*e) + 2*(c^2 + d^2)*e^(2*I*f*x + 2*I*e)), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**2,x)

[Out]

Timed out

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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}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{2}}\,{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))^2,x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^m/(d*tan(f*x + e) + c)^2, x)