3.791 \(\int \frac{(d \cot (e+f x))^n}{a+i a \tan (e+f x)} \, dx\)

Optimal. Leaf size=157 \[ -\frac{i n (d \cot (e+f x))^{n+2} \, _2F_1\left (1,\frac{n+2}{2};\frac{n+4}{2};-\cot ^2(e+f x)\right )}{2 a d^2 f (n+2)}+\frac{(n+1) (d \cot (e+f x))^{n+3} \, _2F_1\left (1,\frac{n+3}{2};\frac{n+5}{2};-\cot ^2(e+f x)\right )}{2 a d^3 f (n+3)}-\frac{(d \cot (e+f x))^{n+2}}{2 d^2 f (a \cot (e+f x)+i a)} \]

[Out]

-(d*Cot[e + f*x])^(2 + n)/(2*d^2*f*(I*a + a*Cot[e + f*x])) - ((I/2)*n*(d*Cot[e + f*x])^(2 + n)*Hypergeometric2
F1[1, (2 + n)/2, (4 + n)/2, -Cot[e + f*x]^2])/(a*d^2*f*(2 + n)) + ((1 + n)*(d*Cot[e + f*x])^(3 + n)*Hypergeome
tric2F1[1, (3 + n)/2, (5 + n)/2, -Cot[e + f*x]^2])/(2*a*d^3*f*(3 + n))

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Rubi [A]  time = 0.254964, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3673, 3552, 3538, 3476, 364} \[ -\frac{i n (d \cot (e+f x))^{n+2} \, _2F_1\left (1,\frac{n+2}{2};\frac{n+4}{2};-\cot ^2(e+f x)\right )}{2 a d^2 f (n+2)}+\frac{(n+1) (d \cot (e+f x))^{n+3} \, _2F_1\left (1,\frac{n+3}{2};\frac{n+5}{2};-\cot ^2(e+f x)\right )}{2 a d^3 f (n+3)}-\frac{(d \cot (e+f x))^{n+2}}{2 d^2 f (a \cot (e+f x)+i a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*Cot[e + f*x])^(2 + n)/(2*d^2*f*(I*a + a*Cot[e + f*x])) - ((I/2)*n*(d*Cot[e + f*x])^(2 + n)*Hypergeometric2
F1[1, (2 + n)/2, (4 + n)/2, -Cot[e + f*x]^2])/(a*d^2*f*(2 + n)) + ((1 + n)*(d*Cot[e + f*x])^(3 + n)*Hypergeome
tric2F1[1, (3 + n)/2, (5 + n)/2, -Cot[e + f*x]^2])/(2*a*d^3*f*(3 + n))

Rule 3673

Int[(cot[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Cot[e + f*x])^(m - n*p)*(b + a*Cot[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rule 3552

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

Rule 3538

Int[((b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*T
an[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Tan[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x] && NeQ
[c^2 + d^2, 0] &&  !IntegerQ[2*m]

Rule 3476

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

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{(d \cot (e+f x))^n}{a+i a \tan (e+f x)} \, dx &=\frac{\int \frac{(d \cot (e+f x))^{1+n}}{i a+a \cot (e+f x)} \, dx}{d}\\ &=-\frac{(d \cot (e+f x))^{2+n}}{2 d^2 f (i a+a \cot (e+f x))}-\frac{\int (d \cot (e+f x))^{1+n} (-i a d n+a d (1+n) \cot (e+f x)) \, dx}{2 a^2 d^2}\\ &=-\frac{(d \cot (e+f x))^{2+n}}{2 d^2 f (i a+a \cot (e+f x))}+\frac{(i n) \int (d \cot (e+f x))^{1+n} \, dx}{2 a d}-\frac{(1+n) \int (d \cot (e+f x))^{2+n} \, dx}{2 a d^2}\\ &=-\frac{(d \cot (e+f x))^{2+n}}{2 d^2 f (i a+a \cot (e+f x))}-\frac{(i n) \operatorname{Subst}\left (\int \frac{x^{1+n}}{d^2+x^2} \, dx,x,d \cot (e+f x)\right )}{2 a f}+\frac{(1+n) \operatorname{Subst}\left (\int \frac{x^{2+n}}{d^2+x^2} \, dx,x,d \cot (e+f x)\right )}{2 a d f}\\ &=-\frac{(d \cot (e+f x))^{2+n}}{2 d^2 f (i a+a \cot (e+f x))}-\frac{i n (d \cot (e+f x))^{2+n} \, _2F_1\left (1,\frac{2+n}{2};\frac{4+n}{2};-\cot ^2(e+f x)\right )}{2 a d^2 f (2+n)}+\frac{(1+n) (d \cot (e+f x))^{3+n} \, _2F_1\left (1,\frac{3+n}{2};\frac{5+n}{2};-\cot ^2(e+f x)\right )}{2 a d^3 f (3+n)}\\ \end{align*}

Mathematica [F]  time = 2.59343, size = 0, normalized size = 0. \[ \int \frac{(d \cot (e+f x))^n}{a+i a \tan (e+f x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.868, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( d\cot \left ( fx+e \right ) \right ) ^{n}}{a+ia\tan \left ( fx+e \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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