3.790 \(\int (d \cot (e+f x))^n (a+i a \tan (e+f x)) \, dx\)

Optimal. Leaf size=37 \[ -\frac {i a (d \cot (e+f x))^n \, _2F_1(1,n;n+1;-i \cot (e+f x))}{f n} \]

[Out]

-I*a*(d*cot(f*x+e))^n*hypergeom([1, n],[1+n],-I*cot(f*x+e))/f/n

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Rubi [A]  time = 0.07, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3673, 3537, 64} \[ -\frac {i a (d \cot (e+f x))^n \, _2F_1(1,n;n+1;-i \cot (e+f x))}{f n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*a*(d*Cot[e + f*x])^n*Hypergeometric2F1[1, n, 1 + n, (-I)*Cot[e + f*x]])/(f*n)

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rule 3537

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

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]

Rubi steps

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

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Mathematica [B]  time = 0.84, size = 166, normalized size = 4.49 \[ -\frac {e^{-i e} 2^{n-1} \left (1+e^{2 i (e+f x)}\right )^{1-n} \left (\frac {i \left (1+e^{2 i (e+f x)}\right )}{-1+e^{2 i (e+f x)}}\right )^{n-1} \cos (e+f x) (a+i a \tan (e+f x)) \, _2F_1\left (1-n,1-n;2-n;\frac {1}{2} \left (1-e^{2 i (e+f x)}\right )\right ) \cot ^{-n}(e+f x) (d \cot (e+f x))^n}{f (n-1) (\cos (f x)+i \sin (f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2^(-1 + n)*(1 + E^((2*I)*(e + f*x)))^(1 - n)*((I*(1 + E^((2*I)*(e + f*x))))/(-1 + E^((2*I)*(e + f*x))))^(-1
+ n)*Cos[e + f*x]*(d*Cot[e + f*x])^n*Hypergeometric2F1[1 - n, 1 - n, 2 - n, (1 - E^((2*I)*(e + f*x)))/2]*(a +
I*a*Tan[e + f*x]))/(E^(I*e)*f*(-1 + n)*Cot[e + f*x]^n*(Cos[f*x] + I*Sin[f*x])))

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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {2 \, a \left (\frac {i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} - 1}\right )^{n} e^{\left (2 i \, f x + 2 i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}, x\right ) \]

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(2*a*((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)/(e^(2*I*f*x + 2
*I*e) + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (f x + e\right ) + a\right )} \left (d \cot \left (f x + e\right )\right )^{n}\,{d x} \]

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

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maple [F]  time = 1.55, size = 0, normalized size = 0.00 \[ \int \left (d \cot \left (f x +e \right )\right )^{n} \left (a +i a \tan \left (f x +e \right )\right )\, dx \]

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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (f x + e\right ) + a\right )} \left (d \cot \left (f x + e\right )\right )^{n}\,{d x} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int {\left (d\,\mathrm {cot}\left (e+f\,x\right )\right )}^n\,\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cot(e + f*x))^n*(a + a*tan(e + f*x)*1i),x)

[Out]

int((d*cot(e + f*x))^n*(a + a*tan(e + f*x)*1i), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ i a \left (\int \left (- i \left (d \cot {\left (e + f x \right )}\right )^{n}\right )\, dx + \int \left (d \cot {\left (e + f x \right )}\right )^{n} \tan {\left (e + f x \right )}\, dx\right ) \]

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]

I*a*(Integral(-I*(d*cot(e + f*x))**n, x) + Integral((d*cot(e + f*x))**n*tan(e + f*x), x))

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