3.789 \(\int (d \cot (e+f x))^n (a+i a \tan (e+f x))^2 \, dx\)

Optimal. Leaf size=72 \[ \frac{a^2 d (d \cot (e+f x))^{n-1}}{f (1-n)}-\frac{2 a^2 d (d \cot (e+f x))^{n-1} \, _2F_1(1,n-1;n;-i \cot (e+f x))}{f (1-n)} \]

[Out]

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

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Rubi [A]  time = 0.177186, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3673, 3543, 3537, 12, 64} \[ \frac{a^2 d (d \cot (e+f x))^{n-1}}{f (1-n)}-\frac{2 a^2 d (d \cot (e+f x))^{n-1} \, _2F_1(1,n-1;n;-i \cot (e+f x))}{f (1-n)} \]

Antiderivative was successfully verified.

[In]

Int[(d*Cot[e + f*x])^n*(a + I*a*Tan[e + f*x])^2,x]

[Out]

(a^2*d*(d*Cot[e + f*x])^(-1 + n))/(f*(1 - n)) - (2*a^2*d*(d*Cot[e + f*x])^(-1 + n)*Hypergeometric2F1[1, -1 + n
, n, (-I)*Cot[e + f*x]])/(f*(1 - 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 3543

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(d^2*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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])))

Rubi steps

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

Mathematica [B]  time = 1.96049, size = 198, normalized size = 2.75 \[ -\frac{e^{-2 i e} \left (1+e^{2 i (e+f x)}\right )^{-n} \left (\frac{i \left (1+e^{2 i (e+f x)}\right )}{-1+e^{2 i (e+f x)}}\right )^{n-1} \cos ^2(e+f x) (a+i a \tan (e+f x))^2 \left (2^n \left (1+e^{2 i (e+f x)}\right ) \, _2F_1\left (1-n,1-n;2-n;\frac{1}{2} \left (1-e^{2 i (e+f x)}\right )\right )-\left (1+e^{2 i (e+f x)}\right )^n\right ) \cot ^{-n}(e+f x) (d \cot (e+f x))^n}{f (n-1) (\cos (f x)+i \sin (f x))^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Cot[e + f*x])^n*(a + I*a*Tan[e + f*x])^2,x]

[Out]

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

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Maple [F]  time = 0.5, size = 0, normalized size = 0. \begin{align*} \int \left ( d\cot \left ( fx+e \right ) \right ) ^{n} \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{2}\, 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))^2,x)

[Out]

int((d*cot(f*x+e))^n*(a+I*a*tan(f*x+e))^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} \left (d \cot \left (f x + e\right )\right )^{n}\,{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))^2,x, algorithm="maxima")

[Out]

integrate((I*a*tan(f*x + e) + a)^2*(d*cot(f*x + e))^n, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{4 \, a^{2} \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 (4 i \, f x + 4 i \, e\right )}}{e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, e^{\left (2 i \, f x + 2 i \, e\right )} + 1}, 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))^2,x, algorithm="fricas")

[Out]

integral(4*a^2*((I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) - 1))^n*e^(4*I*f*x + 4*I*e)/(e^(4*I*f*x +
4*I*e) + 2*e^(2*I*f*x + 2*I*e) + 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int \left (d \cot{\left (e + f x \right )}\right )^{n}\, dx + \int - \left (d \cot{\left (e + f x \right )}\right )^{n} \tan ^{2}{\left (e + f x \right )}\, dx + \int 2 i \left (d \cot{\left (e + f x \right )}\right )^{n} \tan{\left (e + f x \right )}\, dx\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))**2,x)

[Out]

a**2*(Integral((d*cot(e + f*x))**n, x) + Integral(-(d*cot(e + f*x))**n*tan(e + f*x)**2, x) + Integral(2*I*(d*c
ot(e + f*x))**n*tan(e + f*x), x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} \left (d \cot \left (f x + e\right )\right )^{n}\,{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))^2,x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^2*(d*cot(f*x + e))^n, x)