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

Optimal. Leaf size=75 \[ -\frac{a^2 (d \tan (e+f x))^{n+1}}{d f (n+1)}+\frac{2 a^2 (d \tan (e+f x))^{n+1} \, _2F_1(1,n+1;n+2;i \tan (e+f x))}{d f (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.114957, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3543, 3537, 12, 64} \[ -\frac{a^2 (d \tan (e+f x))^{n+1}}{d f (n+1)}+\frac{2 a^2 (d \tan (e+f x))^{n+1} \, _2F_1(1,n+1;n+2;i \tan (e+f x))}{d f (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [B]  time = 1.92924, size = 168, normalized size = 2.24 \[ \frac{e^{-2 i e} 2^{-n} \left (-\frac{i \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}\right )^{n+1} \cos ^2(e+f x) (a+i a \tan (e+f x))^2 \left (-2^n+\left (1+e^{2 i (e+f x)}\right )^{n+1} \, _2F_1\left (n+1,n+1;n+2;\frac{1}{2} \left (1-e^{2 i (e+f x)}\right )\right )\right ) \tan ^{-n}(e+f x) (d \tan (e+f x))^n}{f (n+1) (\cos (f x)+i \sin (f x))^2} \]

Antiderivative was successfully verified.

[In]

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

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Maple [F]  time = 0.452, size = 0, normalized size = 0. \begin{align*} \int \left ( d\tan \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*tan(f*x+e))^n*(a+I*a*tan(f*x+e))^2,x)

[Out]

int((d*tan(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 \tan \left (f x + e\right )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(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*tan(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*tan(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 \tan{\left (e + f x \right )}\right )^{n}\, dx + \int - \left (d \tan{\left (e + f x \right )}\right )^{n} \tan ^{2}{\left (e + f x \right )}\, dx + \int 2 i \left (d \tan{\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*tan(f*x+e))**n*(a+I*a*tan(f*x+e))**2,x)

[Out]

a**2*(Integral((d*tan(e + f*x))**n, x) + Integral(-(d*tan(e + f*x))**n*tan(e + f*x)**2, x) + Integral(2*I*(d*t
an(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 \tan \left (f x + e\right )\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(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*tan(f*x + e))^n, x)