3.1175 \(\int (a+i a \tan (e+f x))^2 (c+d \tan (e+f x))^n \, dx\)

Optimal. Leaf size=95 \[ -\frac{a^2 (c+d \tan (e+f x))^{n+1}}{d f (n+1)}+\frac{2 a^2 (c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{c+d \tan (e+f x)}{c-i d}\right )}{f (n+1) (d+i c)} \]

[Out]

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

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Rubi [A]  time = 0.127351, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {3543, 3537, 68} \[ -\frac{a^2 (c+d \tan (e+f x))^{n+1}}{d f (n+1)}+\frac{2 a^2 (c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{c+d \tan (e+f x)}{c-i d}\right )}{f (n+1) (d+i c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [F]  time = 4.51852, size = 0, normalized size = 0. \[ \int (a+i a \tan (e+f x))^2 (c+d \tan (e+f x))^n \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.275, size = 0, normalized size = 0. \begin{align*} \int \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{2} \left ( c+d\tan \left ( fx+e \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a+I*a*tan(f*x+e))^2*(c+d*tan(f*x+e))^n,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 ) + c\right )}^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2*(c+d*tan(f*x+e))^n,x, algorithm="maxima")

[Out]

integrate((I*a*tan(f*x + e) + a)^2*(d*tan(f*x + e) + c)^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{{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + 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((a+I*a*tan(f*x+e))^2*(c+d*tan(f*x+e))^n,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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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 ) + c\right )}^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^2*(c+d*tan(f*x+e))^n,x, algorithm="giac")

[Out]

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