3.1176 \(\int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^n \, dx\)

Optimal. Leaf size=61 \[ \frac{a (c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{c+d \tan (e+f x)}{c-i d}\right )}{f (n+1) (d+i c)} \]

[Out]

(a*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 = 0.0728441, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {3537, 68} \[ \frac{a (c+d \tan (e+f x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{c+d \tan (e+f x)}{c-i d}\right )}{f (n+1) (d+i c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*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 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)) (c+d \tan (e+f x))^n \, dx &=\frac{\left (i a^2\right ) \operatorname{Subst}\left (\int \frac{\left (c-\frac{i d x}{a}\right )^n}{-a^2+a x} \, dx,x,i a \tan (e+f x)\right )}{f}\\ &=\frac{a \, _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 = 1.83593, size = 0, normalized size = 0. \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^n \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.291, size = 0, normalized size = 0. \begin{align*} \int \left ( a+ia\tan \left ( fx+e \right ) \right ) \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))*(c+d*tan(f*x+e))^n,x)

[Out]

int((a+I*a*tan(f*x+e))*(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 )}{\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))*(c+d*tan(f*x+e))^n,x, algorithm="maxima")

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(Integral(I*(c + d*tan(e + f*x))**n*tan(e + f*x), x) + Integral((c + d*tan(e + f*x))**n, 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 )}{\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))*(c+d*tan(f*x+e))^n,x, algorithm="giac")

[Out]

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