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*hypergeom([1, 1+n],[2+n],(c+d*tan(f*x+e))/(c-I*d))*(c+d*tan(f*x+e))^(1+n)/(I*c+d)/f/(1+n)

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Rubi [A]  time = 0.07, antiderivative size = 61, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [F]  time = 1.96, size = 0, normalized size = 0.00 \[ \int (a+i a \tan (e+f x)) (c+d \tan (e+f x))^n \, dx \]

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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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {2 \, a \left (\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{n} e^{\left (2 i \, f x + 2 i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}, x\right ) \]

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

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)

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maple [F]  time = 1.31, size = 0, normalized size = 0.00 \[ \int \left (a +i a \tan \left (f x +e \right )\right ) \left (c +d \tan \left (f x +e \right )\right )^{n}\, dx \]

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

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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