3.489 \(\int (e \sec (c+d x))^{2-n} (a+i a \tan (c+d x))^n \, dx\)

Optimal. Leaf size=113 \[ \frac{i a 2^{\frac{n}{2}+1} (1+i \tan (c+d x))^{-n/2} (a+i a \tan (c+d x))^{n-1} (e \sec (c+d x))^{2-n} \text{Hypergeometric2F1}\left (\frac{2-n}{2},-\frac{n}{2},\frac{4-n}{2},\frac{1}{2} (1-i \tan (c+d x))\right )}{d (2-n)} \]

[Out]

(I*2^(1 + n/2)*a*Hypergeometric2F1[(2 - n)/2, -n/2, (4 - n)/2, (1 - I*Tan[c + d*x])/2]*(e*Sec[c + d*x])^(2 - n
)*(a + I*a*Tan[c + d*x])^(-1 + n))/(d*(2 - n)*(1 + I*Tan[c + d*x])^(n/2))

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Rubi [A]  time = 0.183201, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3505, 3523, 70, 69} \[ \frac{i a 2^{\frac{n}{2}+1} (1+i \tan (c+d x))^{-n/2} (a+i a \tan (c+d x))^{n-1} (e \sec (c+d x))^{2-n} \text{Hypergeometric2F1}\left (\frac{2-n}{2},-\frac{n}{2},\frac{4-n}{2},\frac{1}{2} (1-i \tan (c+d x))\right )}{d (2-n)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*2^(1 + n/2)*a*Hypergeometric2F1[(2 - n)/2, -n/2, (4 - n)/2, (1 - I*Tan[c + d*x])/2]*(e*Sec[c + d*x])^(2 - n
)*(a + I*a*Tan[c + d*x])^(-1 + n))/(d*(2 - n)*(1 + I*Tan[c + d*x])^(n/2))

Rule 3505

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*S
ec[e + f*x])^m/((a + b*Tan[e + f*x])^(m/2)*(a - b*Tan[e + f*x])^(m/2)), Int[(a + b*Tan[e + f*x])^(m/2 + n)*(a
- b*Tan[e + f*x])^(m/2), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]

Rule 3523

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

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

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

Mathematica [A]  time = 12.5509, size = 112, normalized size = 0.99 \[ \frac{4 e^2 (\cos (2 c)-i \sin (2 c)) (\tan (d x)+i) (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n} \text{Hypergeometric2F1}\left (2,1-\frac{n}{2},2-\frac{n}{2},-\cos (2 (c+d x))+i \sin (2 (c+d x))\right )}{d (n-2) (-1-i \tan (d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.967, size = 0, normalized size = 0. \begin{align*} \int \left ( e\sec \left ( dx+c \right ) \right ) ^{2-n} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [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((e*sec(d*x+c))^(2-n)*(a+I*a*tan(d*x+c))^n,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (\frac{2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{n} \left (\frac{2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{-n + 2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(2-n)*(a+I*a*tan(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^n*(2*e*e^(I*d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1))
^(-n + 2), 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((e*sec(d*x+c))**(2-n)*(a+I*a*tan(d*x+c))**n,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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