3.490 \(\int (e \sec (c+d x))^{3-n} (a+i a \tan (c+d x))^n \, dx\)

Optimal. Leaf size=121 \[ \frac{i a 2^{\frac{n+3}{2}} (1+i \tan (c+d x))^{\frac{1}{2} (-n-1)} (a+i a \tan (c+d x))^{n-1} (e \sec (c+d x))^{3-n} \text{Hypergeometric2F1}\left (\frac{1}{2} (-n-1),\frac{3-n}{2},\frac{5-n}{2},\frac{1}{2} (1-i \tan (c+d x))\right )}{d (3-n)} \]

[Out]

(I*2^((3 + n)/2)*a*Hypergeometric2F1[(-1 - n)/2, (3 - n)/2, (5 - n)/2, (1 - I*Tan[c + d*x])/2]*(e*Sec[c + d*x]
)^(3 - n)*(1 + I*Tan[c + d*x])^((-1 - n)/2)*(a + I*a*Tan[c + d*x])^(-1 + n))/(d*(3 - n))

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Rubi [A]  time = 0.235446, antiderivative size = 121, 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} \[ \frac{i a 2^{\frac{n+3}{2}} (1+i \tan (c+d x))^{\frac{1}{2} (-n-1)} (a+i a \tan (c+d x))^{n-1} (e \sec (c+d x))^{3-n} \text{Hypergeometric2F1}\left (\frac{1}{2} (-n-1),\frac{3-n}{2},\frac{5-n}{2},\frac{1}{2} (1-i \tan (c+d x))\right )}{d (3-n)} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sec[c + d*x])^(3 - n)*(a + I*a*Tan[c + d*x])^n,x]

[Out]

(I*2^((3 + n)/2)*a*Hypergeometric2F1[(-1 - n)/2, (3 - n)/2, (5 - n)/2, (1 - I*Tan[c + d*x])/2]*(e*Sec[c + d*x]
)^(3 - n)*(1 + I*Tan[c + d*x])^((-1 - n)/2)*(a + I*a*Tan[c + d*x])^(-1 + n))/(d*(3 - n))

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))^{3-n} (a+i a \tan (c+d x))^n \, dx &=\left ((e \sec (c+d x))^{3-n} (a-i a \tan (c+d x))^{\frac{1}{2} (-3+n)} (a+i a \tan (c+d x))^{\frac{1}{2} (-3+n)}\right ) \int (a-i a \tan (c+d x))^{\frac{3-n}{2}} (a+i a \tan (c+d x))^{\frac{3-n}{2}+n} \, dx\\ &=\frac{\left (a^2 (e \sec (c+d x))^{3-n} (a-i a \tan (c+d x))^{\frac{1}{2} (-3+n)} (a+i a \tan (c+d x))^{\frac{1}{2} (-3+n)}\right ) \operatorname{Subst}\left (\int (a-i a x)^{-1+\frac{3-n}{2}} (a+i a x)^{-1+\frac{3-n}{2}+n} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\left (2^{\frac{1}{2}+\frac{n}{2}} a^2 (e \sec (c+d x))^{3-n} (a-i a \tan (c+d x))^{\frac{1}{2} (-3+n)} (a+i a \tan (c+d x))^{\frac{1}{2}+\frac{1}{2} (-3+n)+\frac{n}{2}} \left (\frac{a+i a \tan (c+d x)}{a}\right )^{-\frac{1}{2}-\frac{n}{2}}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{i x}{2}\right )^{-1+\frac{3-n}{2}+n} (a-i a x)^{-1+\frac{3-n}{2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{i 2^{\frac{3+n}{2}} a \, _2F_1\left (\frac{1}{2} (-1-n),\frac{3-n}{2};\frac{5-n}{2};\frac{1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{3-n} (1+i \tan (c+d x))^{\frac{1}{2} (-1-n)} (a+i a \tan (c+d x))^{-1+n}}{d (3-n)}\\ \end{align*}

Mathematica [A]  time = 11.0644, size = 116, normalized size = 0.96 \[ \frac{8 e^3 (\tan (d x)+i) \sec (d x) (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n} \text{Hypergeometric2F1}\left (3,\frac{3-n}{2},\frac{5-n}{2},-\cos (2 (c+d x))+i \sin (2 (c+d x))\right )}{d (n-3) (\cos (c)+i \sin (c))^3 (\tan (d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(3 - n)*(a + I*a*Tan[c + d*x])^n,x]

[Out]

(8*e^3*Hypergeometric2F1[3, (3 - n)/2, (5 - n)/2, -Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]]*Sec[d*x]*(I + Tan[d*
x])*(a + I*a*Tan[c + d*x])^n)/(d*(-3 + n)*(e*Sec[c + d*x])^n*(Cos[c] + I*Sin[c])^3*(-I + Tan[d*x])^2)

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Maple [F]  time = 0.886, size = 0, normalized size = 0. \begin{align*} \int \left ( e\sec \left ( dx+c \right ) \right ) ^{3-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))^(3-n)*(a+I*a*tan(d*x+c))^n,x)

[Out]

int((e*sec(d*x+c))^(3-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))^(3-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 + 3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(3-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 + 3), 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))**(3-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 + 3}{\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))^(3-n)*(a+I*a*tan(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(-n + 3)*(I*a*tan(d*x + c) + a)^n, x)