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

Optimal. Leaf size=113 \[ \frac {i a 2^{\frac {n}{2}+1} (1+i \tan (c+d x))^{-n/2} (a+i a \tan (c+d x))^{n-1} (e \sec (c+d x))^{2-n} \, _2F_1\left (\frac {2-n}{2},-\frac {n}{2};\frac {4-n}{2};\frac {1}{2} (1-i \tan (c+d x))\right )}{d (2-n)} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 113, normalized size of antiderivative = 1.00, 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 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]))

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 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]

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*}

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Mathematica [A]  time = 13.62, size = 112, normalized size = 0.99 \[ \frac {4 e^2 (\cos (2 c)-i \sin (2 c)) (\tan (d x)+i) (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-n} \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};i \sin (2 (c+d x))-\cos (2 (c+d x))\right )}{d (n-2) (-1-i \tan (d x))} \]

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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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ \frac {{\left (\left (\frac {2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{-n + 2} {\left (i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (i \, d n x + i \, c n + n \log \left (\frac {2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right ) + n \log \left (\frac {a}{e}\right )\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} {\rm integral}\left (\frac {1}{2} \, {\left (n e^{\left (2 i \, d x + 2 i \, c\right )} + n\right )} \left (\frac {2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{-n + 2} e^{\left (i \, d n x + i \, c n - 2 i \, d x + n \log \left (\frac {2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right ) + n \log \left (\frac {a}{e}\right ) - 2 i \, c\right )}, x\right )\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, d} \]

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]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \sec \left (d x + c\right )\right )^{-n + 2} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}\,{d x} \]

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)

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maple [F]  time = 2.05, size = 0, normalized size = 0.00 \[ \int \left (e \sec \left (d x +c \right )\right )^{2-n} \left (a +i a \tan \left (d x +c \right )\right )^{n}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{2-n}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e/cos(c + d*x))^(2 - n)*(a + a*tan(c + d*x)*1i)^n,x)

[Out]

int((e/cos(c + d*x))^(2 - n)*(a + a*tan(c + d*x)*1i)^n, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \sec {\left (c + d x \right )}\right )^{2 - n} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n}\, dx \]

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]

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

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