3.492 \(\int (e \sec (c+d x))^{5-2 n} (a+i a \tan (c+d x))^n \, dx\)

Optimal. Leaf size=97 \[ -\frac {i 2^{\frac {5}{2}-n} (1-i \tan (c+d x))^{n-\frac {5}{2}} (a+i a \tan (c+d x))^n (e \sec (c+d x))^{5-2 n} \, _2F_1\left (\frac {5}{2},\frac {1}{2} (2 n-3);\frac {7}{2};\frac {1}{2} (i \tan (c+d x)+1)\right )}{5 d} \]

[Out]

-1/5*I*2^(5/2-n)*hypergeom([5/2, -3/2+n],[7/2],1/2+1/2*I*tan(d*x+c))*(e*sec(d*x+c))^(5-2*n)*(1-I*tan(d*x+c))^(
-5/2+n)*(a+I*a*tan(d*x+c))^n/d

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Rubi [A]  time = 0.21, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3505, 3523, 7, 70, 69} \[ -\frac {i 2^{\frac {5}{2}-n} (1-i \tan (c+d x))^{n-\frac {5}{2}} (a+i a \tan (c+d x))^n (e \sec (c+d x))^{5-2 n} \text {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2} (2 n-3),\frac {7}{2},\frac {1}{2} (1+i \tan (c+d x))\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/5)*2^(5/2 - n)*Hypergeometric2F1[5/2, (-3 + 2*n)/2, 7/2, (1 + I*Tan[c + d*x])/2]*(e*Sec[c + d*x])^(5 - 2*
n)*(1 - I*Tan[c + d*x])^(-5/2 + n)*(a + I*a*Tan[c + d*x])^n)/d

Rule 7

Int[(u_.)*(Px_)^(p_), x_Symbol] :> Int[u*Px^Simplify[p], x] /; PolyQ[Px, x] &&  !RationalQ[p] && FreeQ[p, x] &
& RationalQ[Simplify[p]]

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))^{5-2 n} (a+i a \tan (c+d x))^n \, dx &=\left ((e \sec (c+d x))^{5-2 n} (a-i a \tan (c+d x))^{\frac {1}{2} (-5+2 n)} (a+i a \tan (c+d x))^{\frac {1}{2} (-5+2 n)}\right ) \int (a-i a \tan (c+d x))^{\frac {1}{2} (5-2 n)} (a+i a \tan (c+d x))^{\frac {1}{2} (5-2 n)+n} \, dx\\ &=\frac {\left (a^2 (e \sec (c+d x))^{5-2 n} (a-i a \tan (c+d x))^{\frac {1}{2} (-5+2 n)} (a+i a \tan (c+d x))^{\frac {1}{2} (-5+2 n)}\right ) \operatorname {Subst}\left (\int (a-i a x)^{-1+\frac {1}{2} (5-2 n)} (a+i a x)^{-1+\frac {1}{2} (5-2 n)+n} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\left (a^2 (e \sec (c+d x))^{5-2 n} (a-i a \tan (c+d x))^{\frac {1}{2} (-5+2 n)} (a+i a \tan (c+d x))^{\frac {1}{2} (-5+2 n)}\right ) \operatorname {Subst}\left (\int (a-i a x)^{-1+\frac {1}{2} (5-2 n)} (a+i a x)^{3/2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\left (2^{\frac {3}{2}-n} a^3 (e \sec (c+d x))^{5-2 n} (a-i a \tan (c+d x))^{\frac {1}{2}-n+\frac {1}{2} (-5+2 n)} \left (\frac {a-i a \tan (c+d x)}{a}\right )^{-\frac {1}{2}+n} (a+i a \tan (c+d x))^{\frac {1}{2} (-5+2 n)}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{2}-\frac {i x}{2}\right )^{-1+\frac {1}{2} (5-2 n)} (a+i a x)^{3/2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {i 2^{\frac {5}{2}-n} \, _2F_1\left (\frac {5}{2},\frac {1}{2} (-3+2 n);\frac {7}{2};\frac {1}{2} (1+i \tan (c+d x))\right ) (e \sec (c+d x))^{5-2 n} (1-i \tan (c+d x))^{-\frac {5}{2}+n} (a+i a \tan (c+d x))^n}{5 d}\\ \end {align*}

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Mathematica [A]  time = 13.80, size = 166, normalized size = 1.71 \[ -\frac {i 2^{5-n} e^{5 i (c+d x)} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-n} \left (1+e^{2 i (c+d x)}\right )^{-n} \, _2F_1\left (\frac {5}{2},5-n;\frac {7}{2};-e^{2 i (c+d x)}\right ) \sec ^{n-5}(c+d x) (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n (e \sec (c+d x))^{5-2 n}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/5*I)*2^(5 - n)*E^((5*I)*(c + d*x))*(E^(I*d*x))^n*Hypergeometric2F1[5/2, 5 - n, 7/2, -E^((2*I)*(c + d*x))]
*Sec[c + d*x]^(-5 + n)*(e*Sec[c + d*x])^(5 - 2*n)*(a + I*a*Tan[c + d*x])^n)/(d*(E^(I*(c + d*x))/(1 + E^((2*I)*
(c + d*x))))^n*(1 + E^((2*I)*(c + d*x)))^n*(Cos[d*x] + I*Sin[d*x])^n)

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fricas [F]  time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (\frac {2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{-2 \, n + 5} e^{\left (i \, d n x + i \, c n + n \log \left (\frac {2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right ) + n \log \left (\frac {a}{e}\right )\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(5-2*n)*(a+I*a*tan(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((2*e*e^(I*d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1))^(-2*n + 5)*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)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \sec \left (d x + c\right )\right )^{-2 \, n + 5} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(5-2*n)*(a+I*a*tan(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((e*sec(d*x + c))^(-2*n + 5)*(I*a*tan(d*x + c) + a)^n, x)

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maple [F]  time = 2.12, size = 0, normalized size = 0.00 \[ \int \left (e \sec \left (d x +c \right )\right )^{5-2 n} \left (a +i a \tan \left (d x +c \right )\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(5-2*n)*(a+I*a*tan(d*x+c))^n,x)

[Out]

int((e*sec(d*x+c))^(5-2*n)*(a+I*a*tan(d*x+c))^n,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \sec \left (d x + c\right )\right )^{-2 \, n + 5} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(5-2*n)*(a+I*a*tan(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((e*sec(d*x + c))^(-2*n + 5)*(I*a*tan(d*x + c) + a)^n, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{5-2\,n}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e/cos(c + d*x))^(5 - 2*n)*(a + a*tan(c + d*x)*1i)^n,x)

[Out]

int((e/cos(c + d*x))^(5 - 2*n)*(a + a*tan(c + d*x)*1i)^n, x)

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sympy [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))**(5-2*n)*(a+I*a*tan(d*x+c))**n,x)

[Out]

Timed out

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