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

Optimal. Leaf size=96 \[ \frac {i a 2^{n+\frac {9}{4}} (e \sec (c+d x))^{5/2} (1+i \tan (c+d x))^{-n-\frac {1}{4}} (a+i a \tan (c+d x))^{n-1} \, _2F_1\left (\frac {5}{4},-n-\frac {1}{4};\frac {9}{4};\frac {1}{2} (1-i \tan (c+d x))\right )}{5 d} \]

[Out]

1/5*I*2^(9/4+n)*a*hypergeom([5/4, -1/4-n],[9/4],1/2-1/2*I*tan(d*x+c))*(e*sec(d*x+c))^(5/2)*(1+I*tan(d*x+c))^(-
1/4-n)*(a+I*a*tan(d*x+c))^(-1+n)/d

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Rubi [A]  time = 0.20, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3505, 3523, 70, 69} \[ \frac {i a 2^{n+\frac {9}{4}} (e \sec (c+d x))^{5/2} (1+i \tan (c+d x))^{-n-\frac {1}{4}} (a+i a \tan (c+d x))^{n-1} \text {Hypergeometric2F1}\left (\frac {5}{4},-n-\frac {1}{4},\frac {9}{4},\frac {1}{2} (1-i \tan (c+d x))\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/5)*2^(9/4 + n)*a*Hypergeometric2F1[5/4, -1/4 - n, 9/4, (1 - I*Tan[c + d*x])/2]*(e*Sec[c + d*x])^(5/2)*(1 +
I*Tan[c + d*x])^(-1/4 - n)*(a + I*a*Tan[c + d*x])^(-1 + n))/d

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} (a+i a \tan (c+d x))^n \, dx &=\frac {(e \sec (c+d x))^{5/2} \int (a-i a \tan (c+d x))^{5/4} (a+i a \tan (c+d x))^{\frac {5}{4}+n} \, dx}{(a-i a \tan (c+d x))^{5/4} (a+i a \tan (c+d x))^{5/4}}\\ &=\frac {\left (a^2 (e \sec (c+d x))^{5/2}\right ) \operatorname {Subst}\left (\int \sqrt [4]{a-i a x} (a+i a x)^{\frac {1}{4}+n} \, dx,x,\tan (c+d x)\right )}{d (a-i a \tan (c+d x))^{5/4} (a+i a \tan (c+d x))^{5/4}}\\ &=\frac {\left (2^{\frac {1}{4}+n} a^2 (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x))^{-1+n} \left (\frac {a+i a \tan (c+d x)}{a}\right )^{-\frac {1}{4}-n}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{2}+\frac {i x}{2}\right )^{\frac {1}{4}+n} \sqrt [4]{a-i a x} \, dx,x,\tan (c+d x)\right )}{d (a-i a \tan (c+d x))^{5/4}}\\ &=\frac {i 2^{\frac {9}{4}+n} a \, _2F_1\left (\frac {5}{4},-\frac {1}{4}-n;\frac {9}{4};\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{5/2} (1+i \tan (c+d x))^{-\frac {1}{4}-n} (a+i a \tan (c+d x))^{-1+n}}{5 d}\\ \end {align*}

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Mathematica [A]  time = 9.06, size = 156, normalized size = 1.62 \[ -\frac {i 2^{n+\frac {7}{2}} e^{i (c+d x)} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{n+\frac {3}{2}} (e \sec (c+d x))^{5/2} \, _2F_1\left (-\frac {1}{4},1;n+\frac {9}{4};-e^{2 i (c+d x)}\right ) \sec ^{-n-\frac {5}{2}}(c+d x) (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n}{d (4 n+5)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*2^(7/2 + n)*E^(I*(c + d*x))*(E^(I*d*x))^n*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(3/2 + n)*Hypergeo
metric2F1[-1/4, 1, 9/4 + n, -E^((2*I)*(c + d*x))]*Sec[c + d*x]^(-5/2 - n)*(e*Sec[c + d*x])^(5/2)*(a + I*a*Tan[
c + d*x])^n)/(d*(5 + 4*n)*(Cos[d*x] + I*Sin[d*x])^n)

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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {4 \, \sqrt {2} e^{2} \left (\frac {2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{n} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {5}{2} i \, d x + \frac {5}{2} i \, c\right )}}{e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(4*sqrt(2)*e^2*(2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^n*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))
*e^(5/2*I*d*x + 5/2*I*c)/(e^(4*I*d*x + 4*I*c) + 2*e^(2*I*d*x + 2*I*c) + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.85, size = 0, normalized size = 0.00 \[ \int \left (e \sec \left (d x +c \right )\right )^{\frac {5}{2}} \left (a +i a \tan \left (d x +c \right )\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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