3.479 \(\int \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^n \, dx\)

Optimal. Leaf size=94 \[ \frac{i a 2^{n+\frac{5}{4}} \sqrt{e \sec (c+d x)} (1+i \tan (c+d x))^{\frac{3}{4}-n} (a+i a \tan (c+d x))^{n-1} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{3}{4}-n,\frac{5}{4},\frac{1}{2} (1-i \tan (c+d x))\right )}{d} \]

[Out]

(I*2^(5/4 + n)*a*Hypergeometric2F1[1/4, 3/4 - n, 5/4, (1 - I*Tan[c + d*x])/2]*Sqrt[e*Sec[c + d*x]]*(1 + I*Tan[
c + d*x])^(3/4 - n)*(a + I*a*Tan[c + d*x])^(-1 + n))/d

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Rubi [A]  time = 0.174919, antiderivative size = 94, normalized size of antiderivative = 1., 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} \[ \frac{i a 2^{n+\frac{5}{4}} \sqrt{e \sec (c+d x)} (1+i \tan (c+d x))^{\frac{3}{4}-n} (a+i a \tan (c+d x))^{n-1} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{3}{4}-n,\frac{5}{4},\frac{1}{2} (1-i \tan (c+d x))\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*2^(5/4 + n)*a*Hypergeometric2F1[1/4, 3/4 - n, 5/4, (1 - I*Tan[c + d*x])/2]*Sqrt[e*Sec[c + d*x]]*(1 + I*Tan[
c + d*x])^(3/4 - n)*(a + I*a*Tan[c + d*x])^(-1 + n))/d

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 \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^n \, dx &=\frac{\sqrt{e \sec (c+d x)} \int \sqrt [4]{a-i a \tan (c+d x)} (a+i a \tan (c+d x))^{\frac{1}{4}+n} \, dx}{\sqrt [4]{a-i a \tan (c+d x)} \sqrt [4]{a+i a \tan (c+d x)}}\\ &=\frac{\left (a^2 \sqrt{e \sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{-\frac{3}{4}+n}}{(a-i a x)^{3/4}} \, dx,x,\tan (c+d x)\right )}{d \sqrt [4]{a-i a \tan (c+d x)} \sqrt [4]{a+i a \tan (c+d x)}}\\ &=\frac{\left (2^{-\frac{3}{4}+n} a^2 \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^{-1+n} \left (\frac{a+i a \tan (c+d x)}{a}\right )^{\frac{3}{4}-n}\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{1}{2}+\frac{i x}{2}\right )^{-\frac{3}{4}+n}}{(a-i a x)^{3/4}} \, dx,x,\tan (c+d x)\right )}{d \sqrt [4]{a-i a \tan (c+d x)}}\\ &=\frac{i 2^{\frac{5}{4}+n} a \, _2F_1\left (\frac{1}{4},\frac{3}{4}-n;\frac{5}{4};\frac{1}{2} (1-i \tan (c+d x))\right ) \sqrt{e \sec (c+d x)} (1+i \tan (c+d x))^{\frac{3}{4}-n} (a+i a \tan (c+d x))^{-1+n}}{d}\\ \end{align*}

Mathematica [A]  time = 8.34891, size = 156, normalized size = 1.66 \[ -\frac{i 2^{n+\frac{3}{2}} e^{i (c+d x)} \left (e^{i d x}\right )^n \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{n-\frac{1}{2}} \sqrt{e \sec (c+d x)} \sec ^{-n-\frac{1}{2}}(c+d x) (\cos (d x)+i \sin (d x))^{-n} \text{Hypergeometric2F1}\left (\frac{3}{4},1,n+\frac{5}{4},-e^{2 i (c+d x)}\right ) (a+i a \tan (c+d x))^n}{d (4 n+1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*2^(3/2 + n)*E^(I*(c + d*x))*(E^(I*d*x))^n*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(-1/2 + n)*Hyperge
ometric2F1[3/4, 1, 5/4 + n, -E^((2*I)*(c + d*x))]*Sec[c + d*x]^(-1/2 - n)*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c
+ d*x])^n)/(d*(1 + 4*n)*(Cos[d*x] + I*Sin[d*x])^n)

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Maple [F]  time = 0.194, size = 0, normalized size = 0. \begin{align*} \int \sqrt{e\sec \left ( dx+c \right ) } \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))^(1/2)*(a+I*a*tan(d*x+c))^n,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{e \sec \left (d x + c\right )}{\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))^(1/2)*(a+I*a*tan(d*x+c))^n,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^n,x, algorithm="fricas")

[Out]

integral(sqrt(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^(1/
2*I*d*x + 1/2*I*c), 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))**(1/2)*(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 \sqrt{e \sec \left (d x + c\right )}{\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))^(1/2)*(a+I*a*tan(d*x+c))^n,x, algorithm="giac")

[Out]

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