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

Optimal. Leaf size=91 \[ -\frac{i 2^{n+\frac{3}{4}} (1+i \tan (c+d x))^{\frac{1}{4}-n} (a+i a \tan (c+d x))^n \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{5}{4}-n,\frac{3}{4},\frac{1}{2} (1-i \tan (c+d x))\right )}{d \sqrt{e \sec (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.178157, antiderivative size = 91, 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} \[ -\frac{i 2^{n+\frac{3}{4}} (1+i \tan (c+d x))^{\frac{1}{4}-n} (a+i a \tan (c+d x))^n \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{5}{4}-n,\frac{3}{4},\frac{1}{2} (1-i \tan (c+d x))\right )}{d \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 9.53452, size = 129, normalized size = 1.42 \[ -\frac{i 2^{n+\frac{1}{2}} e^{i (c+d x)} \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{n-\frac{3}{2}} \sec ^{\frac{1}{2}-n}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{5}{4},n+\frac{3}{4},-e^{2 i (c+d x)}\right ) (a+i a \tan (c+d x))^n}{d (4 n-1) \sqrt{e \sec (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.199, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{n}{\frac{1}{\sqrt{e\sec \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\sqrt{e \sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a \left (i \tan{\left (c + d x \right )} + 1\right )\right )^{n}}{\sqrt{e \sec{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(I*tan(c + d*x) + 1))**n/sqrt(e*sec(c + d*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\sqrt{e \sec \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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