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

Optimal. Leaf size=96 \[ \frac{6 a E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{2 i a}{5 d (e \sec (c+d x))^{5/2}}+\frac{2 a \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}} \]

[Out]

(((-2*I)/5)*a)/(d*(e*Sec[c + d*x])^(5/2)) + (6*a*EllipticE[(c + d*x)/2, 2])/(5*d*e^2*Sqrt[Cos[c + d*x]]*Sqrt[e
*Sec[c + d*x]]) + (2*a*Sin[c + d*x])/(5*d*e*(e*Sec[c + d*x])^(3/2))

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Rubi [A]  time = 0.0652379, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3486, 3769, 3771, 2639} \[ \frac{6 a E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}-\frac{2 i a}{5 d (e \sec (c+d x))^{5/2}}+\frac{2 a \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-2*I)/5)*a)/(d*(e*Sec[c + d*x])^(5/2)) + (6*a*EllipticE[(c + d*x)/2, 2])/(5*d*e^2*Sqrt[Cos[c + d*x]]*Sqrt[e
*Sec[c + d*x]]) + (2*a*Sin[c + d*x])/(5*d*e*(e*Sec[c + d*x])^(3/2))

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \frac{a+i a \tan (c+d x)}{(e \sec (c+d x))^{5/2}} \, dx &=-\frac{2 i a}{5 d (e \sec (c+d x))^{5/2}}+a \int \frac{1}{(e \sec (c+d x))^{5/2}} \, dx\\ &=-\frac{2 i a}{5 d (e \sec (c+d x))^{5/2}}+\frac{2 a \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}+\frac{(3 a) \int \frac{1}{\sqrt{e \sec (c+d x)}} \, dx}{5 e^2}\\ &=-\frac{2 i a}{5 d (e \sec (c+d x))^{5/2}}+\frac{2 a \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}+\frac{(3 a) \int \sqrt{\cos (c+d x)} \, dx}{5 e^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}\\ &=-\frac{2 i a}{5 d (e \sec (c+d x))^{5/2}}+\frac{6 a E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}}+\frac{2 a \sin (c+d x)}{5 d e (e \sec (c+d x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.75465, size = 99, normalized size = 1.03 \[ -\frac{a (\tan (c+d x)-i) \left (-2 \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 i \sin (2 (c+d x))+2 \cos (2 (c+d x))+2\right )}{5 d e^2 \sqrt{e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(2 + 2*Cos[2*(c + d*x)] - 2*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c +
d*x))] - (3*I)*Sin[2*(c + d*x)])*(-I + Tan[c + d*x]))/(5*d*e^2*Sqrt[e*Sec[c + d*x]])

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Maple [B]  time = 0.191, size = 339, normalized size = 3.5 \begin{align*} -{\frac{2\,a}{5\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}} \left ( 3\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -3\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +i\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3\,i\sin \left ( dx+c \right ){\it EllipticE} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}-3\,i\sin \left ( dx+c \right ){\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-3\,\cos \left ( dx+c \right ) \right ) \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*a/d*(3*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c)
,I)*cos(d*x+c)*sin(d*x+c)-3*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x+
c)-1)/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)+I*sin(d*x+c)*cos(d*x+c)^3+3*I*sin(d*x+c)*EllipticE(I*(cos(d*x+c)-1)/
sin(d*x+c),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-3*I*sin(d*x+c)*EllipticF(I*(cos(d*x+c
)-1)/sin(d*x+c),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+cos(d*x+c)^4+2*cos(d*x+c)^2-3*co
s(d*x+c))/sin(d*x+c)/cos(d*x+c)^3/(e/cos(d*x+c))^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{i \, a \tan \left (d x + c\right ) + a}{\left (e \sec \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{2}{\left (-i \, a e^{\left (5 i \, d x + 5 i \, c\right )} + i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 8 i \, a e^{\left (3 i \, d x + 3 i \, c\right )} - 4 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 7 i \, a e^{\left (i \, d x + i \, c\right )} - 5 i \, a\right )} \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 )} + 10 \,{\left (d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - d e^{3} e^{\left (i \, d x + i \, c\right )}\right )}{\rm integral}\left (\frac{\sqrt{2}{\left (-3 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 6 i \, a e^{\left (i \, d x + i \, c\right )} - 3 i \, a\right )} \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 )}}{5 \,{\left (d e^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{3} e^{\left (i \, d x + i \, c\right )}\right )}}, x\right )}{10 \,{\left (d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - d e^{3} e^{\left (i \, d x + i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(sqrt(2)*(-I*a*e^(5*I*d*x + 5*I*c) + I*a*e^(4*I*d*x + 4*I*c) - 8*I*a*e^(3*I*d*x + 3*I*c) - 4*I*a*e^(2*I*d
*x + 2*I*c) - 7*I*a*e^(I*d*x + I*c) - 5*I*a)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 10*(d
*e^3*e^(2*I*d*x + 2*I*c) - d*e^3*e^(I*d*x + I*c))*integral(1/5*sqrt(2)*(-3*I*a*e^(2*I*d*x + 2*I*c) - 6*I*a*e^(
I*d*x + I*c) - 3*I*a)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)/(d*e^3*e^(3*I*d*x + 3*I*c) - 2
*d*e^3*e^(2*I*d*x + 2*I*c) + d*e^3*e^(I*d*x + I*c)), x))/(d*e^3*e^(2*I*d*x + 2*I*c) - d*e^3*e^(I*d*x + I*c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int \frac{1}{\left (e \sec{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx + \int \frac{i \tan{\left (c + d x \right )}}{\left (e \sec{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(Integral((e*sec(c + d*x))**(-5/2), x) + Integral(I*tan(c + d*x)/(e*sec(c + d*x))**(5/2), x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{i \, a \tan \left (d x + c\right ) + a}{\left (e \sec \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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