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]

-2/5*I*a/d/(e*sec(d*x+c))^(5/2)+2/5*a*sin(d*x+c)/d/e/(e*sec(d*x+c))^(3/2)+6/5*a*(cos(1/2*d*x+1/2*c)^2)^(1/2)/c
os(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/e^2/cos(d*x+c)^(1/2)/(e*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 96, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [C]  time = 0.80, size = 99, normalized size = 1.03 \[ -\frac {a (\tan (c+d x)-i) \left (-2 \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )-3 i \sin (2 (c+d x))+2 \cos (2 (c+d x))+2\right )}{5 d e^2 \sqrt {e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(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]))/(d*e^2*Sqrt[e*Sec[c + d*x]])

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ \frac {\sqrt {2} {\left (-i \, a e^{\left (5 i \, d x + 5 i \, c\right )} + i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 8 i \, a e^{\left (3 i \, d x + 3 i \, c\right )} - 4 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 7 i \, a e^{\left (i \, d x + i \, c\right )} - 5 i \, a\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 10 \, {\left (d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - d e^{3} e^{\left (i \, d x + i \, c\right )}\right )} {\rm integral}\left (\frac {\sqrt {2} {\left (-3 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 6 i \, a e^{\left (i \, d x + i \, c\right )} - 3 i \, a\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{5 \, {\left (d e^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 2 \, d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{3} e^{\left (i \, d x + i \, c\right )}\right )}}, x\right )}{10 \, {\left (d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - d e^{3} e^{\left (i \, d x + i \, c\right )}\right )}} \]

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

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)

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maple [B]  time = 0.82, size = 341, normalized size = 3.55 \[ \frac {2 a \left (3 i \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-3 i \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+3 i \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-3 i \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-i \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )-\left (\cos ^{4}\left (d x +c \right )\right )-2 \left (\cos ^{2}\left (d x +c \right )\right )+3 \cos \left (d x +c \right )\right )}{5 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3} \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}} \]

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*sin(d*x+c)*cos(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*EllipticF(I*(-1+
cos(d*x+c))/sin(d*x+c),I)-3*I*sin(d*x+c)*cos(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)+3*I*sin(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)-3*I*sin(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*
x+c)))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)-I*sin(d*x+c)*cos(d*x+c)^3-cos(d*x+c)^4-2*cos(d*x+c)^2+3
*cos(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {i \, a \tan \left (d x + c\right ) + a}{\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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]

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

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