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

Optimal. Leaf size=94 \[ \frac {2 a e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{3 d}+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}+\frac {2 a e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 94, 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, 3768, 3771, 2641} \[ \frac {2 a e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{3 d}+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}+\frac {2 a e \sin (c+d x) (e \sec (c+d x))^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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 (e \sec (c+d x))^{5/2} (a+i a \tan (c+d x)) \, dx &=\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}+a \int (e \sec (c+d x))^{5/2} \, dx\\ &=\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}+\frac {2 a e (e \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {1}{3} \left (a e^2\right ) \int \sqrt {e \sec (c+d x)} \, dx\\ &=\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}+\frac {2 a e (e \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac {1}{3} \left (a e^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 a e^2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{3 d}+\frac {2 i a (e \sec (c+d x))^{5/2}}{5 d}+\frac {2 a e (e \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.55, size = 57, normalized size = 0.61 \[ \frac {a (e \sec (c+d x))^{5/2} \left (5 \sin (2 (c+d x))+10 \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+6 i\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(e*Sec[c + d*x])^(5/2)*(6*I + 10*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 5*Sin[2*(c + d*x)]))/(15*d)

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ \frac {\sqrt {2} {\left (-10 i \, a e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 24 i \, a e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 10 i \, a e^{2}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} {\rm integral}\left (-\frac {i \, \sqrt {2} a e^{2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}}{3 \, d}, x\right )}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

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)),x, algorithm="fricas")

[Out]

1/15*(sqrt(2)*(-10*I*a*e^2*e^(4*I*d*x + 4*I*c) + 24*I*a*e^2*e^(2*I*d*x + 2*I*c) + 10*I*a*e^2)*sqrt(e/(e^(2*I*d
*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 15*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*integral(
-1/3*I*sqrt(2)*a*e^2*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2*I*d*x - 1/2*I*c)/d, x))/(d*e^(4*I*d*x + 4*I*c)
+ 2*d*e^(2*I*d*x + 2*I*c) + d)

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

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)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.83, size = 192, normalized size = 2.04 \[ \frac {2 a \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+5 \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 i\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}{15 d \sin \left (d x +c \right )^{4}} \]

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)),x)

[Out]

2/15*a/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^2*(5*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*co
s(d*x+c)^3*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+5*I*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)*cos(d*x+c)^2*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+5*cos(d*x+c)*sin(d*x+c)+3*I)*(e/cos(d*x+c))^(5/2)
/sin(d*x+c)^4

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

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)),x, algorithm="maxima")

[Out]

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

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

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),x)

[Out]

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

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

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)),x)

[Out]

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

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