3.190 \(\int \frac {a+i a \tan (c+d x)}{(e \sec (c+d x))^{3/2}} \, dx\)

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

[Out]

-2/3*I*a/d/(e*sec(d*x+c))^(3/2)+2/3*a*sin(d*x+c)/d/e/(e*sec(d*x+c))^(1/2)+2/3*a*(cos(1/2*d*x+1/2*c)^2)^(1/2)/c
os(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/e^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, 2641} \[ \frac {2 a \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}-\frac {2 i a}{3 d (e \sec (c+d x))^{3/2}}+\frac {2 a \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])/(e*Sec[c + d*x])^(3/2),x]

[Out]

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

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

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Mathematica [A]  time = 0.42, size = 62, normalized size = 0.65 \[ \frac {2 a \left (\sin (c+d x)-i \cos (c+d x)+\frac {F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{\sqrt {\cos (c+d x)}}\right )}{3 d e \sqrt {e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])/(e*Sec[c + d*x])^(3/2),x]

[Out]

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

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ \frac {3 \, d e^{2} {\rm integral}\left (-\frac {i \, \sqrt {2} a \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}}{3 \, d e^{2}}, x\right ) + \sqrt {2} {\left (-i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{3 \, d e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))/(e*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {i \, a \tan \left (d x + c\right ) + a}{\left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))/(e*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)/(e*sec(d*x + c))^(3/2), x)

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maple [A]  time = 0.78, size = 170, normalized size = 1.77 \[ \frac {2 a \left (i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right )+i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-i \left (\cos ^{2}\left (d x +c \right )\right )+\cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{2} \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))/(e*sec(d*x+c))^(3/2),x)

[Out]

2/3*a/d*(I*(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
)*cos(d*x+c)+I*(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)-I*cos(d*x+c)^2+cos(d*x+c)*sin(d*x+c))/cos(d*x+c)^2/(e/cos(d*x+c))^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {i \, a \tan \left (d x + c\right ) + a}{\left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))/(e*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)/(e*sec(d*x + c))^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)/(e/cos(c + d*x))^(3/2),x)

[Out]

int((a + a*tan(c + d*x)*1i)/(e/cos(c + d*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ i a \left (\int \left (- \frac {i}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\right )\, dx + \int \frac {\tan {\left (c + d x \right )}}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))/(e*sec(d*x+c))**(3/2),x)

[Out]

I*a*(Integral(-I/(e*sec(c + d*x))**(3/2), x) + Integral(tan(c + d*x)/(e*sec(c + d*x))**(3/2), x))

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