3.3.30 \(\int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))} \, dx\) [230]

Optimal. Leaf size=114 \[ \frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{21 a d e^2}+\frac {10 \sin (c+d x)}{21 a d e \sqrt {e \sec (c+d x)}}+\frac {2 i}{7 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))} \]

[Out]

10/21*sin(d*x+c)/a/d/e/(e*sec(d*x+c))^(1/2)+10/21*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(si
n(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*(e*sec(d*x+c))^(1/2)/a/d/e^2+2/7*I/d/(e*sec(d*x+c))^(3/2)/(a+I*a*ta
n(d*x+c))

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Rubi [A]
time = 0.08, antiderivative size = 114, normalized size of antiderivative = 1.00, 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 = {3583, 3854, 3856, 2720} \begin {gather*} \frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{21 a d e^2}+\frac {10 \sin (c+d x)}{21 a d e \sqrt {e \sec (c+d x)}}+\frac {2 i}{7 d (a+i a \tan (c+d x)) (e \sec (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])),x]

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(21*a*d*e^2) + (10*Sin[c + d*x])/(21*a*
d*e*Sqrt[e*Sec[c + d*x]]) + ((2*I)/7)/(d*(e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x]))

Rule 2720

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

Rule 3583

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(d*
Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/(b*f*(m + 2*n))), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), Int[(d*Sec[
e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2*n]

Rule 3854

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 3856

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 {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))} \, dx &=\frac {2 i}{7 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))}+\frac {5 \int \frac {1}{(e \sec (c+d x))^{3/2}} \, dx}{7 a}\\ &=\frac {10 \sin (c+d x)}{21 a d e \sqrt {e \sec (c+d x)}}+\frac {2 i}{7 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))}+\frac {5 \int \sqrt {e \sec (c+d x)} \, dx}{21 a e^2}\\ &=\frac {10 \sin (c+d x)}{21 a d e \sqrt {e \sec (c+d x)}}+\frac {2 i}{7 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))}+\frac {\left (5 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 a e^2}\\ &=\frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{21 a d e^2}+\frac {10 \sin (c+d x)}{21 a d e \sqrt {e \sec (c+d x)}}+\frac {2 i}{7 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 0.59, size = 125, normalized size = 1.10 \begin {gather*} -\frac {\sec ^3(c+d x) \left (-14 \cos (c+d x)+2 \cos (3 (c+d x))+20 i \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (\cos (c+d x)+i \sin (c+d x))+5 i \sin (c+d x)+5 i \sin (3 (c+d x))\right )}{42 a d (e \sec (c+d x))^{3/2} (-i+\tan (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])),x]

[Out]

-1/42*(Sec[c + d*x]^3*(-14*Cos[c + d*x] + 2*Cos[3*(c + d*x)] + (20*I)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2
, 2]*(Cos[c + d*x] + I*Sin[c + d*x]) + (5*I)*Sin[c + d*x] + (5*I)*Sin[3*(c + d*x)]))/(a*d*(e*Sec[c + d*x])^(3/
2)*(-I + Tan[c + d*x]))

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Maple [A]
time = 0.77, size = 218, normalized size = 1.91

method	result	size
default	\(\frac {2 \cos \left (d x +c \right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (1+\cos \left (d x +c \right )\right )^{2} \left (3 i \left (\cos ^{4}\left (d x +c \right )\right )+5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+3 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+5 \sin \left (d x +c \right ) \cos \left (d x +c \right )\right )}{21 a d \,e^{3} \sin \left (d x +c \right )^{4}}\)	\(218\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

2/21/a/d*cos(d*x+c)*(e/cos(d*x+c))^(3/2)*(-1+cos(d*x+c))^2*(1+cos(d*x+c))^2*(3*I*cos(d*x+c)^4+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)*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)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)+3*sin(d*x+c)*co
s(d*x+c)^3+5*sin(d*x+c)*cos(d*x+c))/e^3/sin(d*x+c)^4

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 106, normalized size = 0.93 \begin {gather*} \frac {{\left (-40 i \, \sqrt {2} e^{\left (4 i \, d x + 4 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \frac {\sqrt {2} {\left (-7 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 9 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 19 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{\sqrt {e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-4 i \, d x - 4 i \, c - \frac {3}{2}\right )}}{84 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/84*(-40*I*sqrt(2)*e^(4*I*d*x + 4*I*c)*weierstrassPInverse(-4, 0, e^(I*d*x + I*c)) + sqrt(2)*(-7*I*e^(6*I*d*x
+ 6*I*c) + 9*I*e^(4*I*d*x + 4*I*c) + 19*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(1/2*I*d*x + 1/2*I*c)/sqrt(e^(2*I*d*x
+ 2*I*c) + 1))*e^(-4*I*d*x - 4*I*c - 3/2)/(a*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {i \int \frac {1}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan {\left (c + d x \right )} - i \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))**(3/2)/(a+I*a*tan(d*x+c)),x)

[Out]

-I*Integral(1/((e*sec(c + d*x))**(3/2)*tan(c + d*x) - I*(e*sec(c + d*x))**(3/2)), x)/a

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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