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

Optimal. Leaf size=152 \[ \frac {20 i e^2}{77 d \left (a^3+i a^3 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}+\frac {10 e \sin (c+d x)}{77 a^3 d \sqrt {e \sec (c+d x)}}+\frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{77 a^3 d}+\frac {2 i \sqrt {e \sec (c+d x)}}{11 d (a+i a \tan (c+d x))^3} \]

[Out]

10/77*e*sin(d*x+c)/a^3/d/(e*sec(d*x+c))^(1/2)+10/77*(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)/a^3/d+2/11*I*(e*sec(d*x+c))^(1/2)/d/(a+I*a*t
an(d*x+c))^3+20/77*I*e^2/d/(e*sec(d*x+c))^(3/2)/(a^3+I*a^3*tan(d*x+c))

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Rubi [A]  time = 0.14, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3502, 3500, 3769, 3771, 2641} \[ \frac {20 i e^2}{77 d \left (a^3+i a^3 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}+\frac {10 e \sin (c+d x)}{77 a^3 d \sqrt {e \sec (c+d x)}}+\frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \sec (c+d x)}}{77 a^3 d}+\frac {2 i \sqrt {e \sec (c+d x)}}{11 d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(77*a^3*d) + (10*e*Sin[c + d*x])/(77*a^
3*d*Sqrt[e*Sec[c + d*x]]) + (((2*I)/11)*Sqrt[e*Sec[c + d*x]])/(d*(a + I*a*Tan[c + d*x])^3) + (((20*I)/77)*e^2)
/(d*(e*Sec[c + d*x])^(3/2)*(a^3 + I*a^3*Tan[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 3500

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

Rule 3502

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 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 {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^3} \, dx &=\frac {2 i \sqrt {e \sec (c+d x)}}{11 d (a+i a \tan (c+d x))^3}+\frac {5 \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^2} \, dx}{11 a}\\ &=\frac {2 i \sqrt {e \sec (c+d x)}}{11 d (a+i a \tan (c+d x))^3}+\frac {20 i e^2}{77 d (e \sec (c+d x))^{3/2} \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\left (15 e^2\right ) \int \frac {1}{(e \sec (c+d x))^{3/2}} \, dx}{77 a^3}\\ &=\frac {10 e \sin (c+d x)}{77 a^3 d \sqrt {e \sec (c+d x)}}+\frac {2 i \sqrt {e \sec (c+d x)}}{11 d (a+i a \tan (c+d x))^3}+\frac {20 i e^2}{77 d (e \sec (c+d x))^{3/2} \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {5 \int \sqrt {e \sec (c+d x)} \, dx}{77 a^3}\\ &=\frac {10 e \sin (c+d x)}{77 a^3 d \sqrt {e \sec (c+d x)}}+\frac {2 i \sqrt {e \sec (c+d x)}}{11 d (a+i a \tan (c+d x))^3}+\frac {20 i e^2}{77 d (e \sec (c+d x))^{3/2} \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\left (5 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{77 a^3}\\ &=\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)}}{77 a^3 d}+\frac {10 e \sin (c+d x)}{77 a^3 d \sqrt {e \sec (c+d x)}}+\frac {2 i \sqrt {e \sec (c+d x)}}{11 d (a+i a \tan (c+d x))^3}+\frac {20 i e^2}{77 d (e \sec (c+d x))^{3/2} \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.52, size = 129, normalized size = 0.85 \[ \frac {i \sec ^3(c+d x) \sqrt {e \sec (c+d x)} \left (-15 \sin (c+d x)-15 \sin (3 (c+d x))+46 i \cos (c+d x)+22 i \cos (3 (c+d x))+20 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))\right )}{154 a^3 d (\tan (c+d x)-i)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/154)*Sec[c + d*x]^3*Sqrt[e*Sec[c + d*x]]*((46*I)*Cos[c + d*x] + (22*I)*Cos[3*(c + d*x)] - 15*Sin[c + d*x]
+ 20*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x)]) - 15*Sin[3*(c + d*x)
]))/(a^3*d*(-I + Tan[c + d*x])^3)

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ \frac {{\left (308 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} {\rm integral}\left (-\frac {5 i \, \sqrt {2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}}{77 \, a^{3} d}, x\right ) + \sqrt {2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (37 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 61 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 31 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{308 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/308*(308*a^3*d*e^(6*I*d*x + 6*I*c)*integral(-5/77*I*sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2*I*d*x
- 1/2*I*c)/(a^3*d), x) + sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(37*I*e^(6*I*d*x + 6*I*c) + 61*I*e^(4*I*d*x
+ 4*I*c) + 31*I*e^(2*I*d*x + 2*I*c) + 7*I)*e^(1/2*I*d*x + 1/2*I*c))*e^(-6*I*d*x - 6*I*c)/(a^3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.45, size = 236, normalized size = 1.55 \[ \frac {2 \sqrt {\frac {e}{\cos \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right )^{2} \left (1+\cos \left (d x +c \right )\right )^{2} \left (28 i \left (\cos ^{6}\left (d x +c \right )\right )+28 \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )-11 i \left (\cos ^{4}\left (d x +c \right )\right )+5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \cos \left (d x +c \right )+5 i \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+3 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+5 \cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{77 a^{3} d \sin \left (d x +c \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/77/a^3/d*(e/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(1+cos(d*x+c))^2*(28*I*cos(d*x+c)^6+28*cos(d*x+c)^5*sin(d*x+
c)-11*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)*EllipticF(I*(-1+cos(d*x+c)
)/sin(d*x+c),I)*cos(d*x+c)+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*cos(d*x+c)^3*sin(d*x+c)+5*cos(d*x+c)*sin(d*x+c))/sin(d*x+c)^4

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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