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*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]))

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Rubi [A]  time = 0.139877, antiderivative size = 152, normalized size of antiderivative = 1., 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 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 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 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]

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]

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*}

Mathematica [A]  time = 0.425284, 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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Maple [A]  time = 0.39, size = 236, normalized size = 1.6 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}{77\,d{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}\sqrt{{\frac{e}{\cos \left ( dx+c \right ) }}} \left ( 28\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}+28\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +5\,i\cos \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) -11\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+5\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( dx+c \right ) -1 \right ) }{\sin \left ( dx+c \right ) }},i \right ) +3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +5\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) } \end{align*}

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)*(cos(d*x+c)-1)^2*(cos(d*x+c)+1)^2*(28*I*cos(d*x+c)^6+28*cos(d*x+c)^5*sin(d*x+c
)+5*I*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x
+c),I)-11*I*cos(d*x+c)^4+5*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x+c
)-1)/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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

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]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{e \sec \left (d x + c\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

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)