3.244 \(\int \frac{1}{(e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=181 \[ \frac{2 \sin (c+d x)}{7 a^2 d e^3 \sqrt{e \sec (c+d x)}}+\frac{4 i e^2}{15 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{11/2}}+\frac{2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{7 a^2 d e^4}+\frac{2 e \sin (c+d x)}{15 a^2 d (e \sec (c+d x))^{9/2}}+\frac{6 \sin (c+d x)}{35 a^2 d e (e \sec (c+d x))^{5/2}} \]

[Out]

(2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(7*a^2*d*e^4) + (2*e*Sin[c + d*x])/(15*a
^2*d*(e*Sec[c + d*x])^(9/2)) + (6*Sin[c + d*x])/(35*a^2*d*e*(e*Sec[c + d*x])^(5/2)) + (2*Sin[c + d*x])/(7*a^2*
d*e^3*Sqrt[e*Sec[c + d*x]]) + (((4*I)/15)*e^2)/(d*(e*Sec[c + d*x])^(11/2)*(a^2 + I*a^2*Tan[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.130622, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3500, 3769, 3771, 2641} \[ \frac{2 \sin (c+d x)}{7 a^2 d e^3 \sqrt{e \sec (c+d x)}}+\frac{4 i e^2}{15 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{11/2}}+\frac{2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{7 a^2 d e^4}+\frac{2 e \sin (c+d x)}{15 a^2 d (e \sec (c+d x))^{9/2}}+\frac{6 \sin (c+d x)}{35 a^2 d e (e \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Sec[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(7*a^2*d*e^4) + (2*e*Sin[c + d*x])/(15*a
^2*d*(e*Sec[c + d*x])^(9/2)) + (6*Sin[c + d*x])/(35*a^2*d*e*(e*Sec[c + d*x])^(5/2)) + (2*Sin[c + d*x])/(7*a^2*
d*e^3*Sqrt[e*Sec[c + d*x]]) + (((4*I)/15)*e^2)/(d*(e*Sec[c + d*x])^(11/2)*(a^2 + I*a^2*Tan[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 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{1}{(e \sec (c+d x))^{7/2} (a+i a \tan (c+d x))^2} \, dx &=\frac{4 i e^2}{15 d (e \sec (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (11 e^2\right ) \int \frac{1}{(e \sec (c+d x))^{11/2}} \, dx}{15 a^2}\\ &=\frac{2 e \sin (c+d x)}{15 a^2 d (e \sec (c+d x))^{9/2}}+\frac{4 i e^2}{15 d (e \sec (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{3 \int \frac{1}{(e \sec (c+d x))^{7/2}} \, dx}{5 a^2}\\ &=\frac{2 e \sin (c+d x)}{15 a^2 d (e \sec (c+d x))^{9/2}}+\frac{6 \sin (c+d x)}{35 a^2 d e (e \sec (c+d x))^{5/2}}+\frac{4 i e^2}{15 d (e \sec (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{3 \int \frac{1}{(e \sec (c+d x))^{3/2}} \, dx}{7 a^2 e^2}\\ &=\frac{2 e \sin (c+d x)}{15 a^2 d (e \sec (c+d x))^{9/2}}+\frac{6 \sin (c+d x)}{35 a^2 d e (e \sec (c+d x))^{5/2}}+\frac{2 \sin (c+d x)}{7 a^2 d e^3 \sqrt{e \sec (c+d x)}}+\frac{4 i e^2}{15 d (e \sec (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\int \sqrt{e \sec (c+d x)} \, dx}{7 a^2 e^4}\\ &=\frac{2 e \sin (c+d x)}{15 a^2 d (e \sec (c+d x))^{9/2}}+\frac{6 \sin (c+d x)}{35 a^2 d e (e \sec (c+d x))^{5/2}}+\frac{2 \sin (c+d x)}{7 a^2 d e^3 \sqrt{e \sec (c+d x)}}+\frac{4 i e^2}{15 d (e \sec (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{e \sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{7 a^2 e^4}\\ &=\frac{2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \sec (c+d x)}}{7 a^2 d e^4}+\frac{2 e \sin (c+d x)}{15 a^2 d (e \sec (c+d x))^{9/2}}+\frac{6 \sin (c+d x)}{35 a^2 d e (e \sec (c+d x))^{5/2}}+\frac{2 \sin (c+d x)}{7 a^2 d e^3 \sqrt{e \sec (c+d x)}}+\frac{4 i e^2}{15 d (e \sec (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.873756, size = 151, normalized size = 0.83 \[ -\frac{(e \sec (c+d x))^{5/2} \left (-17 \sin (2 (c+d x))+128 \sin (4 (c+d x))+11 \sin (6 (c+d x))+228 i \cos (2 (c+d x))-72 i \cos (4 (c+d x))-4 i \cos (6 (c+d x))+480 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+296 i\right )}{1680 a^2 d e^6 (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Sec[c + d*x])^(7/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

-((e*Sec[c + d*x])^(5/2)*(296*I + (228*I)*Cos[2*(c + d*x)] - (72*I)*Cos[4*(c + d*x)] - (4*I)*Cos[6*(c + d*x)]
+ 480*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) - 17*Sin[2*(c + d*x
)] + 128*Sin[4*(c + d*x)] + 11*Sin[6*(c + d*x)]))/(1680*a^2*d*e^6*(-I + Tan[c + d*x])^2)

________________________________________________________________________________________

Maple [A]  time = 0.492, size = 252, normalized size = 1.4 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{105\,{a}^{2}d{e}^{7} \left ( \sin \left ( dx+c \right ) \right ) ^{4}} \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{7}{2}}} \left ( 14\,i \left ( \cos \left ( dx+c \right ) \right ) ^{8}+14\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{7}+7\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +15\,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 ) +15\,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 ) +9\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +15\,\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(1/(e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x)

[Out]

2/105/a^2/d*(e/cos(d*x+c))^(7/2)*(cos(d*x+c)-1)^2*(cos(d*x+c)+1)^2*cos(d*x+c)^3*(14*I*cos(d*x+c)^8+14*sin(d*x+
c)*cos(d*x+c)^7+7*cos(d*x+c)^5*sin(d*x+c)+15*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)+15*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)+9*cos(d*x+c)^3*sin(d*x+c)+15*cos(d*x+c)*sin(d*x+c))/e^7/sin(d*x+c)^
4

________________________________________________________________________________________

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(1/(e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (3360 \, a^{2} d e^{4} e^{\left (8 i \, d x + 8 i \, c\right )}{\rm integral}\left (-\frac{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 )}}{7 \, a^{2} d e^{4}}, x\right ) + \sqrt{2} \sqrt{\frac{e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-15 i \, e^{\left (12 i \, d x + 12 i \, c\right )} - 200 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 245 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 592 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 211 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 56 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 (-8 i \, d x - 8 i \, c\right )}}{3360 \, a^{2} d e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3360*(3360*a^2*d*e^4*e^(8*I*d*x + 8*I*c)*integral(-1/7*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^2*d*e^4), x) + sqrt(2)*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-15*I*e^(12*I*d*x + 12*I*c) - 200
*I*e^(10*I*d*x + 10*I*c) + 245*I*e^(8*I*d*x + 8*I*c) + 592*I*e^(6*I*d*x + 6*I*c) + 211*I*e^(4*I*d*x + 4*I*c) +
56*I*e^(2*I*d*x + 2*I*c) + 7*I)*e^(1/2*I*d*x + 1/2*I*c))*e^(-8*I*d*x - 8*I*c)/(a^2*d*e^4)

________________________________________________________________________________________

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(1/(e*sec(d*x+c))**(7/2)/(a+I*a*tan(d*x+c))**2,x)

[Out]

Exception raised: AttributeError

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (e \sec \left (d x + c\right )\right )^{\frac{7}{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(1/((e*sec(d*x + c))^(7/2)*(I*a*tan(d*x + c) + a)^2), x)