\(\int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx\) [671]

Optimal result

Integrand size = 28, antiderivative size = 126 \[ \int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx=\frac {10 \cos ^{\frac {9}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d (e \cos (c+d x))^{9/2}}+\frac {10 \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (e \cos (c+d x))^{9/2}}-\frac {4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{9/2} \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

10/3*cos(d*x+c)^(9/2)*(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))/a^
2/d/(e*cos(d*x+c))^(9/2)+10/3*cos(d*x+c)^3*sin(d*x+c)/a^2/d/(e*cos(d*x+c))^(9/2)-4*I*cos(d*x+c)^2/d/(e*cos(d*x
+c))^(9/2)/(a^2+I*a^2*tan(d*x+c))

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3596, 3581, 3853, 3856, 2720} \[ \int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx=\frac {10 \cos ^{\frac {9}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d (e \cos (c+d x))^{9/2}}+\frac {10 \sin (c+d x) \cos ^3(c+d x)}{3 a^2 d (e \cos (c+d x))^{9/2}}-\frac {4 i \cos ^2(c+d x)}{d \left (a^2+i a^2 \tan (c+d x)\right ) (e \cos (c+d x))^{9/2}} \]

[In]

Int[1/((e*Cos[c + d*x])^(9/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(10*Cos[c + d*x]^(9/2)*EllipticF[(c + d*x)/2, 2])/(3*a^2*d*(e*Cos[c + d*x])^(9/2)) + (10*Cos[c + d*x]^3*Sin[c
+ d*x])/(3*a^2*d*(e*Cos[c + d*x])^(9/2)) - ((4*I)*Cos[c + d*x]^2)/(d*(e*Cos[c + d*x])^(9/2)*(a^2 + I*a^2*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 3581

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 3596

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] &&  !IntegerQ[m]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[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*} \text {integral}& = \frac {\int \frac {(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^2} \, dx}{(e \cos (c+d x))^{9/2} (e \sec (c+d x))^{9/2}} \\ & = -\frac {4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{9/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (5 e^2\right ) \int (e \sec (c+d x))^{5/2} \, dx}{a^2 (e \cos (c+d x))^{9/2} (e \sec (c+d x))^{9/2}} \\ & = \frac {10 \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (e \cos (c+d x))^{9/2}}-\frac {4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{9/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (5 e^4\right ) \int \sqrt {e \sec (c+d x)} \, dx}{3 a^2 (e \cos (c+d x))^{9/2} (e \sec (c+d x))^{9/2}} \\ & = \frac {10 \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (e \cos (c+d x))^{9/2}}-\frac {4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{9/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (5 \cos ^{\frac {9}{2}}(c+d x)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 a^2 (e \cos (c+d x))^{9/2}} \\ & = \frac {10 \cos ^{\frac {9}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d (e \cos (c+d x))^{9/2}}+\frac {10 \cos ^3(c+d x) \sin (c+d x)}{3 a^2 d (e \cos (c+d x))^{9/2}}-\frac {4 i \cos ^2(c+d x)}{d (e \cos (c+d x))^{9/2} \left (a^2+i a^2 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.57 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.53 \[ \int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx=\frac {2 \left (-6 i \cos (c+d x)+5 \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\sin (c+d x)\right )}{3 a^2 d e^3 (e \cos (c+d x))^{3/2}} \]

[In]

Integrate[1/((e*Cos[c + d*x])^(9/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(2*((-6*I)*Cos[c + d*x] + 5*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] - Sin[c + d*x]))/(3*a^2*d*e^3*(e*Cos[
c + d*x])^(3/2))

Maple [A] (verified)

Time = 4.56 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.65

method	result	size
default	\(\frac {-\frac {20 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-8 i \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {10 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}}{3}+4 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{4} d}\)	\(208\)

[In]

int(1/(e*cos(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

2/3/(2*sin(1/2*d*x+1/2*c)^2-1)/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/e^4*(-10*(sin(1/2*d*
x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2-
12*I*sin(1/2*d*x+1/2*c)^3+2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+5*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipti
cF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)+6*I*sin(1/2*d*x+1/2*c))/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.08 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.21 \[ \int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx=-\frac {2 \, {\left (2 \, \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} {\left (5 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + 7 i \, e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + 5 \, {\left (i \, \sqrt {2} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, \sqrt {2} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, \sqrt {2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )}}{3 \, {\left (a^{2} d e^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{2} d e^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d e^{5}\right )}} \]

[In]

integrate(1/(e*cos(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-2/3*(2*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*(5*I*e^(3*I*d*x + 3*I*c) + 7*I*e^(I*d*x + I*c))*e^(-1/2*I*d*
x - 1/2*I*c) + 5*(I*sqrt(2)*e^(4*I*d*x + 4*I*c) + 2*I*sqrt(2)*e^(2*I*d*x + 2*I*c) + I*sqrt(2))*sqrt(e)*weierst
rassPInverse(-4, 0, e^(I*d*x + I*c)))/(a^2*d*e^5*e^(4*I*d*x + 4*I*c) + 2*a^2*d*e^5*e^(2*I*d*x + 2*I*c) + a^2*d
*e^5)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(1/(e*cos(d*x+c))**(9/2)/(a+I*a*tan(d*x+c))**2,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(1/(e*cos(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx=\int { \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {9}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}} \,d x } \]

[In]

integrate(1/(e*cos(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(1/((e*cos(d*x + c))^(9/2)*(I*a*tan(d*x + c) + a)^2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{9/2} (a+i a \tan (c+d x))^2} \, dx=\int \frac {1}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{9/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \]

[In]

int(1/((e*cos(c + d*x))^(9/2)*(a + a*tan(c + d*x)*1i)^2),x)

[Out]

int(1/((e*cos(c + d*x))^(9/2)*(a + a*tan(c + d*x)*1i)^2), x)