\(\int \frac {1}{(e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \, dx\) [672]

Optimal result

Integrand size = 28, antiderivative size = 164 \[ \int \frac {1}{(e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \, dx=-\frac {14 \cos ^{\frac {11}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d (e \cos (c+d x))^{11/2}}+\frac {14 \cos ^3(c+d x) \sin (c+d x)}{15 a^2 d (e \cos (c+d x))^{11/2}}+\frac {14 \cos ^5(c+d x) \sin (c+d x)}{5 a^2 d (e \cos (c+d x))^{11/2}}-\frac {4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )} \]

[Out]

-14/5*cos(d*x+c)^(11/2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/
a^2/d/(e*cos(d*x+c))^(11/2)+14/15*cos(d*x+c)^3*sin(d*x+c)/a^2/d/(e*cos(d*x+c))^(11/2)+14/5*cos(d*x+c)^5*sin(d*
x+c)/a^2/d/(e*cos(d*x+c))^(11/2)-4/3*I*cos(d*x+c)^2/d/(e*cos(d*x+c))^(11/2)/(a^2+I*a^2*tan(d*x+c))

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3596, 3581, 3853, 3856, 2719} \[ \int \frac {1}{(e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \, dx=-\frac {14 \cos ^{\frac {11}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d (e \cos (c+d x))^{11/2}}+\frac {14 \sin (c+d x) \cos ^5(c+d x)}{5 a^2 d (e \cos (c+d x))^{11/2}}+\frac {14 \sin (c+d x) \cos ^3(c+d x)}{15 a^2 d (e \cos (c+d x))^{11/2}}-\frac {4 i \cos ^2(c+d x)}{3 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \cos (c+d x))^{11/2}} \]

[In]

Int[1/((e*Cos[c + d*x])^(11/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(-14*Cos[c + d*x]^(11/2)*EllipticE[(c + d*x)/2, 2])/(5*a^2*d*(e*Cos[c + d*x])^(11/2)) + (14*Cos[c + d*x]^3*Sin
[c + d*x])/(15*a^2*d*(e*Cos[c + d*x])^(11/2)) + (14*Cos[c + d*x]^5*Sin[c + d*x])/(5*a^2*d*(e*Cos[c + d*x])^(11
/2)) - (((4*I)/3)*Cos[c + d*x]^2)/(d*(e*Cos[c + d*x])^(11/2)*(a^2 + I*a^2*Tan[c + d*x]))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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))^{11/2}}{(a+i a \tan (c+d x))^2} \, dx}{(e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}} \\ & = -\frac {4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (7 e^2\right ) \int (e \sec (c+d x))^{7/2} \, dx}{3 a^2 (e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}} \\ & = \frac {14 \cos ^3(c+d x) \sin (c+d x)}{15 a^2 d (e \cos (c+d x))^{11/2}}-\frac {4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (7 e^4\right ) \int (e \sec (c+d x))^{3/2} \, dx}{5 a^2 (e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}} \\ & = \frac {14 \cos ^3(c+d x) \sin (c+d x)}{15 a^2 d (e \cos (c+d x))^{11/2}}+\frac {14 \cos ^5(c+d x) \sin (c+d x)}{5 a^2 d (e \cos (c+d x))^{11/2}}-\frac {4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {\left (7 e^6\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{5 a^2 (e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}} \\ & = \frac {14 \cos ^3(c+d x) \sin (c+d x)}{15 a^2 d (e \cos (c+d x))^{11/2}}+\frac {14 \cos ^5(c+d x) \sin (c+d x)}{5 a^2 d (e \cos (c+d x))^{11/2}}-\frac {4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {\left (7 \cos ^{\frac {11}{2}}(c+d x)\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 a^2 (e \cos (c+d x))^{11/2}} \\ & = -\frac {14 \cos ^{\frac {11}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d (e \cos (c+d x))^{11/2}}+\frac {14 \cos ^3(c+d x) \sin (c+d x)}{15 a^2 d (e \cos (c+d x))^{11/2}}+\frac {14 \cos ^5(c+d x) \sin (c+d x)}{5 a^2 d (e \cos (c+d x))^{11/2}}-\frac {4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 5.21 (sec) , antiderivative size = 414, normalized size of antiderivative = 2.52 \[ \int \frac {1}{(e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \, dx=\frac {\cos ^3(c+d x) (\cos (d x)+i \sin (d x))^2 \left (7 \cos (c) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c)))-\frac {7}{2} (3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \cot (c) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}+\frac {1}{6} \csc (c) \sqrt {\sec ^2(c)} \sec ^2(c+d x) (\cos (2 c)+i \sin (2 c)) (36 \cos (d x)+27 \cos (2 c+d x)+21 \cos (2 c+3 d x)+20 i \sin (d x)-20 i \sin (2 c+d x)) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}+7 i \left (2 \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (c) \sin (d x+\arctan (\tan (c)))-(3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}\right )+\frac {7}{2} \left (-2 \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (c) \sin (d x+\arctan (\tan (c)))+(3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}\right ) \tan (c)\right )}{5 d (e \cos (c+d x))^{11/2} \sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\arctan (\tan (c)))} (a+i a \tan (c+d x))^2} \]

[In]

Integrate[1/((e*Cos[c + d*x])^(11/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(Cos[c + d*x]^3*(Cos[d*x] + I*Sin[d*x])^2*(7*Cos[c]*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Ta
n[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]] - (7*(3*Cos[c - d*x - ArcTan[Tan[c]]] + Cos[c + d*x + ArcTan[Tan[c]]])*Cot
[c]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2])/2 + (Csc[c]*Sqrt[Sec[c]^2]*Sec[c + d*x]^2*(Cos[2*c] + I*Sin[2*c])*(36*C
os[d*x] + 27*Cos[2*c + d*x] + 21*Cos[2*c + 3*d*x] + (20*I)*Sin[d*x] - (20*I)*Sin[2*c + d*x])*Sqrt[Sin[d*x + Ar
cTan[Tan[c]]]^2])/6 + (7*I)*(2*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[c]*Sin[
d*x + ArcTan[Tan[c]]] - (3*Cos[c - d*x - ArcTan[Tan[c]]] + Cos[c + d*x + ArcTan[Tan[c]]])*Sqrt[Sin[d*x + ArcTa
n[Tan[c]]]^2]) + (7*(-2*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[c]*Sin[d*x + A
rcTan[Tan[c]]] + (3*Cos[c - d*x - ArcTan[Tan[c]]] + Cos[c + d*x + ArcTan[Tan[c]]])*Sqrt[Sin[d*x + ArcTan[Tan[c
]]]^2])*Tan[c])/2))/(5*d*(e*Cos[c + d*x])^(11/2)*Sqrt[Sec[c]^2]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2]*(a + I*a*Tan
[c + d*x])^2)

Maple [A] (verified)

Time = 7.55 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.96

method	result	size
default	\(\frac {\frac {112 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}-\frac {56 E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {112 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}+\frac {56 E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {8 i \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {14 E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}}{5}-\frac {4 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )}{3}}{\left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{5} d}\)	\(321\)

[In]

int(1/(e*cos(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

2/15/(4*sin(1/2*d*x+1/2*c)^4-4*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^5*(168*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-84*EllipticE(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)^(1/2)*sin(1/2*d*x+1/2*c)^4-168*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)
+84*EllipticE(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)^(1/2)*sin(1/
2*d*x+1/2*c)^2+20*I*sin(1/2*d*x+1/2*c)^3+36*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-21*EllipticE(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)^(1/2)-10*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 = 198, normalized size of antiderivative = 1.21 \[ \int \frac {1}{(e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \, dx=-\frac {2 \, {\left (2 \, \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} {\left (21 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 56 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 47 i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + 21 \, {\left (i \, \sqrt {2} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 i \, \sqrt {2} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, \sqrt {2} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, \sqrt {2}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{15 \, {\left (a^{2} d e^{6} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{2} d e^{6} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{2} d e^{6} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d e^{6}\right )}} \]

[In]

integrate(1/(e*cos(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-2/15*(2*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*(21*I*e^(6*I*d*x + 6*I*c) + 56*I*e^(4*I*d*x + 4*I*c) + 47*I
*e^(2*I*d*x + 2*I*c))*e^(-1/2*I*d*x - 1/2*I*c) + 21*(I*sqrt(2)*e^(6*I*d*x + 6*I*c) + 3*I*sqrt(2)*e^(4*I*d*x +
4*I*c) + 3*I*sqrt(2)*e^(2*I*d*x + 2*I*c) + I*sqrt(2))*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0
, e^(I*d*x + I*c))))/(a^2*d*e^6*e^(6*I*d*x + 6*I*c) + 3*a^2*d*e^6*e^(4*I*d*x + 4*I*c) + 3*a^2*d*e^6*e^(2*I*d*x
+ 2*I*c) + a^2*d*e^6)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(1/(e*cos(d*x+c))**(11/2)/(a+I*a*tan(d*x+c))**2,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(1/(e*cos(d*x+c))^(11/2)/(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]

\[ \int \frac {1}{(e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \, dx=\int { \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {11}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}} \,d x } \]

[In]

integrate(1/(e*cos(d*x+c))^(11/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(1/((e*cos(d*x + c))^(11/2)*(I*a*tan(d*x + c) + a)^2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \, dx=\int \frac {1}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{11/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \]

[In]

int(1/((e*cos(c + d*x))^(11/2)*(a + a*tan(c + d*x)*1i)^2),x)

[Out]

int(1/((e*cos(c + d*x))^(11/2)*(a + a*tan(c + d*x)*1i)^2), x)