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

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

[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]
time = 0.14, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3596, 3581, 3853, 3856, 2719} \begin {gather*} -\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}} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {1}{(e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \, dx &=\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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.10, size = 406, normalized size = 2.48 \begin {gather*} \frac {2 \sqrt {2} e^{3 i c+2 i d x} \cos ^{\frac {7}{2}}(c+d x) \csc (c) \left (-42 \sqrt {2-2 i e^{i (c+d x)}} \sqrt {e^{i (c+d x)} \left (-i+e^{i (c+d x)}\right )} \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\text {ArcSin}\left (\sqrt {-i \cos (c+d x)+\sin (c+d x)}\right )\right |-1\right )+42 \sqrt {2-2 i e^{i (c+d x)}} \sqrt {e^{i (c+d x)} \left (-i+e^{i (c+d x)}\right )} \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\text {ArcSin}\left (\sqrt {-i \cos (c+d x)+\sin (c+d x)}\right )\right |-1\right )-\frac {1}{2} e^{-2 i c} \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (\left (-1+e^{2 i c}\right ) \left (47+56 e^{2 i (c+d x)}+21 e^{4 i (c+d x)}\right )+7 \left (1+e^{2 i (c+d x)}\right )^{5/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )\right )\right ) (\cos (2 c)+i \sin (2 c)) (\cos (d x)+i \sin (d x))^2}{15 d \left (1+e^{2 i (c+d x)}\right )^3 (e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[2]*E^((3*I)*c + (2*I)*d*x)*Cos[c + d*x]^(7/2)*Csc[c]*(-42*Sqrt[2 - (2*I)*E^(I*(c + d*x))]*Sqrt[E^(I*(c
+ d*x))*(-I + E^(I*(c + d*x)))]*Cos[c + d*x]^(5/2)*EllipticE[ArcSin[Sqrt[(-I)*Cos[c + d*x] + Sin[c + d*x]]],
-1] + 42*Sqrt[2 - (2*I)*E^(I*(c + d*x))]*Sqrt[E^(I*(c + d*x))*(-I + E^(I*(c + d*x)))]*Cos[c + d*x]^(5/2)*Ellip
ticF[ArcSin[Sqrt[(-I)*Cos[c + d*x] + Sin[c + d*x]]], -1] - (Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]*((
-1 + E^((2*I)*c))*(47 + 56*E^((2*I)*(c + d*x)) + 21*E^((4*I)*(c + d*x))) + 7*(1 + E^((2*I)*(c + d*x)))^(5/2)*H
ypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/(2*E^((2*I)*c)))*(Cos[2*c] + I*Sin[2*c])*(Cos[d*x] + I
*Sin[d*x])^2)/(15*d*(1 + E^((2*I)*(c + d*x)))^3*(e*Cos[c + d*x])^(11/2)*(a + I*a*Tan[c + d*x])^2)

________________________________________________________________________________________

Maple [A]
time = 2.56, size = 321, normalized size = 1.96

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

Verification of antiderivative is not currently implemented for this CAS.

[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*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-84*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(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)^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))*(sin(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+20*I*sin(1/2*d*x+1/2*c)^3+36*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-21*(sin(1/2*d*x+1/2*c)^2)^
(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-10*I*sin(1/2*d*x+1/2*c))/d

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.09, size = 186, normalized size = 1.13 \begin {gather*} -\frac {2 \, {\left (2 \, \sqrt {\frac {1}{2}} {\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 )} \sqrt {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 )} + 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 )} {\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^{\frac {11}{2}} + a^{2} d e^{\left (6 i \, d x + 6 i \, c + \frac {11}{2}\right )} + 3 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c + \frac {11}{2}\right )} + 3 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c + \frac {11}{2}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)*(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))*sqrt(e^(2*
I*d*x + 2*I*c) + 1)*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))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, e^(I*d*x
+ I*c))))/(a^2*d*e^(11/2) + a^2*d*e^(6*I*d*x + 6*I*c + 11/2) + 3*a^2*d*e^(4*I*d*x + 4*I*c + 11/2) + 3*a^2*d*e
^(2*I*d*x + 2*I*c + 11/2))

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________