3.4.32 \(\int \cos ^{11}(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\) [332]

Optimal. Leaf size=342 \[ \frac {195 i a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{1024 \sqrt {2} d}+\frac {65 i a^4 \cos (c+d x)}{512 d \sqrt {a+i a \tan (c+d x)}}+\frac {39 i a^4 \cos ^3(c+d x)}{448 d \sqrt {a+i a \tan (c+d x)}}-\frac {195 i a^3 \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{1024 d}-\frac {13 i a^3 \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{128 d}-\frac {13 i a^3 \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{168 d}-\frac {65 i a^2 \cos ^7(c+d x) (a+i a \tan (c+d x))^{3/2}}{924 d}-\frac {5 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^{5/2}}{66 d}-\frac {i \cos ^{11}(c+d x) (a+i a \tan (c+d x))^{7/2}}{11 d} \]

[Out]

195/2048*I*a^(7/2)*arctanh(1/2*sec(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d*2^(1/2)+65/512*I*a^4*cos
(d*x+c)/d/(a+I*a*tan(d*x+c))^(1/2)+39/448*I*a^4*cos(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(1/2)-195/1024*I*a^3*cos(d*x
+c)*(a+I*a*tan(d*x+c))^(1/2)/d-13/128*I*a^3*cos(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)/d-13/168*I*a^3*cos(d*x+c)^5*
(a+I*a*tan(d*x+c))^(1/2)/d-65/924*I*a^2*cos(d*x+c)^7*(a+I*a*tan(d*x+c))^(3/2)/d-5/66*I*a*cos(d*x+c)^9*(a+I*a*t
an(d*x+c))^(5/2)/d-1/11*I*cos(d*x+c)^11*(a+I*a*tan(d*x+c))^(7/2)/d

________________________________________________________________________________________

Rubi [A]
time = 0.40, antiderivative size = 342, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3578, 3583, 3571, 3570, 212} \begin {gather*} \frac {195 i a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{1024 \sqrt {2} d}+\frac {39 i a^4 \cos ^3(c+d x)}{448 d \sqrt {a+i a \tan (c+d x)}}+\frac {65 i a^4 \cos (c+d x)}{512 d \sqrt {a+i a \tan (c+d x)}}-\frac {13 i a^3 \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{168 d}-\frac {13 i a^3 \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{128 d}-\frac {195 i a^3 \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{1024 d}-\frac {65 i a^2 \cos ^7(c+d x) (a+i a \tan (c+d x))^{3/2}}{924 d}-\frac {i \cos ^{11}(c+d x) (a+i a \tan (c+d x))^{7/2}}{11 d}-\frac {5 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^{5/2}}{66 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^11*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(((195*I)/1024)*a^(7/2)*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*d) + ((
(65*I)/512)*a^4*Cos[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((39*I)/448)*a^4*Cos[c + d*x]^3)/(d*Sqrt[a + I
*a*Tan[c + d*x]]) - (((195*I)/1024)*a^3*Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d - (((13*I)/128)*a^3*Cos[c +
d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d - (((13*I)/168)*a^3*Cos[c + d*x]^5*Sqrt[a + I*a*Tan[c + d*x]])/d - (((65
*I)/924)*a^2*Cos[c + d*x]^7*(a + I*a*Tan[c + d*x])^(3/2))/d - (((5*I)/66)*a*Cos[c + d*x]^9*(a + I*a*Tan[c + d*
x])^(5/2))/d - ((I/11)*Cos[c + d*x]^11*(a + I*a*Tan[c + d*x])^(7/2))/d

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3570

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2*(a/(b*f)), Subst[
Int[1/(2 - a*x^2), x], x, Sec[e + f*x]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 + b^
2, 0]

Rule 3571

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*
Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/(a*f*m)), x] + Dist[a/(2*d^2), Int[(d*Sec[e + f*x])^(m + 2)*(a + b*Tan
[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && EqQ[m/2 + n, 0] && GtQ[n, 0]

Rule 3578

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*S
ec[e + f*x])^m*((a + b*Tan[e + f*x])^n/(a*f*m)), x] + Dist[a*((m + n)/(m*d^2)), Int[(d*Sec[e + f*x])^(m + 2)*(
a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 0] && LtQ[m, -
1] && IntegersQ[2*m, 2*n]

Rule 3583

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]

Rubi steps

\begin {align*} \int \cos ^{11}(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx &=-\frac {i \cos ^{11}(c+d x) (a+i a \tan (c+d x))^{7/2}}{11 d}+\frac {1}{22} (15 a) \int \cos ^9(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\\ &=-\frac {5 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^{5/2}}{66 d}-\frac {i \cos ^{11}(c+d x) (a+i a \tan (c+d x))^{7/2}}{11 d}+\frac {1}{132} \left (65 a^2\right ) \int \cos ^7(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\\ &=-\frac {65 i a^2 \cos ^7(c+d x) (a+i a \tan (c+d x))^{3/2}}{924 d}-\frac {5 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^{5/2}}{66 d}-\frac {i \cos ^{11}(c+d x) (a+i a \tan (c+d x))^{7/2}}{11 d}+\frac {1}{168} \left (65 a^3\right ) \int \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\\ &=-\frac {13 i a^3 \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{168 d}-\frac {65 i a^2 \cos ^7(c+d x) (a+i a \tan (c+d x))^{3/2}}{924 d}-\frac {5 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^{5/2}}{66 d}-\frac {i \cos ^{11}(c+d x) (a+i a \tan (c+d x))^{7/2}}{11 d}+\frac {1}{112} \left (39 a^4\right ) \int \frac {\cos ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\\ &=\frac {39 i a^4 \cos ^3(c+d x)}{448 d \sqrt {a+i a \tan (c+d x)}}-\frac {13 i a^3 \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{168 d}-\frac {65 i a^2 \cos ^7(c+d x) (a+i a \tan (c+d x))^{3/2}}{924 d}-\frac {5 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^{5/2}}{66 d}-\frac {i \cos ^{11}(c+d x) (a+i a \tan (c+d x))^{7/2}}{11 d}+\frac {1}{128} \left (39 a^3\right ) \int \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\\ &=\frac {39 i a^4 \cos ^3(c+d x)}{448 d \sqrt {a+i a \tan (c+d x)}}-\frac {13 i a^3 \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{128 d}-\frac {13 i a^3 \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{168 d}-\frac {65 i a^2 \cos ^7(c+d x) (a+i a \tan (c+d x))^{3/2}}{924 d}-\frac {5 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^{5/2}}{66 d}-\frac {i \cos ^{11}(c+d x) (a+i a \tan (c+d x))^{7/2}}{11 d}+\frac {1}{256} \left (65 a^4\right ) \int \frac {\cos (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\\ &=\frac {65 i a^4 \cos (c+d x)}{512 d \sqrt {a+i a \tan (c+d x)}}+\frac {39 i a^4 \cos ^3(c+d x)}{448 d \sqrt {a+i a \tan (c+d x)}}-\frac {13 i a^3 \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{128 d}-\frac {13 i a^3 \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{168 d}-\frac {65 i a^2 \cos ^7(c+d x) (a+i a \tan (c+d x))^{3/2}}{924 d}-\frac {5 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^{5/2}}{66 d}-\frac {i \cos ^{11}(c+d x) (a+i a \tan (c+d x))^{7/2}}{11 d}+\frac {\left (195 a^3\right ) \int \cos (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx}{1024}\\ &=\frac {65 i a^4 \cos (c+d x)}{512 d \sqrt {a+i a \tan (c+d x)}}+\frac {39 i a^4 \cos ^3(c+d x)}{448 d \sqrt {a+i a \tan (c+d x)}}-\frac {195 i a^3 \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{1024 d}-\frac {13 i a^3 \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{128 d}-\frac {13 i a^3 \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{168 d}-\frac {65 i a^2 \cos ^7(c+d x) (a+i a \tan (c+d x))^{3/2}}{924 d}-\frac {5 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^{5/2}}{66 d}-\frac {i \cos ^{11}(c+d x) (a+i a \tan (c+d x))^{7/2}}{11 d}+\frac {\left (195 a^4\right ) \int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{2048}\\ &=\frac {65 i a^4 \cos (c+d x)}{512 d \sqrt {a+i a \tan (c+d x)}}+\frac {39 i a^4 \cos ^3(c+d x)}{448 d \sqrt {a+i a \tan (c+d x)}}-\frac {195 i a^3 \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{1024 d}-\frac {13 i a^3 \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{128 d}-\frac {13 i a^3 \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{168 d}-\frac {65 i a^2 \cos ^7(c+d x) (a+i a \tan (c+d x))^{3/2}}{924 d}-\frac {5 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^{5/2}}{66 d}-\frac {i \cos ^{11}(c+d x) (a+i a \tan (c+d x))^{7/2}}{11 d}+\frac {\left (195 i a^4\right ) \text {Subst}\left (\int \frac {1}{2-a x^2} \, dx,x,\frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}}\right )}{1024 d}\\ &=\frac {195 i a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{1024 \sqrt {2} d}+\frac {65 i a^4 \cos (c+d x)}{512 d \sqrt {a+i a \tan (c+d x)}}+\frac {39 i a^4 \cos ^3(c+d x)}{448 d \sqrt {a+i a \tan (c+d x)}}-\frac {195 i a^3 \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{1024 d}-\frac {13 i a^3 \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{128 d}-\frac {13 i a^3 \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{168 d}-\frac {65 i a^2 \cos ^7(c+d x) (a+i a \tan (c+d x))^{3/2}}{924 d}-\frac {5 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^{5/2}}{66 d}-\frac {i \cos ^{11}(c+d x) (a+i a \tan (c+d x))^{7/2}}{11 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 6.98, size = 194, normalized size = 0.57 \begin {gather*} -\frac {i a^3 e^{-5 i (c+d x)} \left (-462-7161 e^{2 i (c+d x)}+47413 e^{4 i (c+d x)}+78800 e^{6 i (c+d x)}+38512 e^{8 i (c+d x)}+19552 e^{10 i (c+d x)}+7184 e^{12 i (c+d x)}+1624 e^{14 i (c+d x)}+168 e^{16 i (c+d x)}-45045 e^{4 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \sqrt {a+i a \tan (c+d x)}}{473088 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^11*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

((-1/473088*I)*a^3*(-462 - 7161*E^((2*I)*(c + d*x)) + 47413*E^((4*I)*(c + d*x)) + 78800*E^((6*I)*(c + d*x)) +
38512*E^((8*I)*(c + d*x)) + 19552*E^((10*I)*(c + d*x)) + 7184*E^((12*I)*(c + d*x)) + 1624*E^((14*I)*(c + d*x))
+ 168*E^((16*I)*(c + d*x)) - 45045*E^((4*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I
)*(c + d*x))]])*Sqrt[a + I*a*Tan[c + d*x]])/(d*E^((5*I)*(c + d*x)))

________________________________________________________________________________________

Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1947 vs. \(2 (281 ) = 562\).
time = 1.80, size = 1948, normalized size = 5.70

method	result	size
default	\(\text {Expression too large to display}\)	\(1948\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^11*(a+I*a*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/484442112/d*(-352321536*I*cos(d*x+c)^22+176160768*I*cos(d*x+c)^21+205520896*I*cos(d*x+c)^20-58720256*I*cos(d
*x+c)^19-2097152*I*cos(d*x+c)^18-2621440*I*cos(d*x+c)^17-3407872*I*cos(d*x+c)^16-4685824*I*cos(d*x+c)^15-70287
36*I*cos(d*x+c)^14+352321536*sin(d*x+c)*cos(d*x+c)^21-176160768*sin(d*x+c)*cos(d*x+c)^20-29360128*sin(d*x+c)*c
os(d*x+c)^19-29360128*sin(d*x+c)*cos(d*x+c)^18+450450*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(21/2)*arctan(1/2*(-2*cos
(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin(d*x+c)*cos(d*x+c)*2^(1/2)-45045*I*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c
)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(21/2)*sin(d*x+c)+314572
80*cos(d*x+c)^17*sin(d*x+c)-34078720*cos(d*x+c)^16*sin(d*x+c)[phone]*cos(d*x+c)^15*sin(d*x+c)+45045*(-2*cos(
d*x+c)/(1+cos(d*x+c)))^(21/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin(d*x+c)*cos(d*x+c)^1
0*2^(1/2)+450450*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(21/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)
)*sin(d*x+c)*cos(d*x+c)^9*2^(1/2)+2027025*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(21/2)*arctan(1/2*(-2*cos(d*x+c)/(1+c
os(d*x+c)))^(1/2)*2^(1/2))*sin(d*x+c)*cos(d*x+c)^8*2^(1/2)+5405400*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(21/2)*arcta
n(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin(d*x+c)*cos(d*x+c)^7*2^(1/2)+9459450*(-2*cos(d*x+c)/(1+
cos(d*x+c)))^(21/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin(d*x+c)*cos(d*x+c)^6*2^(1/2)+1
1351340*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(21/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin(d*x
+c)*cos(d*x+c)^5*2^(1/2)+9459450*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(21/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)
))^(1/2)*2^(1/2))*sin(d*x+c)*cos(d*x+c)^4*2^(1/2)+5405400*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(21/2)*arctan(1/2*(-2
*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin(d*x+c)*cos(d*x+c)^3*2^(1/2)+2027025*(-2*cos(d*x+c)/(1+cos(d*x+c
)))^(21/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin(d*x+c)*cos(d*x+c)^2*2^(1/2)-42172416*s
in(d*x+c)*cos(d*x+c)^14-9459450*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(21/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+
c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)*cos(d*x+c)^6*2^(1/2)-11351340*I*(-2*cos(d*x+c)/(1+cos(d*x
+c)))^(21/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)*cos(d*
x+c)^5*2^(1/2)-9459450*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(21/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2
)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)*cos(d*x+c)^4*2^(1/2)-5405400*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(21/
2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)*cos(d*x+c)^3*2^(
1/2)-2027025*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(21/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+
c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)*cos(d*x+c)^2*2^(1/2)-450450*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(21/2)*arctanh(
1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)*cos(d*x+c)*2^(1/2)-45045*I*
(-2*cos(d*x+c)/(1+cos(d*x+c)))^(21/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2
^(1/2))*sin(d*x+c)*cos(d*x+c)^10*2^(1/2)-450450*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(21/2)*arctanh(1/2*(-2*cos(d*
x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)*cos(d*x+c)^9*2^(1/2)-2027025*I*(-2*cos(d*
x+c)/(1+cos(d*x+c)))^(21/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*si
n(d*x+c)*cos(d*x+c)^8*2^(1/2)-5405400*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(21/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)*cos(d*x+c)^7*2^(1/2)-12300288*I*cos(d*x+c)^13-30750
720*I*cos(d*x+c)^12+92252160*I*cos(d*x+c)^11-61501440*sin(d*x+c)*cos(d*x+c)^12+45045*2^(1/2)*arctan(1/2*(-2*co
s(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(21/2)*sin(d*x+c)[phone]*cos(d*x+c)^1
3*sin(d*x+c)[phone]*sin(d*x+c)*cos(d*x+c)^11)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/(I*sin(d*x+c)+c
os(d*x+c)-1)/cos(d*x+c)^10*a^3

________________________________________________________________________________________

Maxima [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(cos(d*x+c)^11*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [A]
time = 0.44, size = 342, normalized size = 1.00 \begin {gather*} -\frac {{\left (45045 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} d e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {195 \, {\left (-i \, a^{4} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{512 \, d}\right ) - 45045 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} d e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {195 \, {\left (-i \, a^{4} - \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{512 \, d}\right ) - \sqrt {2} {\left (-168 i \, a^{3} e^{\left (16 i \, d x + 16 i \, c\right )} - 1624 i \, a^{3} e^{\left (14 i \, d x + 14 i \, c\right )} - 7184 i \, a^{3} e^{\left (12 i \, d x + 12 i \, c\right )} - 19552 i \, a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} - 38512 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 78800 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 47413 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 7161 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 462 i \, a^{3}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{473088 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^11*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

-1/473088*(45045*sqrt(1/2)*sqrt(-a^7/d^2)*d*e^(4*I*d*x + 4*I*c)*log(-195/512*(-I*a^4 + sqrt(2)*sqrt(1/2)*sqrt(
-a^7/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/d) - 45045*sqrt(1/2)
*sqrt(-a^7/d^2)*d*e^(4*I*d*x + 4*I*c)*log(-195/512*(-I*a^4 - sqrt(2)*sqrt(1/2)*sqrt(-a^7/d^2)*(d*e^(2*I*d*x +
2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/d) - sqrt(2)*(-168*I*a^3*e^(16*I*d*x + 16*I*c)
- 1624*I*a^3*e^(14*I*d*x + 14*I*c) - 7184*I*a^3*e^(12*I*d*x + 12*I*c) - 19552*I*a^3*e^(10*I*d*x + 10*I*c) - 3
8512*I*a^3*e^(8*I*d*x + 8*I*c) - 78800*I*a^3*e^(6*I*d*x + 6*I*c) - 47413*I*a^3*e^(4*I*d*x + 4*I*c) + 7161*I*a^
3*e^(2*I*d*x + 2*I*c) + 462*I*a^3)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-4*I*d*x - 4*I*c)/d

________________________________________________________________________________________

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(cos(d*x+c)**11*(a+I*a*tan(d*x+c))**(7/2),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(cos(d*x+c)^11*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(7/2)*cos(d*x + c)^11, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\cos \left (c+d\,x\right )}^{11}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^11*(a + a*tan(c + d*x)*1i)^(7/2),x)

[Out]

int(cos(c + d*x)^11*(a + a*tan(c + d*x)*1i)^(7/2), x)

________________________________________________________________________________________