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

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

[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

________________________________________________________________________________________

Rubi [A]  time = 0.56461, antiderivative size = 342, normalized size of antiderivative = 1., 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 = {3497, 3502, 3490, 3489, 206} \[ -\frac{65 i a^2 \cos ^7(c+d x) (a+i a \tan (c+d x))^{3/2}}{924 d}-\frac{13 i a^3 \cos ^5(c+d x) \sqrt{a+i a \tan (c+d x)}}{168 d}-\frac{13 i a^3 \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{128 d}+\frac{39 i a^4 \cos ^3(c+d x)}{448 d \sqrt{a+i a \tan (c+d x)}}-\frac{195 i a^3 \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{1024 d}+\frac{65 i a^4 \cos (c+d x)}{512 d \sqrt{a+i a \tan (c+d x)}}+\frac{195 i a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{1024 \sqrt{2} d}-\frac{i \cos ^{11}(c+d x) (a+i a \tan (c+d x))^{7/2}}{11 d}-\frac{5 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^{5/2}}{66 d} \]

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 3497

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*(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 3502

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]

Rule 3490

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 3489

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 206

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

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 ) \operatorname{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 = 6.32674, size = 194, normalized size = 0.57 \[ -\frac{i a^3 e^{-5 i (c+d x)} \left (-7161 e^{2 i (c+d x)}+47413 e^{4 i (c+d x)}+78800 e^{6 i (c+d x)}+38512 e^{8 i (c+d x)}+19552 e^{10 i (c+d x)}+7184 e^{12 i (c+d x)}+1624 e^{14 i (c+d x)}+168 e^{16 i (c+d x)}-45045 e^{4 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )-462\right ) \sqrt{a+i a \tan (c+d x)}}{473088 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/473088)*a^3*(-462 - 7161*E^((2*I)*(c + d*x)) + 47413*E^((4*I)*(c + d*x)) + 78800*E^((6*I)*(c + d*x)) + 38
512*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]  time = 0.816, size = 1948, normalized size = 5.7 \begin{align*} \text{result too large to display} \end{align*}

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)

[Out]

-1/484442112/d*a^3*(5405400*I*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*
(-2*cos(d*x+c)/(cos(d*x+c)+1))^(21/2)*2^(1/2)*sin(d*x+c)*cos(d*x+c)^7-37486592*sin(d*x+c)*cos(d*x+c)^15+421724
16*sin(d*x+c)*cos(d*x+c)^14-49201152*sin(d*x+c)*cos(d*x+c)^13-31457280*sin(d*x+c)*cos(d*x+c)^17+61501440*sin(d
*x+c)*cos(d*x+c)^12-92252160*sin(d*x+c)*cos(d*x+c)^11+11351340*I*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c
)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(21/2)*2^(1/2)*sin(d*x+c)*cos(d*x+c)^5+94594
50*I*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c
)+1))^(21/2)*2^(1/2)*sin(d*x+c)*cos(d*x+c)^4+5405400*I*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2
)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(21/2)*2^(1/2)*sin(d*x+c)*cos(d*x+c)^3+2027025*I*arcta
nh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(21/
2)*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2+450450*I*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c
)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(21/2)*2^(1/2)*sin(d*x+c)*cos(d*x+c)+45045*I*arctanh(1/2*2^(1/2)*
(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(21/2)*2^(1/2)*sin(
d*x+c)*cos(d*x+c)^10+450450*I*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*
(-2*cos(d*x+c)/(cos(d*x+c)+1))^(21/2)*2^(1/2)*sin(d*x+c)*cos(d*x+c)^9+2027025*I*arctanh(1/2*2^(1/2)*(-2*cos(d*
x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(21/2)*2^(1/2)*sin(d*x+c)*cos
(d*x+c)^8+9459450*I*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*
x+c)/(cos(d*x+c)+1))^(21/2)*2^(1/2)*sin(d*x+c)*cos(d*x+c)^6-45045*2^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(c
os(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(21/2)*sin(d*x+c)[phone]*I*cos(d*x+c)^22-176160768*I*co
s(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+7028736*I*cos(d*x+c)^14+12300288*I*cos(d*x+c)^13+30750720*I*c
os(d*x+c)^12-92252160*I*cos(d*x+c)^11-352321536*sin(d*x+c)*cos(d*x+c)^21+176160768*sin(d*x+c)*cos(d*x+c)^20+29
360128*sin(d*x+c)*cos(d*x+c)^19+29360128*sin(d*x+c)*cos(d*x+c)^18+34078720*sin(d*x+c)*cos(d*x+c)^16+45045*I*2^
(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+
c)+1))^(21/2)*sin(d*x+c)-45045*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(21/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*
x+c)+1))^(1/2))*2^(1/2)*sin(d*x+c)*cos(d*x+c)^10-450450*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(21/2)*arctan(1/2*2^(1/
2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)*sin(d*x+c)*cos(d*x+c)^9-2027025*(-2*cos(d*x+c)/(cos(d*x+c)+1)
)^(21/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)*sin(d*x+c)*cos(d*x+c)^8-5405400*(-2*
cos(d*x+c)/(cos(d*x+c)+1))^(21/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)*sin(d*x+c)*
cos(d*x+c)^7-9459450*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(21/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(
1/2))*2^(1/2)*sin(d*x+c)*cos(d*x+c)^6-11351340*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(21/2)*arctan(1/2*2^(1/2)*(-2*co
s(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)*sin(d*x+c)*cos(d*x+c)^5-9459450*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(21/2)*
arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)*sin(d*x+c)*cos(d*x+c)^4-5405400*(-2*cos(d*x+c
)/(cos(d*x+c)+1))^(21/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)*sin(d*x+c)*cos(d*x+c
)^3-2027025*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(21/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(
1/2)*sin(d*x+c)*cos(d*x+c)^2-450450*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(21/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(c
os(d*x+c)+1))^(1/2))*2^(1/2)*sin(d*x+c)*cos(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/(I*sin(d*x+
c)+cos(d*x+c)-1)/cos(d*x+c)^10

________________________________________________________________________________________

Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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 = 2.26877, size = 1149, normalized size = 3.36 \begin{align*} -\frac{{\left (45045 \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{7}}{d^{2}}} d e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac{{\left (390 i \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{7}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} + 195 \, \sqrt{2}{\left (a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{195 \, a^{3}}\right ) - 45045 \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{7}}{d^{2}}} d e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac{{\left (-390 i \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{7}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} + 195 \, \sqrt{2}{\left (a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{195 \, a^{3}}\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}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{473088 \, d} \end{align*}

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^(5*I*d*x + 5*I*c)*log(1/195*(390*I*sqrt(1/2)*sqrt(-a^7/d^2)*d*e^
(I*d*x + I*c) + 195*sqrt(2)*(a^3*e^(2*I*d*x + 2*I*c) + a^3)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))
*e^(-I*d*x - I*c)/a^3) - 45045*sqrt(1/2)*sqrt(-a^7/d^2)*d*e^(5*I*d*x + 5*I*c)*log(1/195*(-390*I*sqrt(1/2)*sqrt
(-a^7/d^2)*d*e^(I*d*x + I*c) + 195*sqrt(2)*(a^3*e^(2*I*d*x + 2*I*c) + a^3)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e
^(I*d*x + I*c))*e^(-I*d*x - I*c)/a^3) - sqrt(2)*(-168*I*a^3*e^(16*I*d*x + 16*I*c) - 1624*I*a^3*e^(14*I*d*x + 1
4*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) - 38512*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^(I*d*x + I*c))*e^(-5*I*d*x - 5*I*c)/d

________________________________________________________________________________________

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

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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out