3.292 \(\int \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=223 \[ -\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {21 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{80 d}+\frac {9 i a \cos ^3(c+d x)}{40 d \sqrt {a+i a \tan (c+d x)}}-\frac {63 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{128 d}+\frac {21 i a \cos (c+d x)}{64 d \sqrt {a+i a \tan (c+d x)}}+\frac {63 i \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{128 \sqrt {2} d} \]

[Out]

63/256*I*arctanh(1/2*sec(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*a^(1/2)/d*2^(1/2)+21/64*I*a*cos(d*x+
c)/d/(a+I*a*tan(d*x+c))^(1/2)+9/40*I*a*cos(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(1/2)-63/128*I*cos(d*x+c)*(a+I*a*tan(
d*x+c))^(1/2)/d-21/80*I*cos(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)/d-1/5*I*cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(1/2)/d

________________________________________________________________________________________

Rubi [A]  time = 0.39, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ -\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {21 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{80 d}+\frac {9 i a \cos ^3(c+d x)}{40 d \sqrt {a+i a \tan (c+d x)}}-\frac {63 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{128 d}+\frac {21 i a \cos (c+d x)}{64 d \sqrt {a+i a \tan (c+d x)}}+\frac {63 i \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{128 \sqrt {2} d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(((63*I)/128)*Sqrt[a]*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*d) + (((2
1*I)/64)*a*Cos[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((9*I)/40)*a*Cos[c + d*x]^3)/(d*Sqrt[a + I*a*Tan[c
+ d*x]]) - (((63*I)/128)*Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d - (((21*I)/80)*Cos[c + d*x]^3*Sqrt[a + I*a
*Tan[c + d*x]])/d - ((I/5)*Cos[c + d*x]^5*Sqrt[a + I*a*Tan[c + d*x]])/d

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])

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 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 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]

Rubi steps

\begin {align*} \int \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx &=-\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {1}{10} (9 a) \int \frac {\cos ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\\ &=\frac {9 i a \cos ^3(c+d x)}{40 d \sqrt {a+i a \tan (c+d x)}}-\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {63}{80} \int \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\\ &=\frac {9 i a \cos ^3(c+d x)}{40 d \sqrt {a+i a \tan (c+d x)}}-\frac {21 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{80 d}-\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {1}{32} (21 a) \int \frac {\cos (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\\ &=\frac {21 i a \cos (c+d x)}{64 d \sqrt {a+i a \tan (c+d x)}}+\frac {9 i a \cos ^3(c+d x)}{40 d \sqrt {a+i a \tan (c+d x)}}-\frac {21 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{80 d}-\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {63}{128} \int \cos (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\\ &=\frac {21 i a \cos (c+d x)}{64 d \sqrt {a+i a \tan (c+d x)}}+\frac {9 i a \cos ^3(c+d x)}{40 d \sqrt {a+i a \tan (c+d x)}}-\frac {63 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{128 d}-\frac {21 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{80 d}-\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {1}{256} (63 a) \int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\\ &=\frac {21 i a \cos (c+d x)}{64 d \sqrt {a+i a \tan (c+d x)}}+\frac {9 i a \cos ^3(c+d x)}{40 d \sqrt {a+i a \tan (c+d x)}}-\frac {63 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{128 d}-\frac {21 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{80 d}-\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {(63 i a) \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 )}{128 d}\\ &=\frac {63 i \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{128 \sqrt {2} d}+\frac {21 i a \cos (c+d x)}{64 d \sqrt {a+i a \tan (c+d x)}}+\frac {9 i a \cos ^3(c+d x)}{40 d \sqrt {a+i a \tan (c+d x)}}-\frac {63 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{128 d}-\frac {21 i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{80 d}-\frac {i \cos ^5(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.64, size = 152, normalized size = 0.68 \[ -\frac {i e^{-5 i (c+d x)} \left (-95 e^{2 i (c+d x)}+203 e^{4 i (c+d x)}+344 e^{6 i (c+d x)}+64 e^{8 i (c+d x)}+8 e^{10 i (c+d x)}-315 e^{4 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )-10\right ) \sqrt {a+i a \tan (c+d x)}}{1280 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

((-1/1280*I)*(-10 - 95*E^((2*I)*(c + d*x)) + 203*E^((4*I)*(c + d*x)) + 344*E^((6*I)*(c + d*x)) + 64*E^((8*I)*(
c + d*x)) + 8*E^((10*I)*(c + d*x)) - 315*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)))

________________________________________________________________________________________

fricas [A]  time = 0.51, size = 267, normalized size = 1.20 \[ \frac {{\left (315 \, \sqrt {\frac {1}{2}} d \sqrt {-\frac {a}{d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac {63 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {a}{d^{2}}} + i \, a\right )} e^{\left (-i \, d x - i \, c\right )}}{64 \, d}\right ) - 315 \, \sqrt {\frac {1}{2}} d \sqrt {-\frac {a}{d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {63 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {a}{d^{2}}} - i \, a\right )} e^{\left (-i \, d x - i \, c\right )}}{64 \, d}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-8 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 64 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 344 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 203 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 95 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 10 i\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{1280 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/1280*(315*sqrt(1/2)*d*sqrt(-a/d^2)*e^(4*I*d*x + 4*I*c)*log(63/64*(sqrt(2)*sqrt(1/2)*(d*e^(2*I*d*x + 2*I*c) +
d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-a/d^2) + I*a)*e^(-I*d*x - I*c)/d) - 315*sqrt(1/2)*d*sqrt(-a/d^2)*e
^(4*I*d*x + 4*I*c)*log(-63/64*(sqrt(2)*sqrt(1/2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))
*sqrt(-a/d^2) - I*a)*e^(-I*d*x - I*c)/d) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-8*I*e^(10*I*d*x + 10*I*
c) - 64*I*e^(8*I*d*x + 8*I*c) - 344*I*e^(6*I*d*x + 6*I*c) - 203*I*e^(4*I*d*x + 4*I*c) + 95*I*e^(2*I*d*x + 2*I*
c) + 10*I))*e^(-4*I*d*x - 4*I*c)/d

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {i \, a \tan \left (d x + c\right ) + a} \cos \left (d x + c\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(I*a*tan(d*x + c) + a)*cos(d*x + c)^5, x)

________________________________________________________________________________________

maple [B]  time = 1.32, size = 913, normalized size = 4.09 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

-1/20480/d*(1344*I*cos(d*x+c)^7-315*cos(d*x+c)^4*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arctan(1/2*(-
2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)+512*I*cos(d*x+c)^9-1260*cos(d*x+c)^3*sin(d*x+c)*(-2*cos(d*
x+c)/(1+cos(d*x+c)))^(9/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)+3360*I*cos(d*x+c)^
6-1890*cos(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)*2^(1/2))*2^(1/2)-10080*I*cos(d*x+c)^5-1260*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*ar
ctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)+1890*I*cos(d*x+c)^2*sin(d*x+c)*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)))^(9/2)*2^(1/2)-3
15*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*2^(1/2)*sin(d
*x+c)+768*I*cos(d*x+c)^8+315*I*cos(d*x+c)^4*sin(d*x+c)*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)))^(9/2)*2^(1/2)-4096*sin(d*x+c)*cos(d*x+c)^9+1260*I*cos(
d*x+c)^3*sin(d*x+c)*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)))^(9/2)*2^(1/2)+4608*sin(d*x+c)*cos(d*x+c)^8+1260*I*cos(d*x+c)*sin(d*x+c)*arctanh(1/2*(-2*c
os(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)))^(9/2)*2^(1/2)-53
76*sin(d*x+c)*cos(d*x+c)^7+4096*I*cos(d*x+c)^10+6720*sin(d*x+c)*cos(d*x+c)^6+315*I*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)))^(9/2)*2^(1/2)*sin(d*x+c)-1
0080*cos(d*x+c)^5*sin(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)^4

________________________________________________________________________________________

maxima [B]  time = 2.00, size = 2215, normalized size = 9.93 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-1/5120*((cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x
+ 4*c)))^2 + 2*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)^(3/4)*((60*I*sqrt(2)*cos(4*d*x + 4*c
) + 60*sqrt(2)*sin(4*d*x + 4*c) + 160*I*sqrt(2))*cos(3/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x +
4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)) - 20*(3*sqrt(2)*cos(4*d*x + 4*c) - 3*I*sqr
t(2)*sin(4*d*x + 4*c) + 8*sqrt(2))*sin(3/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1
/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)))*sqrt(a) + (cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x
+ 4*c)))^2 + sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(
4*d*x + 4*c))) + 1)^(1/4)*((32*I*sqrt(2)*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 32*I*sqrt(2)
*sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 64*I*sqrt(2)*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4
*d*x + 4*c))) + 32*I*sqrt(2))*cos(5/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*ar
ctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)) + (-100*I*sqrt(2)*cos(4*d*x + 4*c) - 400*I*sqrt(2)*cos(1/2*ar
ctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) - 100*sqrt(2)*sin(4*d*x + 4*c) - 400*sqrt(2)*sin(1/2*arctan2(sin(4*
d*x + 4*c), cos(4*d*x + 4*c))) + 960*I*sqrt(2))*cos(1/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x +
4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)) - 32*(sqrt(2)*cos(1/2*arctan2(sin(4*d*x + 4
*c), cos(4*d*x + 4*c)))^2 + sqrt(2)*sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*sqrt(2)*cos(1/2
*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + sqrt(2))*sin(5/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos
(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)) + (100*sqrt(2)*cos(4*d*x + 4*c) +
400*sqrt(2)*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) - 100*I*sqrt(2)*sin(4*d*x + 4*c) - 400*I*sqrt
(2)*sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) - 960*sqrt(2))*sin(1/2*arctan2(sin(1/2*arctan2(sin(4*
d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)))*sqrt(a) + (630*sqr
t(2)*arctan2((cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4
*d*x + 4*c)))^2 + 2*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)^(1/4)*sin(1/2*arctan2(sin(1/2*ar
ctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)), (cos(1
/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2
*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)^(1/4)*cos(1/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4
*c), cos(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)) + 1) - 630*sqrt(2)*arctan2
((cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))
)^2 + 2*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)^(1/4)*sin(1/2*arctan2(sin(1/2*arctan2(sin(4*
d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)), (cos(1/2*arctan2(s
in(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*arc
tan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)^(1/4)*cos(1/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d
*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)) - 1) - 315*I*sqrt(2)*log(sqrt(cos(1/2*
arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*co
s(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)*cos(1/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(
4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1))^2 + sqrt(cos(1/2*arctan2(sin(4*d*x
+ 4*c), cos(4*d*x + 4*c)))^2 + sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*arctan2(sin(
4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)*sin(1/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), co
s(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1))^2 + 2*(cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x +
4*c)))^2 + sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*
d*x + 4*c))) + 1)^(1/4)*cos(1/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*arctan2(
sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)) + 1) + 315*I*sqrt(2)*log(sqrt(cos(1/2*arctan2(sin(4*d*x + 4*c), cos
(4*d*x + 4*c)))^2 + sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*arctan2(sin(4*d*x + 4*c
), cos(4*d*x + 4*c))) + 1)*cos(1/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*arcta
n2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1))^2 + sqrt(cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2
+ sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c
))) + 1)*sin(1/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c
), cos(4*d*x + 4*c))) + 1))^2 - 2*(cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + sin(1/2*arctan2(si
n(4*d*x + 4*c), cos(4*d*x + 4*c)))^2 + 2*cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))) + 1)^(1/4)*cos(1
/2*arctan2(sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))), cos(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x +
4*c))) + 1)) + 1))*sqrt(a))/d

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^5\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^(1/2),x)

[Out]

int(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^(1/2), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________