3.317 \(\int \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\)
Optimal. Leaf size=159 \[ \frac {i a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{4 \sqrt {2} d}-\frac {i a^2 \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac {i a \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{6 d} \]
[Out]
1/8*I*a^(5/2)*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)-1/4*I*a^2*cos(d*x+c)*
(a+I*a*tan(d*x+c))^(1/2)/d-1/6*I*a*cos(d*x+c)^3*(a+I*a*tan(d*x+c))^(3/2)/d-1/5*I*cos(d*x+c)^5*(a+I*a*tan(d*x+c
))^(5/2)/d
________________________________________________________________________________________
Rubi [A] time = 0.22, antiderivative size = 159, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used =
{3490, 3489, 206} \[ -\frac {i a^2 \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{4 \sqrt {2} d}-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac {i a \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{6 d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^(5/2),x]
[Out]
((I/4)*a^(5/2)*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*d) - ((I/4)*a^2*
Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d - ((I/6)*a*Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^(3/2))/d - ((I/5)*
Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^(5/2))/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]
Rubi steps
\begin {align*} \int \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx &=-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}+\frac {1}{2} a \int \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\\ &=-\frac {i a \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{6 d}-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}+\frac {1}{4} a^2 \int \cos (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\\ &=-\frac {i a^2 \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {i a \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{6 d}-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}+\frac {1}{8} a^3 \int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\\ &=-\frac {i a^2 \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {i a \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{6 d}-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}+\frac {\left (i a^3\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 )}{4 d}\\ &=\frac {i a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{4 \sqrt {2} d}-\frac {i a^2 \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {i a \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{6 d}-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 1.12, size = 118, normalized size = 0.74 \[ -\frac {i a^2 e^{-i (c+d x)} \left (34 e^{2 i (c+d x)}+14 e^{4 i (c+d x)}+3 e^{6 i (c+d x)}-15 \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )+23\right ) \sqrt {a+i a \tan (c+d x)}}{120 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^(5/2),x]
[Out]
((-1/120*I)*a^2*(23 + 34*E^((2*I)*(c + d*x)) + 14*E^((4*I)*(c + d*x)) + 3*E^((6*I)*(c + d*x)) - 15*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^(I*(c + d*x)))
________________________________________________________________________________________
fricas [B] time = 0.73, size = 244, normalized size = 1.53 \[ \frac {15 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} d \log \left (\frac {{\left (i \, a^{3} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, d}\right ) - 15 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} d \log \left (\frac {{\left (i \, a^{3} - \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, d}\right ) + \sqrt {2} {\left (-3 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 14 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 34 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 23 i \, a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{120 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
1/120*(15*sqrt(1/2)*sqrt(-a^5/d^2)*d*log(1/2*(I*a^3 + sqrt(2)*sqrt(1/2)*sqrt(-a^5/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) - 15*sqrt(1/2)*sqrt(-a^5/d^2)*d*log(1/2*(I*a^3 - s
qrt(2)*sqrt(1/2)*sqrt(-a^5/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)*(-3*I*a^2*e^(6*I*d*x + 6*I*c) - 14*I*a^2*e^(4*I*d*x + 4*I*c) - 34*I*a^2*e^(2*I*d*x + 2*I*c) - 2
3*I*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/d
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{5}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((I*a*tan(d*x + c) + a)^(5/2)*cos(d*x + c)^5, x)
________________________________________________________________________________________
maple [B] time = 1.46, size = 916, normalized size = 5.76 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(5/2),x)
[Out]
-1/1920/d*(-768*I*cos(d*x+c)^9+64*I*cos(d*x+c)^7-15*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)+60*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)*cos(d*x+c)-
60*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)+90*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*co
s(d*x+c)/(1+cos(d*x+c)))^(9/2)*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2-90*cos(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(1+co
s(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)+160*I*cos(d*x+c)^6-60*cos(d*
x+c)*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)-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)+15*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)))^(9/2)*sin(d*x+c)-480*I*cos(d*x+c)^5-1536*sin(d*x+c)*cos(d*x+c)^9+60*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)*cos(d*x+c)^3+768*sin(d*x+c)*cos(d*x+c)^8-512*I*cos(d*x+c)^8-256*sin(d*x+c)*cos(d*x+c)^7+1536*I*c
os(d*x+c)^10+320*sin(d*x+c)*cos(d*x+c)^6+15*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)*cos(d*x+c)^4-480*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*a^2
________________________________________________________________________________________
maxima [B] time = 1.10, size = 1075, normalized size = 6.76 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
-1/480*(20*(I*sqrt(2)*a^2*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - sqrt(2)*a^2*sin(3/2*arcta
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1
)^(3/4)*sqrt(a) - (-60*I*sqrt(2)*a^2*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 60*sqrt(2)*a^2
*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (-12*I*sqrt(2)*a^2*cos(2*d*x + 2*c)^2 - 12*I*sqrt(
2)*a^2*sin(2*d*x + 2*c)^2 - 24*I*sqrt(2)*a^2*cos(2*d*x + 2*c) - 12*I*sqrt(2)*a^2)*cos(5/2*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c) + 1)) + 12*(sqrt(2)*a^2*cos(2*d*x + 2*c)^2 + sqrt(2)*a^2*sin(2*d*x + 2*c)^2 + 2*sqrt(2)
*a^2*cos(2*d*x + 2*c) + sqrt(2)*a^2)*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*(cos(2*d*x + 2*
c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + (30*sqrt(2)*a^2*arctan2((cos(2*d*x + 2*c)^
2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)
), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c) + 1)) + 1) - 30*sqrt(2)*a^2*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x +
2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*
c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 1) - 15*I*sqrt
(2)*a^2*log(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(1/2*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c) + 1))^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(1
/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d
*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) + 15*I*sqrt(2)*a^2*log(sqrt
(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x
+ 2*c) + 1))^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(1/2*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 - 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(
1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1))*sqrt(a))/d
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^5\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^(5/2),x)
[Out]
int(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^(5/2), 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))**(5/2),x)
[Out]
Timed out
________________________________________________________________________________________