Integrand size = 24, antiderivative size = 88 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {i a^3 \cos ^3(c+d x)}{15 d}+\frac {a^3 \sin (c+d x)}{5 d}-\frac {a^3 \sin ^3(c+d x)}{15 d}-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d} \] Output:
-1/15*I*a^3*cos(d*x+c)^3/d+1/5*a^3*sin(d*x+c)/d-1/15*a^3*sin(d*x+c)^3/d-2/ 5*I*a*cos(d*x+c)^5*(a+I*a*tan(d*x+c))^2/d
Time = 0.66 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.65 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^3 (-i \cos (2 (c+d x))+\sin (2 (c+d x))) \left (\cos (c+d x) \left (4+25 \sqrt {\cos ^2(c+d x)}\right )+\left (4+3 \sqrt {\cos ^2(c+d x)}\right ) \cos (3 (c+d x))-i \left (\left (4+5 \sqrt {\cos ^2(c+d x)}\right ) \sin (c+d x)+\left (4-3 \sqrt {\cos ^2(c+d x)}\right ) \sin (3 (c+d x))\right )\right )}{60 d \sqrt {\cos ^2(c+d x)}} \] Input:
Integrate[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^3,x]
Output:
(a^3*((-I)*Cos[2*(c + d*x)] + Sin[2*(c + d*x)])*(Cos[c + d*x]*(4 + 25*Sqrt [Cos[c + d*x]^2]) + (4 + 3*Sqrt[Cos[c + d*x]^2])*Cos[3*(c + d*x)] - I*((4 + 5*Sqrt[Cos[c + d*x]^2])*Sin[c + d*x] + (4 - 3*Sqrt[Cos[c + d*x]^2])*Sin[ 3*(c + d*x)])))/(60*d*Sqrt[Cos[c + d*x]^2])
Time = 0.43 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {3042, 3977, 3042, 3967, 3042, 3113, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \cos ^5(c+d x) (a+i a \tan (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^3}{\sec (c+d x)^5}dx\) |
\(\Big \downarrow \) 3977 |
\(\displaystyle \frac {1}{5} a^2 \int \cos ^3(c+d x) (i \tan (c+d x) a+a)dx-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} a^2 \int \frac {i \tan (c+d x) a+a}{\sec (c+d x)^3}dx-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}\) |
\(\Big \downarrow \) 3967 |
\(\displaystyle \frac {1}{5} a^2 \left (a \int \cos ^3(c+d x)dx-\frac {i a \cos ^3(c+d x)}{3 d}\right )-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} a^2 \left (a \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {i a \cos ^3(c+d x)}{3 d}\right )-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}\) |
\(\Big \downarrow \) 3113 |
\(\displaystyle \frac {1}{5} a^2 \left (-\frac {a \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}-\frac {i a \cos ^3(c+d x)}{3 d}\right )-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} a^2 \left (-\frac {a \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}-\frac {i a \cos ^3(c+d x)}{3 d}\right )-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}\) |
Input:
Int[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^3,x]
Output:
(a^2*(((-1/3*I)*a*Cos[c + d*x]^3)/d - (a*(-Sin[c + d*x] + Sin[c + d*x]^3/3 ))/d))/5 - (((2*I)/5)*a*Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^2)/d
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a Int[(d *Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] || NeQ[a^2 + b^2, 0])
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x _)])^(n_), x_Symbol] :> Simp[2*b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^( n - 1)/(f*m)), x] - Simp[b^2*((m + 2*n - 2)/(d^2*m)) Int[(d*Sec[e + f*x]) ^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILtQ[m, 0] & & LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]
Time = 24.55 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.64
method | result | size |
risch | \(-\frac {i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}}{20 d}-\frac {i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{6 d}-\frac {i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{4 d}\) | \(56\) |
derivativedivides | \(\frac {-i a^{3} \left (-\frac {\cos \left (d x +c \right )^{3} \sin \left (d x +c \right )^{2}}{5}-\frac {2 \cos \left (d x +c \right )^{3}}{15}\right )-3 a^{3} \left (-\frac {\cos \left (d x +c \right )^{4} \sin \left (d x +c \right )}{5}+\frac {\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{15}\right )-\frac {3 i a^{3} \cos \left (d x +c \right )^{5}}{5}+\frac {a^{3} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\) | \(126\) |
default | \(\frac {-i a^{3} \left (-\frac {\cos \left (d x +c \right )^{3} \sin \left (d x +c \right )^{2}}{5}-\frac {2 \cos \left (d x +c \right )^{3}}{15}\right )-3 a^{3} \left (-\frac {\cos \left (d x +c \right )^{4} \sin \left (d x +c \right )}{5}+\frac {\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{15}\right )-\frac {3 i a^{3} \cos \left (d x +c \right )^{5}}{5}+\frac {a^{3} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\) | \(126\) |
Input:
int(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
-1/20*I/d*a^3*exp(5*I*(d*x+c))-1/6*I/d*a^3*exp(3*I*(d*x+c))-1/4*I/d*a^3*ex p(I*(d*x+c))
Time = 0.08 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.55 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {-3 i \, a^{3} e^{\left (5 i \, d x + 5 i \, c\right )} - 10 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 15 i \, a^{3} e^{\left (i \, d x + i \, c\right )}}{60 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
Output:
1/60*(-3*I*a^3*e^(5*I*d*x + 5*I*c) - 10*I*a^3*e^(3*I*d*x + 3*I*c) - 15*I*a ^3*e^(I*d*x + I*c))/d
Time = 0.25 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.32 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^3 \, dx=\begin {cases} \frac {- 24 i a^{3} d^{2} e^{5 i c} e^{5 i d x} - 80 i a^{3} d^{2} e^{3 i c} e^{3 i d x} - 120 i a^{3} d^{2} e^{i c} e^{i d x}}{480 d^{3}} & \text {for}\: d^{3} \neq 0 \\x \left (\frac {a^{3} e^{5 i c}}{4} + \frac {a^{3} e^{3 i c}}{2} + \frac {a^{3} e^{i c}}{4}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**5*(a+I*a*tan(d*x+c))**3,x)
Output:
Piecewise(((-24*I*a**3*d**2*exp(5*I*c)*exp(5*I*d*x) - 80*I*a**3*d**2*exp(3 *I*c)*exp(3*I*d*x) - 120*I*a**3*d**2*exp(I*c)*exp(I*d*x))/(480*d**3), Ne(d **3, 0)), (x*(a**3*exp(5*I*c)/4 + a**3*exp(3*I*c)/2 + a**3*exp(I*c)/4), Tr ue))
Time = 0.04 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.19 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {9 i \, a^{3} \cos \left (d x + c\right )^{5} + i \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{3} - 3 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a^{3} - {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{3}}{15 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
Output:
-1/15*(9*I*a^3*cos(d*x + c)^5 + I*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^ 3 - 3*(3*sin(d*x + c)^5 - 5*sin(d*x + c)^3)*a^3 - (3*sin(d*x + c)^5 - 10*s in(d*x + c)^3 + 15*sin(d*x + c))*a^3)/d
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 929 vs. \(2 (74) = 148\).
Time = 0.45 (sec) , antiderivative size = 929, normalized size of antiderivative = 10.56 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^3 \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
Output:
1/7680*(1785*a^3*e^(8*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 7140*a^3 *e^(6*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*c) + 1) + 7140*a^3*e^(2*I*d*x - 2* I*c)*log(I*e^(I*d*x + I*c) + 1) + 10710*a^3*e^(4*I*d*x)*log(I*e^(I*d*x + I *c) + 1) + 1785*a^3*e^(-4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 1530*a^3*e^(8* I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) - 1) + 6120*a^3*e^(6*I*d*x + 2*I*c)*l og(I*e^(I*d*x + I*c) - 1) + 6120*a^3*e^(2*I*d*x - 2*I*c)*log(I*e^(I*d*x + I*c) - 1) + 9180*a^3*e^(4*I*d*x)*log(I*e^(I*d*x + I*c) - 1) + 1530*a^3*e^( -4*I*c)*log(I*e^(I*d*x + I*c) - 1) - 1785*a^3*e^(8*I*d*x + 4*I*c)*log(-I*e ^(I*d*x + I*c) + 1) - 7140*a^3*e^(6*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 7140*a^3*e^(2*I*d*x - 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 10710*a^ 3*e^(4*I*d*x)*log(-I*e^(I*d*x + I*c) + 1) - 1785*a^3*e^(-4*I*c)*log(-I*e^( I*d*x + I*c) + 1) - 1530*a^3*e^(8*I*d*x + 4*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 6120*a^3*e^(6*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 6120*a^3*e ^(2*I*d*x - 2*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 9180*a^3*e^(4*I*d*x)*log( -I*e^(I*d*x + I*c) - 1) - 1530*a^3*e^(-4*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 255*a^3*e^(8*I*d*x + 4*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 1020*a^3*e^(6* I*d*x + 2*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 1020*a^3*e^(2*I*d*x - 2*I*c)* log(I*e^(I*d*x) + e^(-I*c)) - 1530*a^3*e^(4*I*d*x)*log(I*e^(I*d*x) + e^(-I *c)) - 255*a^3*e^(-4*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 255*a^3*e^(8*I*d*x + 4*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 1020*a^3*e^(6*I*d*x + 2*I*c)*l...
Time = 0.66 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.48 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {2\,a^3\,\left (15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,30{}\mathrm {i}-40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,20{}\mathrm {i}+7\right )}{15\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,5{}\mathrm {i}-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,10{}\mathrm {i}+5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )} \] Input:
int(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^3,x)
Output:
(2*a^3*(tan(c/2 + (d*x)/2)^3*30i - 40*tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d* x)/2)*20i + 15*tan(c/2 + (d*x)/2)^4 + 7))/(15*d*(5*tan(c/2 + (d*x)/2) - ta n(c/2 + (d*x)/2)^2*10i - 10*tan(c/2 + (d*x)/2)^3 + tan(c/2 + (d*x)/2)^4*5i + tan(c/2 + (d*x)/2)^5 + 1i))
Time = 0.20 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.94 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^{3} \left (-12 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} i +19 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} i -7 \cos \left (d x +c \right ) i +12 \sin \left (d x +c \right )^{5}-25 \sin \left (d x +c \right )^{3}+15 \sin \left (d x +c \right )+7 i \right )}{15 d} \] Input:
int(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^3,x)
Output:
(a**3*( - 12*cos(c + d*x)*sin(c + d*x)**4*i + 19*cos(c + d*x)*sin(c + d*x) **2*i - 7*cos(c + d*x)*i + 12*sin(c + d*x)**5 - 25*sin(c + d*x)**3 + 15*si n(c + d*x) + 7*i))/(15*d)