Integrand size = 24, antiderivative size = 78 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^4 \text {arctanh}(\sin (c+d x))}{d}-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}+\frac {2 i \cos (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d} \] Output:
a^4*arctanh(sin(d*x+c))/d-2/3*I*a*cos(d*x+c)^3*(a+I*a*tan(d*x+c))^3/d+2*I* cos(d*x+c)*(a^4+I*a^4*tan(d*x+c))/d
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(246\) vs. \(2(78)=156\).
Time = 0.86 (sec) , antiderivative size = 246, normalized size of antiderivative = 3.15 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^4 \left (-3 \cos (4 c) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+3 \cos (4 c) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 \cos (3 d x) \sin (c)+6 \cos (d x) \sin (3 c)+3 i \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (4 c)-3 i \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (4 c)+\cos (3 c) (6 i \cos (d x)-6 \sin (d x))+6 i \sin (3 c) \sin (d x)-2 i \sin (c) \sin (3 d x)+2 \cos (c) (-i \cos (3 d x)+\sin (3 d x))\right ) (\cos (c+d x)+i \sin (c+d x))^4}{3 d (\cos (d x)+i \sin (d x))^4} \] Input:
Integrate[Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^4,x]
Output:
(a^4*(-3*Cos[4*c]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 3*Cos[4*c]*Lo g[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 2*Cos[3*d*x]*Sin[c] + 6*Cos[d*x]* Sin[3*c] + (3*I)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[4*c] - (3*I) *Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[4*c] + Cos[3*c]*((6*I)*Cos[d *x] - 6*Sin[d*x]) + (6*I)*Sin[3*c]*Sin[d*x] - (2*I)*Sin[c]*Sin[3*d*x] + 2* Cos[c]*((-I)*Cos[3*d*x] + Sin[3*d*x]))*(Cos[c + d*x] + I*Sin[c + d*x])^4)/ (3*d*(Cos[d*x] + I*Sin[d*x])^4)
Time = 0.42 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3977, 3042, 3977, 3042, 4257}
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.
\(\displaystyle \int \cos ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^4}{\sec (c+d x)^3}dx\) |
\(\Big \downarrow \) 3977 |
\(\displaystyle -a^2 \int \cos (c+d x) (i \tan (c+d x) a+a)^2dx-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -a^2 \int \frac {(i \tan (c+d x) a+a)^2}{\sec (c+d x)}dx-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}\) |
\(\Big \downarrow \) 3977 |
\(\displaystyle -a^2 \left (a^2 (-\int \sec (c+d x)dx)-\frac {2 i \cos (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d}\right )-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -a^2 \left (a^2 \left (-\int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )-\frac {2 i \cos (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d}\right )-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle -a^2 \left (-\frac {a^2 \text {arctanh}(\sin (c+d x))}{d}-\frac {2 i \cos (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{d}\right )-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}\) |
Input:
Int[Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^4,x]
Output:
(((-2*I)/3)*a*Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^3)/d - a^2*(-((a^2*Arc Tanh[Sin[c + d*x]])/d) - ((2*I)*Cos[c + d*x]*(a^2 + I*a^2*Tan[c + d*x]))/d )
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]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 12.56 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01
method | result | size |
risch | \(-\frac {2 i a^{4} {\mathrm e}^{3 i \left (d x +c \right )}}{3 d}+\frac {2 i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\) | \(79\) |
derivativedivides | \(\frac {a^{4} \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+\frac {4 i a^{4} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}-2 a^{4} \sin \left (d x +c \right )^{3}-\frac {4 i a^{4} \cos \left (d x +c \right )^{3}}{3}+\frac {a^{4} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(113\) |
default | \(\frac {a^{4} \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+\frac {4 i a^{4} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}-2 a^{4} \sin \left (d x +c \right )^{3}-\frac {4 i a^{4} \cos \left (d x +c \right )^{3}}{3}+\frac {a^{4} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(113\) |
Input:
int(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
-2/3*I*a^4/d*exp(3*I*(d*x+c))+2*I*a^4/d*exp(I*(d*x+c))+a^4/d*ln(exp(I*(d*x +c))+I)-a^4/d*ln(exp(I*(d*x+c))-I)
Time = 0.09 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.87 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {-2 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c\right )} + 6 i \, a^{4} e^{\left (i \, d x + i \, c\right )} + 3 \, a^{4} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 3 \, a^{4} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{3 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")
Output:
1/3*(-2*I*a^4*e^(3*I*d*x + 3*I*c) + 6*I*a^4*e^(I*d*x + I*c) + 3*a^4*log(e^ (I*d*x + I*c) + I) - 3*a^4*log(e^(I*d*x + I*c) - I))/d
Time = 0.27 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.40 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^{4} \left (- \log {\left (e^{i d x} - i e^{- i c} \right )} + \log {\left (e^{i d x} + i e^{- i c} \right )}\right )}{d} + \begin {cases} \frac {- 2 i a^{4} d e^{3 i c} e^{3 i d x} + 6 i a^{4} d e^{i c} e^{i d x}}{3 d^{2}} & \text {for}\: d^{2} \neq 0 \\x \left (2 a^{4} e^{3 i c} - 2 a^{4} e^{i c}\right ) & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**3*(a+I*a*tan(d*x+c))**4,x)
Output:
a**4*(-log(exp(I*d*x) - I*exp(-I*c)) + log(exp(I*d*x) + I*exp(-I*c)))/d + Piecewise(((-2*I*a**4*d*exp(3*I*c)*exp(3*I*d*x) + 6*I*a**4*d*exp(I*c)*exp( I*d*x))/(3*d**2), Ne(d**2, 0)), (x*(2*a**4*exp(3*I*c) - 2*a**4*exp(I*c)), True))
Time = 0.04 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.55 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {8 i \, a^{4} \cos \left (d x + c\right )^{3} + 12 \, a^{4} \sin \left (d x + c\right )^{3} + 8 i \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{4} + {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a^{4} + 2 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{4}}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")
Output:
-1/6*(8*I*a^4*cos(d*x + c)^3 + 12*a^4*sin(d*x + c)^3 + 8*I*(cos(d*x + c)^3 - 3*cos(d*x + c))*a^4 + (2*sin(d*x + c)^3 - 3*log(sin(d*x + c) + 1) + 3*l og(sin(d*x + c) - 1) + 6*sin(d*x + c))*a^4 + 2*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^4)/d
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1299 vs. \(2 (68) = 136\).
Time = 0.59 (sec) , antiderivative size = 1299, normalized size of antiderivative = 16.65 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^4,x, algorithm="giac")
Output:
1/768*(1110*a^4*e^(12*I*d*x + 6*I*c)*log(I*e^(I*d*x + I*c) + 1) + 6660*a^4 *e^(10*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 16650*a^4*e^(8*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*c) + 1) + 16650*a^4*e^(4*I*d*x - 2*I*c)*log(I*e^ (I*d*x + I*c) + 1) + 6660*a^4*e^(2*I*d*x - 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 22200*a^4*e^(6*I*d*x)*log(I*e^(I*d*x + I*c) + 1) + 1110*a^4*e^(-6*I*c )*log(I*e^(I*d*x + I*c) + 1) + 1875*a^4*e^(12*I*d*x + 6*I*c)*log(I*e^(I*d* x + I*c) - 1) + 11250*a^4*e^(10*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) - 1) + 28125*a^4*e^(8*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*c) - 1) + 28125*a^4*e^( 4*I*d*x - 2*I*c)*log(I*e^(I*d*x + I*c) - 1) + 11250*a^4*e^(2*I*d*x - 4*I*c )*log(I*e^(I*d*x + I*c) - 1) + 37500*a^4*e^(6*I*d*x)*log(I*e^(I*d*x + I*c) - 1) + 1875*a^4*e^(-6*I*c)*log(I*e^(I*d*x + I*c) - 1) - 1110*a^4*e^(12*I* d*x + 6*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 6660*a^4*e^(10*I*d*x + 4*I*c)*l og(-I*e^(I*d*x + I*c) + 1) - 16650*a^4*e^(8*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 16650*a^4*e^(4*I*d*x - 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 6660*a^4*e^(2*I*d*x - 4*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 22200*a^4*e^(6 *I*d*x)*log(-I*e^(I*d*x + I*c) + 1) - 1110*a^4*e^(-6*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 1875*a^4*e^(12*I*d*x + 6*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 11250*a^4*e^(10*I*d*x + 4*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 28125*a^4*e^( 8*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 28125*a^4*e^(4*I*d*x - 2*I* c)*log(-I*e^(I*d*x + I*c) - 1) - 11250*a^4*e^(2*I*d*x - 4*I*c)*log(-I*e...
Time = 0.66 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.13 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {2\,a^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {\frac {8\,a^4}{3}-a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,8{}\mathrm {i}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,3{}\mathrm {i}+3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )} \] Input:
int(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i)^4,x)
Output:
(2*a^4*atanh(tan(c/2 + (d*x)/2)))/d - ((8*a^4)/3 - a^4*tan(c/2 + (d*x)/2)* 8i)/(d*(3*tan(c/2 + (d*x)/2) - tan(c/2 + (d*x)/2)^2*3i - tan(c/2 + (d*x)/2 )^3 + 1i))
Time = 0.19 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.97 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^{4} \left (8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} i +4 \cos \left (d x +c \right ) i -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-8 \sin \left (d x +c \right )^{3}-4 i \right )}{3 d} \] Input:
int(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^4,x)
Output:
(a**4*(8*cos(c + d*x)*sin(c + d*x)**2*i + 4*cos(c + d*x)*i - 3*log(tan((c + d*x)/2) - 1) + 3*log(tan((c + d*x)/2) + 1) - 8*sin(c + d*x)**3 - 4*i))/( 3*d)