Integrand size = 22, antiderivative size = 101 \[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {12 \sin (c+d x)}{35 a^3 d}-\frac {4 \sin ^3(c+d x)}{35 a^3 d}+\frac {i \cos (c+d x)}{7 d (a+i a \tan (c+d x))^3}+\frac {8 i \cos ^3(c+d x)}{35 d \left (a^3+i a^3 \tan (c+d x)\right )} \] Output:
12/35*sin(d*x+c)/a^3/d-4/35*sin(d*x+c)^3/a^3/d+1/7*I*cos(d*x+c)/d/(a+I*a*t an(d*x+c))^3+8/35*I*cos(d*x+c)^3/d/(a^3+I*a^3*tan(d*x+c))
Time = 0.36 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.75 \[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\sec ^3(c+d x) (35+84 \cos (2 (c+d x))-15 \cos (4 (c+d x))+56 i \sin (2 (c+d x))-20 i \sin (4 (c+d x)))}{280 a^3 d (-i+\tan (c+d x))^3} \] Input:
Integrate[Cos[c + d*x]/(a + I*a*Tan[c + d*x])^3,x]
Output:
-1/280*(Sec[c + d*x]^3*(35 + 84*Cos[2*(c + d*x)] - 15*Cos[4*(c + d*x)] + ( 56*I)*Sin[2*(c + d*x)] - (20*I)*Sin[4*(c + d*x)]))/(a^3*d*(-I + Tan[c + d* x])^3)
Time = 0.44 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {3042, 3983, 3042, 3981, 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.
\(\displaystyle \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sec (c+d x) (a+i a \tan (c+d x))^3}dx\) |
\(\Big \downarrow \) 3983 |
\(\displaystyle \frac {4 \int \frac {\cos (c+d x)}{(i \tan (c+d x) a+a)^2}dx}{7 a}+\frac {i \cos (c+d x)}{7 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 \int \frac {1}{\sec (c+d x) (i \tan (c+d x) a+a)^2}dx}{7 a}+\frac {i \cos (c+d x)}{7 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 3981 |
\(\displaystyle \frac {4 \left (\frac {3 \int \cos ^3(c+d x)dx}{5 a^2}+\frac {2 i \cos ^3(c+d x)}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{7 a}+\frac {i \cos (c+d x)}{7 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 \left (\frac {3 \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx}{5 a^2}+\frac {2 i \cos ^3(c+d x)}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{7 a}+\frac {i \cos (c+d x)}{7 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 3113 |
\(\displaystyle \frac {4 \left (-\frac {3 \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{5 a^2 d}+\frac {2 i \cos ^3(c+d x)}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{7 a}+\frac {i \cos (c+d x)}{7 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 \left (-\frac {3 \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{5 a^2 d}+\frac {2 i \cos ^3(c+d x)}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}\right )}{7 a}+\frac {i \cos (c+d x)}{7 d (a+i a \tan (c+d x))^3}\) |
Input:
Int[Cos[c + d*x]/(a + I*a*Tan[c + d*x])^3,x]
Output:
((I/7)*Cos[c + d*x])/(d*(a + I*a*Tan[c + d*x])^3) + (4*((-3*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/(5*a^2*d) + (((2*I)/5)*Cos[c + d*x]^3)/(d*(a^2 + I*a ^2*Tan[c + d*x]))))/(7*a)
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 _)])^(n_), x_Symbol] :> Simp[2*d^2*(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + f*x])^(n + 1)/(b*f*(m + 2*n))), x] - Simp[d^2*((m - 2)/(b^2*(m + 2*n))) Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[ {a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (IntegersQ[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]
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] + Simp[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]
Time = 0.83 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.84
method | result | size |
risch | \(\frac {i {\mathrm e}^{-3 i \left (d x +c \right )}}{8 a^{3} d}+\frac {i {\mathrm e}^{-5 i \left (d x +c \right )}}{20 a^{3} d}+\frac {i {\mathrm e}^{-7 i \left (d x +c \right )}}{112 a^{3} d}+\frac {3 i \cos \left (d x +c \right )}{16 a^{3} d}+\frac {5 \sin \left (d x +c \right )}{16 a^{3} d}\) | \(85\) |
derivativedivides | \(\frac {\frac {4 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {9 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {17 i}{4 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {8}{7 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {38}{5 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {15}{2 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {15}{8 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {2}{16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+16 i}}{a^{3} d}\) | \(141\) |
default | \(\frac {\frac {4 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {9 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {17 i}{4 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {8}{7 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {38}{5 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {15}{2 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {15}{8 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {2}{16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+16 i}}{a^{3} d}\) | \(141\) |
Input:
int(cos(d*x+c)/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/8*I/a^3/d*exp(-3*I*(d*x+c))+1/20*I/a^3/d*exp(-5*I*(d*x+c))+1/112*I/a^3/d *exp(-7*I*(d*x+c))+3/16*I/a^3/d*cos(d*x+c)+5/16*sin(d*x+c)/a^3/d
Time = 0.09 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.62 \[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\left (-35 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 140 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 70 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 28 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{560 \, a^{3} d} \] Input:
integrate(cos(d*x+c)/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
Output:
1/560*(-35*I*e^(8*I*d*x + 8*I*c) + 140*I*e^(6*I*d*x + 6*I*c) + 70*I*e^(4*I *d*x + 4*I*c) + 28*I*e^(2*I*d*x + 2*I*c) + 5*I)*e^(-7*I*d*x - 7*I*c)/(a^3* d)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (87) = 174\).
Time = 0.28 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.95 \[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\begin {cases} \frac {\left (- 71680 i a^{12} d^{4} e^{17 i c} e^{i d x} + 286720 i a^{12} d^{4} e^{15 i c} e^{- i d x} + 143360 i a^{12} d^{4} e^{13 i c} e^{- 3 i d x} + 57344 i a^{12} d^{4} e^{11 i c} e^{- 5 i d x} + 10240 i a^{12} d^{4} e^{9 i c} e^{- 7 i d x}\right ) e^{- 16 i c}}{1146880 a^{15} d^{5}} & \text {for}\: a^{15} d^{5} e^{16 i c} \neq 0 \\\frac {x \left (e^{8 i c} + 4 e^{6 i c} + 6 e^{4 i c} + 4 e^{2 i c} + 1\right ) e^{- 7 i c}}{16 a^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)/(a+I*a*tan(d*x+c))**3,x)
Output:
Piecewise(((-71680*I*a**12*d**4*exp(17*I*c)*exp(I*d*x) + 286720*I*a**12*d* *4*exp(15*I*c)*exp(-I*d*x) + 143360*I*a**12*d**4*exp(13*I*c)*exp(-3*I*d*x) + 57344*I*a**12*d**4*exp(11*I*c)*exp(-5*I*d*x) + 10240*I*a**12*d**4*exp(9 *I*c)*exp(-7*I*d*x))*exp(-16*I*c)/(1146880*a**15*d**5), Ne(a**15*d**5*exp( 16*I*c), 0)), (x*(exp(8*I*c) + 4*exp(6*I*c) + 6*exp(4*I*c) + 4*exp(2*I*c) + 1)*exp(-7*I*c)/(16*a**3), True))
Exception generated. \[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(cos(d*x+c)/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
Time = 0.33 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.18 \[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {\frac {35}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right )}} + \frac {525 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 1960 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4025 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4480 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3143 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1176 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 243}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{7}}}{280 \, d} \] Input:
integrate(cos(d*x+c)/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
Output:
1/280*(35/(a^3*(tan(1/2*d*x + 1/2*c) + I)) + (525*tan(1/2*d*x + 1/2*c)^6 - 1960*I*tan(1/2*d*x + 1/2*c)^5 - 4025*tan(1/2*d*x + 1/2*c)^4 + 4480*I*tan( 1/2*d*x + 1/2*c)^3 + 3143*tan(1/2*d*x + 1/2*c)^2 - 1176*I*tan(1/2*d*x + 1/ 2*c) - 243)/(a^3*(tan(1/2*d*x + 1/2*c) - I)^7))/d
Time = 3.05 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.33 \[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\left (35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,105{}\mathrm {i}-175\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,105{}\mathrm {i}-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,77{}\mathrm {i}+43\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-13{}\mathrm {i}\right )\,2{}\mathrm {i}}{35\,a^3\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )\,{\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^7} \] Input:
int(cos(c + d*x)/(a + a*tan(c + d*x)*1i)^3,x)
Output:
-((43*tan(c/2 + (d*x)/2) + tan(c/2 + (d*x)/2)^2*77i - 7*tan(c/2 + (d*x)/2) ^3 + tan(c/2 + (d*x)/2)^4*105i - 175*tan(c/2 + (d*x)/2)^5 - tan(c/2 + (d*x )/2)^6*105i + 35*tan(c/2 + (d*x)/2)^7 - 13i)*2i)/(35*a^3*d*(tan(c/2 + (d*x )/2) + 1i)*(tan(c/2 + (d*x)/2)*1i + 1)^7)
\[ \int \frac {\cos (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\int \frac {\cos \left (d x +c \right )}{\tan \left (d x +c \right )^{3} i +3 \tan \left (d x +c \right )^{2}-3 \tan \left (d x +c \right ) i -1}d x}{a^{3}} \] Input:
int(cos(d*x+c)/(a+I*a*tan(d*x+c))^3,x)
Output:
( - int(cos(c + d*x)/(tan(c + d*x)**3*i + 3*tan(c + d*x)**2 - 3*tan(c + d* x)*i - 1),x))/a**3