Integrand size = 24, antiderivative size = 120 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {5 a^4 \sin (c+d x)}{21 d}-\frac {10 a^4 \sin ^3(c+d x)}{63 d}+\frac {a^4 \sin ^5(c+d x)}{21 d}-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^3}{9 d}-\frac {2 i \cos ^7(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{21 d} \] Output:
5/21*a^4*sin(d*x+c)/d-10/63*a^4*sin(d*x+c)^3/d+1/21*a^4*sin(d*x+c)^5/d-2/9 *I*a*cos(d*x+c)^9*(a+I*a*tan(d*x+c))^3/d-2/21*I*cos(d*x+c)^7*(a^4+I*a^4*ta n(d*x+c))/d
Time = 0.81 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.80 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^4 (-i \cos (4 (c+d x))+\sin (4 (c+d x))) \left (168 \cos (c+d x) \sqrt {\cos ^2(c+d x)}+4 \left (16+45 \sqrt {\cos ^2(c+d x)}\right ) \cos (3 (c+d x))+64 \cos (5 (c+d x))-28 \sqrt {\cos ^2(c+d x)} \cos (5 (c+d x))-42 i \sqrt {\cos ^2(c+d x)} \sin (c+d x)-64 i \sin (3 (c+d x))-135 i \sqrt {\cos ^2(c+d x)} \sin (3 (c+d x))-64 i \sin (5 (c+d x))+35 i \sqrt {\cos ^2(c+d x)} \sin (5 (c+d x))\right )}{1008 d \sqrt {\cos ^2(c+d x)}} \] Input:
Integrate[Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^4,x]
Output:
(a^4*((-I)*Cos[4*(c + d*x)] + Sin[4*(c + d*x)])*(168*Cos[c + d*x]*Sqrt[Cos [c + d*x]^2] + 4*(16 + 45*Sqrt[Cos[c + d*x]^2])*Cos[3*(c + d*x)] + 64*Cos[ 5*(c + d*x)] - 28*Sqrt[Cos[c + d*x]^2]*Cos[5*(c + d*x)] - (42*I)*Sqrt[Cos[ c + d*x]^2]*Sin[c + d*x] - (64*I)*Sin[3*(c + d*x)] - (135*I)*Sqrt[Cos[c + d*x]^2]*Sin[3*(c + d*x)] - (64*I)*Sin[5*(c + d*x)] + (35*I)*Sqrt[Cos[c + d *x]^2]*Sin[5*(c + d*x)]))/(1008*d*Sqrt[Cos[c + d*x]^2])
Time = 0.47 (sec) , antiderivative size = 119, 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, 3977, 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.
| \(\displaystyle \int \cos ^9(c+d x) (a+i a \tan (c+d x))^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^4}{\sec (c+d x)^9}dx\) |
| \(\Big \downarrow \) 3977 |
| \(\displaystyle \frac {1}{3} a^2 \int \cos ^7(c+d x) (i \tan (c+d x) a+a)^2dx-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^3}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} a^2 \int \frac {(i \tan (c+d x) a+a)^2}{\sec (c+d x)^7}dx-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^3}{9 d}\) |
| \(\Big \downarrow \) 3977 |
| \(\displaystyle \frac {1}{3} a^2 \left (\frac {5}{7} a^2 \int \cos ^5(c+d x)dx-\frac {2 i \cos ^7(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{7 d}\right )-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^3}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} a^2 \left (\frac {5}{7} a^2 \int \sin \left (c+d x+\frac {\pi }{2}\right )^5dx-\frac {2 i \cos ^7(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{7 d}\right )-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^3}{9 d}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {1}{3} a^2 \left (-\frac {5 a^2 \int \left (\sin ^4(c+d x)-2 \sin ^2(c+d x)+1\right )d(-\sin (c+d x))}{7 d}-\frac {2 i \cos ^7(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{7 d}\right )-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^3}{9 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} a^2 \left (-\frac {5 a^2 \left (-\frac {1}{5} \sin ^5(c+d x)+\frac {2}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{7 d}-\frac {2 i \cos ^7(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{7 d}\right )-\frac {2 i a \cos ^9(c+d x) (a+i a \tan (c+d x))^3}{9 d}\) |
Input:
Int[Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^4,x]
Output:
(((-2*I)/9)*a*Cos[c + d*x]^9*(a + I*a*Tan[c + d*x])^3)/d + (a^2*((-5*a^2*( -Sin[c + d*x] + (2*Sin[c + d*x]^3)/3 - Sin[c + d*x]^5/5))/(7*d) - (((2*I)/ 7)*Cos[c + d*x]^7*(a^2 + I*a^2*Tan[c + d*x]))/d))/3
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*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 = 306.94 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.86
| method | result | size |
| risch | \(-\frac {i a^{4} {\mathrm e}^{9 i \left (d x +c \right )}}{288 d}-\frac {5 i a^{4} {\mathrm e}^{7 i \left (d x +c \right )}}{224 d}-\frac {i a^{4} {\mathrm e}^{5 i \left (d x +c \right )}}{16 d}-\frac {5 i a^{4} {\mathrm e}^{3 i \left (d x +c \right )}}{48 d}-\frac {i a^{4} \cos \left (d x +c \right )}{8 d}+\frac {3 a^{4} \sin \left (d x +c \right )}{16 d}\) | \(103\) |
| derivativedivides | \(\frac {a^{4} \left (-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{6}}{9}-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{6}}{21}+\frac {\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{105}\right )-4 i a^{4} \left (-\frac {\cos \left (d x +c \right )^{7} \sin \left (d x +c \right )^{2}}{9}-\frac {2 \cos \left (d x +c \right )^{7}}{63}\right )-6 a^{4} \left (-\frac {\cos \left (d x +c \right )^{8} \sin \left (d x +c \right )}{9}+\frac {\left (\frac {16}{5}+\cos \left (d x +c \right )^{6}+\frac {6 \cos \left (d x +c \right )^{4}}{5}+\frac {8 \cos \left (d x +c \right )^{2}}{5}\right ) \sin \left (d x +c \right )}{63}\right )-\frac {4 i a^{4} \cos \left (d x +c \right )^{9}}{9}+\frac {a^{4} \left (\frac {128}{35}+\cos \left (d x +c \right )^{8}+\frac {8 \cos \left (d x +c \right )^{6}}{7}+\frac {48 \cos \left (d x +c \right )^{4}}{35}+\frac {64 \cos \left (d x +c \right )^{2}}{35}\right ) \sin \left (d x +c \right )}{9}}{d}\) | \(233\) |
| default | \(\frac {a^{4} \left (-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{6}}{9}-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{6}}{21}+\frac {\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{105}\right )-4 i a^{4} \left (-\frac {\cos \left (d x +c \right )^{7} \sin \left (d x +c \right )^{2}}{9}-\frac {2 \cos \left (d x +c \right )^{7}}{63}\right )-6 a^{4} \left (-\frac {\cos \left (d x +c \right )^{8} \sin \left (d x +c \right )}{9}+\frac {\left (\frac {16}{5}+\cos \left (d x +c \right )^{6}+\frac {6 \cos \left (d x +c \right )^{4}}{5}+\frac {8 \cos \left (d x +c \right )^{2}}{5}\right ) \sin \left (d x +c \right )}{63}\right )-\frac {4 i a^{4} \cos \left (d x +c \right )^{9}}{9}+\frac {a^{4} \left (\frac {128}{35}+\cos \left (d x +c \right )^{8}+\frac {8 \cos \left (d x +c \right )^{6}}{7}+\frac {48 \cos \left (d x +c \right )^{4}}{35}+\frac {64 \cos \left (d x +c \right )^{2}}{35}\right ) \sin \left (d x +c \right )}{9}}{d}\) | \(233\) |
Input:
int(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
-1/288*I*a^4/d*exp(9*I*(d*x+c))-5/224*I*a^4/d*exp(7*I*(d*x+c))-1/16*I*a^4/ d*exp(5*I*(d*x+c))-5/48*I*a^4/d*exp(3*I*(d*x+c))-1/8*I*a^4/d*cos(d*x+c)+3/ 16*a^4*sin(d*x+c)/d
Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.75 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {{\left (-7 i \, a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} - 45 i \, a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} - 126 i \, a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - 210 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 315 i \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + 63 i \, a^{4}\right )} e^{\left (-i \, d x - i \, c\right )}}{2016 \, d} \] Input:
integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")
Output:
1/2016*(-7*I*a^4*e^(10*I*d*x + 10*I*c) - 45*I*a^4*e^(8*I*d*x + 8*I*c) - 12 6*I*a^4*e^(6*I*d*x + 6*I*c) - 210*I*a^4*e^(4*I*d*x + 4*I*c) - 315*I*a^4*e^ (2*I*d*x + 2*I*c) + 63*I*a^4)*e^(-I*d*x - I*c)/d
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (107) = 214\).
Time = 0.40 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.90 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^4 \, dx=\begin {cases} \frac {\left (- 176160768 i a^{4} d^{5} e^{10 i c} e^{9 i d x} - 1132462080 i a^{4} d^{5} e^{8 i c} e^{7 i d x} - 3170893824 i a^{4} d^{5} e^{6 i c} e^{5 i d x} - 5284823040 i a^{4} d^{5} e^{4 i c} e^{3 i d x} - 7927234560 i a^{4} d^{5} e^{2 i c} e^{i d x} + 1585446912 i a^{4} d^{5} e^{- i d x}\right ) e^{- i c}}{50734301184 d^{6}} & \text {for}\: d^{6} e^{i c} \neq 0 \\\frac {x \left (a^{4} e^{10 i c} + 5 a^{4} e^{8 i c} + 10 a^{4} e^{6 i c} + 10 a^{4} e^{4 i c} + 5 a^{4} e^{2 i c} + a^{4}\right ) e^{- i c}}{32} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**9*(a+I*a*tan(d*x+c))**4,x)
Output:
Piecewise(((-176160768*I*a**4*d**5*exp(10*I*c)*exp(9*I*d*x) - 1132462080*I *a**4*d**5*exp(8*I*c)*exp(7*I*d*x) - 3170893824*I*a**4*d**5*exp(6*I*c)*exp (5*I*d*x) - 5284823040*I*a**4*d**5*exp(4*I*c)*exp(3*I*d*x) - 7927234560*I* a**4*d**5*exp(2*I*c)*exp(I*d*x) + 1585446912*I*a**4*d**5*exp(-I*d*x))*exp( -I*c)/(50734301184*d**6), Ne(d**6*exp(I*c), 0)), (x*(a**4*exp(10*I*c) + 5* a**4*exp(8*I*c) + 10*a**4*exp(6*I*c) + 10*a**4*exp(4*I*c) + 5*a**4*exp(2*I *c) + a**4)*exp(-I*c)/32, True))
Time = 0.24 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.51 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {140 i \, a^{4} \cos \left (d x + c\right )^{9} + 20 i \, {\left (7 \, \cos \left (d x + c\right )^{9} - 9 \, \cos \left (d x + c\right )^{7}\right )} a^{4} - {\left (35 \, \sin \left (d x + c\right )^{9} - 90 \, \sin \left (d x + c\right )^{7} + 63 \, \sin \left (d x + c\right )^{5}\right )} a^{4} - 6 \, {\left (35 \, \sin \left (d x + c\right )^{9} - 135 \, \sin \left (d x + c\right )^{7} + 189 \, \sin \left (d x + c\right )^{5} - 105 \, \sin \left (d x + c\right )^{3}\right )} a^{4} - {\left (35 \, \sin \left (d x + c\right )^{9} - 180 \, \sin \left (d x + c\right )^{7} + 378 \, \sin \left (d x + c\right )^{5} - 420 \, \sin \left (d x + c\right )^{3} + 315 \, \sin \left (d x + c\right )\right )} a^{4}}{315 \, d} \] Input:
integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")
Output:
-1/315*(140*I*a^4*cos(d*x + c)^9 + 20*I*(7*cos(d*x + c)^9 - 9*cos(d*x + c) ^7)*a^4 - (35*sin(d*x + c)^9 - 90*sin(d*x + c)^7 + 63*sin(d*x + c)^5)*a^4 - 6*(35*sin(d*x + c)^9 - 135*sin(d*x + c)^7 + 189*sin(d*x + c)^5 - 105*sin (d*x + c)^3)*a^4 - (35*sin(d*x + c)^9 - 180*sin(d*x + c)^7 + 378*sin(d*x + c)^5 - 420*sin(d*x + c)^3 + 315*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. 1409 vs. \(2 (102) = 204\).
Time = 0.46 (sec) , antiderivative size = 1409, normalized size of antiderivative = 11.74 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^4 \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^4,x, algorithm="giac")
Output:
1/516096*(435267*a^4*e^(13*I*d*x + 7*I*c)*log(I*e^(I*d*x + I*c) + 1) + 261 1602*a^4*e^(11*I*d*x + 5*I*c)*log(I*e^(I*d*x + I*c) + 1) + 6529005*a^4*e^( 9*I*d*x + 3*I*c)*log(I*e^(I*d*x + I*c) + 1) + 8705340*a^4*e^(7*I*d*x + I*c )*log(I*e^(I*d*x + I*c) + 1) + 6529005*a^4*e^(5*I*d*x - I*c)*log(I*e^(I*d* x + I*c) + 1) + 2611602*a^4*e^(3*I*d*x - 3*I*c)*log(I*e^(I*d*x + I*c) + 1) + 435267*a^4*e^(I*d*x - 5*I*c)*log(I*e^(I*d*x + I*c) + 1) + 427896*a^4*e^ (13*I*d*x + 7*I*c)*log(I*e^(I*d*x + I*c) - 1) + 2567376*a^4*e^(11*I*d*x + 5*I*c)*log(I*e^(I*d*x + I*c) - 1) + 6418440*a^4*e^(9*I*d*x + 3*I*c)*log(I* e^(I*d*x + I*c) - 1) + 8557920*a^4*e^(7*I*d*x + I*c)*log(I*e^(I*d*x + I*c) - 1) + 6418440*a^4*e^(5*I*d*x - I*c)*log(I*e^(I*d*x + I*c) - 1) + 2567376 *a^4*e^(3*I*d*x - 3*I*c)*log(I*e^(I*d*x + I*c) - 1) + 427896*a^4*e^(I*d*x - 5*I*c)*log(I*e^(I*d*x + I*c) - 1) - 435267*a^4*e^(13*I*d*x + 7*I*c)*log( -I*e^(I*d*x + I*c) + 1) - 2611602*a^4*e^(11*I*d*x + 5*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 6529005*a^4*e^(9*I*d*x + 3*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 8705340*a^4*e^(7*I*d*x + I*c)*log(-I*e^(I*d*x + I*c) + 1) - 6529005*a^4 *e^(5*I*d*x - I*c)*log(-I*e^(I*d*x + I*c) + 1) - 2611602*a^4*e^(3*I*d*x - 3*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 435267*a^4*e^(I*d*x - 5*I*c)*log(-I*e ^(I*d*x + I*c) + 1) - 427896*a^4*e^(13*I*d*x + 7*I*c)*log(-I*e^(I*d*x + I* c) - 1) - 2567376*a^4*e^(11*I*d*x + 5*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 6 418440*a^4*e^(9*I*d*x + 3*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 8557920*a^...
Time = 1.88 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.21 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {2\,a^4\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {89\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}-\frac {55\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{4}+\frac {55\,\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{4}-\frac {355\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )}{16}+\frac {35\,\sin \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )}{16}-\frac {\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,21{}\mathrm {i}}{2}+\frac {\cos \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )\,21{}\mathrm {i}}{2}-\frac {\cos \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,87{}\mathrm {i}}{4}+\frac {\cos \left (\frac {9\,c}{2}+\frac {9\,d\,x}{2}\right )\,7{}\mathrm {i}}{4}\right )}{63\,d\,\left (\cos \left (4\,c+4\,d\,x\right )-\sin \left (4\,c+4\,d\,x\right )\,1{}\mathrm {i}\right )} \] Input:
int(cos(c + d*x)^9*(a + a*tan(c + d*x)*1i)^4,x)
Output:
(2*a^4*cos(c/2 + (d*x)/2)*((cos((5*c)/2 + (5*d*x)/2)*21i)/2 - (cos((3*c)/2 + (3*d*x)/2)*21i)/2 - (cos((7*c)/2 + (7*d*x)/2)*87i)/4 + (cos((9*c)/2 + ( 9*d*x)/2)*7i)/4 + (89*sin(c/2 + (d*x)/2))/8 - (55*sin((3*c)/2 + (3*d*x)/2) )/4 + (55*sin((5*c)/2 + (5*d*x)/2))/4 - (355*sin((7*c)/2 + (7*d*x)/2))/16 + (35*sin((9*c)/2 + (9*d*x)/2))/16))/(63*d*(cos(4*c + 4*d*x) - sin(4*c + 4 *d*x)*1i))
Time = 0.22 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.14 \[ \int \cos ^9(c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {a^{4} \left (-56 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{8} i +188 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} i -228 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} i +116 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} i -20 \cos \left (d x +c \right ) i +56 \sin \left (d x +c \right )^{9}-216 \sin \left (d x +c \right )^{7}+315 \sin \left (d x +c \right )^{5}-210 \sin \left (d x +c \right )^{3}+63 \sin \left (d x +c \right )+20 i \right )}{63 d} \] Input:
int(cos(d*x+c)^9*(a+I*a*tan(d*x+c))^4,x)
Output:
(a**4*( - 56*cos(c + d*x)*sin(c + d*x)**8*i + 188*cos(c + d*x)*sin(c + d*x )**6*i - 228*cos(c + d*x)*sin(c + d*x)**4*i + 116*cos(c + d*x)*sin(c + d*x )**2*i - 20*cos(c + d*x)*i + 56*sin(c + d*x)**9 - 216*sin(c + d*x)**7 + 31 5*sin(c + d*x)**5 - 210*sin(c + d*x)**3 + 63*sin(c + d*x) + 20*i))/(63*d)