Integrand size = 31, antiderivative size = 99 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^n \, dx=\frac {4 i a^3 (c-i c \tan (e+f x))^n}{f n}-\frac {4 i a^3 (c-i c \tan (e+f x))^{1+n}}{c f (1+n)}+\frac {i a^3 (c-i c \tan (e+f x))^{2+n}}{c^2 f (2+n)} \] Output:
4*I*a^3*(c-I*c*tan(f*x+e))^n/f/n-4*I*a^3*(c-I*c*tan(f*x+e))^(1+n)/c/f/(1+n )+I*a^3*(c-I*c*tan(f*x+e))^(2+n)/c^2/f/(2+n)
Time = 0.47 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.76 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^n \, dx=-\frac {i a^3 (c-i c \tan (e+f x))^n \left (-8-5 n-n^2-2 i n (3+n) \tan (e+f x)+n (1+n) \tan ^2(e+f x)\right )}{f n (1+n) (2+n)} \] Input:
Integrate[(a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^n,x]
Output:
((-I)*a^3*(c - I*c*Tan[e + f*x])^n*(-8 - 5*n - n^2 - (2*I)*n*(3 + n)*Tan[e + f*x] + n*(1 + n)*Tan[e + f*x]^2))/(f*n*(1 + n)*(2 + n))
Time = 0.39 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.86, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3042, 4005, 3042, 3968, 53, 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.
\(\displaystyle \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^ndx\) |
\(\Big \downarrow \) 4005 |
\(\displaystyle a^3 c^3 \int \sec ^6(e+f x) (c-i c \tan (e+f x))^{n-3}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^3 c^3 \int \sec (e+f x)^6 (c-i c \tan (e+f x))^{n-3}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle \frac {i a^3 \int (c-i c \tan (e+f x))^{n-1} (i \tan (e+f x) c+c)^2d(-i c \tan (e+f x))}{c^2 f}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {i a^3 \int \left (4 c^2 (c-i c \tan (e+f x))^{n-1}-4 c (c-i c \tan (e+f x))^n+(c-i c \tan (e+f x))^{n+1}\right )d(-i c \tan (e+f x))}{c^2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i a^3 \left (\frac {4 c^2 (c-i c \tan (e+f x))^n}{n}-\frac {4 c (c-i c \tan (e+f x))^{n+1}}{n+1}+\frac {(c-i c \tan (e+f x))^{n+2}}{n+2}\right )}{c^2 f}\) |
Input:
Int[(a + I*a*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^n,x]
Output:
(I*a^3*((4*c^2*(c - I*c*Tan[e + f*x])^n)/n - (4*c*(c - I*c*Tan[e + f*x])^( 1 + n))/(1 + n) + (c - I*c*Tan[e + f*x])^(2 + n)/(2 + n)))/(c^2*f)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(a^(m - 2)*b*f) Subst[Int[(a - x)^(m/2 - 1)*(a + x )^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] && !(IGtQ[n, 0] && (LtQ[ m, 0] || GtQ[m, n]))
Time = 0.81 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.30
method | result | size |
derivativedivides | \(\frac {i a^{3} \left (n^{2}+5 n +8\right ) {\mathrm e}^{n \ln \left (c -i c \tan \left (f x +e \right )\right )}}{\left (1+n \right ) f n \left (2+n \right )}-\frac {i a^{3} \tan \left (f x +e \right )^{2} {\mathrm e}^{n \ln \left (c -i c \tan \left (f x +e \right )\right )}}{f \left (2+n \right )}-\frac {2 a^{3} \left (3+n \right ) \tan \left (f x +e \right ) {\mathrm e}^{n \ln \left (c -i c \tan \left (f x +e \right )\right )}}{f \left (1+n \right ) \left (2+n \right )}\) | \(129\) |
default | \(\frac {i a^{3} \left (n^{2}+5 n +8\right ) {\mathrm e}^{n \ln \left (c -i c \tan \left (f x +e \right )\right )}}{\left (1+n \right ) f n \left (2+n \right )}-\frac {i a^{3} \tan \left (f x +e \right )^{2} {\mathrm e}^{n \ln \left (c -i c \tan \left (f x +e \right )\right )}}{f \left (2+n \right )}-\frac {2 a^{3} \left (3+n \right ) \tan \left (f x +e \right ) {\mathrm e}^{n \ln \left (c -i c \tan \left (f x +e \right )\right )}}{f \left (1+n \right ) \left (2+n \right )}\) | \(129\) |
norman | \(\frac {i a^{3} \left (n^{2}+5 n +8\right ) {\mathrm e}^{n \ln \left (c -i c \tan \left (f x +e \right )\right )}}{\left (1+n \right ) f n \left (2+n \right )}-\frac {i a^{3} \tan \left (f x +e \right )^{2} {\mathrm e}^{n \ln \left (c -i c \tan \left (f x +e \right )\right )}}{f \left (2+n \right )}-\frac {2 a^{3} \left (3+n \right ) \tan \left (f x +e \right ) {\mathrm e}^{n \ln \left (c -i c \tan \left (f x +e \right )\right )}}{f \left (1+n \right ) \left (2+n \right )}\) | \(129\) |
risch | \(\text {Expression too large to display}\) | \(947\) |
Input:
int((a+I*a*tan(f*x+e))^3*(c-I*c*tan(f*x+e))^n,x,method=_RETURNVERBOSE)
Output:
I*a^3*(n^2+5*n+8)/(1+n)/f/n/(2+n)*exp(n*ln(c-I*c*tan(f*x+e)))-I/f/(2+n)*a^ 3*tan(f*x+e)^2*exp(n*ln(c-I*c*tan(f*x+e)))-2*a^3*(3+n)/f/(1+n)/(2+n)*tan(f *x+e)*exp(n*ln(c-I*c*tan(f*x+e)))
Time = 0.11 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.52 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^n \, dx=-\frac {4 \, {\left (-2 i \, a^{3} + {\left (-i \, a^{3} n^{2} - 3 i \, a^{3} n - 2 i \, a^{3}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (-i \, a^{3} n - 2 i \, a^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \left (\frac {2 \, c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{n}}{f n^{3} + 3 \, f n^{2} + 2 \, f n + {\left (f n^{3} + 3 \, f n^{2} + 2 \, f n\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (f n^{3} + 3 \, f n^{2} + 2 \, f n\right )} e^{\left (2 i \, f x + 2 i \, e\right )}} \] Input:
integrate((a+I*a*tan(f*x+e))^3*(c-I*c*tan(f*x+e))^n,x, algorithm="fricas")
Output:
-4*(-2*I*a^3 + (-I*a^3*n^2 - 3*I*a^3*n - 2*I*a^3)*e^(4*I*f*x + 4*I*e) + 2* (-I*a^3*n - 2*I*a^3)*e^(2*I*f*x + 2*I*e))*(2*c/(e^(2*I*f*x + 2*I*e) + 1))^ n/(f*n^3 + 3*f*n^2 + 2*f*n + (f*n^3 + 3*f*n^2 + 2*f*n)*e^(4*I*f*x + 4*I*e) + 2*(f*n^3 + 3*f*n^2 + 2*f*n)*e^(2*I*f*x + 2*I*e))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 979 vs. \(2 (80) = 160\).
Time = 0.82 (sec) , antiderivative size = 979, normalized size of antiderivative = 9.89 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^n \, dx=\text {Too large to display} \] Input:
integrate((a+I*a*tan(f*x+e))**3*(c-I*c*tan(f*x+e))**n,x)
Output:
Piecewise((x*(I*a*tan(e) + a)**3*(-I*c*tan(e) + c)**n, Eq(f, 0)), (2*a**3* f*x*tan(e + f*x)**2/(2*c**2*f*tan(e + f*x)**2 + 4*I*c**2*f*tan(e + f*x) - 2*c**2*f) + 4*I*a**3*f*x*tan(e + f*x)/(2*c**2*f*tan(e + f*x)**2 + 4*I*c**2 *f*tan(e + f*x) - 2*c**2*f) - 2*a**3*f*x/(2*c**2*f*tan(e + f*x)**2 + 4*I*c **2*f*tan(e + f*x) - 2*c**2*f) + I*a**3*log(tan(e + f*x)**2 + 1)*tan(e + f *x)**2/(2*c**2*f*tan(e + f*x)**2 + 4*I*c**2*f*tan(e + f*x) - 2*c**2*f) - 2 *a**3*log(tan(e + f*x)**2 + 1)*tan(e + f*x)/(2*c**2*f*tan(e + f*x)**2 + 4* I*c**2*f*tan(e + f*x) - 2*c**2*f) - I*a**3*log(tan(e + f*x)**2 + 1)/(2*c** 2*f*tan(e + f*x)**2 + 4*I*c**2*f*tan(e + f*x) - 2*c**2*f) - 8*a**3*tan(e + f*x)/(2*c**2*f*tan(e + f*x)**2 + 4*I*c**2*f*tan(e + f*x) - 2*c**2*f) - 4* I*a**3/(2*c**2*f*tan(e + f*x)**2 + 4*I*c**2*f*tan(e + f*x) - 2*c**2*f), Eq (n, -2)), (-4*a**3*f*x*tan(e + f*x)/(c*f*tan(e + f*x) + I*c*f) - 4*I*a**3* f*x/(c*f*tan(e + f*x) + I*c*f) - 2*I*a**3*log(tan(e + f*x)**2 + 1)*tan(e + f*x)/(c*f*tan(e + f*x) + I*c*f) + 2*a**3*log(tan(e + f*x)**2 + 1)/(c*f*ta n(e + f*x) + I*c*f) + a**3*tan(e + f*x)**2/(c*f*tan(e + f*x) + I*c*f) + 5* a**3/(c*f*tan(e + f*x) + I*c*f), Eq(n, -1)), (4*a**3*x + 2*I*a**3*log(tan( e + f*x)**2 + 1)/f - I*a**3*tan(e + f*x)**2/(2*f) - 3*a**3*tan(e + f*x)/f, Eq(n, 0)), (-I*a**3*n**2*(-I*c*tan(e + f*x) + c)**n*tan(e + f*x)**2/(f*n* *3 + 3*f*n**2 + 2*f*n) - 2*a**3*n**2*(-I*c*tan(e + f*x) + c)**n*tan(e + f* x)/(f*n**3 + 3*f*n**2 + 2*f*n) + I*a**3*n**2*(-I*c*tan(e + f*x) + c)**n...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 547 vs. \(2 (87) = 174\).
Time = 0.29 (sec) , antiderivative size = 547, normalized size of antiderivative = 5.53 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^n \, dx =\text {Too large to display} \] Input:
integrate((a+I*a*tan(f*x+e))^3*(c-I*c*tan(f*x+e))^n,x, algorithm="maxima")
Output:
(2^(n + 3)*a^3*c^n*cos(n*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - I*2^(n + 3)*a^3*c^n*sin(n*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1 )) + 8*(a^3*c^n*n + 2*a^3*c^n)*2^n*cos(-2*f*x + n*arctan2(sin(2*f*x + 2*e) , cos(2*f*x + 2*e) + 1) - 2*e) + 4*(a^3*c^n*n^2 + 3*a^3*c^n*n + 2*a^3*c^n) *2^n*cos(-4*f*x + n*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) - 4*e) + 8*(-I*a^3*c^n*n - 2*I*a^3*c^n)*2^n*sin(-2*f*x + n*arctan2(sin(2*f*x + 2 *e), cos(2*f*x + 2*e) + 1) - 2*e) + 4*(-I*a^3*c^n*n^2 - 3*I*a^3*c^n*n - 2* I*a^3*c^n)*2^n*sin(-4*f*x + n*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) - 4*e))/(((-I*n^3 - 3*I*n^2 - 2*I*n)*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/2*n)*cos(4*f*x + 4*e) + (n^3 + 3*n^2 + 2*n)*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^ (1/2*n)*sin(4*f*x + 4*e) + (-I*n^3 - 3*I*n^2 - 2*(I*n^3 + 3*I*n^2 + 2*I*n) *cos(2*f*x + 2*e) + 2*(n^3 + 3*n^2 + 2*n)*sin(2*f*x + 2*e) - 2*I*n)*(cos(2 *f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/2*n))*f)
\[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^n \, dx=\int { {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{n} \,d x } \] Input:
integrate((a+I*a*tan(f*x+e))^3*(c-I*c*tan(f*x+e))^n,x, algorithm="giac")
Output:
integrate((I*a*tan(f*x + e) + a)^3*(-I*c*tan(f*x + e) + c)^n, x)
Time = 1.55 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.32 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^n \, dx=\frac {2\,a^3\,{\left (\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}\right )}^n\,\left (n\,7{}\mathrm {i}+\cos \left (2\,e+2\,f\,x\right )\,16{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,4{}\mathrm {i}-2\,n^2\,\sin \left (2\,e+2\,f\,x\right )-n^2\,\sin \left (4\,e+4\,f\,x\right )+n\,\cos \left (2\,e+2\,f\,x\right )\,10{}\mathrm {i}+n\,\cos \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}-6\,n\,\sin \left (2\,e+2\,f\,x\right )-3\,n\,\sin \left (4\,e+4\,f\,x\right )+n^2\,1{}\mathrm {i}+n^2\,\cos \left (2\,e+2\,f\,x\right )\,2{}\mathrm {i}+n^2\,\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+12{}\mathrm {i}\right )}{f\,n\,\left (4\,\cos \left (2\,e+2\,f\,x\right )+\cos \left (4\,e+4\,f\,x\right )+3\right )\,\left (n^2+3\,n+2\right )} \] Input:
int((a + a*tan(e + f*x)*1i)^3*(c - c*tan(e + f*x)*1i)^n,x)
Output:
(2*a^3*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^n*(n*7i + cos(2*e + 2*f*x)*16i + cos(4*e + 4*f*x)*4i - 2*n^2*sin(2* e + 2*f*x) - n^2*sin(4*e + 4*f*x) + n*cos(2*e + 2*f*x)*10i + n*cos(4*e + 4 *f*x)*3i - 6*n*sin(2*e + 2*f*x) - 3*n*sin(4*e + 4*f*x) + n^2*1i + n^2*cos( 2*e + 2*f*x)*2i + n^2*cos(4*e + 4*f*x)*1i + 12i))/(f*n*(4*cos(2*e + 2*f*x) + cos(4*e + 4*f*x) + 3)*(3*n + n^2 + 2))
Time = 0.20 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.94 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^n \, dx=\frac {\left (-\tan \left (f x +e \right ) c i +c \right )^{n} a^{3} \left (-\tan \left (f x +e \right )^{2} i \,n^{2}-\tan \left (f x +e \right )^{2} i n -2 \tan \left (f x +e \right ) n^{2}-6 \tan \left (f x +e \right ) n +i \,n^{2}+5 i n +8 i \right )}{f n \left (n^{2}+3 n +2\right )} \] Input:
int((a+I*a*tan(f*x+e))^3*(c-I*c*tan(f*x+e))^n,x)
Output:
(( - tan(e + f*x)*c*i + c)**n*a**3*( - tan(e + f*x)**2*i*n**2 - tan(e + f* x)**2*i*n - 2*tan(e + f*x)*n**2 - 6*tan(e + f*x)*n + i*n**2 + 5*i*n + 8*i) )/(f*n*(n**2 + 3*n + 2))