Integrand size = 30, antiderivative size = 98 \[ \int (e \sec (c+d x))^{4-2 n} (a+i a \tan (c+d x))^n \, dx=\frac {2 i a^2 (e \sec (c+d x))^{4-2 n} (a+i a \tan (c+d x))^{-2+n}}{d \left (6-5 n+n^2\right )}+\frac {i a (e \sec (c+d x))^{4-2 n} (a+i a \tan (c+d x))^{-1+n}}{d (3-n)} \] Output:
2*I*a^2*(e*sec(d*x+c))^(4-2*n)*(a+I*a*tan(d*x+c))^(-2+n)/d/(n^2-5*n+6)+I*a *(e*sec(d*x+c))^(4-2*n)*(a+I*a*tan(d*x+c))^(-1+n)/d/(3-n)
Time = 1.99 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.93 \[ \int (e \sec (c+d x))^{4-2 n} (a+i a \tan (c+d x))^n \, dx=\frac {e^4 \sec ^2(c+d x) (e \sec (c+d x))^{-2 n} (\cos (2 (c+d x))-i \sin (2 (c+d x))) (a+i a \tan (c+d x))^n (-i (-4+n)+(-2+n) \tan (c+d x))}{d (-3+n) (-2+n)} \] Input:
Integrate[(e*Sec[c + d*x])^(4 - 2*n)*(a + I*a*Tan[c + d*x])^n,x]
Output:
(e^4*Sec[c + d*x]^2*(Cos[2*(c + d*x)] - I*Sin[2*(c + d*x)])*(a + I*a*Tan[c + d*x])^n*((-I)*(-4 + n) + (-2 + n)*Tan[c + d*x]))/(d*(-3 + n)*(-2 + n)*( e*Sec[c + d*x])^(2*n))
Time = 0.43 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3042, 3975, 3042, 3974}
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.
\(\displaystyle \int (a+i a \tan (c+d x))^n (e \sec (c+d x))^{4-2 n} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+i a \tan (c+d x))^n (e \sec (c+d x))^{4-2 n}dx\) |
\(\Big \downarrow \) 3975 |
\(\displaystyle \frac {2 a \int (e \sec (c+d x))^{4-2 n} (i \tan (c+d x) a+a)^{n-1}dx}{3-n}+\frac {i a (a+i a \tan (c+d x))^{n-1} (e \sec (c+d x))^{4-2 n}}{d (3-n)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 a \int (e \sec (c+d x))^{4-2 n} (i \tan (c+d x) a+a)^{n-1}dx}{3-n}+\frac {i a (a+i a \tan (c+d x))^{n-1} (e \sec (c+d x))^{4-2 n}}{d (3-n)}\) |
\(\Big \downarrow \) 3974 |
\(\displaystyle \frac {2 i a^2 (a+i a \tan (c+d x))^{n-2} (e \sec (c+d x))^{4-2 n}}{d (2-n) (3-n)}+\frac {i a (a+i a \tan (c+d x))^{n-1} (e \sec (c+d x))^{4-2 n}}{d (3-n)}\) |
Input:
Int[(e*Sec[c + d*x])^(4 - 2*n)*(a + I*a*Tan[c + d*x])^n,x]
Output:
((2*I)*a^2*(e*Sec[c + d*x])^(4 - 2*n)*(a + I*a*Tan[c + d*x])^(-2 + n))/(d* (2 - n)*(3 - n)) + (I*a*(e*Sec[c + d*x])^(4 - 2*n)*(a + I*a*Tan[c + d*x])^ (-1 + n))/(d*(3 - n))
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] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] && EqQ[Simplify[m/2 + n - 1], 0]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_), x_Symbol] :> Simp[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] + Simp[a*((m + 2*n - 2)/(m + n - 1)) Int[(d*Se c[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] && IGtQ[Simplify[m/2 + n - 1], 0] && !Inte gerQ[n]
\[\int \left (e \sec \left (d x +c \right )\right )^{4-2 n} \left (a +i a \tan \left (d x +c \right )\right )^{n}d x\]
Input:
int((e*sec(d*x+c))^(4-2*n)*(a+I*a*tan(d*x+c))^n,x)
Output:
int((e*sec(d*x+c))^(4-2*n)*(a+I*a*tan(d*x+c))^n,x)
Time = 0.09 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.37 \[ \int (e \sec (c+d x))^{4-2 n} (a+i a \tan (c+d x))^n \, dx=\frac {{\left ({\left (-i \, n + 3 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-i \, n + 4 i\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} \left (\frac {2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{-2 \, n + 4} e^{\left (i \, d n x + i \, c n - 4 i \, d x + n \log \left (\frac {2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right ) + n \log \left (\frac {a}{e}\right ) - 4 i \, c\right )}}{2 \, {\left (d n^{2} - 5 \, d n + 6 \, d\right )}} \] Input:
integrate((e*sec(d*x+c))^(4-2*n)*(a+I*a*tan(d*x+c))^n,x, algorithm="fricas ")
Output:
1/2*((-I*n + 3*I)*e^(4*I*d*x + 4*I*c) + (-I*n + 4*I)*e^(2*I*d*x + 2*I*c) + I)*(2*e*e^(I*d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1))^(-2*n + 4)*e^(I*d*n*x + I*c*n - 4*I*d*x + n*log(2*e*e^(I*d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1)) + n*log(a/e) - 4*I*c)/(d*n^2 - 5*d*n + 6*d)
\[ \int (e \sec (c+d x))^{4-2 n} (a+i a \tan (c+d x))^n \, dx=\int \left (e \sec {\left (c + d x \right )}\right )^{4 - 2 n} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n}\, dx \] Input:
integrate((e*sec(d*x+c))**(4-2*n)*(a+I*a*tan(d*x+c))**n,x)
Output:
Integral((e*sec(c + d*x))**(4 - 2*n)*(I*a*(tan(c + d*x) - I))**n, x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 595 vs. \(2 (88) = 176\).
Time = 0.37 (sec) , antiderivative size = 595, normalized size of antiderivative = 6.07 \[ \int (e \sec (c+d x))^{4-2 n} (a+i a \tan (c+d x))^n \, dx =\text {Too large to display} \] Input:
integrate((e*sec(d*x+c))^(4-2*n)*(a+I*a*tan(d*x+c))^n,x, algorithm="maxima ")
Output:
8*((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/2 *n)*a^n*e^4*cos(n*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + I*(co s(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/2*n)*a^ n*e^4*sin(n*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - (a^n*e^4*n - 3*a^n*e^4)*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/2*n)*cos(2*d*x + n*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1 ) + 2*c) + (-I*a^n*e^4*n + 3*I*a^n*e^4)*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/2*n)*sin(2*d*x + n*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1) + 2*c))/(((-I*e^(2*n)*n^2 + 5*I*e^(2*n)*n - 6 *I*e^(2*n))*2^n*cos(6*d*x + 6*c) - 3*(I*e^(2*n)*n^2 - 5*I*e^(2*n)*n + 6*I* e^(2*n))*2^n*cos(4*d*x + 4*c) - 3*(I*e^(2*n)*n^2 - 5*I*e^(2*n)*n + 6*I*e^( 2*n))*2^n*cos(2*d*x + 2*c) + (e^(2*n)*n^2 - 5*e^(2*n)*n + 6*e^(2*n))*2^n*s in(6*d*x + 6*c) + 3*(e^(2*n)*n^2 - 5*e^(2*n)*n + 6*e^(2*n))*2^n*sin(4*d*x + 4*c) + 3*(e^(2*n)*n^2 - 5*e^(2*n)*n + 6*e^(2*n))*2^n*sin(2*d*x + 2*c) + (-I*e^(2*n)*n^2 + 5*I*e^(2*n)*n - 6*I*e^(2*n))*2^n)*d)
\[ \int (e \sec (c+d x))^{4-2 n} (a+i a \tan (c+d x))^n \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{-2 \, n + 4} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \] Input:
integrate((e*sec(d*x+c))^(4-2*n)*(a+I*a*tan(d*x+c))^n,x, algorithm="giac")
Output:
integrate((e*sec(d*x + c))^(-2*n + 4)*(I*a*tan(d*x + c) + a)^n, x)
Time = 2.95 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.78 \[ \int (e \sec (c+d x))^{4-2 n} (a+i a \tan (c+d x))^n \, dx=\frac {4\,e^4\,{\left (\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}\right )}^n\,\left (4\,\sin \left (2\,c+2\,d\,x\right )+\cos \left (2\,c+2\,d\,x\right )\,4{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,1{}\mathrm {i}-n\,1{}\mathrm {i}+\sin \left (4\,c+4\,d\,x\right )-n\,\cos \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}-n\,\sin \left (2\,c+2\,d\,x\right )+3{}\mathrm {i}\right )}{d\,{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{2\,n}\,\left (4\,\cos \left (2\,c+2\,d\,x\right )+\cos \left (4\,c+4\,d\,x\right )+3\right )\,\left (n^2-5\,n+6\right )} \] Input:
int((e/cos(c + d*x))^(4 - 2*n)*(a + a*tan(c + d*x)*1i)^n,x)
Output:
(4*e^4*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^n*(cos(2*c + 2*d*x)*4i - n*1i + cos(4*c + 4*d*x)*1i + 4*sin(2*c + 2 *d*x) + sin(4*c + 4*d*x) - n*cos(2*c + 2*d*x)*1i - n*sin(2*c + 2*d*x) + 3i ))/(d*(e/cos(c + d*x))^(2*n)*(4*cos(2*c + 2*d*x) + cos(4*c + 4*d*x) + 3)*( n^2 - 5*n + 6))
\[ \int (e \sec (c+d x))^{4-2 n} (a+i a \tan (c+d x))^n \, dx=\frac {\left (\int \frac {\left (\tan \left (d x +c \right ) a i +a \right )^{n} \sec \left (d x +c \right )^{4}}{\sec \left (d x +c \right )^{2 n}}d x \right ) e^{4}}{e^{2 n}} \] Input:
int((e*sec(d*x+c))^(4-2*n)*(a+I*a*tan(d*x+c))^n,x)
Output:
(int(((tan(c + d*x)*a*i + a)**n*sec(c + d*x)**4)/sec(c + d*x)**(2*n),x)*e* *4)/e**(2*n)