Integrand size = 24, antiderivative size = 55 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {2 i (a+i a \tan (c+d x))^7}{7 a^2 d}+\frac {i (a+i a \tan (c+d x))^8}{8 a^3 d} \] Output:
-2/7*I*(a+I*a*tan(d*x+c))^7/a^2/d+1/8*I*(a+I*a*tan(d*x+c))^8/a^3/d
Time = 0.35 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {i a^5 (-i+\tan (c+d x))^7 (9 i+7 \tan (c+d x))}{56 d} \] Input:
Integrate[Sec[c + d*x]^4*(a + I*a*Tan[c + d*x])^5,x]
Output:
((I/56)*a^5*(-I + Tan[c + d*x])^7*(9*I + 7*Tan[c + d*x]))/d
Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3968, 49, 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.
\(\displaystyle \int \sec ^4(c+d x) (a+i a \tan (c+d x))^5 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x)^4 (a+i a \tan (c+d x))^5dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i \int (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^6d(i a \tan (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {i \int \left (2 a (i \tan (c+d x) a+a)^6-(i \tan (c+d x) a+a)^7\right )d(i a \tan (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i \left (\frac {2}{7} a (a+i a \tan (c+d x))^7-\frac {1}{8} (a+i a \tan (c+d x))^8\right )}{a^3 d}\) |
Input:
Int[Sec[c + d*x]^4*(a + I*a*Tan[c + d*x])^5,x]
Output:
((-I)*((2*a*(a + I*a*Tan[c + d*x])^7)/7 - (a + I*a*Tan[c + d*x])^8/8))/(a^ 3*d)
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}, x] && IGtQ[m, 0] && IGtQ[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]
Time = 59.60 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.65
method | result | size |
risch | \(\frac {32 i a^{5} \left (28 \,{\mathrm e}^{12 i \left (d x +c \right )}+56 \,{\mathrm e}^{10 i \left (d x +c \right )}+70 \,{\mathrm e}^{8 i \left (d x +c \right )}+56 \,{\mathrm e}^{6 i \left (d x +c \right )}+28 \,{\mathrm e}^{4 i \left (d x +c \right )}+8 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{7 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{8}}\) | \(91\) |
derivativedivides | \(\frac {i a^{5} \left (\frac {\sin \left (d x +c \right )^{6}}{8 \cos \left (d x +c \right )^{8}}+\frac {\sin \left (d x +c \right )^{6}}{24 \cos \left (d x +c \right )^{6}}\right )+5 a^{5} \left (\frac {\sin \left (d x +c \right )^{5}}{7 \cos \left (d x +c \right )^{7}}+\frac {2 \sin \left (d x +c \right )^{5}}{35 \cos \left (d x +c \right )^{5}}\right )-10 i a^{5} \left (\frac {\sin \left (d x +c \right )^{4}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{4}}{12 \cos \left (d x +c \right )^{4}}\right )-10 a^{5} \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )+\frac {5 i a^{5}}{4 \cos \left (d x +c \right )^{4}}-a^{5} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(213\) |
default | \(\frac {i a^{5} \left (\frac {\sin \left (d x +c \right )^{6}}{8 \cos \left (d x +c \right )^{8}}+\frac {\sin \left (d x +c \right )^{6}}{24 \cos \left (d x +c \right )^{6}}\right )+5 a^{5} \left (\frac {\sin \left (d x +c \right )^{5}}{7 \cos \left (d x +c \right )^{7}}+\frac {2 \sin \left (d x +c \right )^{5}}{35 \cos \left (d x +c \right )^{5}}\right )-10 i a^{5} \left (\frac {\sin \left (d x +c \right )^{4}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{4}}{12 \cos \left (d x +c \right )^{4}}\right )-10 a^{5} \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )+\frac {5 i a^{5}}{4 \cos \left (d x +c \right )^{4}}-a^{5} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(213\) |
Input:
int(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^5,x,method=_RETURNVERBOSE)
Output:
32/7*I*a^5*(28*exp(12*I*(d*x+c))+56*exp(10*I*(d*x+c))+70*exp(8*I*(d*x+c))+ 56*exp(6*I*(d*x+c))+28*exp(4*I*(d*x+c))+8*exp(2*I*(d*x+c))+1)/d/(exp(2*I*( d*x+c))+1)^8
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (43) = 86\).
Time = 0.08 (sec) , antiderivative size = 191, normalized size of antiderivative = 3.47 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {32 \, {\left (-28 i \, a^{5} e^{\left (12 i \, d x + 12 i \, c\right )} - 56 i \, a^{5} e^{\left (10 i \, d x + 10 i \, c\right )} - 70 i \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} - 56 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} - 28 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} - 8 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{5}\right )}}{7 \, {\left (d e^{\left (16 i \, d x + 16 i \, c\right )} + 8 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 28 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 56 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 70 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 56 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 28 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:
integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^5,x, algorithm="fricas")
Output:
-32/7*(-28*I*a^5*e^(12*I*d*x + 12*I*c) - 56*I*a^5*e^(10*I*d*x + 10*I*c) - 70*I*a^5*e^(8*I*d*x + 8*I*c) - 56*I*a^5*e^(6*I*d*x + 6*I*c) - 28*I*a^5*e^( 4*I*d*x + 4*I*c) - 8*I*a^5*e^(2*I*d*x + 2*I*c) - I*a^5)/(d*e^(16*I*d*x + 1 6*I*c) + 8*d*e^(14*I*d*x + 14*I*c) + 28*d*e^(12*I*d*x + 12*I*c) + 56*d*e^( 10*I*d*x + 10*I*c) + 70*d*e^(8*I*d*x + 8*I*c) + 56*d*e^(6*I*d*x + 6*I*c) + 28*d*e^(4*I*d*x + 4*I*c) + 8*d*e^(2*I*d*x + 2*I*c) + d)
\[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^5 \, dx=i a^{5} \left (\int \left (- i \sec ^{4}{\left (c + d x \right )}\right )\, dx + \int 5 \tan {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \left (- 10 \tan ^{3}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\right )\, dx + \int \tan ^{5}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int 10 i \tan ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \left (- 5 i \tan ^{4}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\right )\, dx\right ) \] Input:
integrate(sec(d*x+c)**4*(a+I*a*tan(d*x+c))**5,x)
Output:
I*a**5*(Integral(-I*sec(c + d*x)**4, x) + Integral(5*tan(c + d*x)*sec(c + d*x)**4, x) + Integral(-10*tan(c + d*x)**3*sec(c + d*x)**4, x) + Integral( tan(c + d*x)**5*sec(c + d*x)**4, x) + Integral(10*I*tan(c + d*x)**2*sec(c + d*x)**4, x) + Integral(-5*I*tan(c + d*x)**4*sec(c + d*x)**4, x))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (43) = 86\).
Time = 0.04 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.96 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {-7 i \, a^{5} \tan \left (d x + c\right )^{8} - 40 \, a^{5} \tan \left (d x + c\right )^{7} + 84 i \, a^{5} \tan \left (d x + c\right )^{6} + 56 \, a^{5} \tan \left (d x + c\right )^{5} + 70 i \, a^{5} \tan \left (d x + c\right )^{4} + 168 \, a^{5} \tan \left (d x + c\right )^{3} - 140 i \, a^{5} \tan \left (d x + c\right )^{2} - 56 \, a^{5} \tan \left (d x + c\right )}{56 \, d} \] Input:
integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^5,x, algorithm="maxima")
Output:
-1/56*(-7*I*a^5*tan(d*x + c)^8 - 40*a^5*tan(d*x + c)^7 + 84*I*a^5*tan(d*x + c)^6 + 56*a^5*tan(d*x + c)^5 + 70*I*a^5*tan(d*x + c)^4 + 168*a^5*tan(d*x + c)^3 - 140*I*a^5*tan(d*x + c)^2 - 56*a^5*tan(d*x + c))/d
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (43) = 86\).
Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.96 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {-7 i \, a^{5} \tan \left (d x + c\right )^{8} - 40 \, a^{5} \tan \left (d x + c\right )^{7} + 84 i \, a^{5} \tan \left (d x + c\right )^{6} + 56 \, a^{5} \tan \left (d x + c\right )^{5} + 70 i \, a^{5} \tan \left (d x + c\right )^{4} + 168 \, a^{5} \tan \left (d x + c\right )^{3} - 140 i \, a^{5} \tan \left (d x + c\right )^{2} - 56 \, a^{5} \tan \left (d x + c\right )}{56 \, d} \] Input:
integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^5,x, algorithm="giac")
Output:
-1/56*(-7*I*a^5*tan(d*x + c)^8 - 40*a^5*tan(d*x + c)^7 + 84*I*a^5*tan(d*x + c)^6 + 56*a^5*tan(d*x + c)^5 + 70*I*a^5*tan(d*x + c)^4 + 168*a^5*tan(d*x + c)^3 - 140*I*a^5*tan(d*x + c)^2 - 56*a^5*tan(d*x + c))/d
Time = 0.41 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.75 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {a^5\,\sin \left (c+d\,x\right )\,\left (56\,{\cos \left (c+d\,x\right )}^7+{\cos \left (c+d\,x\right )}^6\,\sin \left (c+d\,x\right )\,140{}\mathrm {i}-168\,{\cos \left (c+d\,x\right )}^5\,{\sin \left (c+d\,x\right )}^2-{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^3\,70{}\mathrm {i}-56\,{\cos \left (c+d\,x\right )}^3\,{\sin \left (c+d\,x\right )}^4-{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^5\,84{}\mathrm {i}+40\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^6+{\sin \left (c+d\,x\right )}^7\,7{}\mathrm {i}\right )}{56\,d\,{\cos \left (c+d\,x\right )}^8} \] Input:
int((a + a*tan(c + d*x)*1i)^5/cos(c + d*x)^4,x)
Output:
(a^5*sin(c + d*x)*(40*cos(c + d*x)*sin(c + d*x)^6 + cos(c + d*x)^6*sin(c + d*x)*140i + 56*cos(c + d*x)^7 + sin(c + d*x)^7*7i - cos(c + d*x)^2*sin(c + d*x)^5*84i - 56*cos(c + d*x)^3*sin(c + d*x)^4 - cos(c + d*x)^4*sin(c + d *x)^3*70i - 168*cos(c + d*x)^5*sin(c + d*x)^2))/(56*d*cos(c + d*x)^8)
Time = 0.16 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.82 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {\sin \left (d x +c \right ) a^{5} \left (-128 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}+448 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-336 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+56 \cos \left (d x +c \right )-119 \sin \left (d x +c \right )^{7} i +476 \sin \left (d x +c \right )^{5} i -490 \sin \left (d x +c \right )^{3} i +140 \sin \left (d x +c \right ) i \right )}{56 d \left (\sin \left (d x +c \right )^{8}-4 \sin \left (d x +c \right )^{6}+6 \sin \left (d x +c \right )^{4}-4 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^5,x)
Output:
(sin(c + d*x)*a**5*( - 128*cos(c + d*x)*sin(c + d*x)**6 + 448*cos(c + d*x) *sin(c + d*x)**4 - 336*cos(c + d*x)*sin(c + d*x)**2 + 56*cos(c + d*x) - 11 9*sin(c + d*x)**7*i + 476*sin(c + d*x)**5*i - 490*sin(c + d*x)**3*i + 140* sin(c + d*x)*i))/(56*d*(sin(c + d*x)**8 - 4*sin(c + d*x)**6 + 6*sin(c + d* x)**4 - 4*sin(c + d*x)**2 + 1))