Integrand size = 24, antiderivative size = 55 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {2 i (a+i a \tan (c+d x))^5}{5 a^2 d}+\frac {i (a+i a \tan (c+d x))^6}{6 a^3 d} \] Output:
-2/5*I*(a+I*a*tan(d*x+c))^5/a^2/d+1/6*I*(a+I*a*tan(d*x+c))^6/a^3/d
Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.62 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^3 (7-5 i \tan (c+d x)) (-i+\tan (c+d x))^5}{30 d} \] Input:
Integrate[Sec[c + d*x]^4*(a + I*a*Tan[c + d*x])^3,x]
Output:
(a^3*(7 - (5*I)*Tan[c + d*x])*(-I + Tan[c + d*x])^5)/(30*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.
\(\displaystyle \int \sec ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x)^4 (a+i a \tan (c+d x))^3dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i \int (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^4d(i a \tan (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {i \int \left (2 a (i \tan (c+d x) a+a)^4-(i \tan (c+d x) a+a)^5\right )d(i a \tan (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i \left (\frac {2}{5} a (a+i a \tan (c+d x))^5-\frac {1}{6} (a+i a \tan (c+d x))^6\right )}{a^3 d}\) |
Input:
Int[Sec[c + d*x]^4*(a + I*a*Tan[c + d*x])^3,x]
Output:
((-I)*((2*a*(a + I*a*Tan[c + d*x])^5)/5 - (a + I*a*Tan[c + d*x])^6/6))/(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 = 15.39 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.25
method | result | size |
risch | \(\frac {32 i a^{3} \left (15 \,{\mathrm e}^{8 i \left (d x +c \right )}+20 \,{\mathrm e}^{6 i \left (d x +c \right )}+15 \,{\mathrm e}^{4 i \left (d x +c \right )}+6 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}\) | \(69\) |
derivativedivides | \(\frac {-i a^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{4}}{12 \cos \left (d x +c \right )^{4}}\right )-3 a^{3} \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )+\frac {3 i a^{3}}{4 \cos \left (d x +c \right )^{4}}-a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(128\) |
default | \(\frac {-i a^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{4}}{12 \cos \left (d x +c \right )^{4}}\right )-3 a^{3} \left (\frac {\sin \left (d x +c \right )^{3}}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \sin \left (d x +c \right )^{3}}{15 \cos \left (d x +c \right )^{3}}\right )+\frac {3 i a^{3}}{4 \cos \left (d x +c \right )^{4}}-a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(128\) |
Input:
int(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
32/15*I*a^3*(15*exp(8*I*(d*x+c))+20*exp(6*I*(d*x+c))+15*exp(4*I*(d*x+c))+6 *exp(2*I*(d*x+c))+1)/d/(exp(2*I*(d*x+c))+1)^6
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (43) = 86\).
Time = 0.07 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.53 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {32 \, {\left (-15 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 20 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 15 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 6 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3}\right )}}{15 \, {\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:
integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
Output:
-32/15*(-15*I*a^3*e^(8*I*d*x + 8*I*c) - 20*I*a^3*e^(6*I*d*x + 6*I*c) - 15* I*a^3*e^(4*I*d*x + 4*I*c) - 6*I*a^3*e^(2*I*d*x + 2*I*c) - I*a^3)/(d*e^(12* I*d*x + 12*I*c) + 6*d*e^(10*I*d*x + 10*I*c) + 15*d*e^(8*I*d*x + 8*I*c) + 2 0*d*e^(6*I*d*x + 6*I*c) + 15*d*e^(4*I*d*x + 4*I*c) + 6*d*e^(2*I*d*x + 2*I* c) + d)
\[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=- i a^{3} \left (\int i \sec ^{4}{\left (c + d x \right )}\, dx + \int \left (- 3 \tan {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\right )\, dx + \int \tan ^{3}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \left (- 3 i \tan ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\right )\, dx\right ) \] Input:
integrate(sec(d*x+c)**4*(a+I*a*tan(d*x+c))**3,x)
Output:
-I*a**3*(Integral(I*sec(c + d*x)**4, x) + Integral(-3*tan(c + d*x)*sec(c + d*x)**4, x) + Integral(tan(c + d*x)**3*sec(c + d*x)**4, x) + Integral(-3* I*tan(c + d*x)**2*sec(c + d*x)**4, x))
Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.49 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {5 i \, a^{3} \tan \left (d x + c\right )^{6} + 18 \, a^{3} \tan \left (d x + c\right )^{5} - 15 i \, a^{3} \tan \left (d x + c\right )^{4} + 20 \, a^{3} \tan \left (d x + c\right )^{3} - 45 i \, a^{3} \tan \left (d x + c\right )^{2} - 30 \, a^{3} \tan \left (d x + c\right )}{30 \, d} \] Input:
integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
Output:
-1/30*(5*I*a^3*tan(d*x + c)^6 + 18*a^3*tan(d*x + c)^5 - 15*I*a^3*tan(d*x + c)^4 + 20*a^3*tan(d*x + c)^3 - 45*I*a^3*tan(d*x + c)^2 - 30*a^3*tan(d*x + c))/d
Time = 0.19 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.49 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {5 i \, a^{3} \tan \left (d x + c\right )^{6} + 18 \, a^{3} \tan \left (d x + c\right )^{5} - 15 i \, a^{3} \tan \left (d x + c\right )^{4} + 20 \, a^{3} \tan \left (d x + c\right )^{3} - 45 i \, a^{3} \tan \left (d x + c\right )^{2} - 30 \, a^{3} \tan \left (d x + c\right )}{30 \, d} \] Input:
integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
Output:
-1/30*(5*I*a^3*tan(d*x + c)^6 + 18*a^3*tan(d*x + c)^5 - 15*I*a^3*tan(d*x + c)^4 + 20*a^3*tan(d*x + c)^3 - 45*I*a^3*tan(d*x + c)^2 - 30*a^3*tan(d*x + c))/d
Time = 0.35 (sec) , antiderivative size = 114, normalized size of antiderivative = 2.07 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {a^3\,\sin \left (c+d\,x\right )\,\left (-30\,{\cos \left (c+d\,x\right )}^5-{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )\,45{}\mathrm {i}+20\,{\cos \left (c+d\,x\right )}^3\,{\sin \left (c+d\,x\right )}^2-{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^3\,15{}\mathrm {i}+18\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^4+{\sin \left (c+d\,x\right )}^5\,5{}\mathrm {i}\right )}{30\,d\,{\cos \left (c+d\,x\right )}^6} \] Input:
int((a + a*tan(c + d*x)*1i)^3/cos(c + d*x)^4,x)
Output:
-(a^3*sin(c + d*x)*(18*cos(c + d*x)*sin(c + d*x)^4 - cos(c + d*x)^4*sin(c + d*x)*45i - 30*cos(c + d*x)^5 + sin(c + d*x)^5*5i - cos(c + d*x)^2*sin(c + d*x)^3*15i + 20*cos(c + d*x)^3*sin(c + d*x)^2))/(30*d*cos(c + d*x)^6)
Time = 0.16 (sec) , antiderivative size = 118, normalized size of antiderivative = 2.15 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {\sin \left (d x +c \right ) a^{3} \left (-32 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+80 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-30 \cos \left (d x +c \right )-25 \sin \left (d x +c \right )^{5} i +75 \sin \left (d x +c \right )^{3} i -45 \sin \left (d x +c \right ) i \right )}{30 d \left (\sin \left (d x +c \right )^{6}-3 \sin \left (d x +c \right )^{4}+3 \sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^3,x)
Output:
(sin(c + d*x)*a**3*( - 32*cos(c + d*x)*sin(c + d*x)**4 + 80*cos(c + d*x)*s in(c + d*x)**2 - 30*cos(c + d*x) - 25*sin(c + d*x)**5*i + 75*sin(c + d*x)* *3*i - 45*sin(c + d*x)*i))/(30*d*(sin(c + d*x)**6 - 3*sin(c + d*x)**4 + 3* sin(c + d*x)**2 - 1))