\(\int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) [132]

Optimal result

Integrand size = 24, antiderivative size = 55 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {2 i (a-i a \tan (c+d x))^5}{5 a^8 d}-\frac {i (a-i a \tan (c+d x))^6}{6 a^9 d} \]

[Out]

2/5*I*(a-I*a*tan(d*x+c))^5/a^8/d-1/6*I*(a-I*a*tan(d*x+c))^6/a^9/d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3568, 45} \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {2 i (a-i a \tan (c+d x))^5}{5 a^8 d}-\frac {i (a-i a \tan (c+d x))^6}{6 a^9 d} \]

[In]

Int[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(((2*I)/5)*(a - I*a*Tan[c + d*x])^5)/(a^8*d) - ((I/6)*(a - I*a*Tan[c + d*x])^6)/(a^9*d)

Rule 45

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] && NeQ[b*c - a*d, 0] && 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])

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {i \text {Subst}\left (\int (a-x)^4 (a+x) \, dx,x,i a \tan (c+d x)\right )}{a^9 d} \\ & = -\frac {i \text {Subst}\left (\int \left (2 a (a-x)^4-(a-x)^5\right ) \, dx,x,i a \tan (c+d x)\right )}{a^9 d} \\ & = \frac {2 i (a-i a \tan (c+d x))^5}{5 a^8 d}-\frac {i (a-i a \tan (c+d x))^6}{6 a^9 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.62 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {(7+5 i \tan (c+d x)) (i+\tan (c+d x))^5}{30 a^3 d} \]

[In]

Integrate[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^3,x]

[Out]

((7 + (5*I)*Tan[c + d*x])*(I + Tan[c + d*x])^5)/(30*a^3*d)

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65

method	result	size
risch	\(\frac {32 i \left (6 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{15 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}\)	\(36\)
derivativedivides	\(-\frac {i \left (i \tan \left (d x +c \right )-\frac {\left (\tan ^{6}\left (d x +c \right )\right )}{6}-\frac {3 i \left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{2}-\frac {2 i \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {3 \left (\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{a^{3} d}\)	\(72\)
default	\(-\frac {i \left (i \tan \left (d x +c \right )-\frac {\left (\tan ^{6}\left (d x +c \right )\right )}{6}-\frac {3 i \left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{2}-\frac {2 i \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {3 \left (\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{a^{3} d}\)	\(72\)

[In]

int(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

32/15*I*(6*exp(2*I*(d*x+c))+1)/d/a^3/(exp(2*I*(d*x+c))+1)^6

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (43) = 86\).

Time = 0.24 (sec) , antiderivative size = 112, normalized size of antiderivative = 2.04 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {32 \, {\left (-6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{15 \, {\left (a^{3} d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \]

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-32/15*(-6*I*e^(2*I*d*x + 2*I*c) - I)/(a^3*d*e^(12*I*d*x + 12*I*c) + 6*a^3*d*e^(10*I*d*x + 10*I*c) + 15*a^3*d*
e^(8*I*d*x + 8*I*c) + 20*a^3*d*e^(6*I*d*x + 6*I*c) + 15*a^3*d*e^(4*I*d*x + 4*I*c) + 6*a^3*d*e^(2*I*d*x + 2*I*c
) + a^3*d)

Sympy [F]

\[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {i \int \frac {\sec ^{10}{\left (c + d x \right )}}{\tan ^{3}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} + i}\, dx}{a^{3}} \]

[In]

integrate(sec(d*x+c)**10/(a+I*a*tan(d*x+c))**3,x)

[Out]

I*Integral(sec(c + d*x)**10/(tan(c + d*x)**3 - 3*I*tan(c + d*x)**2 - 3*tan(c + d*x) + I), x)/a**3

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.22 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {5 i \, \tan \left (d x + c\right )^{6} - 18 \, \tan \left (d x + c\right )^{5} - 15 i \, \tan \left (d x + c\right )^{4} - 20 \, \tan \left (d x + c\right )^{3} - 45 i \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right )}{30 \, a^{3} d} \]

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/30*(5*I*tan(d*x + c)^6 - 18*tan(d*x + c)^5 - 15*I*tan(d*x + c)^4 - 20*tan(d*x + c)^3 - 45*I*tan(d*x + c)^2 +
30*tan(d*x + c))/(a^3*d)

Giac [A] (verification not implemented)

none

Time = 0.64 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.22 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {-5 i \, \tan \left (d x + c\right )^{6} + 18 \, \tan \left (d x + c\right )^{5} + 15 i \, \tan \left (d x + c\right )^{4} + 20 \, \tan \left (d x + c\right )^{3} + 45 i \, \tan \left (d x + c\right )^{2} - 30 \, \tan \left (d x + c\right )}{30 \, a^{3} d} \]

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/30*(-5*I*tan(d*x + c)^6 + 18*tan(d*x + c)^5 + 15*I*tan(d*x + c)^4 + 20*tan(d*x + c)^3 + 45*I*tan(d*x + c)^2
- 30*tan(d*x + c))/(a^3*d)

Mupad [B] (verification not implemented)

Time = 3.79 (sec) , antiderivative size = 114, normalized size of antiderivative = 2.07 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\sin \left (c+d\,x\right )\,\left (-30\,{\cos \left (c+d\,x\right )}^5+{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )\,45{}\mathrm {i}+20\,{\cos \left (c+d\,x\right )}^3\,{\sin \left (c+d\,x\right )}^2+{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^3\,15{}\mathrm {i}+18\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^4-{\sin \left (c+d\,x\right )}^5\,5{}\mathrm {i}\right )}{30\,a^3\,d\,{\cos \left (c+d\,x\right )}^6} \]

[In]

int(1/(cos(c + d*x)^10*(a + a*tan(c + d*x)*1i)^3),x)

[Out]

-(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*a^3*d*cos(c + d*x)^6)