\(\int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) [130]

Optimal result

Integrand size = 24, antiderivative size = 109 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {8 i (a-i a \tan (c+d x))^7}{7 a^{10} d}-\frac {3 i (a-i a \tan (c+d x))^8}{2 a^{11} d}+\frac {2 i (a-i a \tan (c+d x))^9}{3 a^{12} d}-\frac {i (a-i a \tan (c+d x))^{10}}{10 a^{13} d} \]

[Out]

8/7*I*(a-I*a*tan(d*x+c))^7/a^10/d-3/2*I*(a-I*a*tan(d*x+c))^8/a^11/d+2/3*I*(a-I*a*tan(d*x+c))^9/a^12/d-1/10*I*(
a-I*a*tan(d*x+c))^10/a^13/d

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 109, 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} \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {i (a-i a \tan (c+d x))^{10}}{10 a^{13} d}+\frac {2 i (a-i a \tan (c+d x))^9}{3 a^{12} d}-\frac {3 i (a-i a \tan (c+d x))^8}{2 a^{11} d}+\frac {8 i (a-i a \tan (c+d x))^7}{7 a^{10} d} \]

[In]

Int[Sec[c + d*x]^14/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(((8*I)/7)*(a - I*a*Tan[c + d*x])^7)/(a^10*d) - (((3*I)/2)*(a - I*a*Tan[c + d*x])^8)/(a^11*d) + (((2*I)/3)*(a
- I*a*Tan[c + d*x])^9)/(a^12*d) - ((I/10)*(a - I*a*Tan[c + d*x])^10)/(a^13*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)^6 (a+x)^3 \, dx,x,i a \tan (c+d x)\right )}{a^{13} d} \\ & = -\frac {i \text {Subst}\left (\int \left (8 a^3 (a-x)^6-12 a^2 (a-x)^7+6 a (a-x)^8-(a-x)^9\right ) \, dx,x,i a \tan (c+d x)\right )}{a^{13} d} \\ & = \frac {8 i (a-i a \tan (c+d x))^7}{7 a^{10} d}-\frac {3 i (a-i a \tan (c+d x))^8}{2 a^{11} d}+\frac {2 i (a-i a \tan (c+d x))^9}{3 a^{12} d}-\frac {i (a-i a \tan (c+d x))^{10}}{10 a^{13} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.51 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {(i+\tan (c+d x))^7 \left (-44-98 i \tan (c+d x)+77 \tan ^2(c+d x)+21 i \tan ^3(c+d x)\right )}{210 a^3 d} \]

[In]

Integrate[Sec[c + d*x]^14/(a + I*a*Tan[c + d*x])^3,x]

[Out]

((I + Tan[c + d*x])^7*(-44 - (98*I)*Tan[c + d*x] + 77*Tan[c + d*x]^2 + (21*I)*Tan[c + d*x]^3))/(210*a^3*d)

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.53

method	result	size
risch	\(\frac {128 i \left (120 \,{\mathrm e}^{6 i \left (d x +c \right )}+45 \,{\mathrm e}^{4 i \left (d x +c \right )}+10 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{105 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{10}}\)	\(58\)
derivativedivides	\(-\frac {i \left (i \tan \left (d x +c \right )-\frac {\left (\tan ^{10}\left (d x +c \right )\right )}{10}-\frac {i \left (\tan ^{9}\left (d x +c \right )\right )}{3}-\frac {8 i \left (\tan ^{7}\left (d x +c \right )\right )}{7}+\tan ^{6}\left (d x +c \right )-\frac {6 i \left (\tan ^{5}\left (d x +c \right )\right )}{5}+2 \left (\tan ^{4}\left (d x +c \right )\right )+\frac {3 \left (\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{a^{3} d}\)	\(91\)
default	\(-\frac {i \left (i \tan \left (d x +c \right )-\frac {\left (\tan ^{10}\left (d x +c \right )\right )}{10}-\frac {i \left (\tan ^{9}\left (d x +c \right )\right )}{3}-\frac {8 i \left (\tan ^{7}\left (d x +c \right )\right )}{7}+\tan ^{6}\left (d x +c \right )-\frac {6 i \left (\tan ^{5}\left (d x +c \right )\right )}{5}+2 \left (\tan ^{4}\left (d x +c \right )\right )+\frac {3 \left (\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{a^{3} d}\)	\(91\)

[In]

int(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

128/105*I*(120*exp(6*I*(d*x+c))+45*exp(4*I*(d*x+c))+10*exp(2*I*(d*x+c))+1)/d/a^3/(exp(2*I*(d*x+c))+1)^10

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. 194 vs. \(2 (85) = 170\).

Time = 0.25 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.78 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {128 \, {\left (-120 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 45 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 10 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{105 \, {\left (a^{3} d e^{\left (20 i \, d x + 20 i \, c\right )} + 10 \, a^{3} d e^{\left (18 i \, d x + 18 i \, c\right )} + 45 \, a^{3} d e^{\left (16 i \, d x + 16 i \, c\right )} + 120 \, a^{3} d e^{\left (14 i \, d x + 14 i \, c\right )} + 210 \, a^{3} d e^{\left (12 i \, d x + 12 i \, c\right )} + 252 \, a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 210 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 120 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 45 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 10 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \]

[In]

integrate(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-128/105*(-120*I*e^(6*I*d*x + 6*I*c) - 45*I*e^(4*I*d*x + 4*I*c) - 10*I*e^(2*I*d*x + 2*I*c) - I)/(a^3*d*e^(20*I
*d*x + 20*I*c) + 10*a^3*d*e^(18*I*d*x + 18*I*c) + 45*a^3*d*e^(16*I*d*x + 16*I*c) + 120*a^3*d*e^(14*I*d*x + 14*
I*c) + 210*a^3*d*e^(12*I*d*x + 12*I*c) + 252*a^3*d*e^(10*I*d*x + 10*I*c) + 210*a^3*d*e^(8*I*d*x + 8*I*c) + 120
*a^3*d*e^(6*I*d*x + 6*I*c) + 45*a^3*d*e^(4*I*d*x + 4*I*c) + 10*a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)

Sympy [F]

\[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {i \int \frac {\sec ^{14}{\left (c + d x \right )}}{\tan ^{3}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} + i}\, dx}{a^{3}} \]

[In]

integrate(sec(d*x+c)**14/(a+I*a*tan(d*x+c))**3,x)

[Out]

I*Integral(sec(c + d*x)**14/(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 = 0.22 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.80 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {-21 i \, \tan \left (d x + c\right )^{10} + 70 \, \tan \left (d x + c\right )^{9} + 240 \, \tan \left (d x + c\right )^{7} + 210 i \, \tan \left (d x + c\right )^{6} + 252 \, \tan \left (d x + c\right )^{5} + 420 i \, \tan \left (d x + c\right )^{4} + 315 i \, \tan \left (d x + c\right )^{2} - 210 \, \tan \left (d x + c\right )}{210 \, a^{3} d} \]

[In]

integrate(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/210*(-21*I*tan(d*x + c)^10 + 70*tan(d*x + c)^9 + 240*tan(d*x + c)^7 + 210*I*tan(d*x + c)^6 + 252*tan(d*x +
c)^5 + 420*I*tan(d*x + c)^4 + 315*I*tan(d*x + c)^2 - 210*tan(d*x + c))/(a^3*d)

Giac [A] (verification not implemented)

none

Time = 0.68 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.80 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {-21 i \, \tan \left (d x + c\right )^{10} + 70 \, \tan \left (d x + c\right )^{9} + 240 \, \tan \left (d x + c\right )^{7} + 210 i \, \tan \left (d x + c\right )^{6} + 252 \, \tan \left (d x + c\right )^{5} + 420 i \, \tan \left (d x + c\right )^{4} + 315 i \, \tan \left (d x + c\right )^{2} - 210 \, \tan \left (d x + c\right )}{210 \, a^{3} d} \]

[In]

integrate(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/210*(-21*I*tan(d*x + c)^10 + 70*tan(d*x + c)^9 + 240*tan(d*x + c)^7 + 210*I*tan(d*x + c)^6 + 252*tan(d*x +
c)^5 + 420*I*tan(d*x + c)^4 + 315*I*tan(d*x + c)^2 - 210*tan(d*x + c))/(a^3*d)

Mupad [B] (verification not implemented)

Time = 4.54 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.09 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\cos \left (c+d\,x\right )}^{10}\,84{}\mathrm {i}+128\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^9+64\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^7+48\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^5+40\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^3-{\cos \left (c+d\,x\right )}^2\,105{}\mathrm {i}-70\,\sin \left (c+d\,x\right )\,\cos \left (c+d\,x\right )+21{}\mathrm {i}}{210\,a^3\,d\,{\cos \left (c+d\,x\right )}^{10}} \]

[In]

int(1/(cos(c + d*x)^14*(a + a*tan(c + d*x)*1i)^3),x)

[Out]

(40*cos(c + d*x)^3*sin(c + d*x) - 70*cos(c + d*x)*sin(c + d*x) + 48*cos(c + d*x)^5*sin(c + d*x) + 64*cos(c + d
*x)^7*sin(c + d*x) + 128*cos(c + d*x)^9*sin(c + d*x) - cos(c + d*x)^2*105i + cos(c + d*x)^10*84i + 21i)/(210*a
^3*d*cos(c + d*x)^10)