\(\int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\) [114]

Optimal result

Integrand size = 24, antiderivative size = 82 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {4 i (a-i a \tan (c+d x))^5}{5 a^7 d}-\frac {2 i (a-i a \tan (c+d x))^6}{3 a^8 d}+\frac {i (a-i a \tan (c+d x))^7}{7 a^9 d} \]

[Out]

4/5*I*(a-I*a*tan(d*x+c))^5/a^7/d-2/3*I*(a-I*a*tan(d*x+c))^6/a^8/d+1/7*I*(a-I*a*tan(d*x+c))^7/a^9/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 82, 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} \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {i (a-i a \tan (c+d x))^7}{7 a^9 d}-\frac {2 i (a-i a \tan (c+d x))^6}{3 a^8 d}+\frac {4 i (a-i a \tan (c+d x))^5}{5 a^7 d} \]

[In]

Int[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(((4*I)/5)*(a - I*a*Tan[c + d*x])^5)/(a^7*d) - (((2*I)/3)*(a - I*a*Tan[c + d*x])^6)/(a^8*d) + ((I/7)*(a - I*a*
Tan[c + d*x])^7)/(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)^2 \, dx,x,i a \tan (c+d x)\right )}{a^9 d} \\ & = -\frac {i \text {Subst}\left (\int \left (4 a^2 (a-x)^4-4 a (a-x)^5+(a-x)^6\right ) \, dx,x,i a \tan (c+d x)\right )}{a^9 d} \\ & = \frac {4 i (a-i a \tan (c+d x))^5}{5 a^7 d}-\frac {2 i (a-i a \tan (c+d x))^6}{3 a^8 d}+\frac {i (a-i a \tan (c+d x))^7}{7 a^9 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.54 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {(i+\tan (c+d x))^5 \left (-29-40 i \tan (c+d x)+15 \tan ^2(c+d x)\right )}{105 a^2 d} \]

[In]

Integrate[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^2,x]

[Out]

-1/105*((I + Tan[c + d*x])^5*(-29 - (40*I)*Tan[c + d*x] + 15*Tan[c + d*x]^2))/(a^2*d)

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.57

method	result	size
risch	\(\frac {128 i \left (21 \,{\mathrm e}^{4 i \left (d x +c \right )}+7 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{105 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{7}}\)	\(47\)
derivativedivides	\(\frac {\tan \left (d x +c \right )-\frac {\left (\tan ^{7}\left (d x +c \right )\right )}{7}-\frac {i \left (\tan ^{6}\left (d x +c \right )\right )}{3}-\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-i \left (\tan ^{4}\left (d x +c \right )\right )+\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-i \left (\tan ^{2}\left (d x +c \right )\right )}{a^{2} d}\)	\(78\)
default	\(\frac {\tan \left (d x +c \right )-\frac {\left (\tan ^{7}\left (d x +c \right )\right )}{7}-\frac {i \left (\tan ^{6}\left (d x +c \right )\right )}{3}-\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-i \left (\tan ^{4}\left (d x +c \right )\right )+\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-i \left (\tan ^{2}\left (d x +c \right )\right )}{a^{2} d}\)	\(78\)

[In]

int(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

128/105*I*(21*exp(4*I*(d*x+c))+7*exp(2*I*(d*x+c))+1)/d/a^2/(exp(2*I*(d*x+c))+1)^7

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. 138 vs. \(2 (64) = 128\).

Time = 0.24 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.68 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {128 \, {\left (-21 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 7 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{105 \, {\left (a^{2} d e^{\left (14 i \, d x + 14 i \, c\right )} + 7 \, a^{2} d e^{\left (12 i \, d x + 12 i \, c\right )} + 21 \, a^{2} d e^{\left (10 i \, d x + 10 i \, c\right )} + 35 \, a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 35 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 21 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 7 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \]

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-128/105*(-21*I*e^(4*I*d*x + 4*I*c) - 7*I*e^(2*I*d*x + 2*I*c) - I)/(a^2*d*e^(14*I*d*x + 14*I*c) + 7*a^2*d*e^(1
2*I*d*x + 12*I*c) + 21*a^2*d*e^(10*I*d*x + 10*I*c) + 35*a^2*d*e^(8*I*d*x + 8*I*c) + 35*a^2*d*e^(6*I*d*x + 6*I*
c) + 21*a^2*d*e^(4*I*d*x + 4*I*c) + 7*a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)

Sympy [F]

\[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=- \frac {\int \frac {\sec ^{10}{\left (c + d x \right )}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx}{a^{2}} \]

[In]

integrate(sec(d*x+c)**10/(a+I*a*tan(d*x+c))**2,x)

[Out]

-Integral(sec(c + d*x)**10/(tan(c + d*x)**2 - 2*I*tan(c + d*x) - 1), x)/a**2

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.94 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {15 \, \tan \left (d x + c\right )^{7} + 35 i \, \tan \left (d x + c\right )^{6} + 21 \, \tan \left (d x + c\right )^{5} + 105 i \, \tan \left (d x + c\right )^{4} - 35 \, \tan \left (d x + c\right )^{3} + 105 i \, \tan \left (d x + c\right )^{2} - 105 \, \tan \left (d x + c\right )}{105 \, a^{2} d} \]

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/105*(15*tan(d*x + c)^7 + 35*I*tan(d*x + c)^6 + 21*tan(d*x + c)^5 + 105*I*tan(d*x + c)^4 - 35*tan(d*x + c)^3
+ 105*I*tan(d*x + c)^2 - 105*tan(d*x + c))/(a^2*d)

Giac [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.94 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=-\frac {15 \, \tan \left (d x + c\right )^{7} + 35 i \, \tan \left (d x + c\right )^{6} + 21 \, \tan \left (d x + c\right )^{5} + 105 i \, \tan \left (d x + c\right )^{4} - 35 \, \tan \left (d x + c\right )^{3} + 105 i \, \tan \left (d x + c\right )^{2} - 105 \, \tan \left (d x + c\right )}{105 \, a^{2} d} \]

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/105*(15*tan(d*x + c)^7 + 35*I*tan(d*x + c)^6 + 21*tan(d*x + c)^5 + 105*I*tan(d*x + c)^4 - 35*tan(d*x + c)^3
+ 105*I*tan(d*x + c)^2 - 105*tan(d*x + c))/(a^2*d)

Mupad [B] (verification not implemented)

Time = 3.91 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.13 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx=\frac {{\cos \left (c+d\,x\right )}^7\,35{}\mathrm {i}+64\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^6+32\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^4+24\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^2-\cos \left (c+d\,x\right )\,35{}\mathrm {i}-15\,\sin \left (c+d\,x\right )}{105\,a^2\,d\,{\cos \left (c+d\,x\right )}^7} \]

[In]

int(1/(cos(c + d*x)^10*(a + a*tan(c + d*x)*1i)^2),x)

[Out]

(24*cos(c + d*x)^2*sin(c + d*x) - 15*sin(c + d*x) - cos(c + d*x)*35i + 32*cos(c + d*x)^4*sin(c + d*x) + 64*cos
(c + d*x)^6*sin(c + d*x) + cos(c + d*x)^7*35i)/(105*a^2*d*cos(c + d*x)^7)