\(\int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [356]

Optimal result

Integrand size = 26, antiderivative size = 73 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {8 i a^2 \sec ^7(c+d x)}{63 d (a+i a \tan (c+d x))^{7/2}}+\frac {2 i a \sec ^7(c+d x)}{9 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

8/63*I*a^2*sec(d*x+c)^7/d/(a+I*a*tan(d*x+c))^(7/2)+2/9*I*a*sec(d*x+c)^7/d/(a+I*a*tan(d*x+c))^(5/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3575, 3574} \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {8 i a^2 \sec ^7(c+d x)}{63 d (a+i a \tan (c+d x))^{7/2}}+\frac {2 i a \sec ^7(c+d x)}{9 d (a+i a \tan (c+d x))^{5/2}} \]

[In]

Int[Sec[c + d*x]^7/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((8*I)/63)*a^2*Sec[c + d*x]^7)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + (((2*I)/9)*a*Sec[c + d*x]^7)/(d*(a + I*a*Ta
n[c + d*x])^(5/2))

Rule 3574

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[2*b*(
d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*m)), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2
, 0] && EqQ[Simplify[m/2 + n - 1], 0]

Rule 3575

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*
Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] + Dist[a*((m + 2*n - 2)/(m + n - 1)), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]
&& IGtQ[Simplify[m/2 + n - 1], 0] &&  !IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \frac {2 i a \sec ^7(c+d x)}{9 d (a+i a \tan (c+d x))^{5/2}}+\frac {1}{9} (4 a) \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx \\ & = \frac {8 i a^2 \sec ^7(c+d x)}{63 d (a+i a \tan (c+d x))^{7/2}}+\frac {2 i a \sec ^7(c+d x)}{9 d (a+i a \tan (c+d x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.10 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.10 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 \sec ^5(c+d x) (i \cos (2 (c+d x))+\sin (2 (c+d x))) (-11 i+7 \tan (c+d x))}{63 a d (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \]

[In]

Integrate[Sec[c + d*x]^7/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(2*Sec[c + d*x]^5*(I*Cos[2*(c + d*x)] + Sin[2*(c + d*x)])*(-11*I + 7*Tan[c + d*x]))/(63*a*d*(-I + Tan[c + d*x]
)*Sqrt[a + I*a*Tan[c + d*x]])

Maple [F(-1)]

Timed out.

\[\int \frac {\sec ^{7}\left (d x +c \right )}{\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}d x\]

[In]

int(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

int(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^(3/2),x)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.40 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {32 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-9 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i\right )}}{63 \, {\left (a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \]

[In]

integrate(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-32/63*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-9*I*e^(2*I*d*x + 2*I*c) - 2*I)/(a^2*d*e^(8*I*d*x + 8*I*c) +
4*a^2*d*e^(6*I*d*x + 6*I*c) + 6*a^2*d*e^(4*I*d*x + 4*I*c) + 4*a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)

Sympy [F]

\[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {\sec ^{7}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(sec(d*x+c)**7/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Integral(sec(c + d*x)**7/(I*a*(tan(c + d*x) - I))**(3/2), x)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 488 vs. \(2 (57) = 114\).

Time = 0.37 (sec) , antiderivative size = 488, normalized size of antiderivative = 6.68 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (-11 i \, \sqrt {a} - \frac {30 \, \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {12 i \, \sqrt {a} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {86 \, \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {9 i \, \sqrt {a} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {108 \, \sqrt {a} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {108 \, \sqrt {a} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {9 i \, \sqrt {a} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {86 \, \sqrt {a} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac {12 i \, \sqrt {a} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac {30 \, \sqrt {a} \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} + \frac {11 i \, \sqrt {a} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}\right )} {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {3}{2}} {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}^{\frac {3}{2}}}{63 \, {\left (a^{2} - \frac {6 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {15 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {20 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {15 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {6 \, a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {a^{2} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}\right )} d {\left (-\frac {2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}^{\frac {3}{2}}} \]

[In]

integrate(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-2/63*(-11*I*sqrt(a) - 30*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 12*I*sqrt(a)*sin(d*x + c)^2/(cos(d*x + c)
+ 1)^2 - 86*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 9*I*sqrt(a)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 10
8*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 108*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 9*I*sqrt(a)*
sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 86*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 12*I*sqrt(a)*sin(d*x +
c)^10/(cos(d*x + c) + 1)^10 - 30*sqrt(a)*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 11*I*sqrt(a)*sin(d*x + c)^12/
(cos(d*x + c) + 1)^12)*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(3/2)*(sin(d*x + c)/(cos(d*x + c) + 1) - 1)^(3/2)
/((a^2 - 6*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 20*a^2*sin(d
*x + c)^6/(cos(d*x + c) + 1)^6 + 15*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 6*a^2*sin(d*x + c)^10/(cos(d*x +
c) + 1)^10 + a^2*sin(d*x + c)^12/(cos(d*x + c) + 1)^12)*d*(-2*I*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c
)^2/(cos(d*x + c) + 1)^2 - 1)^(3/2))

Giac [F]

\[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{7}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(sec(d*x+c)^7/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^7/(I*a*tan(d*x + c) + a)^(3/2), x)

Mupad [B] (verification not implemented)

Time = 7.74 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.25 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {32\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,9{}\mathrm {i}+2{}\mathrm {i}\right )\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}}{63\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4} \]

[In]

int(1/(cos(c + d*x)^7*(a + a*tan(c + d*x)*1i)^(3/2)),x)

[Out]

(32*exp(- c*1i - d*x*1i)*(exp(c*2i + d*x*2i)*9i + 2i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x
*2i) + 1))^(1/2))/(63*a^2*d*(exp(c*2i + d*x*2i) + 1)^4)