\(\int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\) [379]

Optimal result

Integrand size = 26, antiderivative size = 113 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{a^4 d}+\frac {8 i (a+i a \tan (c+d x))^{3/2}}{a^5 d}-\frac {12 i (a+i a \tan (c+d x))^{5/2}}{5 a^6 d}+\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a^7 d} \]

[Out]

-16*I*(a+I*a*tan(d*x+c))^(1/2)/a^4/d+8*I*(a+I*a*tan(d*x+c))^(3/2)/a^5/d-12/5*I*(a+I*a*tan(d*x+c))^(5/2)/a^6/d+
2/7*I*(a+I*a*tan(d*x+c))^(7/2)/a^7/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3568, 45} \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a^7 d}-\frac {12 i (a+i a \tan (c+d x))^{5/2}}{5 a^6 d}+\frac {8 i (a+i a \tan (c+d x))^{3/2}}{a^5 d}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{a^4 d} \]

[In]

Int[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

((-16*I)*Sqrt[a + I*a*Tan[c + d*x]])/(a^4*d) + ((8*I)*(a + I*a*Tan[c + d*x])^(3/2))/(a^5*d) - (((12*I)/5)*(a +
I*a*Tan[c + d*x])^(5/2))/(a^6*d) + (((2*I)/7)*(a + I*a*Tan[c + d*x])^(7/2))/(a^7*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 \frac {(a-x)^3}{\sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{a^7 d} \\ & = -\frac {i \text {Subst}\left (\int \left (\frac {8 a^3}{\sqrt {a+x}}-12 a^2 \sqrt {a+x}+6 a (a+x)^{3/2}-(a+x)^{5/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^7 d} \\ & = -\frac {16 i \sqrt {a+i a \tan (c+d x)}}{a^4 d}+\frac {8 i (a+i a \tan (c+d x))^{3/2}}{a^5 d}-\frac {12 i (a+i a \tan (c+d x))^{5/2}}{5 a^6 d}+\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a^7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.54 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {2 \sqrt {a+i a \tan (c+d x)} \left (-177 i-71 \tan (c+d x)+27 i \tan ^2(c+d x)+5 \tan ^3(c+d x)\right )}{35 a^4 d} \]

[In]

Integrate[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(2*Sqrt[a + I*a*Tan[c + d*x]]*(-177*I - 71*Tan[c + d*x] + (27*I)*Tan[c + d*x]^2 + 5*Tan[c + d*x]^3))/(35*a^4*d
)

Maple [A] (verified)

Time = 1.32 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {6 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+4 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}-8 a^{3} \sqrt {a +i a \tan \left (d x +c \right )}\right )}{d \,a^{7}}\)	\(82\)
default	\(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {6 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+4 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}-8 a^{3} \sqrt {a +i a \tan \left (d x +c \right )}\right )}{d \,a^{7}}\)	\(82\)

[In]

int(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.05 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {16 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (16 i \, e^{\left (7 i \, d x + 7 i \, c\right )} + 56 i \, e^{\left (5 i \, d x + 5 i \, c\right )} + 70 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + 35 i \, e^{\left (i \, d x + i \, c\right )}\right )}}{35 \, {\left (a^{4} d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )}} \]

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

-16/35*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(16*I*e^(7*I*d*x + 7*I*c) + 56*I*e^(5*I*d*x + 5*I*c) + 70*I*e
^(3*I*d*x + 3*I*c) + 35*I*e^(I*d*x + I*c))/(a^4*d*e^(6*I*d*x + 6*I*c) + 3*a^4*d*e^(4*I*d*x + 4*I*c) + 3*a^4*d*
e^(2*I*d*x + 2*I*c) + a^4*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**8/(a+I*a*tan(d*x+c))**(7/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.67 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {2 i \, {\left (5 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 42 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a + 140 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2} - 280 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{3}\right )}}{35 \, a^{7} d} \]

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

2/35*I*(5*(I*a*tan(d*x + c) + a)^(7/2) - 42*(I*a*tan(d*x + c) + a)^(5/2)*a + 140*(I*a*tan(d*x + c) + a)^(3/2)*
a^2 - 280*sqrt(I*a*tan(d*x + c) + a)*a^3)/(a^7*d)

Giac [F]

\[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{8}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^8/(I*a*tan(d*x + c) + a)^(7/2), x)

Mupad [B] (verification not implemented)

Time = 7.49 (sec) , antiderivative size = 242, normalized size of antiderivative = 2.14 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,256{}\mathrm {i}}{35\,a^4\,d}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,128{}\mathrm {i}}{35\,a^4\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,96{}\mathrm {i}}{35\,a^4\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,16{}\mathrm {i}}{7\,a^4\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3} \]

[In]

int(1/(cos(c + d*x)^8*(a + a*tan(c + d*x)*1i)^(7/2)),x)

[Out]

- ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*256i)/(35*a^4*d) - ((a - (a*(exp(c
*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*128i)/(35*a^4*d*(exp(c*2i + d*x*2i) + 1)) - ((a - (
a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*96i)/(35*a^4*d*(exp(c*2i + d*x*2i) + 1)^2)
- ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*16i)/(7*a^4*d*(exp(c*2i + d*x*2i)
+ 1)^3)