Integrand size = 26, antiderivative size = 113 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{a^4 d}+\frac {8 i (a+i a \tan (c+d x))^{3/2}}{a^5 d}-\frac {12 i (a+i a \tan (c+d x))^{5/2}}{5 a^6 d}+\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a^7 d} \] Output:
-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
Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.54 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {2 \sqrt {a+i a \tan (c+d x)} \left (-177 i-71 \tan (c+d x)+27 i \tan ^2(c+d x)+5 \tan ^3(c+d x)\right )}{35 a^4 d} \] Input:
Integrate[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^(7/2),x]
Output:
(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)
Time = 0.29 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3042, 3968, 53, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^8}{(a+i a \tan (c+d x))^{7/2}}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i \int \frac {(a-i a \tan (c+d x))^3}{\sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\frac {i \int \left (\frac {8 a^3}{\sqrt {i \tan (c+d x) a+a}}-12 \sqrt {i \tan (c+d x) a+a} a^2+6 (i \tan (c+d x) a+a)^{3/2} a-(i \tan (c+d x) a+a)^{5/2}\right )d(i a \tan (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i \left (16 a^3 \sqrt {a+i a \tan (c+d x)}-8 a^2 (a+i a \tan (c+d x))^{3/2}-\frac {2}{7} (a+i a \tan (c+d x))^{7/2}+\frac {12}{5} a (a+i a \tan (c+d x))^{5/2}\right )}{a^7 d}\) |
Input:
Int[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^(7/2),x]
Output:
((-I)*(16*a^3*Sqrt[a + I*a*Tan[c + d*x]] - 8*a^2*(a + I*a*Tan[c + d*x])^(3 /2) + (12*a*(a + I*a*Tan[c + d*x])^(5/2))/5 - (2*(a + I*a*Tan[c + d*x])^(7 /2))/7))/(a^7*d)
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] && 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])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[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]
Time = 1.02 (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 | \(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {6 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+4 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}-8 a^{3} \sqrt {a +i a \tan \left (d x +c \right )}\right )}{d \,a^{7}}\) | \(82\) |
Input:
int(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
Output:
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))
Time = 0.08 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.05 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {16 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (16 i \, e^{\left (7 i \, d x + 7 i \, c\right )} + 56 i \, e^{\left (5 i \, d x + 5 i \, c\right )} + 70 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + 35 i \, e^{\left (i \, d x + i \, c\right )}\right )}}{35 \, {\left (a^{4} d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )}} \] Input:
integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")
Output:
-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)
Timed out. \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**8/(a+I*a*tan(d*x+c))**(7/2),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.67 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {2 i \, {\left (5 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 42 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a + 140 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2} - 280 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{3}\right )}}{35 \, a^{7} d} \] Input:
integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")
Output:
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)
Exception generated. \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad Argument TypeError: Bad Argument TypeError: Bad Argument TypeDone
Time = 4.65 (sec) , antiderivative size = 242, normalized size of antiderivative = 2.14 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,256{}\mathrm {i}}{35\,a^4\,d}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,128{}\mathrm {i}}{35\,a^4\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,96{}\mathrm {i}}{35\,a^4\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,16{}\mathrm {i}}{7\,a^4\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3} \] Input:
int(1/(cos(c + d*x)^8*(a + a*tan(c + d*x)*1i)^(7/2)),x)
Output:
- ((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)
\[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {too large to display} \] Input:
int(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(7/2),x)
Output:
(sqrt(a)*( - 16*sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**8*i - 13*int(( - sq rt(tan(c + d*x)*i + 1)*sec(c + d*x)**8*tan(c + d*x)**4)/(tan(c + d*x)**5*i + 3*tan(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*tan(c + d*x)**2 - 3*tan(c + d*x)*i - 1),x)*tan(c + d*x)**2*d - 13*int(( - sqrt(tan(c + d*x)*i + 1)*se c(c + d*x)**8*tan(c + d*x)**4)/(tan(c + d*x)**5*i + 3*tan(c + d*x)**4 - 2* tan(c + d*x)**3*i + 2*tan(c + d*x)**2 - 3*tan(c + d*x)*i - 1),x)*d + 36*in t(( - sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**8*tan(c + d*x)**2)/(tan(c + d *x)**5*i + 3*tan(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*tan(c + d*x)**2 - 3 *tan(c + d*x)*i - 1),x)*tan(c + d*x)**2*d + 36*int(( - sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**8*tan(c + d*x)**2)/(tan(c + d*x)**5*i + 3*tan(c + d*x) **4 - 2*tan(c + d*x)**3*i + 2*tan(c + d*x)**2 - 3*tan(c + d*x)*i - 1),x)*d - 10*int(( - sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**8*tan(c + d*x))/(tan( c + d*x)**5*i + 3*tan(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*tan(c + d*x)** 2 - 3*tan(c + d*x)*i - 1),x)*tan(c + d*x)**2*d*i - 10*int(( - sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**8*tan(c + d*x))/(tan(c + d*x)**5*i + 3*tan(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*tan(c + d*x)**2 - 3*tan(c + d*x)*i - 1), x)*d*i + int(( - sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**8)/(tan(c + d*x)** 5*i + 3*tan(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*tan(c + d*x)**2 - 3*tan( c + d*x)*i - 1),x)*tan(c + d*x)**2*d + int(( - sqrt(tan(c + d*x)*i + 1)*se c(c + d*x)**8)/(tan(c + d*x)**5*i + 3*tan(c + d*x)**4 - 2*tan(c + d*x)*...