Integrand size = 26, antiderivative size = 117 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {16 i (a+i a \tan (c+d x))^{3/2}}{3 a^4 d}+\frac {24 i (a+i a \tan (c+d x))^{5/2}}{5 a^5 d}-\frac {12 i (a+i a \tan (c+d x))^{7/2}}{7 a^6 d}+\frac {2 i (a+i a \tan (c+d x))^{9/2}}{9 a^7 d} \] Output:
-16/3*I*(a+I*a*tan(d*x+c))^(3/2)/a^4/d+24/5*I*(a+I*a*tan(d*x+c))^(5/2)/a^5 /d-12/7*I*(a+I*a*tan(d*x+c))^(7/2)/a^6/d+2/9*I*(a+I*a*tan(d*x+c))^(9/2)/a^ 7/d
Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.62 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2 (1+i \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \left (-319 i-321 \tan (c+d x)+165 i \tan ^2(c+d x)+35 \tan ^3(c+d x)\right )}{315 a^3 d} \] Input:
Integrate[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^(5/2),x]
Output:
(2*(1 + I*Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]]*(-319*I - 321*Tan[c + d *x] + (165*I)*Tan[c + d*x]^2 + 35*Tan[c + d*x]^3))/(315*a^3*d)
Time = 0.29 (sec) , antiderivative size = 102, 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.
\(\displaystyle \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^8}{(a+i a \tan (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i \int (a-i a \tan (c+d x))^3 \sqrt {i \tan (c+d x) a+a}d(i a \tan (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\frac {i \int \left (-(i \tan (c+d x) a+a)^{7/2}+6 a (i \tan (c+d x) a+a)^{5/2}-12 a^2 (i \tan (c+d x) a+a)^{3/2}+8 a^3 \sqrt {i \tan (c+d x) a+a}\right )d(i a \tan (c+d x))}{a^7 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i \left (\frac {16}{3} a^3 (a+i a \tan (c+d x))^{3/2}-\frac {24}{5} a^2 (a+i a \tan (c+d x))^{5/2}-\frac {2}{9} (a+i a \tan (c+d x))^{9/2}+\frac {12}{7} a (a+i a \tan (c+d x))^{7/2}\right )}{a^7 d}\) |
Input:
Int[Sec[c + d*x]^8/(a + I*a*Tan[c + d*x])^(5/2),x]
Output:
((-I)*((16*a^3*(a + I*a*Tan[c + d*x])^(3/2))/3 - (24*a^2*(a + I*a*Tan[c + d*x])^(5/2))/5 + (12*a*(a + I*a*Tan[c + d*x])^(7/2))/7 - (2*(a + I*a*Tan[c + d*x])^(9/2))/9))/(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 = 1.05 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.70
method | result | size |
derivativedivides | \(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{9}-\frac {6 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}+\frac {12 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {8 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}\right )}{d \,a^{7}}\) | \(82\) |
default | \(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{9}-\frac {6 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}+\frac {12 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {8 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}\right )}{d \,a^{7}}\) | \(82\) |
Input:
int(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
2*I/d/a^7*(1/9*(a+I*a*tan(d*x+c))^(9/2)-6/7*a*(a+I*a*tan(d*x+c))^(7/2)+12/ 5*a^2*(a+I*a*tan(d*x+c))^(5/2)-8/3*a^3*(a+I*a*tan(d*x+c))^(3/2))
Time = 0.09 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.15 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {32 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (16 i \, e^{\left (9 i \, d x + 9 i \, c\right )} + 72 i \, e^{\left (7 i \, d x + 7 i \, c\right )} + 126 i \, e^{\left (5 i \, d x + 5 i \, c\right )} + 105 i \, e^{\left (3 i \, d x + 3 i \, c\right )}\right )}}{315 \, {\left (a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \] Input:
integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
Output:
-32/315*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(16*I*e^(9*I*d*x + 9*I*c ) + 72*I*e^(7*I*d*x + 7*I*c) + 126*I*e^(5*I*d*x + 5*I*c) + 105*I*e^(3*I*d* x + 3*I*c))/(a^3*d*e^(8*I*d*x + 8*I*c) + 4*a^3*d*e^(6*I*d*x + 6*I*c) + 6*a ^3*d*e^(4*I*d*x + 4*I*c) + 4*a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)
\[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\sec ^{8}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(sec(d*x+c)**8/(a+I*a*tan(d*x+c))**(5/2),x)
Output:
Integral(sec(c + d*x)**8/(I*a*(tan(c + d*x) - I))**(5/2), x)
Time = 0.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.65 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2 i \, {\left (35 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 270 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a + 756 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - 840 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3}\right )}}{315 \, a^{7} d} \] Input:
integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
Output:
2/315*I*(35*(I*a*tan(d*x + c) + a)^(9/2) - 270*(I*a*tan(d*x + c) + a)^(7/2 )*a + 756*(I*a*tan(d*x + c) + a)^(5/2)*a^2 - 840*(I*a*tan(d*x + c) + a)^(3 /2)*a^3)/(a^7*d)
Exception generated. \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(5/2),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 TypeDone
Time = 4.07 (sec) , antiderivative size = 306, normalized size of antiderivative = 2.62 \[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,512{}\mathrm {i}}{315\,a^3\,d}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,256{}\mathrm {i}}{315\,a^3\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,64{}\mathrm {i}}{105\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,32{}\mathrm {i}}{63\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,32{}\mathrm {i}}{9\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4} \] Input:
int(1/(cos(c + d*x)^8*(a + a*tan(c + d*x)*1i)^(5/2)),x)
Output:
((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)* 32i)/(9*a^3*d*(exp(c*2i + d*x*2i) + 1)^4) - ((a - (a*(exp(c*2i + d*x*2i)*1 i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*256i)/(315*a^3*d*(exp(c*2i + d *x*2i) + 1)) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2 i) + 1))^(1/2)*64i)/(105*a^3*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)*32i)/(63*a^3* d*(exp(c*2i + d*x*2i) + 1)^3) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/ (exp(c*2i + d*x*2i) + 1))^(1/2)*512i)/(315*a^3*d)
\[ \int \frac {\sec ^8(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\int \frac {\sec \left (d x +c \right )^{8}}{\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right ) i +1}}d x}{\sqrt {a}\, a^{2}} \] Input:
int(sec(d*x+c)^8/(a+I*a*tan(d*x+c))^(5/2),x)
Output:
( - int(sec(c + d*x)**8/(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 2*sqrt (tan(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + d*x)*i + 1)),x))/(sqrt( a)*a**2)