Integrand size = 26, antiderivative size = 146 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {32 i (a+i a \tan (c+d x))^{5/2}}{5 a^5 d}+\frac {64 i (a+i a \tan (c+d x))^{7/2}}{7 a^6 d}-\frac {16 i (a+i a \tan (c+d x))^{9/2}}{3 a^7 d}+\frac {16 i (a+i a \tan (c+d x))^{11/2}}{11 a^8 d}-\frac {2 i (a+i a \tan (c+d x))^{13/2}}{13 a^9 d} \]
-32/5*I*(a+I*a*tan(d*x+c))^(5/2)/a^5/d+64/7*I*(a+I*a*tan(d*x+c))^(7/2)/a^6 /d-16/3*I*(a+I*a*tan(d*x+c))^(9/2)/a^7/d+16/11*I*(a+I*a*tan(d*x+c))^(11/2) /a^8/d-2/13*I*(a+I*a*tan(d*x+c))^(13/2)/a^9/d
Time = 0.40 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.58 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2 (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)} \left (9683 i+16700 \tan (c+d x)-14210 i \tan ^2(c+d x)-6300 \tan ^3(c+d x)+1155 i \tan ^4(c+d x)\right )}{15015 a^3 d} \]
(2*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]]*(9683*I + 16700*Tan[c + d*x] - (14210*I)*Tan[c + d*x]^2 - 6300*Tan[c + d*x]^3 + (1155*I)*Tan[c + d*x]^4))/(15015*a^3*d)
Time = 0.29 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.86, 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.
| \(\displaystyle \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)^{10}}{(a+i a \tan (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3968 |
| \(\displaystyle -\frac {i \int (a-i a \tan (c+d x))^4 (i \tan (c+d x) a+a)^{3/2}d(i a \tan (c+d x))}{a^9 d}\) |
| \(\Big \downarrow \) 53 |
| \(\displaystyle -\frac {i \int \left ((i \tan (c+d x) a+a)^{11/2}-8 a (i \tan (c+d x) a+a)^{9/2}+24 a^2 (i \tan (c+d x) a+a)^{7/2}-32 a^3 (i \tan (c+d x) a+a)^{5/2}+16 a^4 (i \tan (c+d x) a+a)^{3/2}\right )d(i a \tan (c+d x))}{a^9 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {i \left (\frac {32}{5} a^4 (a+i a \tan (c+d x))^{5/2}-\frac {64}{7} a^3 (a+i a \tan (c+d x))^{7/2}+\frac {16}{3} a^2 (a+i a \tan (c+d x))^{9/2}+\frac {2}{13} (a+i a \tan (c+d x))^{13/2}-\frac {16}{11} a (a+i a \tan (c+d x))^{11/2}\right )}{a^9 d}\) |
((-I)*((32*a^4*(a + I*a*Tan[c + d*x])^(5/2))/5 - (64*a^3*(a + I*a*Tan[c + d*x])^(7/2))/7 + (16*a^2*(a + I*a*Tan[c + d*x])^(9/2))/3 - (16*a*(a + I*a* Tan[c + d*x])^(11/2))/11 + (2*(a + I*a*Tan[c + d*x])^(13/2))/13))/(a^9*d)
3.4.62.3.1 Defintions of rubi rules used
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 = 1.29 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.69
| method | result | size |
| derivativedivides | \(\frac {2 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {13}{2}}}{13}+\frac {8 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {11}{2}}}{11}-\frac {8 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{3}+\frac {32 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {16 a^{4} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}\right )}{d \,a^{9}}\) | \(101\) |
| default | \(\frac {2 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {13}{2}}}{13}+\frac {8 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {11}{2}}}{11}-\frac {8 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {9}{2}}}{3}+\frac {32 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {16 a^{4} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}\right )}{d \,a^{9}}\) | \(101\) |
2*I/d/a^9*(-1/13*(a+I*a*tan(d*x+c))^(13/2)+8/11*a*(a+I*a*tan(d*x+c))^(11/2 )-8/3*a^2*(a+I*a*tan(d*x+c))^(9/2)+32/7*a^3*(a+I*a*tan(d*x+c))^(7/2)-16/5* a^4*(a+I*a*tan(d*x+c))^(5/2))
Time = 0.27 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.20 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {128 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (128 i \, e^{\left (13 i \, d x + 13 i \, c\right )} + 832 i \, e^{\left (11 i \, d x + 11 i \, c\right )} + 2288 i \, e^{\left (9 i \, d x + 9 i \, c\right )} + 3432 i \, e^{\left (7 i \, d x + 7 i \, c\right )} + 3003 i \, e^{\left (5 i \, d x + 5 i \, c\right )}\right )}}{15015 \, {\left (a^{3} d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \]
-128/15015*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(128*I*e^(13*I*d*x + 13*I*c) + 832*I*e^(11*I*d*x + 11*I*c) + 2288*I*e^(9*I*d*x + 9*I*c) + 3432* I*e^(7*I*d*x + 7*I*c) + 3003*I*e^(5*I*d*x + 5*I*c))/(a^3*d*e^(12*I*d*x + 1 2*I*c) + 6*a^3*d*e^(10*I*d*x + 10*I*c) + 15*a^3*d*e^(8*I*d*x + 8*I*c) + 20 *a^3*d*e^(6*I*d*x + 6*I*c) + 15*a^3*d*e^(4*I*d*x + 4*I*c) + 6*a^3*d*e^(2*I *d*x + 2*I*c) + a^3*d)
Timed out. \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.64 \[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {2 i \, {\left (1155 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {13}{2}} - 10920 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a + 40040 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a^{2} - 68640 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{3} + 48048 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{4}\right )}}{15015 \, a^{9} d} \]
-2/15015*I*(1155*(I*a*tan(d*x + c) + a)^(13/2) - 10920*(I*a*tan(d*x + c) + a)^(11/2)*a + 40040*(I*a*tan(d*x + c) + a)^(9/2)*a^2 - 68640*(I*a*tan(d*x + c) + a)^(7/2)*a^3 + 48048*(I*a*tan(d*x + c) + a)^(5/2)*a^4)/(a^9*d)
\[ \int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{10}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Time = 10.15 (sec) , antiderivative size = 434, normalized size of antiderivative = 2.97 \[ \int \frac {\sec ^{10}(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}}\,16384{}\mathrm {i}}{15015\,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}}\,8192{}\mathrm {i}}{15015\,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}}\,2048{}\mathrm {i}}{5005\,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}}\,1024{}\mathrm {i}}{3003\,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}}\,128{}\mathrm {i}}{429\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}+\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}}\,1792{}\mathrm {i}}{143\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^5}-\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}}{13\,a^3\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^6} \]
((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)* 1792i)/(143*a^3*d*(exp(c*2i + d*x*2i) + 1)^5) - ((a - (a*(exp(c*2i + d*x*2 i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*8192i)/(15015*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*2i) + 1))^(1/2)*2048i)/(5005*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)*102 4i)/(3003*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)*128i)/(429*a^3*d*(exp(c*2i + d*x*2i) + 1)^4) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d *x*2i) + 1))^(1/2)*16384i)/(15015*a^3*d) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*128i)/(13*a^3*d*(exp(c*2i + d*x *2i) + 1)^6)