Integrand size = 30, antiderivative size = 206 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2 i}{13 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}}+\frac {16 i}{117 a d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}+\frac {32 i}{195 a^2 d (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}+\frac {256 i \sqrt {e \sec (c+d x)}}{585 a^2 d e^2 \sqrt {a+i a \tan (c+d x)}}-\frac {128 i \sqrt {a+i a \tan (c+d x)}}{585 a^3 d (e \sec (c+d x))^{3/2}} \] Output:
2/13*I/d/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(5/2)+16/117*I/a/d/(e*sec (d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(3/2)+32/195*I/a^2/d/(e*sec(d*x+c))^(3/2 )/(a+I*a*tan(d*x+c))^(1/2)+256/585*I*(e*sec(d*x+c))^(1/2)/a^2/d/e^2/(a+I*a *tan(d*x+c))^(1/2)-128/585*I*(a+I*a*tan(d*x+c))^(1/2)/a^3/d/(e*sec(d*x+c)) ^(3/2)
Time = 1.49 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.52 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sec ^4(c+d x) (-351 i-1300 i \cos (2 (c+d x))+75 i \cos (4 (c+d x))+1040 \sin (2 (c+d x))-120 \sin (4 (c+d x)))}{2340 a^2 d (e \sec (c+d x))^{3/2} (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \] Input:
Integrate[1/((e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]
Output:
(Sec[c + d*x]^4*(-351*I - (1300*I)*Cos[2*(c + d*x)] + (75*I)*Cos[4*(c + d* x)] + 1040*Sin[2*(c + d*x)] - 120*Sin[4*(c + d*x)]))/(2340*a^2*d*(e*Sec[c + d*x])^(3/2)*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])
Time = 1.03 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3983, 3042, 3983, 3042, 3983, 3042, 3978, 3042, 3969}
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.
\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^{5/2} (e \sec (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+i a \tan (c+d x))^{5/2} (e \sec (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 3983 |
\(\displaystyle \frac {8 \int \frac {1}{(e \sec (c+d x))^{3/2} (i \tan (c+d x) a+a)^{3/2}}dx}{13 a}+\frac {2 i}{13 d (a+i a \tan (c+d x))^{5/2} (e \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8 \int \frac {1}{(e \sec (c+d x))^{3/2} (i \tan (c+d x) a+a)^{3/2}}dx}{13 a}+\frac {2 i}{13 d (a+i a \tan (c+d x))^{5/2} (e \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3983 |
\(\displaystyle \frac {8 \left (\frac {2 \int \frac {1}{(e \sec (c+d x))^{3/2} \sqrt {i \tan (c+d x) a+a}}dx}{3 a}+\frac {2 i}{9 d (a+i a \tan (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}\right )}{13 a}+\frac {2 i}{13 d (a+i a \tan (c+d x))^{5/2} (e \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8 \left (\frac {2 \int \frac {1}{(e \sec (c+d x))^{3/2} \sqrt {i \tan (c+d x) a+a}}dx}{3 a}+\frac {2 i}{9 d (a+i a \tan (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}\right )}{13 a}+\frac {2 i}{13 d (a+i a \tan (c+d x))^{5/2} (e \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3983 |
\(\displaystyle \frac {8 \left (\frac {2 \left (\frac {4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{(e \sec (c+d x))^{3/2}}dx}{5 a}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}\right )}{3 a}+\frac {2 i}{9 d (a+i a \tan (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}\right )}{13 a}+\frac {2 i}{13 d (a+i a \tan (c+d x))^{5/2} (e \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8 \left (\frac {2 \left (\frac {4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{(e \sec (c+d x))^{3/2}}dx}{5 a}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}\right )}{3 a}+\frac {2 i}{9 d (a+i a \tan (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}\right )}{13 a}+\frac {2 i}{13 d (a+i a \tan (c+d x))^{5/2} (e \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3978 |
\(\displaystyle \frac {8 \left (\frac {2 \left (\frac {4 \left (\frac {2 a \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}dx}{3 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{3 d (e \sec (c+d x))^{3/2}}\right )}{5 a}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}\right )}{3 a}+\frac {2 i}{9 d (a+i a \tan (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}\right )}{13 a}+\frac {2 i}{13 d (a+i a \tan (c+d x))^{5/2} (e \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {8 \left (\frac {2 \left (\frac {4 \left (\frac {2 a \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}dx}{3 e^2}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{3 d (e \sec (c+d x))^{3/2}}\right )}{5 a}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}\right )}{3 a}+\frac {2 i}{9 d (a+i a \tan (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}\right )}{13 a}+\frac {2 i}{13 d (a+i a \tan (c+d x))^{5/2} (e \sec (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3969 |
\(\displaystyle \frac {8 \left (\frac {2 \left (\frac {4 \left (\frac {4 i a \sqrt {e \sec (c+d x)}}{3 d e^2 \sqrt {a+i a \tan (c+d x)}}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{3 d (e \sec (c+d x))^{3/2}}\right )}{5 a}+\frac {2 i}{5 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}\right )}{3 a}+\frac {2 i}{9 d (a+i a \tan (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}\right )}{13 a}+\frac {2 i}{13 d (a+i a \tan (c+d x))^{5/2} (e \sec (c+d x))^{3/2}}\) |
Input:
Int[1/((e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]
Output:
((2*I)/13)/(d*(e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)) + (8*(( (2*I)/9)/(d*(e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^(3/2)) + (2*(((2 *I)/5)/(d*(e*Sec[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) + (4*((((4*I) /3)*a*Sqrt[e*Sec[c + d*x]])/(d*e^2*Sqrt[a + I*a*Tan[c + d*x]]) - (((2*I)/3 )*Sqrt[a + I*a*Tan[c + d*x]])/(d*(e*Sec[c + d*x])^(3/2))))/(5*a)))/(3*a))) /(13*a)
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/ (a*f*m)), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] && EqQ [Simplify[m + n], 0]
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/( a*f*m)), x] + Simp[a*((m + n)/(m*d^2)) Int[(d*Sec[e + f*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b ^2, 0] && GtQ[n, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_), x_Symbol] :> Simp[a*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/ (b*f*(m + 2*n))), x] + Simp[Simplify[m + n]/(a*(m + 2*n)) Int[(d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x ] && EqQ[a^2 + b^2, 0] && LtQ[n, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2* n]
Time = 6.61 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.56
method | result | size |
default | \(\frac {\frac {2 i \sin \left (d x +c \right ) \left (-120 \cos \left (d x +c \right )^{2}+320\right )}{585}-\frac {10 \cos \left (d x +c \right )^{3}}{39}+\frac {160 \cos \left (d x +c \right )}{117}-\frac {256 \sec \left (d x +c \right )}{585}}{d \left (2 \cos \left (d x +c \right ) \sin \left (d x +c \right )-2 i \cos \left (d x +c \right )^{2}+i\right ) \sqrt {e \sec \left (d x +c \right )}\, e \,a^{2} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\) | \(115\) |
Input:
int(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOS E)
Output:
2/585/d/(2*cos(d*x+c)*sin(d*x+c)-2*I*cos(d*x+c)^2+I)/(e*sec(d*x+c))^(1/2)/ e/a^2/(a*(1+I*tan(d*x+c)))^(1/2)*(I*sin(d*x+c)*(-120*cos(d*x+c)^2+320)-75* cos(d*x+c)^3+400*cos(d*x+c)-128*sec(d*x+c))
Time = 0.09 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.54 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-195 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 2145 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 3042 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 962 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 305 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 45 i\right )} e^{\left (-\frac {13}{2} i \, d x - \frac {13}{2} i \, c\right )}}{4680 \, a^{3} d e^{2}} \] Input:
integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fr icas")
Output:
1/4680*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1)) *(-195*I*e^(10*I*d*x + 10*I*c) + 2145*I*e^(8*I*d*x + 8*I*c) + 3042*I*e^(6* I*d*x + 6*I*c) + 962*I*e^(4*I*d*x + 4*I*c) + 305*I*e^(2*I*d*x + 2*I*c) + 4 5*I)*e^(-13/2*I*d*x - 13/2*I*c)/(a^3*d*e^2)
Timed out. \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(e*sec(d*x+c))**(3/2)/(a+I*a*tan(d*x+c))**(5/2),x)
Output:
Timed out
Time = 0.25 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.10 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {45 i \, \cos \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right ) + 260 i \, \cos \left (\frac {9}{13} \, \arctan \left (\sin \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right ), \cos \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right )\right )\right ) + 702 i \, \cos \left (\frac {5}{13} \, \arctan \left (\sin \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right ), \cos \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right )\right )\right ) - 195 i \, \cos \left (\frac {3}{13} \, \arctan \left (\sin \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right ), \cos \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right )\right )\right ) + 2340 i \, \cos \left (\frac {1}{13} \, \arctan \left (\sin \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right ), \cos \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right )\right )\right ) + 45 \, \sin \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right ) + 260 \, \sin \left (\frac {9}{13} \, \arctan \left (\sin \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right ), \cos \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right )\right )\right ) + 702 \, \sin \left (\frac {5}{13} \, \arctan \left (\sin \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right ), \cos \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right )\right )\right ) + 195 \, \sin \left (\frac {3}{13} \, \arctan \left (\sin \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right ), \cos \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right )\right )\right ) + 2340 \, \sin \left (\frac {1}{13} \, \arctan \left (\sin \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right ), \cos \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right )\right )\right )}{4680 \, a^{\frac {5}{2}} d e^{\frac {3}{2}}} \] Input:
integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="ma xima")
Output:
1/4680*(45*I*cos(13/2*d*x + 13/2*c) + 260*I*cos(9/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) + 702*I*cos(5/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) - 195*I*cos(3/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) + 2340*I*cos(1/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) + 45*sin(13/2*d*x + 13/2*c) + 260*sin (9/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) + 702*sin(5 /13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) + 195*sin(3/1 3*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) + 2340*sin(1/13 *arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))))/(a^(5/2)*d*e^(3 /2))
Exception generated. \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="gi ac")
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 = 1.95 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.66 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,507{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,260{}\mathrm {i}+\cos \left (6\,c+6\,d\,x\right )\,45{}\mathrm {i}+897\,\sin \left (2\,c+2\,d\,x\right )+260\,\sin \left (4\,c+4\,d\,x\right )+45\,\sin \left (6\,c+6\,d\,x\right )+2340{}\mathrm {i}\right )}{4680\,a^2\,d\,e^2\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}} \] Input:
int(1/((e/cos(c + d*x))^(3/2)*(a + a*tan(c + d*x)*1i)^(5/2)),x)
Output:
((e/cos(c + d*x))^(1/2)*(cos(2*c + 2*d*x)*507i + cos(4*c + 4*d*x)*260i + c os(6*c + 6*d*x)*45i + 897*sin(2*c + 2*d*x) + 260*sin(4*c + 4*d*x) + 45*sin (6*c + 6*d*x) + 2340i))/(4680*a^2*d*e^2*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2))
\[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\int \frac {1}{\sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right ) \tan \left (d x +c \right )^{2}-2 \sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right ) \tan \left (d x +c \right ) i -\sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \sec \left (d x +c \right )}d x}{\sqrt {e}\, \sqrt {a}\, a^{2} e} \] Input:
int(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x)
Output:
( - int(1/(sqrt(sec(c + d*x))*sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)*tan(c + d*x)**2 - 2*sqrt(sec(c + d*x))*sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)*tan (c + d*x)*i - sqrt(sec(c + d*x))*sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)),x) )/(sqrt(e)*sqrt(a)*a**2*e)