Integrand size = 30, antiderivative size = 81 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{7/2}} \, dx=-\frac {4 i a (a+i a \tan (c+d x))^{3/2}}{21 d e^2 (e \sec (c+d x))^{3/2}}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{7 d (e \sec (c+d x))^{7/2}} \] Output:
-4/21*I*a*(a+I*a*tan(d*x+c))^(3/2)/d/e^2/(e*sec(d*x+c))^(3/2)-2/7*I*(a+I*a *tan(d*x+c))^(5/2)/d/(e*sec(d*x+c))^(7/2)
Time = 1.35 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.14 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{7/2}} \, dx=-\frac {2 a^2 (\cos (2 (c+2 d x))+i \sin (2 (c+2 d x))) (5 i+2 \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{21 d e^2 (e \sec (c+d x))^{3/2} (\cos (d x)+i \sin (d x))^2} \] Input:
Integrate[(a + I*a*Tan[c + d*x])^(5/2)/(e*Sec[c + d*x])^(7/2),x]
Output:
(-2*a^2*(Cos[2*(c + 2*d*x)] + I*Sin[2*(c + 2*d*x)])*(5*I + 2*Tan[c + d*x]) *Sqrt[a + I*a*Tan[c + d*x]])/(21*d*e^2*(e*Sec[c + d*x])^(3/2)*(Cos[d*x] + I*Sin[d*x])^2)
Time = 0.43 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {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.
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{7/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{7/2}}dx\) |
\(\Big \downarrow \) 3978 |
\(\displaystyle \frac {2 a \int \frac {(i \tan (c+d x) a+a)^{3/2}}{(e \sec (c+d x))^{3/2}}dx}{7 e^2}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{7 d (e \sec (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 a \int \frac {(i \tan (c+d x) a+a)^{3/2}}{(e \sec (c+d x))^{3/2}}dx}{7 e^2}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{7 d (e \sec (c+d x))^{7/2}}\) |
\(\Big \downarrow \) 3969 |
\(\displaystyle -\frac {4 i a (a+i a \tan (c+d x))^{3/2}}{21 d e^2 (e \sec (c+d x))^{3/2}}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{7 d (e \sec (c+d x))^{7/2}}\) |
Input:
Int[(a + I*a*Tan[c + d*x])^(5/2)/(e*Sec[c + d*x])^(7/2),x]
Output:
(((-4*I)/21)*a*(a + I*a*Tan[c + d*x])^(3/2))/(d*e^2*(e*Sec[c + d*x])^(3/2) ) - (((2*I)/7)*(a + I*a*Tan[c + d*x])^(5/2))/(d*(e*Sec[c + d*x])^(7/2))
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]
Time = 6.21 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.01
method | result | size |
default | \(-\frac {2 \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (6 i \cos \left (d x +c \right )^{3}-6 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}-i \cos \left (d x +c \right )-2 \sin \left (d x +c \right )\right )}{21 d \sqrt {e \sec \left (d x +c \right )}\, e^{3}}\) | \(82\) |
risch | \(-\frac {i a^{2} \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left (3 \,{\mathrm e}^{3 i \left (d x +c \right )}+7 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{21 e^{3} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\) | \(88\) |
Input:
int((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
Output:
-2/21/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*(6*I*cos(d*x+c)^3-6*sin(d*x+c)*cos( d*x+c)^2-I*cos(d*x+c)-2*sin(d*x+c))/(e*sec(d*x+c))^(1/2)/e^3
Time = 0.07 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.16 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {{\left (-3 i \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} - 10 i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} - 7 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{21 \, d e^{4}} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(7/2),x, algorithm="fric as")
Output:
1/21*(-3*I*a^2*e^(5*I*d*x + 5*I*c) - 10*I*a^2*e^(3*I*d*x + 3*I*c) - 7*I*a^ 2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)/(d*e^4)
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{7/2}} \, dx=\text {Timed out} \] Input:
integrate((a+I*a*tan(d*x+c))**(5/2)/(e*sec(d*x+c))**(7/2),x)
Output:
Timed out
Time = 0.26 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.16 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {{\left (-7 i \, a^{2} \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) - 3 i \, a^{2} \cos \left (\frac {7}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 7 \, a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, a^{2} \sin \left (\frac {7}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )\right )} \sqrt {a}}{21 \, d e^{\frac {7}{2}}} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(7/2),x, algorithm="maxi ma")
Output:
1/21*(-7*I*a^2*cos(3/2*d*x + 3/2*c) - 3*I*a^2*cos(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 7*a^2*sin(3/2*d*x + 3/2*c) + 3*a^2*sin( 7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*sqrt(a)/(d*e^(7/ 2))
Exception generated. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(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 TypeDone
Time = 2.17 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.38 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{7/2}} \, dx=-\frac {a^2\,\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,10{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,3{}\mathrm {i}-10\,\sin \left (2\,c+2\,d\,x\right )-3\,\sin \left (4\,c+4\,d\,x\right )+7{}\mathrm {i}\right )}{42\,d\,e^4} \] Input:
int((a + a*tan(c + d*x)*1i)^(5/2)/(e/cos(c + d*x))^(7/2),x)
Output:
-(a^2*(e/cos(c + d*x))^(1/2)*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2)*(cos(2*c + 2*d*x)*10i + cos(4*c + 4*d*x )*3i - 10*sin(2*c + 2*d*x) - 3*sin(4*c + 4*d*x) + 7i))/(42*d*e^4)
\[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {\sqrt {e}\, \sqrt {a}\, a^{2} \left (-\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{4}}d x \right )+2 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )^{4}}d x \right ) i +\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\sec \left (d x +c \right )^{4}}d x \right )}{e^{4}} \] Input:
int((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(7/2),x)
Output:
(sqrt(e)*sqrt(a)*a**2*( - int((sqrt(sec(c + d*x))*sqrt(tan(c + d*x)*i + 1) *tan(c + d*x)**2)/sec(c + d*x)**4,x) + 2*int((sqrt(sec(c + d*x))*sqrt(tan( c + d*x)*i + 1)*tan(c + d*x))/sec(c + d*x)**4,x)*i + int((sqrt(sec(c + d*x ))*sqrt(tan(c + d*x)*i + 1))/sec(c + d*x)**4,x)))/e**4