Integrand size = 35, antiderivative size = 154 \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {\tan (e+f x)}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {8 \tan (e+f x)}{15 a^2 c^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \] Output:
1/5*tan(f*x+e)/f/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(5/2)+4/15*ta n(f*x+e)/a/c/f/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(3/2)+8/15*tan( f*x+e)/a^2/c^2/f/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(1/2)
Time = 2.77 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.49 \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {(89+28 \cos (2 (e+f x))+3 \cos (4 (e+f x))) \tan (e+f x)}{120 a^2 c^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \] Input:
Integrate[1/((a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(5/2)),x]
Output:
((89 + 28*Cos[2*(e + f*x)] + 3*Cos[4*(e + f*x)])*Tan[e + f*x])/(120*a^2*c^ 2*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])
Time = 0.32 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 4006, 42, 42, 41}
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.
\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 4006 |
\(\displaystyle \frac {a c \int \frac {1}{(i \tan (e+f x) a+a)^{7/2} (c-i c \tan (e+f x))^{7/2}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 42 |
\(\displaystyle \frac {a c \left (\frac {4 \int \frac {1}{(i \tan (e+f x) a+a)^{5/2} (c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{5 a c}+\frac {\tan (e+f x)}{5 a c (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
\(\Big \downarrow \) 42 |
\(\displaystyle \frac {a c \left (\frac {4 \left (\frac {2 \int \frac {1}{(i \tan (e+f x) a+a)^{3/2} (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{3 a c}+\frac {\tan (e+f x)}{3 a c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}\right )}{5 a c}+\frac {\tan (e+f x)}{5 a c (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
\(\Big \downarrow \) 41 |
\(\displaystyle \frac {a c \left (\frac {4 \left (\frac {2 \tan (e+f x)}{3 a^2 c^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}+\frac {\tan (e+f x)}{3 a c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}\right )}{5 a c}+\frac {\tan (e+f x)}{5 a c (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
Input:
Int[1/((a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(5/2)),x]
Output:
(a*c*(Tan[e + f*x]/(5*a*c*(a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f* x])^(5/2)) + (4*(Tan[e + f*x]/(3*a*c*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c *Tan[e + f*x])^(3/2)) + (2*Tan[e + f*x])/(3*a^2*c^2*Sqrt[a + I*a*Tan[e + f *x]]*Sqrt[c - I*c*Tan[e + f*x]])))/(5*a*c)))/f
Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> S imp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d*x]), x] /; FreeQ[{a, b, c, d}, x] && Eq Q[b*c + a*d, 0]
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(- x)*(a + b*x)^(m + 1)*((c + d*x)^(m + 1)/(2*a*c*(m + 1))), x] + Simp[(2*m + 3)/(2*a*c*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(m + 1), x], x] /; Fre eQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && ILtQ[m + 3/2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(c/f) Subst[Int[(a + b*x)^(m - 1)*( c + d*x)^(n - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]
Time = 0.33 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.68
method | result | size |
derivativedivides | \(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (1+\tan \left (f x +e \right )^{2}\right ) \tan \left (f x +e \right ) \left (8 \tan \left (f x +e \right )^{4}+20 \tan \left (f x +e \right )^{2}+15\right )}{15 f \,a^{3} c^{3} \left (-\tan \left (f x +e \right )+i\right )^{4} \left (i+\tan \left (f x +e \right )\right )^{4}}\) | \(105\) |
default | \(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (1+\tan \left (f x +e \right )^{2}\right ) \tan \left (f x +e \right ) \left (8 \tan \left (f x +e \right )^{4}+20 \tan \left (f x +e \right )^{2}+15\right )}{15 f \,a^{3} c^{3} \left (-\tan \left (f x +e \right )+i\right )^{4} \left (i+\tan \left (f x +e \right )\right )^{4}}\) | \(105\) |
Input:
int(1/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(5/2),x,method=_RETURNVE RBOSE)
Output:
1/15/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)/a^3/c^3*(1+t an(f*x+e)^2)*tan(f*x+e)*(8*tan(f*x+e)^4+20*tan(f*x+e)^2+15)/(-tan(f*x+e)+I )^4/(I+tan(f*x+e))^4
Time = 0.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.72 \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-3 i \, e^{\left (12 i \, f x + 12 i \, e\right )} - 28 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 175 i \, e^{\left (8 i \, f x + 8 i \, e\right )} + 175 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 28 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i\right )} e^{\left (-5 i \, f x - 5 i \, e\right )}}{480 \, a^{3} c^{3} f} \] Input:
integrate(1/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm ="fricas")
Output:
1/480*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))* (-3*I*e^(12*I*f*x + 12*I*e) - 28*I*e^(10*I*f*x + 10*I*e) - 175*I*e^(8*I*f* x + 8*I*e) + 175*I*e^(4*I*f*x + 4*I*e) + 28*I*e^(2*I*f*x + 2*I*e) + 3*I)*e ^(-5*I*f*x - 5*I*e)/(a^3*c^3*f)
\[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\int \frac {1}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/(a+I*a*tan(f*x+e))**(5/2)/(c-I*c*tan(f*x+e))**(5/2),x)
Output:
Integral(1/((I*a*(tan(e + f*x) - I))**(5/2)*(-I*c*(tan(e + f*x) + I))**(5/ 2)), x)
Time = 0.19 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.46 \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {3 \, \sin \left (5 \, f x + 5 \, e\right ) + 25 \, \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (5 \, f x + 5 \, e\right ), \cos \left (5 \, f x + 5 \, e\right )\right )\right ) + 150 \, \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (5 \, f x + 5 \, e\right ), \cos \left (5 \, f x + 5 \, e\right )\right )\right )}{240 \, a^{\frac {5}{2}} c^{\frac {5}{2}} f} \] Input:
integrate(1/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm ="maxima")
Output:
1/240*(3*sin(5*f*x + 5*e) + 25*sin(3/5*arctan2(sin(5*f*x + 5*e), cos(5*f*x + 5*e))) + 150*sin(1/5*arctan2(sin(5*f*x + 5*e), cos(5*f*x + 5*e))))/(a^( 5/2)*c^(5/2)*f)
\[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\int { \frac {1}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm ="giac")
Output:
integrate(1/((I*a*tan(f*x + e) + a)^(5/2)*(-I*c*tan(f*x + e) + c)^(5/2)), x)
Time = 2.91 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,125{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,22{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}+175\,\sin \left (2\,e+2\,f\,x\right )+28\,\sin \left (4\,e+4\,f\,x\right )+3\,\sin \left (6\,e+6\,f\,x\right )-150{}\mathrm {i}\right )}{480\,a^3\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \] Input:
int(1/((a + a*tan(e + f*x)*1i)^(5/2)*(c - c*tan(e + f*x)*1i)^(5/2)),x)
Output:
(((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1)) ^(1/2)*(cos(2*e + 2*f*x)*125i + cos(4*e + 4*f*x)*22i + cos(6*e + 6*f*x)*3i + 175*sin(2*e + 2*f*x) + 28*sin(4*e + 4*f*x) + 3*sin(6*e + 6*f*x) - 150i) )/(480*a^3*c^2*f*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(cos(2* e + 2*f*x) + 1))^(1/2))
Time = 0.21 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.64 \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right ) \left (8 \tan \left (f x +e \right )^{4}+20 \tan \left (f x +e \right )^{2}+15\right )}{15 a^{3} c^{3} f \left (\tan \left (f x +e \right )^{6}+3 \tan \left (f x +e \right )^{4}+3 \tan \left (f x +e \right )^{2}+1\right )} \] Input:
int(1/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(5/2),x)
Output:
(sqrt(c)*sqrt(a)*sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan( e + f*x)*(8*tan(e + f*x)**4 + 20*tan(e + f*x)**2 + 15))/(15*a**3*c**3*f*(t an(e + f*x)**6 + 3*tan(e + f*x)**4 + 3*tan(e + f*x)**2 + 1))