Integrand size = 35, antiderivative size = 136 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{9/2}} \, dx=\frac {i (c-i c \tan (e+f x))^{5/2}}{9 f (a+i a \tan (e+f x))^{9/2}}+\frac {2 i (c-i c \tan (e+f x))^{5/2}}{63 a f (a+i a \tan (e+f x))^{7/2}}+\frac {2 i (c-i c \tan (e+f x))^{5/2}}{315 a^2 f (a+i a \tan (e+f x))^{5/2}} \] Output:
1/9*I*(c-I*c*tan(f*x+e))^(5/2)/f/(a+I*a*tan(f*x+e))^(9/2)+2/63*I*(c-I*c*ta n(f*x+e))^(5/2)/a/f/(a+I*a*tan(f*x+e))^(7/2)+2/315*I*(c-I*c*tan(f*x+e))^(5 /2)/a^2/f/(a+I*a*tan(f*x+e))^(5/2)
Time = 2.77 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.70 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{9/2}} \, dx=\frac {c^2 (i+\tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)} \left (-47 i+14 \tan (e+f x)+2 i \tan ^2(e+f x)\right )}{315 a^4 f (-i+\tan (e+f x))^4 \sqrt {a+i a \tan (e+f x)}} \] Input:
Integrate[(c - I*c*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^(9/2),x]
Output:
(c^2*(I + Tan[e + f*x])^2*Sqrt[c - I*c*Tan[e + f*x]]*(-47*I + 14*Tan[e + f *x] + (2*I)*Tan[e + f*x]^2))/(315*a^4*f*(-I + Tan[e + f*x])^4*Sqrt[a + I*a *Tan[e + f*x]])
Time = 0.32 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 4006, 55, 55, 48}
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.
| \(\displaystyle \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{9/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{9/2}}dx\) |
| \(\Big \downarrow \) 4006 |
| \(\displaystyle \frac {a c \int \frac {(c-i c \tan (e+f x))^{3/2}}{(i \tan (e+f x) a+a)^{11/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle \frac {a c \left (\frac {2 \int \frac {(c-i c \tan (e+f x))^{3/2}}{(i \tan (e+f x) a+a)^{9/2}}d\tan (e+f x)}{9 a}+\frac {i (c-i c \tan (e+f x))^{5/2}}{9 a c (a+i a \tan (e+f x))^{9/2}}\right )}{f}\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle \frac {a c \left (\frac {2 \left (\frac {\int \frac {(c-i c \tan (e+f x))^{3/2}}{(i \tan (e+f x) a+a)^{7/2}}d\tan (e+f x)}{7 a}+\frac {i (c-i c \tan (e+f x))^{5/2}}{7 a c (a+i a \tan (e+f x))^{7/2}}\right )}{9 a}+\frac {i (c-i c \tan (e+f x))^{5/2}}{9 a c (a+i a \tan (e+f x))^{9/2}}\right )}{f}\) |
| \(\Big \downarrow \) 48 |
| \(\displaystyle \frac {a c \left (\frac {2 \left (\frac {i (c-i c \tan (e+f x))^{5/2}}{35 a^2 c (a+i a \tan (e+f x))^{5/2}}+\frac {i (c-i c \tan (e+f x))^{5/2}}{7 a c (a+i a \tan (e+f x))^{7/2}}\right )}{9 a}+\frac {i (c-i c \tan (e+f x))^{5/2}}{9 a c (a+i a \tan (e+f x))^{9/2}}\right )}{f}\) |
Input:
Int[(c - I*c*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^(9/2),x]
Output:
(a*c*(((I/9)*(c - I*c*Tan[e + f*x])^(5/2))/(a*c*(a + I*a*Tan[e + f*x])^(9/ 2)) + (2*(((I/7)*(c - I*c*Tan[e + f*x])^(5/2))/(a*c*(a + I*a*Tan[e + f*x]) ^(7/2)) + ((I/35)*(c - I*c*Tan[e + f*x])^(5/2))/(a^2*c*(a + I*a*Tan[e + f* x])^(5/2))))/(9*a)))/f
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
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 = 99, normalized size of antiderivative = 0.73
| method | result | size |
| derivativedivides | \(-\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{2} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (2 i \tan \left (f x +e \right )^{3}-33 i \tan \left (f x +e \right )+12 \tan \left (f x +e \right )^{2}+47\right )}{315 f \,a^{5} \left (-\tan \left (f x +e \right )+i\right )^{6}}\) | \(99\) |
| default | \(-\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{2} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (2 i \tan \left (f x +e \right )^{3}-33 i \tan \left (f x +e \right )+12 \tan \left (f x +e \right )^{2}+47\right )}{315 f \,a^{5} \left (-\tan \left (f x +e \right )+i\right )^{6}}\) | \(99\) |
Input:
int((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(9/2),x,method=_RETURNVERB OSE)
Output:
-1/315*I/f*(-c*(I*tan(f*x+e)-1))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*c^2/a^5* (1+tan(f*x+e)^2)*(2*I*tan(f*x+e)^3-33*I*tan(f*x+e)+12*tan(f*x+e)^2+47)/(-t an(f*x+e)+I)^6
Time = 0.08 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.73 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{9/2}} \, dx=\frac {{\left (63 i \, c^{2} e^{\left (6 i \, f x + 6 i \, e\right )} + 153 i \, c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 125 i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 35 i \, c^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-9 i \, f x - 9 i \, e\right )}}{1260 \, a^{5} f} \] Input:
integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(9/2),x, algorithm=" fricas")
Output:
1/1260*(63*I*c^2*e^(6*I*f*x + 6*I*e) + 153*I*c^2*e^(4*I*f*x + 4*I*e) + 125 *I*c^2*e^(2*I*f*x + 2*I*e) + 35*I*c^2)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*s qrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(-9*I*f*x - 9*I*e)/(a^5*f)
Timed out. \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{9/2}} \, dx=\text {Timed out} \] Input:
integrate((c-I*c*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e))**(9/2),x)
Output:
Timed out
Time = 0.21 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.09 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{9/2}} \, dx=\frac {{\left (35 i \, c^{2} \cos \left (9 \, f x + 9 \, e\right ) + 90 i \, c^{2} \cos \left (\frac {7}{9} \, \arctan \left (\sin \left (9 \, f x + 9 \, e\right ), \cos \left (9 \, f x + 9 \, e\right )\right )\right ) + 63 i \, c^{2} \cos \left (\frac {5}{9} \, \arctan \left (\sin \left (9 \, f x + 9 \, e\right ), \cos \left (9 \, f x + 9 \, e\right )\right )\right ) + 35 \, c^{2} \sin \left (9 \, f x + 9 \, e\right ) + 90 \, c^{2} \sin \left (\frac {7}{9} \, \arctan \left (\sin \left (9 \, f x + 9 \, e\right ), \cos \left (9 \, f x + 9 \, e\right )\right )\right ) + 63 \, c^{2} \sin \left (\frac {5}{9} \, \arctan \left (\sin \left (9 \, f x + 9 \, e\right ), \cos \left (9 \, f x + 9 \, e\right )\right )\right )\right )} \sqrt {c}}{1260 \, a^{\frac {9}{2}} f} \] Input:
integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(9/2),x, algorithm=" maxima")
Output:
1/1260*(35*I*c^2*cos(9*f*x + 9*e) + 90*I*c^2*cos(7/9*arctan2(sin(9*f*x + 9 *e), cos(9*f*x + 9*e))) + 63*I*c^2*cos(5/9*arctan2(sin(9*f*x + 9*e), cos(9 *f*x + 9*e))) + 35*c^2*sin(9*f*x + 9*e) + 90*c^2*sin(7/9*arctan2(sin(9*f*x + 9*e), cos(9*f*x + 9*e))) + 63*c^2*sin(5/9*arctan2(sin(9*f*x + 9*e), cos (9*f*x + 9*e))))*sqrt(c)/(a^(9/2)*f)
\[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{9/2}} \, dx=\int { \frac {{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {9}{2}}} \,d x } \] Input:
integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(9/2),x, algorithm=" giac")
Output:
integrate((-I*c*tan(f*x + e) + c)^(5/2)/(I*a*tan(f*x + e) + a)^(9/2), x)
Time = 4.21 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.35 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{9/2}} \, dx=\frac {c^2\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (63\,\sin \left (4\,e+4\,f\,x\right )+153\,\sin \left (6\,e+6\,f\,x\right )+125\,\sin \left (8\,e+8\,f\,x\right )+35\,\sin \left (10\,e+10\,f\,x\right )+\cos \left (4\,e+4\,f\,x\right )\,63{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,153{}\mathrm {i}+\cos \left (8\,e+8\,f\,x\right )\,125{}\mathrm {i}+\cos \left (10\,e+10\,f\,x\right )\,35{}\mathrm {i}\right )}{2520\,a^5\,f} \] Input:
int((c - c*tan(e + f*x)*1i)^(5/2)/(a + a*tan(e + f*x)*1i)^(9/2),x)
Output:
(c^2*((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2* f*x) + 1))^(1/2)*(cos(4*e + 4*f*x)*63i + cos(6*e + 6*f*x)*153i + cos(8*e + 8*f*x)*125i + cos(10*e + 10*f*x)*35i + 63*sin(4*e + 4*f*x) + 153*sin(6*e + 6*f*x) + 125*sin(8*e + 8*f*x) + 35*sin(10*e + 10*f*x)))/(2520*a^5*f)
\[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{9/2}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, c^{2} \left (-\left (\int -\frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{6}-4 \tan \left (f x +e \right )^{5} i -5 \tan \left (f x +e \right )^{4}-5 \tan \left (f x +e \right )^{2}+4 \tan \left (f x +e \right ) i +1}d x \right ) i +3 \left (\int -\frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{6}-4 \tan \left (f x +e \right )^{5} i -5 \tan \left (f x +e \right )^{4}-5 \tan \left (f x +e \right )^{2}+4 \tan \left (f x +e \right ) i +1}d x \right ) i -\left (\int -\frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}}{\tan \left (f x +e \right )^{6}-4 \tan \left (f x +e \right )^{5} i -5 \tan \left (f x +e \right )^{4}-5 \tan \left (f x +e \right )^{2}+4 \tan \left (f x +e \right ) i +1}d x \right )-3 \left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{6}-4 \tan \left (f x +e \right )^{5} i -5 \tan \left (f x +e \right )^{4}-5 \tan \left (f x +e \right )^{2}+4 \tan \left (f x +e \right ) i +1}d x \right )\right )}{a^{5}} \] Input:
int((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(9/2),x)
Output:
(sqrt(c)*sqrt(a)*c**2*( - int(( - sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**3)/(tan(e + f*x)**6 - 4*tan(e + f*x)**5*i - 5*t an(e + f*x)**4 - 5*tan(e + f*x)**2 + 4*tan(e + f*x)*i + 1),x)*i + 3*int(( - sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x))/(tan( e + f*x)**6 - 4*tan(e + f*x)**5*i - 5*tan(e + f*x)**4 - 5*tan(e + f*x)**2 + 4*tan(e + f*x)*i + 1),x)*i - int(( - sqrt(tan(e + f*x)*i + 1)*sqrt( - ta n(e + f*x)*i + 1))/(tan(e + f*x)**6 - 4*tan(e + f*x)**5*i - 5*tan(e + f*x) **4 - 5*tan(e + f*x)**2 + 4*tan(e + f*x)*i + 1),x) - 3*int((sqrt(tan(e + f *x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(tan(e + f*x)**6 - 4*tan(e + f*x)**5*i - 5*tan(e + f*x)**4 - 5*tan(e + f*x)**2 + 4*tan(e + f *x)*i + 1),x)))/a**5