Integrand size = 35, antiderivative size = 204 \[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {2 i a^{7/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{5/2} f}-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {2 i a^2 (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 i a^3 \sqrt {a+i a \tan (e+f x)}}{c^2 f \sqrt {c-i c \tan (e+f x)}} \] Output:
2*I*a^(7/2)*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I*c*tan(f*x +e))^(1/2))/c^(5/2)/f-2/5*I*a*(a+I*a*tan(f*x+e))^(5/2)/f/(c-I*c*tan(f*x+e) )^(5/2)+2/3*I*a^2*(a+I*a*tan(f*x+e))^(3/2)/c/f/(c-I*c*tan(f*x+e))^(3/2)-2* I*a^3*(a+I*a*tan(f*x+e))^(1/2)/c^2/f/(c-I*c*tan(f*x+e))^(1/2)
Time = 3.87 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.08 \[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {2 i a^{7/2} \sec ^3(e+f x) \left (\sqrt {a} (10 \cos (e+f x)+3 \cos (3 (e+f x))-20 i \sin (e+f x)+3 i \sin (3 (e+f x))) \sqrt {1-i \tan (e+f x)}-15 \arcsin \left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) (\cos (3 (e+f x))-i \sin (3 (e+f x))) \sqrt {a+i a \tan (e+f x)}\right )}{15 c^2 f \sqrt {1-i \tan (e+f x)} (i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \] Input:
Integrate[(a + I*a*Tan[e + f*x])^(7/2)/(c - I*c*Tan[e + f*x])^(5/2),x]
Output:
(((2*I)/15)*a^(7/2)*Sec[e + f*x]^3*(Sqrt[a]*(10*Cos[e + f*x] + 3*Cos[3*(e + f*x)] - (20*I)*Sin[e + f*x] + (3*I)*Sin[3*(e + f*x)])*Sqrt[1 - I*Tan[e + f*x]] - 15*ArcSin[Sqrt[a + I*a*Tan[e + f*x]]/(Sqrt[2]*Sqrt[a])]*(Cos[3*(e + f*x)] - I*Sin[3*(e + f*x)])*Sqrt[a + I*a*Tan[e + f*x]]))/(c^2*f*Sqrt[1 - I*Tan[e + f*x]]*(I + Tan[e + f*x])^2*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])
Time = 0.35 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4006, 57, 57, 57, 45, 218}
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.
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4006 |
| \(\displaystyle \frac {a c \int \frac {(i \tan (e+f x) a+a)^{5/2}}{(c-i c \tan (e+f x))^{7/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 57 |
| \(\displaystyle \frac {a c \left (-\frac {a \int \frac {(i \tan (e+f x) a+a)^{3/2}}{(c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
| \(\Big \downarrow \) 57 |
| \(\displaystyle \frac {a c \left (-\frac {a \left (-\frac {a \int \frac {\sqrt {i \tan (e+f x) a+a}}{(c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{c}-\frac {2 i (a+i a \tan (e+f x))^{3/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
| \(\Big \downarrow \) 57 |
| \(\displaystyle \frac {a c \left (-\frac {a \left (-\frac {a \left (-\frac {a \int \frac {1}{\sqrt {i \tan (e+f x) a+a} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{c}-\frac {2 i \sqrt {a+i a \tan (e+f x)}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{3/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
| \(\Big \downarrow \) 45 |
| \(\displaystyle \frac {a c \left (-\frac {a \left (-\frac {a \left (-\frac {2 a \int \frac {1}{i a+\frac {i c (i \tan (e+f x) a+a)}{c-i c \tan (e+f x)}}d\frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c-i c \tan (e+f x)}}}{c}-\frac {2 i \sqrt {a+i a \tan (e+f x)}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{3/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {a c \left (-\frac {a \left (-\frac {a \left (\frac {2 i \sqrt {a} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{3/2}}-\frac {2 i \sqrt {a+i a \tan (e+f x)}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{3/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
Input:
Int[(a + I*a*Tan[e + f*x])^(7/2)/(c - I*c*Tan[e + f*x])^(5/2),x]
Output:
(a*c*((((-2*I)/5)*(a + I*a*Tan[e + f*x])^(5/2))/(c*(c - I*c*Tan[e + f*x])^ (5/2)) - (a*((((-2*I)/3)*(a + I*a*Tan[e + f*x])^(3/2))/(c*(c - I*c*Tan[e + f*x])^(3/2)) - (a*(((2*I)*Sqrt[a]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f* x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/c^(3/2) - ((2*I)*Sqrt[a + I*a* Tan[e + f*x]])/(c*Sqrt[c - I*c*Tan[e + f*x]])))/c))/c))/f
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && !GtQ[c, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c , d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 428 vs. \(2 (164 ) = 328\).
Time = 0.35 (sec) , antiderivative size = 429, normalized size of antiderivative = 2.10
| method | result | size |
| derivativedivides | \(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (60 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{3}+15 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{4}-60 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-94 i \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )^{2}-90 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{2}-46 \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3} \sqrt {a c}+26 i \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}+15 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c +74 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{15 f \,c^{3} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i+\tan \left (f x +e \right )\right )^{4} \sqrt {a c}}\) | \(429\) |
| default | \(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (60 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{3}+15 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{4}-60 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-94 i \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )^{2}-90 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{2}-46 \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3} \sqrt {a c}+26 i \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}+15 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c +74 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{15 f \,c^{3} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i+\tan \left (f x +e \right )\right )^{4} \sqrt {a c}}\) | \(429\) |
Input:
int((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(5/2),x,method=_RETURNVERB OSE)
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*(60 *I*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2 ))*a*c*tan(f*x+e)^3+15*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a* c)^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)^4-60*I*ln((a*c*tan(f*x+e)+(a*c*(1+ta n(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)-94*I*(a*c*(1+t an(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^2-90*ln((a*c*tan(f*x+e)+(a*c*(1 +tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)^2-46*(a*c*( 1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^3*(a*c)^(1/2)+26*I*(a*c*(1+tan(f*x+e)^2) )^(1/2)*(a*c)^(1/2)+15*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a* c)^(1/2))/(a*c)^(1/2))*a*c+74*tan(f*x+e)*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c )^(1/2))/(a*c*(1+tan(f*x+e)^2))^(1/2)/(I+tan(f*x+e))^4/(a*c)^(1/2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 382 vs. \(2 (152) = 304\).
Time = 0.11 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.87 \[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {15 \, c^{3} f \sqrt {\frac {a^{7}}{c^{5} f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{3} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (i \, c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, c^{3} f\right )} \sqrt {\frac {a^{7}}{c^{5} f^{2}}}\right )}}{a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}}\right ) - 15 \, c^{3} f \sqrt {\frac {a^{7}}{c^{5} f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{3} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (-i \, c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c^{3} f\right )} \sqrt {\frac {a^{7}}{c^{5} f^{2}}}\right )}}{a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}}\right ) + 4 \, {\left (3 i \, a^{3} e^{\left (7 i \, f x + 7 i \, e\right )} - 2 i \, a^{3} e^{\left (5 i \, f x + 5 i \, e\right )} + 10 i \, a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + 15 i \, a^{3} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{30 \, c^{3} f} \] Input:
integrate((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm=" fricas")
Output:
-1/30*(15*c^3*f*sqrt(a^7/(c^5*f^2))*log(4*(2*(a^3*e^(3*I*f*x + 3*I*e) + a^ 3*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - (I*c^3*f*e^(2*I*f*x + 2*I*e) - I*c^3*f)*sqrt(a^7/(c^5*f^2)) )/(a^3*e^(2*I*f*x + 2*I*e) + a^3)) - 15*c^3*f*sqrt(a^7/(c^5*f^2))*log(4*(2 *(a^3*e^(3*I*f*x + 3*I*e) + a^3*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I* e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - (-I*c^3*f*e^(2*I*f*x + 2*I*e) + I*c^3*f)*sqrt(a^7/(c^5*f^2)))/(a^3*e^(2*I*f*x + 2*I*e) + a^3)) + 4*(3*I *a^3*e^(7*I*f*x + 7*I*e) - 2*I*a^3*e^(5*I*f*x + 5*I*e) + 10*I*a^3*e^(3*I*f *x + 3*I*e) + 15*I*a^3*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))* sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))/(c^3*f)
Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((a+I*a*tan(f*x+e))**(7/2)/(c-I*c*tan(f*x+e))**(5/2),x)
Output:
Timed out
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 440 vs. \(2 (152) = 304\).
Time = 0.27 (sec) , antiderivative size = 440, normalized size of antiderivative = 2.16 \[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx =\text {Too large to display} \] Input:
integrate((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm=" maxima")
Output:
-1/30*(-30*I*a^3*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e ))), sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - 30*I*a^3* arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), -sin(1/2*arc tan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + 12*I*a^3*cos(5/2*arctan2( sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 20*I*a^3*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 60*I*a^3*cos(1/2*arctan2(sin(2*f*x + 2*e), co s(2*f*x + 2*e))) + 15*a^3*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*si n(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - 15*a^3*log(cos(1 /2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2* f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos (2*f*x + 2*e))) + 1) - 12*a^3*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 20*a^3*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 6 0*a^3*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)/(c^(5/ 2)*f)
\[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm=" giac")
Output:
integrate((I*a*tan(f*x + e) + a)^(7/2)/(-I*c*tan(f*x + e) + c)^(5/2), x)
Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \] Input:
int((a + a*tan(e + f*x)*1i)^(7/2)/(c - c*tan(e + f*x)*1i)^(5/2),x)
Output:
int((a + a*tan(e + f*x)*1i)^(7/2)/(c - c*tan(e + f*x)*1i)^(5/2), x)
\[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {a}\, a^{3} \left (-\left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}}{\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}+2 \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right ) i -\sqrt {-\tan \left (f x +e \right ) i +1}}d x \right )+\left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{3}}{\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}+2 \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right ) i -\sqrt {-\tan \left (f x +e \right ) i +1}}d x \right ) i +3 \left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}}{\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}+2 \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right ) i -\sqrt {-\tan \left (f x +e \right ) i +1}}d x \right )-3 \left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )}{\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}+2 \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right ) i -\sqrt {-\tan \left (f x +e \right ) i +1}}d x \right ) i \right )}{\sqrt {c}\, c^{2}} \] Input:
int((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(5/2),x)
Output:
(sqrt(a)*a**3*( - int(sqrt(tan(e + f*x)*i + 1)/(sqrt( - tan(e + f*x)*i + 1 )*tan(e + f*x)**2 + 2*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)*i - sqrt( - tan(e + f*x)*i + 1)),x) + int((sqrt(tan(e + f*x)*i + 1)*tan(e + f*x)**3)/ (sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2 + 2*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)*i - sqrt( - tan(e + f*x)*i + 1)),x)*i + 3*int((sqrt(tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)** 2 + 2*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)*i - sqrt( - tan(e + f*x)*i + 1)),x) - 3*int((sqrt(tan(e + f*x)*i + 1)*tan(e + f*x))/(sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2 + 2*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)*i - sqrt( - tan(e + f*x)*i + 1)),x)*i))/(sqrt(c)*c**2)