Integrand size = 33, antiderivative size = 197 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {i c^{5/2} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a^3 f}+\frac {i c^4 (c-i c \tan (e+f x))^{3/2}}{3 a^3 f (c+i c \tan (e+f x))^3}-\frac {i c^4 \sqrt {c-i c \tan (e+f x)}}{4 a^3 f (c+i c \tan (e+f x))^2}+\frac {i c^4 \sqrt {c-i c \tan (e+f x)}}{16 a^3 f \left (c^2+i c^2 \tan (e+f x)\right )} \] Output:
1/32*I*c^(5/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/ 2)/a^3/f+1/3*I*c^4*(c-I*c*tan(f*x+e))^(3/2)/a^3/f/(c+I*c*tan(f*x+e))^3-1/4 *I*c^4*(c-I*c*tan(f*x+e))^(1/2)/a^3/f/(c+I*c*tan(f*x+e))^2+1/16*I*c^4*(c-I *c*tan(f*x+e))^(1/2)/a^3/f/(c^2+I*c^2*tan(f*x+e))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.99 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.27 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {i \operatorname {Hypergeometric2F1}\left (\frac {5}{2},4,\frac {7}{2},-\frac {1}{2} i (i+\tan (e+f x))\right ) (c-i c \tan (e+f x))^{5/2}}{40 a^3 f} \] Input:
Integrate[(c - I*c*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^3,x]
Output:
((I/40)*Hypergeometric2F1[5/2, 4, 7/2, (-1/2*I)*(I + Tan[e + f*x])]*(c - I *c*Tan[e + f*x])^(5/2))/(a^3*f)
Time = 0.45 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.86, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 4005, 3042, 3968, 51, 51, 52, 73, 219}
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.
| \(\displaystyle \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3}dx\) |
| \(\Big \downarrow \) 4005 |
| \(\displaystyle \frac {\int \cos ^6(e+f x) (c-i c \tan (e+f x))^{11/2}dx}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(c-i c \tan (e+f x))^{11/2}}{\sec (e+f x)^6}dx}{a^3 c^3}\) |
| \(\Big \downarrow \) 3968 |
| \(\displaystyle \frac {i c^4 \int \frac {(c-i c \tan (e+f x))^{3/2}}{(i \tan (e+f x) c+c)^4}d(-i c \tan (e+f x))}{a^3 f}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {i c^4 \left (\frac {(c-i c \tan (e+f x))^{3/2}}{3 (c+i c \tan (e+f x))^3}-\frac {1}{2} \int \frac {\sqrt {c-i c \tan (e+f x)}}{(i \tan (e+f x) c+c)^3}d(-i c \tan (e+f x))\right )}{a^3 f}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {i c^4 \left (\frac {1}{2} \left (\frac {1}{4} \int \frac {1}{\sqrt {c-i c \tan (e+f x)} (i \tan (e+f x) c+c)^2}d(-i c \tan (e+f x))-\frac {\sqrt {c-i c \tan (e+f x)}}{2 (c+i c \tan (e+f x))^2}\right )+\frac {(c-i c \tan (e+f x))^{3/2}}{3 (c+i c \tan (e+f x))^3}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {i c^4 \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {\int \frac {1}{\sqrt {c-i c \tan (e+f x)} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{4 c}+\frac {\sqrt {c-i c \tan (e+f x)}}{2 c (c+i c \tan (e+f x))}\right )-\frac {\sqrt {c-i c \tan (e+f x)}}{2 (c+i c \tan (e+f x))^2}\right )+\frac {(c-i c \tan (e+f x))^{3/2}}{3 (c+i c \tan (e+f x))^3}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {i c^4 \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {\int \frac {1}{c^2 \tan ^2(e+f x)+2 c}d\sqrt {c-i c \tan (e+f x)}}{2 c}+\frac {\sqrt {c-i c \tan (e+f x)}}{2 c (c+i c \tan (e+f x))}\right )-\frac {\sqrt {c-i c \tan (e+f x)}}{2 (c+i c \tan (e+f x))^2}\right )+\frac {(c-i c \tan (e+f x))^{3/2}}{3 (c+i c \tan (e+f x))^3}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {i c^4 \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {\sqrt {c-i c \tan (e+f x)}}{2 c (c+i c \tan (e+f x))}-\frac {i \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2}}\right )}{2 \sqrt {2} c^{3/2}}\right )-\frac {\sqrt {c-i c \tan (e+f x)}}{2 (c+i c \tan (e+f x))^2}\right )+\frac {(c-i c \tan (e+f x))^{3/2}}{3 (c+i c \tan (e+f x))^3}\right )}{a^3 f}\) |
Input:
Int[(c - I*c*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^3,x]
Output:
(I*c^4*((c - I*c*Tan[e + f*x])^(3/2)/(3*(c + I*c*Tan[e + f*x])^3) + (-1/2* Sqrt[c - I*c*Tan[e + f*x]]/(c + I*c*Tan[e + f*x])^2 + (((-1/2*I)*ArcTan[(S qrt[c]*Tan[e + f*x])/Sqrt[2]])/(Sqrt[2]*c^(3/2)) + Sqrt[c - I*c*Tan[e + f* x]]/(2*c*(c + I*c*Tan[e + f*x])))/4)/2))/(a^3*f)
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, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(a^(m - 2)*b*f) Subst[Int[(a - x)^(m/2 - 1)*(a + x )^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] && !(IGtQ[n, 0] && (LtQ[ m, 0] || GtQ[m, n]))
Time = 0.36 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.58
| method | result | size |
| derivativedivides | \(\frac {2 i c^{4} \left (\frac {\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{32 c}+\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{6}-\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, c}{8}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 c^{\frac {3}{2}}}\right )}{f \,a^{3}}\) | \(114\) |
| default | \(\frac {2 i c^{4} \left (\frac {\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{32 c}+\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{6}-\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, c}{8}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 c^{\frac {3}{2}}}\right )}{f \,a^{3}}\) | \(114\) |
Input:
int((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^3,x,method=_RETURNVERBOSE)
Output:
2*I/f/a^3*c^4*(8*(1/256/c*(c-I*c*tan(f*x+e))^(5/2)+1/48*(c-I*c*tan(f*x+e)) ^(3/2)-1/64*(c-I*c*tan(f*x+e))^(1/2)*c)/(c+I*c*tan(f*x+e))^3+1/64/c^(3/2)* 2^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 312 vs. \(2 (152) = 304\).
Time = 0.09 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.58 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {c^{5}}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (i \, c^{3} + \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c^{5}}{a^{6} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a^{3} f}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {c^{5}}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left (i \, c^{3} - \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c^{5}}{a^{6} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a^{3} f}\right ) + \sqrt {2} {\left (-3 i \, c^{2} e^{\left (6 i \, f x + 6 i \, e\right )} - i \, c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 10 i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i \, c^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{96 \, a^{3} f} \] Input:
integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="fric as")
Output:
1/96*(3*sqrt(1/2)*a^3*f*sqrt(-c^5/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/8*( I*c^3 + sqrt(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt(c/(e^(2 *I*f*x + 2*I*e) + 1))*sqrt(-c^5/(a^6*f^2)))*e^(-I*f*x - I*e)/(a^3*f)) - 3* sqrt(1/2)*a^3*f*sqrt(-c^5/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/8*(I*c^3 - sqrt(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-c^5/(a^6*f^2)))*e^(-I*f*x - I*e)/(a^3*f)) + sqrt(2)*(- 3*I*c^2*e^(6*I*f*x + 6*I*e) - I*c^2*e^(4*I*f*x + 4*I*e) + 10*I*c^2*e^(2*I* f*x + 2*I*e) + 8*I*c^2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-6*I*f*x - 6 *I*e)/(a^3*f)
\[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {i \left (\int \frac {c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx + \int \left (- \frac {c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\right )\, dx + \int \left (- \frac {2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\right )\, dx\right )}{a^{3}} \] Input:
integrate((c-I*c*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e))**3,x)
Output:
I*(Integral(c**2*sqrt(-I*c*tan(e + f*x) + c)/(tan(e + f*x)**3 - 3*I*tan(e + f*x)**2 - 3*tan(e + f*x) + I), x) + Integral(-c**2*sqrt(-I*c*tan(e + f*x ) + c)*tan(e + f*x)**2/(tan(e + f*x)**3 - 3*I*tan(e + f*x)**2 - 3*tan(e + f*x) + I), x) + Integral(-2*I*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x )/(tan(e + f*x)**3 - 3*I*tan(e + f*x)**2 - 3*tan(e + f*x) + I), x))/a**3
Time = 0.12 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.97 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {i \, {\left (\frac {3 \, \sqrt {2} c^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3}} + \frac {4 \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} c^{4} + 16 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c^{5} - 12 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} c^{6}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} a^{3} - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{3} c + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{3} c^{2} - 8 \, a^{3} c^{3}}\right )}}{192 \, c f} \] Input:
integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="maxi ma")
Output:
-1/192*I*(3*sqrt(2)*c^(7/2)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(sqrt(2)*sqrt(c) + sqrt(-I*c*tan(f*x + e) + c)))/a^3 + 4*(3*(-I*c*t an(f*x + e) + c)^(5/2)*c^4 + 16*(-I*c*tan(f*x + e) + c)^(3/2)*c^5 - 12*sqr t(-I*c*tan(f*x + e) + c)*c^6)/((-I*c*tan(f*x + e) + c)^3*a^3 - 6*(-I*c*tan (f*x + e) + c)^2*a^3*c + 12*(-I*c*tan(f*x + e) + c)*a^3*c^2 - 8*a^3*c^3))/ (c*f)
Exception generated. \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^3,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 = 2.14 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.92 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {-\frac {c^5\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{4\,a^3\,f}+\frac {c^4\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,1{}\mathrm {i}}{3\,a^3\,f}+\frac {c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,1{}\mathrm {i}}{16\,a^3\,f}}{6\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-12\,c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3+8\,c^3}+\frac {\sqrt {2}\,{\left (-c\right )}^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,1{}\mathrm {i}}{32\,a^3\,f} \] Input:
int((c - c*tan(e + f*x)*1i)^(5/2)/(a + a*tan(e + f*x)*1i)^3,x)
Output:
((c^4*(c - c*tan(e + f*x)*1i)^(3/2)*1i)/(3*a^3*f) - (c^5*(c - c*tan(e + f* x)*1i)^(1/2)*1i)/(4*a^3*f) + (c^3*(c - c*tan(e + f*x)*1i)^(5/2)*1i)/(16*a^ 3*f))/(6*c*(c - c*tan(e + f*x)*1i)^2 - 12*c^2*(c - c*tan(e + f*x)*1i) - (c - c*tan(e + f*x)*1i)^3 + 8*c^3) + (2^(1/2)*(-c)^(5/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*1i)/(32*a^3*f)
\[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {\sqrt {c}\, c^{2} \left (2 \sqrt {-\tan \left (f x +e \right ) i +1}\, i +\left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{4}}{\tan \left (f x +e \right )^{3} i +3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i -1}d x \right ) f -4 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{3} i +3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i -1}d x \right ) f i -5 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{3} i +3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i -1}d x \right ) f +6 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{3} i +3 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i -1}d x \right ) f i \right )}{a^{3} f} \] Input:
int((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^3,x)
Output:
(sqrt(c)*c**2*(2*sqrt( - tan(e + f*x)*i + 1)*i + int((sqrt( - tan(e + f*x) *i + 1)*tan(e + f*x)**4)/(tan(e + f*x)**3*i + 3*tan(e + f*x)**2 - 3*tan(e + f*x)*i - 1),x)*f - 4*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**3)/( tan(e + f*x)**3*i + 3*tan(e + f*x)**2 - 3*tan(e + f*x)*i - 1),x)*f*i - 5*i nt((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(tan(e + f*x)**3*i + 3*ta n(e + f*x)**2 - 3*tan(e + f*x)*i - 1),x)*f + 6*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x))/(tan(e + f*x)**3*i + 3*tan(e + f*x)**2 - 3*tan(e + f*x )*i - 1),x)*f*i))/(a**3*f)