Integrand size = 33, antiderivative size = 146 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx=\frac {3 i c^{5/2} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{4 \sqrt {2} a^2 f}+\frac {i c^3 (c-i c \tan (e+f x))^{3/2}}{2 a^2 f (c+i c \tan (e+f x))^2}-\frac {3 i c^3 \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (c+i c \tan (e+f x))} \]
3/8*I*c^(5/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^2/f* 2^(1/2)+1/2*I*c^3*(c-I*c*tan(f*x+e))^(3/2)/a^2/f/(c+I*c*tan(f*x+e))^2-3/4* I*c^3*(c-I*c*tan(f*x+e))^(1/2)/a^2/f/(c+I*c*tan(f*x+e))
Time = 2.06 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.10 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {i c^{5/2} \sec ^2(e+f x) \left (2 \sqrt {c} (-3+2 \cos (2 (e+f x))-2 i \sin (2 (e+f x)))+3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right ) (\cos (2 (e+f x))+i \sin (2 (e+f x))) \sqrt {c-i c \tan (e+f x)}\right )}{8 a^2 f (-i+\tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}} \]
((-1/8*I)*c^(5/2)*Sec[e + f*x]^2*(2*Sqrt[c]*(-3 + 2*Cos[2*(e + f*x)] - (2* I)*Sin[2*(e + f*x)]) + 3*Sqrt[2]*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[ 2]*Sqrt[c])]*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]]))/(a^2*f*(-I + Tan[e + f*x])^2*Sqrt[c - I*c*Tan[e + f*x]])
Time = 0.41 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.82, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3042, 4005, 3042, 3968, 51, 51, 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.
| \(\displaystyle \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2}dx\) |
| \(\Big \downarrow \) 4005 |
| \(\displaystyle \frac {\int \cos ^4(e+f x) (c-i c \tan (e+f x))^{9/2}dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(c-i c \tan (e+f x))^{9/2}}{\sec (e+f x)^4}dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 3968 |
| \(\displaystyle \frac {i c^3 \int \frac {(c-i c \tan (e+f x))^{3/2}}{(i \tan (e+f x) c+c)^3}d(-i c \tan (e+f x))}{a^2 f}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {i c^3 \left (\frac {(c-i c \tan (e+f x))^{3/2}}{2 (c+i c \tan (e+f x))^2}-\frac {3}{4} \int \frac {\sqrt {c-i c \tan (e+f x)}}{(i \tan (e+f x) c+c)^2}d(-i c \tan (e+f x))\right )}{a^2 f}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {i c^3 \left (\frac {(c-i c \tan (e+f x))^{3/2}}{2 (c+i c \tan (e+f x))^2}-\frac {3}{4} \left (\frac {\sqrt {c-i c \tan (e+f x)}}{c+i c \tan (e+f x)}-\frac {1}{2} \int \frac {1}{\sqrt {c-i c \tan (e+f x)} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))\right )\right )}{a^2 f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {i c^3 \left (\frac {(c-i c \tan (e+f x))^{3/2}}{2 (c+i c \tan (e+f x))^2}-\frac {3}{4} \left (\frac {\sqrt {c-i c \tan (e+f x)}}{c+i c \tan (e+f x)}-\int \frac {1}{c^2 \tan ^2(e+f x)+2 c}d\sqrt {c-i c \tan (e+f x)}\right )\right )}{a^2 f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {i c^3 \left (\frac {(c-i c \tan (e+f x))^{3/2}}{2 (c+i c \tan (e+f x))^2}-\frac {3}{4} \left (\frac {i \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2}}\right )}{\sqrt {2} \sqrt {c}}+\frac {\sqrt {c-i c \tan (e+f x)}}{c+i c \tan (e+f x)}\right )\right )}{a^2 f}\) |
(I*c^3*((c - I*c*Tan[e + f*x])^(3/2)/(2*(c + I*c*Tan[e + f*x])^2) - (3*((I *ArcTan[(Sqrt[c]*Tan[e + f*x])/Sqrt[2]])/(Sqrt[2]*Sqrt[c]) + Sqrt[c - I*c* Tan[e + f*x]]/(c + I*c*Tan[e + f*x])))/4))/(a^2*f)
3.10.72.3.1 Defintions of rubi rules used
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] :> 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 = 0.64 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.65
| method | result | size |
| derivativedivides | \(\frac {2 i c^{3} \left (\frac {\frac {5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{8}-\frac {3 c \sqrt {c -i c \tan \left (f x +e \right )}}{4}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 \sqrt {c}}\right )}{f \,a^{2}}\) | \(95\) |
| default | \(\frac {2 i c^{3} \left (\frac {\frac {5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{8}-\frac {3 c \sqrt {c -i c \tan \left (f x +e \right )}}{4}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 \sqrt {c}}\right )}{f \,a^{2}}\) | \(95\) |
2*I/f/a^2*c^3*(4*(5/32*(c-I*c*tan(f*x+e))^(3/2)-3/16*c*(c-I*c*tan(f*x+e))^ (1/2))/(c+I*c*tan(f*x+e))^2+3/16*2^(1/2)/c^(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. 299 vs. \(2 (111) = 222\).
Time = 0.25 (sec) , antiderivative size = 299, normalized size of antiderivative = 2.05 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {c^{5}}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {3 \, {\left (-i \, c^{3} + \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c^{5}}{a^{4} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a^{2} f}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {c^{5}}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {3 \, {\left (-i \, c^{3} - \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c^{5}}{a^{4} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a^{2} f}\right ) - \sqrt {2} {\left (-3 i \, c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} - i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, c^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{8 \, a^{2} f} \]
-1/8*(3*sqrt(1/2)*a^2*f*sqrt(-c^5/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(-3/2* (-I*c^3 + sqrt(2)*sqrt(1/2)*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(c/(e^ (2*I*f*x + 2*I*e) + 1))*sqrt(-c^5/(a^4*f^2)))*e^(-I*f*x - I*e)/(a^2*f)) - 3*sqrt(1/2)*a^2*f*sqrt(-c^5/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(-3/2*(-I*c^ 3 - sqrt(2)*sqrt(1/2)*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(c/(e^(2*I*f *x + 2*I*e) + 1))*sqrt(-c^5/(a^4*f^2)))*e^(-I*f*x - I*e)/(a^2*f)) - sqrt(2 )*(-3*I*c^2*e^(4*I*f*x + 4*I*e) - I*c^2*e^(2*I*f*x + 2*I*e) + 2*I*c^2)*sqr t(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-4*I*f*x - 4*I*e)/(a^2*f)
\[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx=- \frac {\int \frac {c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx + \int \left (- \frac {c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\right )\, dx + \int \left (- \frac {2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\right )\, dx}{a^{2}} \]
-(Integral(c**2*sqrt(-I*c*tan(e + f*x) + c)/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x) + Integral(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)** 2/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x) + Integral(-2*I*c**2*sqrt(- I*c*tan(e + f*x) + c)*tan(e + f*x)/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1 ), x))/a**2
Time = 0.30 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.05 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {i \, {\left (\frac {3 \, \sqrt {2} c^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{2}} - \frac {4 \, {\left (5 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c^{4} - 6 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} c^{5}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{2} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{2} c + 4 \, a^{2} c^{2}}\right )}}{16 \, c f} \]
-1/16*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^2 - 4*(5*(-I*c*ta n(f*x + e) + c)^(3/2)*c^4 - 6*sqrt(-I*c*tan(f*x + e) + c)*c^5)/((-I*c*tan( f*x + e) + c)^2*a^2 - 4*(-I*c*tan(f*x + e) + c)*a^2*c + 4*a^2*c^2))/(c*f)
\[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx=\int { \frac {{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
Time = 0.30 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.92 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {\frac {c^4\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,3{}\mathrm {i}}{2\,a^2\,f}-\frac {c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,5{}\mathrm {i}}{4\,a^2\,f}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-4\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )+4\,c^2}+\frac {\sqrt {2}\,{\left (-c\right )}^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,3{}\mathrm {i}}{8\,a^2\,f} \]