Integrand size = 33, antiderivative size = 95 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=\frac {i \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{2 \sqrt {2} a f}+\frac {i \sqrt {c-i c \tan (e+f x)}}{2 a f (1+i \tan (e+f x))} \]
1/4*I*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^(1/2)/a/f*2^ (1/2)+1/2*I*(c-I*c*tan(f*x+e))^(1/2)/a/f/(1+I*tan(f*x+e))
Time = 0.76 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=\frac {\sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right ) (1+i \tan (e+f x))+2 \sqrt {c-i c \tan (e+f x)}}{4 a f (-i+\tan (e+f x))} \]
(Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])]*(1 + I*Tan[e + f*x]) + 2*Sqrt[c - I*c*Tan[e + f*x]])/(4*a*f*(-I + Tan[e + f*x ]))
Time = 0.38 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3042, 4005, 3042, 3968, 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.
\(\displaystyle \int \frac {\sqrt {c-i c \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {c-i c \tan (e+f x)}}{a+i a \tan (e+f x)}dx\) |
\(\Big \downarrow \) 4005 |
\(\displaystyle \frac {\int \cos ^2(e+f x) (c-i c \tan (e+f x))^{3/2}dx}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(c-i c \tan (e+f x))^{3/2}}{\sec (e+f x)^2}dx}{a c}\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle \frac {i c^2 \int \frac {1}{\sqrt {c-i c \tan (e+f x)} (i \tan (e+f x) c+c)^2}d(-i c \tan (e+f x))}{a f}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {i c^2 \left (\frac {\int \frac {1}{\sqrt {c-i c \tan (e+f x)} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{4 c}+\frac {\sqrt {c-i c \tan (e+f x)}}{2 c (c+i c \tan (e+f x))}\right )}{a f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {i c^2 \left (\frac {\int \frac {1}{c^2 \tan ^2(e+f x)+2 c}d\sqrt {c-i c \tan (e+f x)}}{2 c}+\frac {\sqrt {c-i c \tan (e+f x)}}{2 c (c+i c \tan (e+f x))}\right )}{a f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {i c^2 \left (\frac {\sqrt {c-i c \tan (e+f x)}}{2 c (c+i c \tan (e+f x))}-\frac {i \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2}}\right )}{2 \sqrt {2} c^{3/2}}\right )}{a f}\) |
(I*c^2*(((-1/2*I)*ArcTan[(Sqrt[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]))))/(a*f)
3.10.59.3.1 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 + 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 = 0.77 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(\frac {2 i c^{2} \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{4 c \left (c +i c \tan \left (f x +e \right )\right )}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 c^{\frac {3}{2}}}\right )}{f a}\) | \(78\) |
default | \(\frac {2 i c^{2} \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{4 c \left (c +i c \tan \left (f x +e \right )\right )}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 c^{\frac {3}{2}}}\right )}{f a}\) | \(78\) |
2*I/f/a*c^2*(1/4*(c-I*c*tan(f*x+e))^(1/2)/c/(c+I*c*tan(f*x+e))+1/8/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. 250 vs. \(2 (70) = 140\).
Time = 0.25 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.63 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=\frac {{\left (\sqrt {\frac {1}{2}} a f \sqrt {-\frac {c}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c}{a^{2} f^{2}}} + i \, c\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) - \sqrt {\frac {1}{2}} a f \sqrt {-\frac {c}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c}{a^{2} f^{2}}} - i \, c\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a f} \]
1/4*(sqrt(1/2)*a*f*sqrt(-c/(a^2*f^2))*e^(2*I*f*x + 2*I*e)*log((sqrt(2)*sqr t(1/2)*(a*f*e^(2*I*f*x + 2*I*e) + a*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*s qrt(-c/(a^2*f^2)) + I*c)*e^(-I*f*x - I*e)/(a*f)) - sqrt(1/2)*a*f*sqrt(-c/( a^2*f^2))*e^(2*I*f*x + 2*I*e)*log(-(sqrt(2)*sqrt(1/2)*(a*f*e^(2*I*f*x + 2* I*e) + a*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-c/(a^2*f^2)) - I*c)*e^ (-I*f*x - I*e)/(a*f)) + sqrt(2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(I*e^(2* I*f*x + 2*I*e) + I))*e^(-2*I*f*x - 2*I*e)/(a*f)
\[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=- \frac {i \int \frac {\sqrt {- i c \tan {\left (e + f x \right )} + c}}{\tan {\left (e + f x \right )} - i}\, dx}{a} \]
Time = 0.30 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.12 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=-\frac {i \, {\left (\frac {\sqrt {2} c^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a} + \frac {4 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} c^{2}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )} a - 2 \, a c}\right )}}{8 \, c f} \]
-1/8*I*(sqrt(2)*c^(3/2)*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 + 4*sqrt(-I*c*tan(f* x + e) + c)*c^2/((-I*c*tan(f*x + e) + c)*a - 2*a*c))/(c*f)
\[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=\int { \frac {\sqrt {-i \, c \tan \left (f x + e\right ) + c}}{i \, a \tan \left (f x + e\right ) + a} \,d x } \]
Time = 0.51 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{a+i a \tan (e+f x)} \, dx=\frac {\sqrt {2}\,\sqrt {-c}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,1{}\mathrm {i}}{4\,a\,f}+\frac {c\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,a\,f\,\left (c+c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )} \]