Integrand size = 33, antiderivative size = 124 \[ \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {3 i \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{4 \sqrt {2} a \sqrt {c} f}-\frac {3 i}{4 a f \sqrt {c-i c \tan (e+f x)}}+\frac {i}{2 a f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \]
3/8*I*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a/f*2^(1/2)/c^ (1/2)-3/4*I/a/f/(c-I*c*tan(f*x+e))^(1/2)+1/2*I/a/f/(c-I*c*tan(f*x+e))^(1/2 )/(1+I*tan(f*x+e))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.65 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.43 \[ \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {i \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2,\frac {1}{2},-\frac {1}{2} i (i+\tan (e+f x))\right )}{2 a f \sqrt {c-i c \tan (e+f x)}} \]
((-1/2*I)*Hypergeometric2F1[-1/2, 2, 1/2, (-1/2*I)*(I + Tan[e + f*x])])/(a *f*Sqrt[c - I*c*Tan[e + f*x]])
Time = 0.39 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.93, 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, 52, 61, 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.
\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}dx\) |
\(\Big \downarrow \) 4005 |
\(\displaystyle \frac {\int \cos ^2(e+f x) \sqrt {c-i c \tan (e+f x)}dx}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sqrt {c-i c \tan (e+f x)}}{\sec (e+f x)^2}dx}{a c}\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle \frac {i c^2 \int \frac {1}{(c-i c \tan (e+f x))^{3/2} (i \tan (e+f x) c+c)^2}d(-i c \tan (e+f x))}{a f}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {i c^2 \left (\frac {3 \int \frac {1}{(c-i c \tan (e+f x))^{3/2} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{4 c}+\frac {1}{2 c \sqrt {c-i c \tan (e+f x)} (c+i c \tan (e+f x))}\right )}{a f}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {i c^2 \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {c-i c \tan (e+f x)} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{2 c}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}\right )}{4 c}+\frac {1}{2 c \sqrt {c-i c \tan (e+f x)} (c+i c \tan (e+f x))}\right )}{a f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {i c^2 \left (\frac {3 \left (\frac {\int \frac {1}{c^2 \tan ^2(e+f x)+2 c}d\sqrt {c-i c \tan (e+f x)}}{c}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}\right )}{4 c}+\frac {1}{2 c \sqrt {c-i c \tan (e+f x)} (c+i c \tan (e+f x))}\right )}{a f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {i c^2 \left (\frac {3 \left (-\frac {i \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2}}\right )}{\sqrt {2} c^{3/2}}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}\right )}{4 c}+\frac {1}{2 c \sqrt {c-i c \tan (e+f x)} (c+i c \tan (e+f x))}\right )}{a f}\) |
(I*c^2*(1/(2*c*Sqrt[c - I*c*Tan[e + f*x]]*(c + I*c*Tan[e + f*x])) + (3*((( -I)*ArcTan[(Sqrt[c]*Tan[e + f*x])/Sqrt[2]])/(Sqrt[2]*c^(3/2)) - 1/(c*Sqrt[ c - I*c*Tan[e + f*x]])))/(4*c)))/(a*f)
3.10.77.3.1 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] :> 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] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
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 = 0.62 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(\frac {2 i c^{2} \left (\frac {\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{2 c +2 i c \tan \left (f x +e \right )}+\frac {3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}}{4 c^{2}}-\frac {1}{4 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f a}\) | \(102\) |
default | \(\frac {2 i c^{2} \left (\frac {\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{2 c +2 i c \tan \left (f x +e \right )}+\frac {3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}}{4 c^{2}}-\frac {1}{4 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f a}\) | \(102\) |
2*I/f/a*c^2*(1/4/c^2*(1/4*(c-I*c*tan(f*x+e))^(1/2)/(1/2*c+1/2*I*c*tan(f*x+ e))+3/4*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/ 2)))-1/4/c^2/(c-I*c*tan(f*x+e))^(1/2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 272 vs. \(2 (91) = 182\).
Time = 0.26 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.19 \[ \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {{\left (-3 i \, \sqrt {\frac {1}{2}} a c f \sqrt {\frac {1}{a^{2} c f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {3 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{2} c f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a f}\right ) + 3 i \, \sqrt {\frac {1}{2}} a c f \sqrt {\frac {1}{a^{2} c f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {3 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{2} c f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-2 i \, e^{\left (4 i \, f x + 4 i \, e\right )} - i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a c f} \]
1/8*(-3*I*sqrt(1/2)*a*c*f*sqrt(1/(a^2*c*f^2))*e^(2*I*f*x + 2*I*e)*log(-3/2 *(sqrt(2)*sqrt(1/2)*(I*a*f*e^(2*I*f*x + 2*I*e) + I*a*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^2*c*f^2)) - I)*e^(-I*f*x - I*e)/(a*f)) + 3*I*sqr t(1/2)*a*c*f*sqrt(1/(a^2*c*f^2))*e^(2*I*f*x + 2*I*e)*log(-3/2*(sqrt(2)*sqr t(1/2)*(-I*a*f*e^(2*I*f*x + 2*I*e) - I*a*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^2*c*f^2)) - I)*e^(-I*f*x - I*e)/(a*f)) + sqrt(2)*sqrt(c/(e^( 2*I*f*x + 2*I*e) + 1))*(-2*I*e^(4*I*f*x + 4*I*e) - I*e^(2*I*f*x + 2*I*e) + I))*e^(-2*I*f*x - 2*I*e)/(a*c*f)
\[ \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \, dx=- \frac {i \int \frac {1}{\sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} - i \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx}{a} \]
-I*Integral(1/(sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) - I*sqrt(-I*c*tan( e + f*x) + c)), x)/a
Time = 0.31 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {i \, {\left (\frac {3 \, \sqrt {2} \sqrt {c} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a} + \frac {4 \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} c - 4 \, c^{2}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a - 2 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} a c}\right )}}{16 \, c f} \]
-1/16*I*(3*sqrt(2)*sqrt(c)*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*(3*(-I*c*tan( f*x + e) + c)*c - 4*c^2)/((-I*c*tan(f*x + e) + c)^(3/2)*a - 2*sqrt(-I*c*ta n(f*x + e) + c)*a*c))/(c*f)
\[ \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \, dx=\int { \frac {1}{{\left (i \, a \tan \left (f x + e\right ) + a\right )} \sqrt {-i \, c \tan \left (f x + e\right ) + c}} \,d x } \]
Time = 6.05 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.91 \[ \int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {\frac {c\,1{}\mathrm {i}}{a\,f}-\frac {\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{4\,a\,f}}{2\,c\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,3{}\mathrm {i}}{8\,a\,\sqrt {-c}\,f} \]