Integrand size = 45, antiderivative size = 110 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}} \, dx=-\frac {2 B \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {a} f}+\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{f \sqrt {a+i a \tan (e+f x)}} \]
-2*B*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^(1/2)/f/a^(1/2)+(I*A-B)*(c-I*c*tan(f*x+e))^(1/2)/f/(a+I*a*tan(f*x+e ))^(1/2)
Time = 3.02 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}} \, dx=-\frac {2 B \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {a} f}+\frac {(A+i B) c (i+\tan (e+f x))}{f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]
(-2*B*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(Sqrt[a]*f) + ((A + I*B)*c*(I + Tan[e + f*x]))/(f*S qrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])
Time = 0.36 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3042, 4071, 87, 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.
| \(\displaystyle \int \frac {\sqrt {c-i c \tan (e+f x)} (A+B \tan (e+f x))}{\sqrt {a+i a \tan (e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {c-i c \tan (e+f x)} (A+B \tan (e+f x))}{\sqrt {a+i a \tan (e+f x)}}dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {A+B \tan (e+f x)}{(i \tan (e+f x) a+a)^{3/2} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {a c \left (\frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{a c \sqrt {a+i a \tan (e+f x)}}-\frac {i B \int \frac {1}{\sqrt {i \tan (e+f x) a+a} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{a}\right )}{f}\) |
| \(\Big \downarrow \) 45 |
| \(\displaystyle \frac {a c \left (\frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{a c \sqrt {a+i a \tan (e+f x)}}-\frac {2 i B \int \frac {1}{i a+\frac {i c (i \tan (e+f x) a+a)}{c-i c \tan (e+f x)}}d\frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c-i c \tan (e+f x)}}}{a}\right )}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {a c \left (\frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{a c \sqrt {a+i a \tan (e+f x)}}-\frac {2 B \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{a^{3/2} \sqrt {c}}\right )}{f}\) |
(a*c*((-2*B*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(a^(3/2)*Sqrt[c]) + ((I*A - B)*Sqrt[c - I*c*Tan[e + f *x]])/(a*c*Sqrt[a + I*a*Tan[e + f*x]])))/f
3.9.32.3.1 Defintions of rubi rules used
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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si mp[a*(c/f) Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x], x , Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, 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. 322 vs. \(2 (89 ) = 178\).
Time = 0.41 (sec) , antiderivative size = 323, normalized size of antiderivative = 2.94
| method | result | size |
| derivativedivides | \(-\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (-2 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )+B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{2}+i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}-B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c -B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{f a \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i-\tan \left (f x +e \right )\right )^{2} \sqrt {a c}}\) | \(323\) |
| default | \(-\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (-2 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )+B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{2}+i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}-B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c -B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{f a \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i-\tan \left (f x +e \right )\right )^{2} \sqrt {a c}}\) | \(323\) |
| parts | \(-\frac {i A \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (1+i \tan \left (f x +e \right )\right )}{f a \left (i-\tan \left (f x +e \right )\right )^{2}}+\frac {i B \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (2 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) \tan \left (f x +e \right ) a c -\ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) \tan \left (f x +e \right )^{2} a c -i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+a c \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right )+\tan \left (f x +e \right ) \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{f a \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i-\tan \left (f x +e \right )\right )^{2} \sqrt {a c}}\) | \(332\) |
int((c-I*c*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(1/2),x,m ethod=_RETURNVERBOSE)
-I/f*(-c*(I*tan(f*x+e)-1))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a*(-2*I*B*ln(( a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*a*c* tan(f*x+e)+B*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/ (a*c)^(1/2))*a*c*tan(f*x+e)^2+I*A*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2) *tan(f*x+e)+I*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)-B*ln((a*c*tan(f*x +e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*a*c-B*(a*c)^(1/ 2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)+A*(a*c)^(1/2)*(a*c*(1+tan(f*x+e )^2))^(1/2))/(a*c*(1+tan(f*x+e)^2))^(1/2)/(I-tan(f*x+e))^2/(a*c)^(1/2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (84) = 168\).
Time = 0.27 (sec) , antiderivative size = 348, normalized size of antiderivative = 3.16 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}} \, dx=\frac {{\left (a f \sqrt {-\frac {B^{2} c}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left (\frac {4 \, {\left (2 \, {\left (B e^{\left (3 i \, f x + 3 i \, e\right )} + B e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} - a f\right )} \sqrt {-\frac {B^{2} c}{a f^{2}}}\right )}}{B e^{\left (2 i \, f x + 2 i \, e\right )} + B}\right ) - a f \sqrt {-\frac {B^{2} c}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left (\frac {4 \, {\left (2 \, {\left (B e^{\left (3 i \, f x + 3 i \, e\right )} + B e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} - a f\right )} \sqrt {-\frac {B^{2} c}{a f^{2}}}\right )}}{B e^{\left (2 i \, f x + 2 i \, e\right )} + B}\right ) - 2 \, {\left ({\left (-i \, A + B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a f} \]
integrate((c-I*c*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(1/ 2),x, algorithm="fricas")
1/2*(a*f*sqrt(-B^2*c/(a*f^2))*e^(I*f*x + I*e)*log(4*(2*(B*e^(3*I*f*x + 3*I *e) + B*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)) + (a*f*e^(2*I*f*x + 2*I*e) - a*f)*sqrt(-B^2*c/(a*f^2))) /(B*e^(2*I*f*x + 2*I*e) + B)) - a*f*sqrt(-B^2*c/(a*f^2))*e^(I*f*x + I*e)*l og(4*(2*(B*e^(3*I*f*x + 3*I*e) + B*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)) - (a*f*e^(2*I*f*x + 2*I*e) - a*f)*sqrt(-B^2*c/(a*f^2)))/(B*e^(2*I*f*x + 2*I*e) + B)) - 2*((-I*A + B)*e ^(2*I*f*x + 2*I*e) - I*A + B)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^ (2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(a*f)
\[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}} \, dx=\int \frac {\sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )} \left (A + B \tan {\left (e + f x \right )}\right )}{\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )}}\, dx \]
Integral(sqrt(-I*c*(tan(e + f*x) + I))*(A + B*tan(e + f*x))/sqrt(I*a*(tan( e + f*x) - I)), x)
Time = 0.39 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.25 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}} \, dx=-\frac {{\left (2 \, B \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + 2 \, B \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (i \, A - B\right )} \cos \left (f x + e\right ) + i \, B \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - i \, B \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (A + i \, B\right )} \sin \left (f x + e\right )\right )} \sqrt {c}}{2 \, \sqrt {a} f} \]
integrate((c-I*c*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(1/ 2),x, algorithm="maxima")
-1/2*(2*B*arctan2(cos(f*x + e), sin(f*x + e) + 1) + 2*B*arctan2(cos(f*x + e), -sin(f*x + e) + 1) - 2*(I*A - B)*cos(f*x + e) + I*B*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - I*B*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e) + 1) - 2*(A + I*B)*sin(f*x + e))*sqrt(c)/(sqrt(a) *f)
\[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {i \, a \tan \left (f x + e\right ) + a}} \,d x } \]
integrate((c-I*c*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(1/ 2),x, algorithm="giac")
Time = 11.56 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.27 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}} \, dx=\frac {A\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{f\,\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}-\frac {4\,B\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\left (\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}{\sqrt {a}\,\left (\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {c}\right )}\right )}{\sqrt {a}\,f}-\frac {4\,B\,\left (\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}{f\,\left (\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {c}\right )\,\left (-\frac {a}{c}+\frac {{\left (\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}^2}{{\left (\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {c}\right )}^2}+\frac {2\,\sqrt {a}\,\left (\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}{\sqrt {c}\,\left (\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {c}\right )}\right )} \]
(A*(c - c*tan(e + f*x)*1i)^(1/2)*1i)/(f*(a + a*tan(e + f*x)*1i)^(1/2)) - ( 4*B*c^(1/2)*atan((c^(1/2)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/(a^(1 /2)*((c - c*tan(e + f*x)*1i)^(1/2) - c^(1/2)))))/(a^(1/2)*f) - (4*B*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/(f*((c - c*tan(e + f*x)*1i)^(1/2) - c ^(1/2))*(((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))^2/((c - c*tan(e + f*x)* 1i)^(1/2) - c^(1/2))^2 - a/c + (2*a^(1/2)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/(c^(1/2)*((c - c*tan(e + f*x)*1i)^(1/2) - c^(1/2)))))