Integrand size = 45, antiderivative size = 92 \[ \int \frac {A+B \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {i A+B}{f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}+\frac {i A \sqrt {c-i c \tan (e+f x)}}{c f \sqrt {a+i a \tan (e+f x)}} \]
(-I*A-B)/f/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(1/2)+I*A*(c-I*c*ta n(f*x+e))^(1/2)/c/f/(a+I*a*tan(f*x+e))^(1/2)
Time = 2.49 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.54 \[ \int \frac {A+B \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {-B+A \tan (e+f x)}{f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]
Time = 0.34 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3042, 4071, 87, 48}
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.
\(\displaystyle \int \frac {A+B \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}dx\) |
\(\Big \downarrow \) 4071 |
\(\displaystyle \frac {a c \int \frac {A+B \tan (e+f x)}{(i \tan (e+f x) a+a)^{3/2} (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {a c \left (\frac {A \int \frac {1}{(i \tan (e+f x) a+a)^{3/2} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{c}-\frac {B+i A}{a c \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}\right )}{f}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {a c \left (\frac {i A \sqrt {c-i c \tan (e+f x)}}{a c^2 \sqrt {a+i a \tan (e+f x)}}-\frac {B+i A}{a c \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}\right )}{f}\) |
(a*c*(-((I*A + B)/(a*c*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x ]])) + (I*A*Sqrt[c - I*c*Tan[e + f*x]])/(a*c^2*Sqrt[a + I*a*Tan[e + f*x]]) ))/f
3.9.33.3.1 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] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
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_.)*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]
Time = 0.47 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-\frac {i A \,{\mathrm e}^{2 i \left (f x +e \right )}+B \,{\mathrm e}^{2 i \left (f x +e \right )}-i A +B}{2 \sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\) | \(92\) |
derivativedivides | \(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (A \tan \left (f x +e \right )^{3}-B \tan \left (f x +e \right )^{2}+A \tan \left (f x +e \right )-B \right )}{f a c \left (i+\tan \left (f x +e \right )\right )^{2} \left (i-\tan \left (f x +e \right )\right )^{2}}\) | \(99\) |
default | \(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (A \tan \left (f x +e \right )^{3}-B \tan \left (f x +e \right )^{2}+A \tan \left (f x +e \right )-B \right )}{f a c \left (i+\tan \left (f x +e \right )\right )^{2} \left (i-\tan \left (f x +e \right )\right )^{2}}\) | \(99\) |
parts | \(\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (1+\tan \left (f x +e \right )^{2}\right ) \tan \left (f x +e \right )}{f a c \left (i-\tan \left (f x +e \right )\right )^{2} \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {B \left (-\tan \left (f x +e \right )^{2}-1\right ) \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}}{f c a \left (i-\tan \left (f x +e \right )\right )^{2} \left (i+\tan \left (f x +e \right )\right )^{2}}\) | \(162\) |
int((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(1/2),x,m ethod=_RETURNVERBOSE)
-1/2/(a*exp(2*I*(f*x+e))/(exp(2*I*(f*x+e))+1))^(1/2)/(exp(2*I*(f*x+e))+1)/ (c/(exp(2*I*(f*x+e))+1))^(1/2)*(I*A*exp(2*I*(f*x+e))+B*exp(2*I*(f*x+e))-I* A+B)/f
Time = 0.25 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.24 \[ \int \frac {A+B \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {{\left ({\left (-i \, A - B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, B e^{\left (3 i \, f x + 3 i \, e\right )} - 2 \, B e^{\left (2 i \, f x + 2 i \, e\right )} + 2 \, B e^{\left (i \, f x + i \, e\right )} + i \, A - B\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-i \, f x - i \, e\right )}}{2 \, a c f} \]
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(1/ 2),x, algorithm="fricas")
1/2*((-I*A - B)*e^(4*I*f*x + 4*I*e) + 2*B*e^(3*I*f*x + 3*I*e) - 2*B*e^(2*I *f*x + 2*I*e) + 2*B*e^(I*f*x + 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*c*f)
\[ \int \frac {A+B \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \, dx=\int \frac {A + B \tan {\left (e + f x \right )}}{\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )} \sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )}}\, dx \]
Integral((A + B*tan(e + f*x))/(sqrt(I*a*(tan(e + f*x) - I))*sqrt(-I*c*(tan (e + f*x) + I))), x)
Time = 0.39 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.35 \[ \int \frac {A+B \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2 \, {\left ({\left (A - i \, B\right )} \cos \left (4 \, f x + 4 \, e\right ) - 2 i \, B \cos \left (2 \, f x + 2 \, e\right ) - {\left (-i \, A - B\right )} \sin \left (4 \, f x + 4 \, e\right ) + 2 \, B \sin \left (2 \, f x + 2 \, e\right ) - A - i \, B\right )} \sqrt {a} \sqrt {c}}{-4 \, {\left (i \, a c \cos \left (3 \, f x + 3 \, e\right ) + i \, a c \cos \left (f x + e\right ) - a c \sin \left (3 \, f x + 3 \, e\right ) - a c \sin \left (f x + e\right )\right )} f} \]
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(1/ 2),x, algorithm="maxima")
-2*((A - I*B)*cos(4*f*x + 4*e) - 2*I*B*cos(2*f*x + 2*e) - (-I*A - B)*sin(4 *f*x + 4*e) + 2*B*sin(2*f*x + 2*e) - A - I*B)*sqrt(a)*sqrt(c)/((-4*I*a*c*c os(3*f*x + 3*e) - 4*I*a*c*cos(f*x + e) + 4*a*c*sin(3*f*x + 3*e) + 4*a*c*si n(f*x + e))*f)
\[ \int \frac {A+B \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \, dx=\int { \frac {B \tan \left (f x + e\right ) + A}{\sqrt {i \, a \tan \left (f x + e\right ) + a} \sqrt {-i \, c \tan \left (f x + e\right ) + c}} \,d x } \]
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(1/ 2),x, algorithm="giac")
Time = 0.80 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.55 \[ \int \frac {A+B \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,1{}\mathrm {i}+B-A\,\cos \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}+B\,\cos \left (2\,e+2\,f\,x\right )-A\,\sin \left (2\,e+2\,f\,x\right )-B\,\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{2\,a\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]