Integrand size = 41, antiderivative size = 60 \[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {2 a (i A+B)}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a B}{c f \sqrt {c-i c \tan (e+f x)}} \]
Time = 2.55 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.95 \[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {2 a \left (\frac {i A+B}{(c-i c \tan (e+f x))^{3/2}}-\frac {3 B}{c \sqrt {c-i c \tan (e+f x)}}\right )}{3 f} \]
(-2*a*((I*A + B)/(c - I*c*Tan[e + f*x])^(3/2) - (3*B)/(c*Sqrt[c - I*c*Tan[ e + f*x]])))/(3*f)
Time = 0.30 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {3042, 4071, 53, 2009}
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.
\(\displaystyle \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 4071 |
\(\displaystyle \frac {a c \int \frac {A+B \tan (e+f x)}{(c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {a c \int \left (\frac {A-i B}{(c-i c \tan (e+f x))^{5/2}}+\frac {i B}{c (c-i c \tan (e+f x))^{3/2}}\right )d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a c \left (\frac {2 B}{c^2 \sqrt {c-i c \tan (e+f x)}}-\frac {2 (B+i A)}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{f}\) |
(a*c*((-2*(I*A + B))/(3*c*(c - I*c*Tan[e + f*x])^(3/2)) + (2*B)/(c^2*Sqrt[ c - I*c*Tan[e + f*x]])))/f
3.8.45.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
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 = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(\frac {2 i a \left (-\frac {c \left (-i B +A \right )}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {i B}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f c}\) | \(53\) |
default | \(\frac {2 i a \left (-\frac {c \left (-i B +A \right )}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {i B}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f c}\) | \(53\) |
risch | \(-\frac {a \left (i A \,{\mathrm e}^{2 i \left (f x +e \right )}+B \,{\mathrm e}^{2 i \left (f x +e \right )}+i A -5 B \right ) \sqrt {2}}{6 c \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\) | \(62\) |
parts | \(\frac {2 i A a c \left (-\frac {1}{4 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{6 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 c^{\frac {5}{2}}}\right )}{f}+\frac {a \left (i A +B \right ) \left (-\frac {1}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {1}{2 c \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 c^{\frac {3}{2}}}\right )}{f}-\frac {2 a B \left (-\frac {3}{4 \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {c}{6 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 \sqrt {c}}\right )}{f c}\) | \(234\) |
Time = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.25 \[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {2} {\left ({\left (-i \, A - B\right )} a e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, {\left (i \, A - 2 \, B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, A + 5 \, B\right )} a\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{6 \, c^{2} f} \]
1/6*sqrt(2)*((-I*A - B)*a*e^(4*I*f*x + 4*I*e) - 2*(I*A - 2*B)*a*e^(2*I*f*x + 2*I*e) + (-I*A + 5*B)*a)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c^2*f)
\[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=i a \left (\int \left (- \frac {i A}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx + \int \frac {A \tan {\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \frac {B \tan ^{2}{\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \left (- \frac {i B \tan {\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\right )\, dx\right ) \]
I*a*(Integral(-I*A/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt (-I*c*tan(e + f*x) + c)), x) + Integral(A*tan(e + f*x)/(-I*c*sqrt(-I*c*tan (e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) + c)), x) + Integra l(B*tan(e + f*x)**2/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqr t(-I*c*tan(e + f*x) + c)), x) + Integral(-I*B*tan(e + f*x)/(-I*c*sqrt(-I*c *tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) + c)), x))
Time = 0.23 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.75 \[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {2 i \, {\left (3 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} B a + {\left (A - i \, B\right )} a c\right )}}{3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c f} \]
\[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
Time = 9.06 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.83 \[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {a\,\sqrt {-\frac {c\,\left (-2\,{\cos \left (e+f\,x\right )}^2+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{2\,{\cos \left (e+f\,x\right )}^2}}\,\left (B\,\left (2\,{\cos \left (2\,e+2\,f\,x\right )}^2-1\right )-4\,B\,\left (2\,{\cos \left (e+f\,x\right )}^2-1\right )-2\,A\,\sin \left (2\,e+2\,f\,x\right )-A\,\sin \left (4\,e+4\,f\,x\right )-5\,B+A\,1{}\mathrm {i}+A\,\left (2\,{\cos \left (e+f\,x\right )}^2-1\right )\,2{}\mathrm {i}-B\,\sin \left (2\,e+2\,f\,x\right )\,4{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+A\,\left (2\,{\cos \left (2\,e+2\,f\,x\right )}^2-1\right )\,1{}\mathrm {i}\right )}{6\,c^2\,f} \]
-(a*(-(c*(sin(2*e + 2*f*x)*1i - 2*cos(e + f*x)^2))/(2*cos(e + f*x)^2))^(1/ 2)*(A*1i - 5*B + A*(2*cos(e + f*x)^2 - 1)*2i - 4*B*(2*cos(e + f*x)^2 - 1) - 2*A*sin(2*e + 2*f*x) - A*sin(4*e + 4*f*x) - B*sin(2*e + 2*f*x)*4i + B*si n(4*e + 4*f*x)*1i + A*(2*cos(2*e + 2*f*x)^2 - 1)*1i + B*(2*cos(2*e + 2*f*x )^2 - 1)))/(6*c^2*f)