Integrand size = 41, antiderivative size = 62 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\frac {2 a (i A+B) (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac {2 a B (c-i c \tan (e+f x))^{5/2}}{5 c f} \]
Time = 1.55 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.87 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\frac {2 a c (i+\tan (e+f x)) (5 A-2 i B+3 B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{15 f} \]
(2*a*c*(I + Tan[e + f*x])*(5*A - (2*I)*B + 3*B*Tan[e + f*x])*Sqrt[c - I*c* Tan[e + f*x]])/(15*f)
Time = 0.30 (sec) , antiderivative size = 63, 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.
\(\displaystyle \int (a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} (A+B \tan (e+f x))dx\) |
\(\Big \downarrow \) 4071 |
\(\displaystyle \frac {a c \int (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {a c \int \left (\frac {i B (c-i c \tan (e+f x))^{3/2}}{c}+(A-i B) \sqrt {c-i c \tan (e+f x)}\right )d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a c \left (\frac {2 (B+i A) (c-i c \tan (e+f x))^{3/2}}{3 c}-\frac {2 B (c-i c \tan (e+f x))^{5/2}}{5 c^2}\right )}{f}\) |
(a*c*((2*(I*A + B)*(c - I*c*Tan[e + f*x])^(3/2))/(3*c) - (2*B*(c - I*c*Tan [e + f*x])^(5/2))/(5*c^2)))/f
3.8.42.3.1 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 = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {2 i a \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {\left (-i B c +c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}\right )}{f c}\) | \(55\) |
default | \(\frac {2 i a \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {\left (-i B c +c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}\right )}{f c}\) | \(55\) |
parts | \(\frac {2 i A a c \left (-\sqrt {c -i c \tan \left (f x +e \right )}+\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f}+\frac {a \left (i A +B \right ) \left (\frac {2 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 c \sqrt {c -i c \tan \left (f x +e \right )}-2 c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f}-\frac {2 a B \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\sqrt {c -i c \tan \left (f x +e \right )}\, c^{2}-c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f c}\) | \(210\) |
Time = 0.26 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.26 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=-\frac {4 \, \sqrt {2} {\left (5 \, {\left (-i \, A - B\right )} a c e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-5 i \, A + B\right )} a c\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{15 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]
-4/15*sqrt(2)*(5*(-I*A - B)*a*c*e^(2*I*f*x + 2*I*e) + (-5*I*A + B)*a*c)*sq rt(c/(e^(2*I*f*x + 2*I*e) + 1))/(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)
\[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=i a \left (\int \left (- i A c \sqrt {- i c \tan {\left (e + f x \right )} + c}\right )\, dx + \int \left (- i A c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- i B c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}\right )\, dx + \int \left (- i B c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )}\right )\, dx\right ) \]
I*a*(Integral(-I*A*c*sqrt(-I*c*tan(e + f*x) + c), x) + Integral(-I*A*c*sqr t(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2, x) + Integral(-I*B*c*sqrt(-I*c*t an(e + f*x) + c)*tan(e + f*x), x) + Integral(-I*B*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3, x))
Time = 0.35 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.77 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\frac {2 i \, {\left (3 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} B a + 5 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (A - i \, B\right )} a c\right )}}{15 \, c f} \]
2/15*I*(3*I*(-I*c*tan(f*x + e) + c)^(5/2)*B*a + 5*(-I*c*tan(f*x + e) + c)^ (3/2)*(A - I*B)*a*c)/(c*f)
\[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\int { {\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \,d x } \]
Time = 10.87 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.60 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\frac {4\,a\,c\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,\left (A\,5{}\mathrm {i}-B+A\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,5{}\mathrm {i}+5\,B\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\right )}{15\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2} \]