Integrand size = 43, antiderivative size = 184 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {(5 i A-B) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{8 \sqrt {2} a c^{3/2} f}-\frac {5 i A-B}{12 a f (c-i c \tan (e+f x))^{3/2}}+\frac {i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {5 i A-B}{8 a c f \sqrt {c-i c \tan (e+f x)}} \]
1/16*(5*I*A-B)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a/c^( 3/2)/f*2^(1/2)+1/8*(-5*I*A+B)/a/c/f/(c-I*c*tan(f*x+e))^(1/2)+1/12*(-5*I*A+ B)/a/f/(c-I*c*tan(f*x+e))^(3/2)+1/2*(I*A-B)/a/f/(1+I*tan(f*x+e))/(c-I*c*ta n(f*x+e))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 4.71 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.56 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {3 (-5 i A+B) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\frac {1}{2} i (i+\tan (e+f x))\right )+2 i \cos (e+f x) ((A+5 i B) \cos (e+f x)+(-5 i A+B) \sin (e+f x))}{24 a c f \sqrt {c-i c \tan (e+f x)}} \]
(3*((-5*I)*A + B)*Hypergeometric2F1[-1/2, 1, 1/2, (-1/2*I)*(I + Tan[e + f* x])] + (2*I)*Cos[e + f*x]*((A + (5*I)*B)*Cos[e + f*x] + ((-5*I)*A + B)*Sin [e + f*x]))/(24*a*c*f*Sqrt[c - I*c*Tan[e + f*x]])
Time = 0.41 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.93, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {3042, 4071, 27, 87, 61, 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.
| \(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {A+B \tan (e+f x)}{a^2 (i \tan (e+f x)+1)^2 (c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {c \int \frac {A+B \tan (e+f x)}{(i \tan (e+f x)+1)^2 (c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{a f}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {c \left (\frac {1}{4} (5 A+i B) \int \frac {1}{(i \tan (e+f x)+1) (c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)+\frac {-B+i A}{2 c (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}\right )}{a f}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {c \left (\frac {1}{4} (5 A+i B) \left (\frac {\int \frac {1}{(i \tan (e+f x)+1) (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{2 c}-\frac {i}{3 c (c-i c \tan (e+f x))^{3/2}}\right )+\frac {-B+i A}{2 c (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}\right )}{a f}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {c \left (\frac {1}{4} (5 A+i B) \left (\frac {\frac {\int \frac {1}{(i \tan (e+f x)+1) \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{2 c}-\frac {i}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {i}{3 c (c-i c \tan (e+f x))^{3/2}}\right )+\frac {-B+i A}{2 c (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}\right )}{a f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {c \left (\frac {1}{4} (5 A+i B) \left (\frac {\frac {i \int \frac {1}{2-\frac {c-i c \tan (e+f x)}{c}}d\sqrt {c-i c \tan (e+f x)}}{c^2}-\frac {i}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {i}{3 c (c-i c \tan (e+f x))^{3/2}}\right )+\frac {-B+i A}{2 c (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}\right )}{a f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {c \left (\frac {1}{4} (5 A+i B) \left (\frac {\frac {i \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} c^{3/2}}-\frac {i}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {i}{3 c (c-i c \tan (e+f x))^{3/2}}\right )+\frac {-B+i A}{2 c (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}\right )}{a f}\) |
(c*((I*A - B)/(2*c*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2)) + (( 5*A + I*B)*((-1/3*I)/(c*(c - I*c*Tan[e + f*x])^(3/2)) + ((I*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*c^(3/2)) - I/(c*Sqrt[c - I*c*Tan[e + f*x]]))/(2*c)))/4))/(a*f)
3.8.69.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_))*((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[(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[((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 = 141, normalized size of antiderivative = 0.77
| method | result | size |
| derivativedivides | \(\frac {2 i c \left (\frac {\frac {\left (\frac {i B}{8}+\frac {A}{8}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\frac {c}{2}+\frac {i c \tan \left (f x +e \right )}{2}}+\frac {\left (\frac {i B}{2}+\frac {5 A}{2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}}{4 c^{2}}-\frac {A}{4 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {-i B +A}{12 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f a}\) | \(141\) |
| default | \(\frac {2 i c \left (\frac {\frac {\left (\frac {i B}{8}+\frac {A}{8}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\frac {c}{2}+\frac {i c \tan \left (f x +e \right )}{2}}+\frac {\left (\frac {i B}{2}+\frac {5 A}{2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}}{4 c^{2}}-\frac {A}{4 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {-i B +A}{12 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f a}\) | \(141\) |
2*I/f/a*c*(1/4/c^2*((1/8*I*B+1/8*A)*(c-I*c*tan(f*x+e))^(1/2)/(1/2*c+1/2*I* c*tan(f*x+e))+1/4*(1/2*I*B+5/2*A)*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*A/(c-I*c*tan(f*x+e))^(1/2)-1/12/c*( A-I*B)/(c-I*c*tan(f*x+e))^(3/2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 388 vs. \(2 (139) = 278\).
Time = 0.26 (sec) , antiderivative size = 388, normalized size of antiderivative = 2.11 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {{\left (3 \, \sqrt {\frac {1}{2}} a c^{2} f \sqrt {-\frac {25 \, A^{2} + 10 i \, A B - B^{2}}{a^{2} c^{3} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a c f e^{\left (2 i \, f x + 2 i \, e\right )} + a c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {25 \, A^{2} + 10 i \, A B - B^{2}}{a^{2} c^{3} f^{2}}} + 5 i \, A - B\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a c f}\right ) - 3 \, \sqrt {\frac {1}{2}} a c^{2} f \sqrt {-\frac {25 \, A^{2} + 10 i \, A B - B^{2}}{a^{2} c^{3} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a c f e^{\left (2 i \, f x + 2 i \, e\right )} + a c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {25 \, A^{2} + 10 i \, A B - B^{2}}{a^{2} c^{3} f^{2}}} - 5 i \, A + B\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a c f}\right ) - \sqrt {2} {\left (2 \, {\left (i \, A + B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 4 \, {\left (4 i \, A + B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - {\left (-11 i \, A - 5 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, A + 3 \, B\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{48 \, a c^{2} f} \]
1/48*(3*sqrt(1/2)*a*c^2*f*sqrt(-(25*A^2 + 10*I*A*B - B^2)/(a^2*c^3*f^2))*e ^(2*I*f*x + 2*I*e)*log(1/4*(sqrt(2)*sqrt(1/2)*(a*c*f*e^(2*I*f*x + 2*I*e) + a*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(25*A^2 + 10*I*A*B - B^2)/ (a^2*c^3*f^2)) + 5*I*A - B)*e^(-I*f*x - I*e)/(a*c*f)) - 3*sqrt(1/2)*a*c^2* f*sqrt(-(25*A^2 + 10*I*A*B - B^2)/(a^2*c^3*f^2))*e^(2*I*f*x + 2*I*e)*log(- 1/4*(sqrt(2)*sqrt(1/2)*(a*c*f*e^(2*I*f*x + 2*I*e) + a*c*f)*sqrt(c/(e^(2*I* f*x + 2*I*e) + 1))*sqrt(-(25*A^2 + 10*I*A*B - B^2)/(a^2*c^3*f^2)) - 5*I*A + B)*e^(-I*f*x - I*e)/(a*c*f)) - sqrt(2)*(2*(I*A + B)*e^(6*I*f*x + 6*I*e) + 4*(4*I*A + B)*e^(4*I*f*x + 4*I*e) - (-11*I*A - 5*B)*e^(2*I*f*x + 2*I*e) - 3*I*A + 3*B)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a* c^2*f)
\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=- \frac {i \left (\int \frac {A}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - i c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \frac {B \tan {\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - i c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx\right )}{a} \]
-I*(Integral(A/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - I*c*sqr t(-I*c*tan(e + f*x) + c)), x) + Integral(B*tan(e + f*x)/(-I*c*sqrt(-I*c*ta n(e + f*x) + c)*tan(e + f*x)**2 - I*c*sqrt(-I*c*tan(e + f*x) + c)), x))/a
Time = 0.29 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.91 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {i \, {\left (\frac {3 \, \sqrt {2} {\left (5 \, A + i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a \sqrt {c}} + \frac {4 \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} {\left (5 \, A + i \, B\right )} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (5 \, A + i \, B\right )} c - 8 \, {\left (A - i \, B\right )} c^{2}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a - 2 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a c}\right )}}{96 \, c f} \]
-1/96*I*(3*sqrt(2)*(5*A + I*B)*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*sqrt(c)) + 4 *(3*(-I*c*tan(f*x + e) + c)^2*(5*A + I*B) - 4*(-I*c*tan(f*x + e) + c)*(5*A + I*B)*c - 8*(A - I*B)*c^2)/((-I*c*tan(f*x + e) + c)^(5/2)*a - 2*(-I*c*ta n(f*x + e) + c)^(3/2)*a*c))/(c*f)
\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=\int { \frac {B \tan \left (f x + e\right ) + A}{{\left (i \, a \tan \left (f x + e\right ) + a\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
Time = 9.70 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.42 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {\frac {B\,c}{3}-\frac {B\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}{6}+\frac {B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2}{8\,c}}{a\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}-2\,a\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\frac {A\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,5{}\mathrm {i}}{6\,a\,f}+\frac {A\,c\,1{}\mathrm {i}}{3\,a\,f}-\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,5{}\mathrm {i}}{8\,a\,c\,f}}{2\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,5{}\mathrm {i}}{16\,a\,{\left (-c\right )}^{3/2}\,f}-\frac {\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{16\,a\,c^{3/2}\,f} \]
((B*c)/3 - (B*(c - c*tan(e + f*x)*1i))/6 + (B*(c - c*tan(e + f*x)*1i)^2)/( 8*c))/(a*f*(c - c*tan(e + f*x)*1i)^(5/2) - 2*a*c*f*(c - c*tan(e + f*x)*1i) ^(3/2)) - ((A*(c - c*tan(e + f*x)*1i)*5i)/(6*a*f) + (A*c*1i)/(3*a*f) - (A* (c - c*tan(e + f*x)*1i)^2*5i)/(8*a*c*f))/(2*c*(c - c*tan(e + f*x)*1i)^(3/2 ) - (c - c*tan(e + f*x)*1i)^(5/2)) + (2^(1/2)*A*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*5i)/(16*a*(-c)^(3/2)*f) - (2^(1/2)*B*at anh((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*c^(1/2))))/(16*a*c^(3/2)*f)