Integrand size = 43, antiderivative size = 245 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {5 (7 i A+5 B) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{128 \sqrt {2} a^3 \sqrt {c} f}-\frac {5 (7 i A+5 B)}{128 a^3 f \sqrt {c-i c \tan (e+f x)}}+\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}+\frac {7 i A+5 B}{48 a^3 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 (7 i A+5 B)}{192 a^3 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \] Output:
5/256*(7*I*A+5*B)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^ (1/2)/a^3/c^(1/2)/f-5/128*(7*I*A+5*B)/a^3/f/(c-I*c*tan(f*x+e))^(1/2)+1/6*( I*A-B)/a^3/f/(1+I*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(1/2)+1/48*(7*I*A+5*B)/ a^3/f/(1+I*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(1/2)+5/192*(7*I*A+5*B)/a^3/f/ (1+I*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2)
Time = 3.01 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.86 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {\sec ^3(e+f x) \left (2 \sqrt {c} ((-125 A+7 i B) \cos (e+f x)+8 (5 A-7 i B) \cos (3 (e+f x))-i (7 A-5 i B) (7 \sin (e+f x)-8 \sin (3 (e+f x))))-15 \sqrt {2} (7 A-5 i B) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right ) (\cos (3 (e+f x))+i \sin (3 (e+f x))) \sqrt {c-i c \tan (e+f x)}\right )}{768 a^3 \sqrt {c} f (-i+\tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \] Input:
Integrate[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])^3*Sqrt[c - I*c*Tan[ e + f*x]]),x]
Output:
(Sec[e + f*x]^3*(2*Sqrt[c]*((-125*A + (7*I)*B)*Cos[e + f*x] + 8*(5*A - (7* I)*B)*Cos[3*(e + f*x)] - I*(7*A - (5*I)*B)*(7*Sin[e + f*x] - 8*Sin[3*(e + f*x)])) - 15*Sqrt[2]*(7*A - (5*I)*B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(S qrt[2]*Sqrt[c])]*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)])*Sqrt[c - I*c*Tan[ e + f*x]]))/(768*a^3*Sqrt[c]*f*(-I + Tan[e + f*x])^3*Sqrt[c - I*c*Tan[e + f*x]])
Time = 0.71 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.93, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.209, Rules used = {3042, 4071, 27, 87, 52, 52, 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.
| \(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {A+B \tan (e+f x)}{a^4 (i \tan (e+f x)+1)^4 (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {c \int \frac {A+B \tan (e+f x)}{(i \tan (e+f x)+1)^4 (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{a^3 f}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {c \left (\frac {1}{12} (7 A-5 i B) \int \frac {1}{(i \tan (e+f x)+1)^3 (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)+\frac {-B+i A}{6 c (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {c \left (\frac {1}{12} (7 A-5 i B) \left (\frac {5}{8} \int \frac {1}{(i \tan (e+f x)+1)^2 (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)+\frac {i}{4 c (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}\right )+\frac {-B+i A}{6 c (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {c \left (\frac {1}{12} (7 A-5 i B) \left (\frac {5}{8} \left (\frac {3}{4} \int \frac {1}{(i \tan (e+f x)+1) (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)+\frac {i}{2 c (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}\right )+\frac {i}{4 c (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}\right )+\frac {-B+i A}{6 c (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {c \left (\frac {1}{12} (7 A-5 i B) \left (\frac {5}{8} \left (\frac {3}{4} \left (\frac {\int \frac {1}{(i \tan (e+f x)+1) \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{2 c}-\frac {i}{c \sqrt {c-i c \tan (e+f x)}}\right )+\frac {i}{2 c (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}\right )+\frac {i}{4 c (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}\right )+\frac {-B+i A}{6 c (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {c \left (\frac {1}{12} (7 A-5 i B) \left (\frac {5}{8} \left (\frac {3}{4} \left (\frac {i \int \frac {1}{2-\frac {c-i c \tan (e+f x)}{c}}d\sqrt {c-i c \tan (e+f x)}}{c^2}-\frac {i}{c \sqrt {c-i c \tan (e+f x)}}\right )+\frac {i}{2 c (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}\right )+\frac {i}{4 c (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}\right )+\frac {-B+i A}{6 c (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {c \left (\frac {1}{12} (7 A-5 i B) \left (\frac {5}{8} \left (\frac {3}{4} \left (\frac {i \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} c^{3/2}}-\frac {i}{c \sqrt {c-i c \tan (e+f x)}}\right )+\frac {i}{2 c (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}\right )+\frac {i}{4 c (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}\right )+\frac {-B+i A}{6 c (1+i \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}}\right )}{a^3 f}\) |
Input:
Int[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])^3*Sqrt[c - I*c*Tan[e + f* x]]),x]
Output:
(c*((I*A - B)/(6*c*(1 + I*Tan[e + f*x])^3*Sqrt[c - I*c*Tan[e + f*x]]) + (( 7*A - (5*I)*B)*((I/4)/(c*(1 + I*Tan[e + f*x])^2*Sqrt[c - I*c*Tan[e + f*x]] ) + (5*((I/2)/(c*(1 + I*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]]) + (3*((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]])))/4))/8))/12))/(a^3*f)
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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
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 = 0.56 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.73
| method | result | size |
| derivativedivides | \(\frac {2 i c^{3} \left (-\frac {-i B +A}{16 c^{3} \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {\frac {8 \left (-\frac {9 i B}{128}+\frac {19 A}{128}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}+8 \left (\frac {7}{24} i B c -\frac {17}{24} c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+8 \left (-\frac {7}{32} i B \,c^{2}+\frac {29}{32} A \,c^{2}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {5 \left (-\frac {5 i B}{8}+\frac {7 A}{8}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}}{16 c^{3}}\right )}{f \,a^{3}}\) | \(178\) |
| default | \(\frac {2 i c^{3} \left (-\frac {-i B +A}{16 c^{3} \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {\frac {8 \left (-\frac {9 i B}{128}+\frac {19 A}{128}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}+8 \left (\frac {7}{24} i B c -\frac {17}{24} c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+8 \left (-\frac {7}{32} i B \,c^{2}+\frac {29}{32} A \,c^{2}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {5 \left (-\frac {5 i B}{8}+\frac {7 A}{8}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}}{16 c^{3}}\right )}{f \,a^{3}}\) | \(178\) |
Input:
int((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(1/2),x,metho d=_RETURNVERBOSE)
Output:
2*I/f/a^3*c^3*(-1/16/c^3*(A-I*B)/(c-I*c*tan(f*x+e))^(1/2)+1/16/c^3*(8*((-9 /128*I*B+19/128*A)*(c-I*c*tan(f*x+e))^(5/2)+(7/24*I*B*c-17/24*c*A)*(c-I*c* tan(f*x+e))^(3/2)+(-7/32*I*B*c^2+29/32*A*c^2)*(c-I*c*tan(f*x+e))^(1/2))/(c +I*c*tan(f*x+e))^3+5/4*(-5/8*I*B+7/8*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))))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 408 vs. \(2 (190) = 380\).
Time = 0.12 (sec) , antiderivative size = 408, normalized size of antiderivative = 1.67 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} c f \sqrt {-\frac {49 \, A^{2} - 70 i \, A B - 25 \, B^{2}}{a^{6} c f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {5 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {49 \, A^{2} - 70 i \, A B - 25 \, B^{2}}{a^{6} c f^{2}}} + 7 i \, A + 5 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a^{3} f}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} c f \sqrt {-\frac {49 \, A^{2} - 70 i \, A B - 25 \, B^{2}}{a^{6} c f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {5 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {49 \, A^{2} - 70 i \, A B - 25 \, B^{2}}{a^{6} c f^{2}}} - 7 i \, A - 5 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{64 \, a^{3} f}\right ) - \sqrt {2} {\left (48 \, {\left (i \, A + B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + 3 \, {\left (-13 i \, A + 9 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} - {\left (125 i \, A + 7 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (-23 i \, A + 11 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, A + 8 \, B\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{768 \, a^{3} c f} \] Input:
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(1/2),x , algorithm="fricas")
Output:
1/768*(15*sqrt(1/2)*a^3*c*f*sqrt(-(49*A^2 - 70*I*A*B - 25*B^2)/(a^6*c*f^2) )*e^(6*I*f*x + 6*I*e)*log(5/64*(sqrt(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*x + 2*I* e) + a^3*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(49*A^2 - 70*I*A*B - 2 5*B^2)/(a^6*c*f^2)) + 7*I*A + 5*B)*e^(-I*f*x - I*e)/(a^3*f)) - 15*sqrt(1/2 )*a^3*c*f*sqrt(-(49*A^2 - 70*I*A*B - 25*B^2)/(a^6*c*f^2))*e^(6*I*f*x + 6*I *e)*log(-5/64*(sqrt(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt( c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(49*A^2 - 70*I*A*B - 25*B^2)/(a^6*c*f^2 )) - 7*I*A - 5*B)*e^(-I*f*x - I*e)/(a^3*f)) - sqrt(2)*(48*(I*A + B)*e^(8*I *f*x + 8*I*e) + 3*(-13*I*A + 9*B)*e^(6*I*f*x + 6*I*e) - (125*I*A + 7*B)*e^ (4*I*f*x + 4*I*e) + 2*(-23*I*A + 11*B)*e^(2*I*f*x + 2*I*e) - 8*I*A + 8*B)* sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-6*I*f*x - 6*I*e)/(a^3*c*f)
\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {i \left (\int \frac {A}{\sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 i \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + i \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \frac {B \tan {\left (e + f x \right )}}{\sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 3 i \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - 3 \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + i \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx\right )}{a^{3}} \] Input:
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))**3/(c-I*c*tan(f*x+e))**(1/2) ,x)
Output:
I*(Integral(A/(sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - 3*I*sqrt(-I*c *tan(e + f*x) + c)*tan(e + f*x)**2 - 3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + I*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(B*tan(e + f*x)/(sqrt (-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - 3*I*sqrt(-I*c*tan(e + f*x) + c)* tan(e + f*x)**2 - 3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + I*sqrt(-I*c *tan(e + f*x) + c)), x))/a**3
Time = 0.12 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.98 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {i \, {\left (\frac {4 \, {\left (15 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} {\left (7 \, A - 5 i \, B\right )} c - 80 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} {\left (7 \, A - 5 i \, B\right )} c^{2} + 132 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (7 \, A - 5 i \, B\right )} c^{3} - 384 \, {\left (A - i \, B\right )} c^{4}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{3} - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{3} c + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} c^{2} - 8 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} a^{3} c^{3}} + \frac {15 \, \sqrt {2} {\left (7 \, A - 5 i \, B\right )} \sqrt {c} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3}}\right )}}{1536 \, c f} \] Input:
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(1/2),x , algorithm="maxima")
Output:
-1/1536*I*(4*(15*(-I*c*tan(f*x + e) + c)^3*(7*A - 5*I*B)*c - 80*(-I*c*tan( f*x + e) + c)^2*(7*A - 5*I*B)*c^2 + 132*(-I*c*tan(f*x + e) + c)*(7*A - 5*I *B)*c^3 - 384*(A - I*B)*c^4)/((-I*c*tan(f*x + e) + c)^(7/2)*a^3 - 6*(-I*c* tan(f*x + e) + c)^(5/2)*a^3*c + 12*(-I*c*tan(f*x + e) + c)^(3/2)*a^3*c^2 - 8*sqrt(-I*c*tan(f*x + e) + c)*a^3*c^3) + 15*sqrt(2)*(7*A - 5*I*B)*sqrt(c) *log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(sqrt(2)*sqrt(c) + s qrt(-I*c*tan(f*x + e) + c)))/a^3)/(c*f)
Exception generated. \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(1/2),x , algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad Argument TypeDone
Time = 6.70 (sec) , antiderivative size = 394, normalized size of antiderivative = 1.61 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {B\,c^3-\frac {25\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3}{128}+\frac {25\,B\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2}{24}-\frac {55\,B\,c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}{32}}{a^3\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}-6\,a^3\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}-8\,a^3\,c^3\,f\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}+12\,a^3\,c^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}+\frac {\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,35{}\mathrm {i}}{128\,a^3\,f}-\frac {A\,c^3\,1{}\mathrm {i}}{a^3\,f}-\frac {A\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,35{}\mathrm {i}}{24\,a^3\,f}+\frac {A\,c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,77{}\mathrm {i}}{32\,a^3\,f}}{6\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}+8\,c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-12\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,35{}\mathrm {i}}{256\,a^3\,\sqrt {-c}\,f}+\frac {25\,\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{256\,a^3\,\sqrt {c}\,f} \] Input:
int((A + B*tan(e + f*x))/((a + a*tan(e + f*x)*1i)^3*(c - c*tan(e + f*x)*1i )^(1/2)),x)
Output:
(B*c^3 - (25*B*(c - c*tan(e + f*x)*1i)^3)/128 + (25*B*c*(c - c*tan(e + f*x )*1i)^2)/24 - (55*B*c^2*(c - c*tan(e + f*x)*1i))/32)/(a^3*f*(c - c*tan(e + f*x)*1i)^(7/2) - 6*a^3*c*f*(c - c*tan(e + f*x)*1i)^(5/2) - 8*a^3*c^3*f*(c - c*tan(e + f*x)*1i)^(1/2) + 12*a^3*c^2*f*(c - c*tan(e + f*x)*1i)^(3/2)) + ((A*(c - c*tan(e + f*x)*1i)^3*35i)/(128*a^3*f) - (A*c^3*1i)/(a^3*f) - (A *c*(c - c*tan(e + f*x)*1i)^2*35i)/(24*a^3*f) + (A*c^2*(c - c*tan(e + f*x)* 1i)*77i)/(32*a^3*f))/(6*c*(c - c*tan(e + f*x)*1i)^(5/2) - (c - c*tan(e + f *x)*1i)^(7/2) + 8*c^3*(c - c*tan(e + f*x)*1i)^(1/2) - 12*c^2*(c - c*tan(e + f*x)*1i)^(3/2)) - (2^(1/2)*A*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2) )/(2*(-c)^(1/2)))*35i)/(256*a^3*(-c)^(1/2)*f) + (25*2^(1/2)*B*atanh((2^(1/ 2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*c^(1/2))))/(256*a^3*c^(1/2)*f)
\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 \sqrt {c-i c \tan (e+f x)}} \, dx=\text {too large to display} \] Input:
int((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(1/2),x)
Output:
(sqrt(c)*(12*sqrt( - tan(e + f*x)*i + 1)*a*i + 2*sqrt( - tan(e + f*x)*i + 1)*b + int(( - sqrt( - tan(e + f*x)*i + 1))/(tan(e + f*x)**5*i + 3*tan(e + f*x)**4 - 2*tan(e + f*x)**3*i + 2*tan(e + f*x)**2 - 3*tan(e + f*x)*i - 1) ,x)*tan(e + f*x)**2*b*f*i + int(( - sqrt( - tan(e + f*x)*i + 1))/(tan(e + f*x)**5*i + 3*tan(e + f*x)**4 - 2*tan(e + f*x)**3*i + 2*tan(e + f*x)**2 - 3*tan(e + f*x)*i - 1),x)*b*f*i - 3*int(( - sqrt( - tan(e + f*x)*i + 1)*tan (e + f*x)**4)/(tan(e + f*x)**5*i + 3*tan(e + f*x)**4 - 2*tan(e + f*x)**3*i + 2*tan(e + f*x)**2 - 3*tan(e + f*x)*i - 1),x)*tan(e + f*x)**2*b*f*i - 3* int(( - sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**4)/(tan(e + f*x)**5*i + 3*tan(e + f*x)**4 - 2*tan(e + f*x)**3*i + 2*tan(e + f*x)**2 - 3*tan(e + f* x)*i - 1),x)*b*f*i + 6*int(( - sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2 )/(tan(e + f*x)**5*i + 3*tan(e + f*x)**4 - 2*tan(e + f*x)**3*i + 2*tan(e + f*x)**2 - 3*tan(e + f*x)*i - 1),x)*tan(e + f*x)**2*b*f*i + 6*int(( - sqrt ( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(tan(e + f*x)**5*i + 3*tan(e + f* x)**4 - 2*tan(e + f*x)**3*i + 2*tan(e + f*x)**2 - 3*tan(e + f*x)*i - 1),x) *b*f*i - 18*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**4)/(tan(e + f*x )**5*i + 3*tan(e + f*x)**4 - 2*tan(e + f*x)**3*i + 2*tan(e + f*x)**2 - 3*t an(e + f*x)*i - 1),x)*tan(e + f*x)**2*a*f - 18*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**4)/(tan(e + f*x)**5*i + 3*tan(e + f*x)**4 - 2*tan(e + f*x)**3*i + 2*tan(e + f*x)**2 - 3*tan(e + f*x)*i - 1),x)*a*f + 48*int((...