Integrand size = 45, antiderivative size = 251 \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=-\frac {2 a^{7/2} B \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{7/2} f}-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac {2 a B (a+i a \tan (e+f x))^{5/2}}{5 c f (c-i c \tan (e+f x))^{5/2}}-\frac {2 a^2 B (a+i a \tan (e+f x))^{3/2}}{3 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a^3 B \sqrt {a+i a \tan (e+f x)}}{c^3 f \sqrt {c-i c \tan (e+f x)}} \]
-2*a^(7/2)*B*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I*c*tan(f* x+e))^(1/2))/c^(7/2)/f+2*a^3*B*(a+I*a*tan(f*x+e))^(1/2)/c^3/f/(c-I*c*tan(f *x+e))^(1/2)-1/7*(I*A+B)*(a+I*a*tan(f*x+e))^(7/2)/f/(c-I*c*tan(f*x+e))^(7/ 2)+2/5*a*B*(a+I*a*tan(f*x+e))^(5/2)/c/f/(c-I*c*tan(f*x+e))^(5/2)-2/3*a^2*B *(a+I*a*tan(f*x+e))^(3/2)/c^2/f/(c-I*c*tan(f*x+e))^(3/2)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(570\) vs. \(2(251)=502\).
Time = 19.88 (sec) , antiderivative size = 570, normalized size of antiderivative = 2.27 \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=-\frac {2 B e^{-i (4 e+f x)} \sqrt {e^{i f x}} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \arctan \left (e^{i (e+f x)}\right ) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{c^3 \sqrt {\frac {c}{1+e^{2 i (e+f x)}}} f \sec ^{\frac {9}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{7/2} (A \cos (e+f x)+B \sin (e+f x))}+\frac {\cos ^4(e+f x) \left (\frac {B \cos (3 e)}{c^4}+\cos (4 f x) \left (-\frac {2 B \cos (e)}{15 c^4}-\frac {2 i B \sin (e)}{15 c^4}\right )+\cos (2 f x) \left (\frac {2 B \cos (e)}{3 c^4}-\frac {2 i B \sin (e)}{3 c^4}\right )-\frac {i B \sin (3 e)}{c^4}+(-5 i A+9 B) \cos (6 f x) \left (\frac {\cos (3 e)}{70 c^4}+\frac {i \sin (3 e)}{70 c^4}\right )+(A-i B) \cos (8 f x) \left (-\frac {i \cos (5 e)}{14 c^4}+\frac {\sin (5 e)}{14 c^4}\right )+\left (\frac {2 i B \cos (e)}{3 c^4}+\frac {2 B \sin (e)}{3 c^4}\right ) \sin (2 f x)+\left (-\frac {2 i B \cos (e)}{15 c^4}+\frac {2 B \sin (e)}{15 c^4}\right ) \sin (4 f x)+(5 A+9 i B) \left (\frac {\cos (3 e)}{70 c^4}+\frac {i \sin (3 e)}{70 c^4}\right ) \sin (6 f x)+(A-i B) \left (\frac {\cos (5 e)}{14 c^4}+\frac {i \sin (5 e)}{14 c^4}\right ) \sin (8 f x)\right ) \sqrt {\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))} \]
Integrate[((a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan [e + f*x])^(7/2),x]
(-2*B*Sqrt[E^(I*f*x)]*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*ArcT an[E^(I*(e + f*x))]*(a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(c^ 3*E^(I*(4*e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*f*Sec[e + f*x]^(9/2) *(Cos[f*x] + I*Sin[f*x])^(7/2)*(A*Cos[e + f*x] + B*Sin[e + f*x])) + (Cos[e + f*x]^4*((B*Cos[3*e])/c^4 + Cos[4*f*x]*((-2*B*Cos[e])/(15*c^4) - (((2*I) /15)*B*Sin[e])/c^4) + Cos[2*f*x]*((2*B*Cos[e])/(3*c^4) - (((2*I)/3)*B*Sin[ e])/c^4) - (I*B*Sin[3*e])/c^4 + ((-5*I)*A + 9*B)*Cos[6*f*x]*(Cos[3*e]/(70* c^4) + ((I/70)*Sin[3*e])/c^4) + (A - I*B)*Cos[8*f*x]*(((-1/14*I)*Cos[5*e]) /c^4 + Sin[5*e]/(14*c^4)) + ((((2*I)/3)*B*Cos[e])/c^4 + (2*B*Sin[e])/(3*c^ 4))*Sin[2*f*x] + ((((-2*I)/15)*B*Cos[e])/c^4 + (2*B*Sin[e])/(15*c^4))*Sin[ 4*f*x] + (5*A + (9*I)*B)*(Cos[3*e]/(70*c^4) + ((I/70)*Sin[3*e])/c^4)*Sin[6 *f*x] + (A - I*B)*(Cos[5*e]/(14*c^4) + ((I/14)*Sin[5*e])/c^4)*Sin[8*f*x])* Sqrt[Sec[e + f*x]*(c*Cos[e + f*x] - I*c*Sin[e + f*x])]*(a + I*a*Tan[e + f* x])^(7/2)*(A + B*Tan[e + f*x]))/(f*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f* x] + B*Sin[e + f*x]))
Time = 0.44 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.178, Rules used = {3042, 4071, 87, 57, 57, 57, 45, 218}
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.
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}}dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {(i \tan (e+f x) a+a)^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{9/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {a c \left (\frac {i B \int \frac {(i \tan (e+f x) a+a)^{5/2}}{(c-i c \tan (e+f x))^{7/2}}d\tan (e+f x)}{c}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{7 a c (c-i c \tan (e+f x))^{7/2}}\right )}{f}\) |
| \(\Big \downarrow \) 57 |
| \(\displaystyle \frac {a c \left (\frac {i B \left (-\frac {a \int \frac {(i \tan (e+f x) a+a)^{3/2}}{(c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{c}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{7 a c (c-i c \tan (e+f x))^{7/2}}\right )}{f}\) |
| \(\Big \downarrow \) 57 |
| \(\displaystyle \frac {a c \left (\frac {i B \left (-\frac {a \left (-\frac {a \int \frac {\sqrt {i \tan (e+f x) a+a}}{(c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{c}-\frac {2 i (a+i a \tan (e+f x))^{3/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{c}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{7 a c (c-i c \tan (e+f x))^{7/2}}\right )}{f}\) |
| \(\Big \downarrow \) 57 |
| \(\displaystyle \frac {a c \left (\frac {i B \left (-\frac {a \left (-\frac {a \left (-\frac {a \int \frac {1}{\sqrt {i \tan (e+f x) a+a} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{c}-\frac {2 i \sqrt {a+i a \tan (e+f x)}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{3/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{c}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{7 a c (c-i c \tan (e+f x))^{7/2}}\right )}{f}\) |
| \(\Big \downarrow \) 45 |
| \(\displaystyle \frac {a c \left (\frac {i B \left (-\frac {a \left (-\frac {a \left (-\frac {2 a \int \frac {1}{i a+\frac {i c (i \tan (e+f x) a+a)}{c-i c \tan (e+f x)}}d\frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c-i c \tan (e+f x)}}}{c}-\frac {2 i \sqrt {a+i a \tan (e+f x)}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{3/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{c}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{7 a c (c-i c \tan (e+f x))^{7/2}}\right )}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {a c \left (\frac {i B \left (-\frac {a \left (-\frac {a \left (\frac {2 i \sqrt {a} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{3/2}}-\frac {2 i \sqrt {a+i a \tan (e+f x)}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{3/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{c}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{7 a c (c-i c \tan (e+f x))^{7/2}}\right )}{f}\) |
(a*c*(-1/7*((I*A + B)*(a + I*a*Tan[e + f*x])^(7/2))/(a*c*(c - I*c*Tan[e + f*x])^(7/2)) + (I*B*((((-2*I)/5)*(a + I*a*Tan[e + f*x])^(5/2))/(c*(c - I*c *Tan[e + f*x])^(5/2)) - (a*((((-2*I)/3)*(a + I*a*Tan[e + f*x])^(3/2))/(c*( c - I*c*Tan[e + f*x])^(3/2)) - (a*(((2*I)*Sqrt[a]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/c^(3/2) - ((2*I )*Sqrt[a + I*a*Tan[e + f*x]])/(c*Sqrt[c - I*c*Tan[e + f*x]])))/c))/c))/c)) /f
3.9.24.3.1 Defintions of rubi rules used
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && !GtQ[c, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[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[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 597 vs. \(2 (206 ) = 412\).
Time = 0.34 (sec) , antiderivative size = 598, normalized size of antiderivative = 2.38
| method | result | size |
| 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 )}\, a^{3} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (-\tan \left (f x +e \right )^{2}+2 i \tan \left (f x +e \right )+1\right )}{7 f \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{5}}-\frac {i B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (525 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) \tan \left (f x +e \right )^{4} a c +105 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) \tan \left (f x +e \right )^{5} a c -1050 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) \tan \left (f x +e \right )^{2} a c -1050 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) \tan \left (f x +e \right )^{3} a c -950 i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3}-337 \tan \left (f x +e \right )^{4} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}+105 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +525 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) \tan \left (f x +e \right ) a c +730 i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+1176 \tan \left (f x +e \right )^{2} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}-167 \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{105 f \,c^{4} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i+\tan \left (f x +e \right )\right )^{5} \sqrt {a c}}\) | \(598\) |
| 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 )}\, a^{3} \left (-105 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{5}+1050 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{3}+337 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{4}+525 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{4}+30 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3}-15 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{4}-525 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-1176 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}-1050 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{2}-950 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3}+30 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+167 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+105 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +730 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+15 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{105 f \,c^{4} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i+\tan \left (f x +e \right )\right )^{5} \sqrt {a c}}\) | \(638\) |
| 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 )}\, a^{3} \left (-105 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{5}+1050 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{3}+337 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{4}+525 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{4}+30 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3}-15 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{4}-525 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-1176 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}-1050 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{2}-950 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3}+30 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+167 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+105 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +730 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+15 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{105 f \,c^{4} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i+\tan \left (f x +e \right )\right )^{5} \sqrt {a c}}\) | \(638\) |
int((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x,m ethod=_RETURNVERBOSE)
1/7*A/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^3/c^4*(1+ tan(f*x+e)^2)*(-tan(f*x+e)^2+2*I*tan(f*x+e)+1)/(I+tan(f*x+e))^5-1/105*I*B/ f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^3/c^4*(525*I*ln ((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*ta n(f*x+e)^4*a*c+105*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^( 1/2))/(a*c)^(1/2))*tan(f*x+e)^5*a*c-1050*I*ln((a*c*tan(f*x+e)+(a*c)^(1/2)* (a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*tan(f*x+e)^2*a*c-1050*ln((a*c*t an(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*tan(f*x+e )^3*a*c-950*I*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^3-337*ta n(f*x+e)^4*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+105*I*ln((a*c*tan(f*x+ e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*a*c+525*ln((a*c* tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*tan(f*x+ e)*a*c+730*I*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)+1176*tan( f*x+e)^2*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)-167*(a*c)^(1/2)*(a*c*(1+ tan(f*x+e)^2))^(1/2))/(a*c*(1+tan(f*x+e)^2))^(1/2)/(I+tan(f*x+e))^5/(a*c)^ (1/2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 434 vs. \(2 (197) = 394\).
Time = 0.28 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.73 \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=\frac {105 \, c^{4} f \sqrt {-\frac {B^{2} a^{7}}{c^{7} f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (B a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + B a^{3} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + {\left (c^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{4} f\right )} \sqrt {-\frac {B^{2} a^{7}}{c^{7} f^{2}}}\right )}}{B a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + B a^{3}}\right ) - 105 \, c^{4} f \sqrt {-\frac {B^{2} a^{7}}{c^{7} f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (B a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + B a^{3} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (c^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{4} f\right )} \sqrt {-\frac {B^{2} a^{7}}{c^{7} f^{2}}}\right )}}{B a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + B a^{3}}\right ) - 2 \, {\left (15 \, {\left (i \, A + B\right )} a^{3} e^{\left (9 i \, f x + 9 i \, e\right )} + 3 \, {\left (5 i \, A - 9 \, B\right )} a^{3} e^{\left (7 i \, f x + 7 i \, e\right )} + 28 \, B a^{3} e^{\left (5 i \, f x + 5 i \, e\right )} - 140 \, B a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} - 210 \, B a^{3} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{210 \, c^{4} f} \]
integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/ 2),x, algorithm="fricas")
1/210*(105*c^4*f*sqrt(-B^2*a^7/(c^7*f^2))*log(4*(2*(B*a^3*e^(3*I*f*x + 3*I *e) + B*a^3*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^( 2*I*f*x + 2*I*e) + 1)) + (c^4*f*e^(2*I*f*x + 2*I*e) - c^4*f)*sqrt(-B^2*a^7 /(c^7*f^2)))/(B*a^3*e^(2*I*f*x + 2*I*e) + B*a^3)) - 105*c^4*f*sqrt(-B^2*a^ 7/(c^7*f^2))*log(4*(2*(B*a^3*e^(3*I*f*x + 3*I*e) + B*a^3*e^(I*f*x + I*e))* sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - (c^4 *f*e^(2*I*f*x + 2*I*e) - c^4*f)*sqrt(-B^2*a^7/(c^7*f^2)))/(B*a^3*e^(2*I*f* x + 2*I*e) + B*a^3)) - 2*(15*(I*A + B)*a^3*e^(9*I*f*x + 9*I*e) + 3*(5*I*A - 9*B)*a^3*e^(7*I*f*x + 7*I*e) + 28*B*a^3*e^(5*I*f*x + 5*I*e) - 140*B*a^3* e^(3*I*f*x + 3*I*e) - 210*B*a^3*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I* e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))/(c^4*f)
Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=\text {Timed out} \]
Time = 0.44 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.98 \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=-\frac {{\left (210 \, B a^{3} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + 210 \, B a^{3} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) - 30 \, {\left (-i \, A - B\right )} a^{3} \cos \left (7 \, f x + 7 \, e\right ) - 84 \, B a^{3} \cos \left (5 \, f x + 5 \, e\right ) + 140 \, B a^{3} \cos \left (3 \, f x + 3 \, e\right ) - 420 \, B a^{3} \cos \left (f x + e\right ) + 105 i \, B a^{3} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - 105 i \, B a^{3} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) - 30 \, {\left (A - i \, B\right )} a^{3} \sin \left (7 \, f x + 7 \, e\right ) - 84 i \, B a^{3} \sin \left (5 \, f x + 5 \, e\right ) + 140 i \, B a^{3} \sin \left (3 \, f x + 3 \, e\right ) - 420 i \, B a^{3} \sin \left (f x + e\right )\right )} \sqrt {a}}{210 \, c^{\frac {7}{2}} f} \]
integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/ 2),x, algorithm="maxima")
-1/210*(210*B*a^3*arctan2(cos(f*x + e), sin(f*x + e) + 1) + 210*B*a^3*arct an2(cos(f*x + e), -sin(f*x + e) + 1) - 30*(-I*A - B)*a^3*cos(7*f*x + 7*e) - 84*B*a^3*cos(5*f*x + 5*e) + 140*B*a^3*cos(3*f*x + 3*e) - 420*B*a^3*cos(f *x + e) + 105*I*B*a^3*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - 105*I*B*a^3*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e) + 1) - 30*(A - I*B)*a^3*sin(7*f*x + 7*e) - 84*I*B*a^3*sin(5*f*x + 5*e) + 14 0*I*B*a^3*sin(3*f*x + 3*e) - 420*I*B*a^3*sin(f*x + e))*sqrt(a)/(c^(7/2)*f)
\[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}}} \,d x } \]
integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/ 2),x, algorithm="giac")
integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^(7/2)/(-I*c*tan(f*x + e) + c)^(7/2), x)
Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}} \,d x \]