Integrand size = 45, antiderivative size = 283 \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {2 a^{7/2} (i A+6 B) \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{5/2} f}-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {2 a (i A+6 B) (a+i a \tan (e+f x))^{5/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 a^2 (i A+6 B) (a+i a \tan (e+f x))^{3/2}}{3 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {a^3 (i A+6 B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c^3 f} \]
2*a^(7/2)*(I*A+6*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^(5/2)/f-a^3*(I*A+6*B)*(a+I*a*tan(f*x+e))^(1/2)*(c-I* c*tan(f*x+e))^(1/2)/c^3/f-2/3*a^2*(I*A+6*B)*(a+I*a*tan(f*x+e))^(3/2)/c^2/f /(c-I*c*tan(f*x+e))^(1/2)-1/5*(I*A+B)*(a+I*a*tan(f*x+e))^(7/2)/f/(c-I*c*ta n(f*x+e))^(5/2)+2/15*a*(I*A+6*B)*(a+I*a*tan(f*x+e))^(5/2)/c/f/(c-I*c*tan(f *x+e))^(3/2)
Time = 19.93 (sec) , antiderivative size = 528, normalized size of antiderivative = 1.87 \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {2 (i A+6 B) e^{-i (4 e+f x)} \sqrt {e^{i f x}} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \arctan \left (e^{i (e+f x)}\right ) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{c^2 \sqrt {\frac {c}{1+e^{2 i (e+f x)}}} f \sec ^{\frac {9}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{7/2} (A \cos (e+f x)+B \sin (e+f x))}+\frac {\cos ^4(e+f x) \left ((A-6 i B) \cos (2 f x) \left (-\frac {2 i \cos (e)}{3 c^3}-\frac {2 \sin (e)}{3 c^3}\right )+(i A+6 B) \cos (4 f x) \left (\frac {2 \cos (e)}{15 c^3}+\frac {2 i \sin (e)}{15 c^3}\right )+(A-6 i B) \left (-\frac {i \cos (3 e)}{c^3}-\frac {\sin (3 e)}{c^3}\right )+(A-i B) \cos (6 f x) \left (-\frac {i \cos (3 e)}{5 c^3}+\frac {\sin (3 e)}{5 c^3}\right )+(A-6 i B) \left (\frac {2 \cos (e)}{3 c^3}-\frac {2 i \sin (e)}{3 c^3}\right ) \sin (2 f x)+(A-6 i B) \left (-\frac {2 \cos (e)}{15 c^3}-\frac {2 i \sin (e)}{15 c^3}\right ) \sin (4 f x)+(A-i B) \left (\frac {\cos (3 e)}{5 c^3}+\frac {i \sin (3 e)}{5 c^3}\right ) \sin (6 f x)\right ) \sqrt {\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))} \]
Integrate[((a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan [e + f*x])^(5/2),x]
(2*(I*A + 6*B)*Sqrt[E^(I*f*x)]*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x )))]*ArcTan[E^(I*(e + f*x))]*(a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f *x]))/(c^2*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*((A - (6*I)*B)*Cos[2*f*x]*((((-2*I)/3)*Cos[e])/c^3 - (2 *Sin[e])/(3*c^3)) + (I*A + 6*B)*Cos[4*f*x]*((2*Cos[e])/(15*c^3) + (((2*I)/ 15)*Sin[e])/c^3) + (A - (6*I)*B)*(((-I)*Cos[3*e])/c^3 - Sin[3*e]/c^3) + (A - I*B)*Cos[6*f*x]*(((-1/5*I)*Cos[3*e])/c^3 + Sin[3*e]/(5*c^3)) + (A - (6* I)*B)*((2*Cos[e])/(3*c^3) - (((2*I)/3)*Sin[e])/c^3)*Sin[2*f*x] + (A - (6*I )*B)*((-2*Cos[e])/(15*c^3) - (((2*I)/15)*Sin[e])/c^3)*Sin[4*f*x] + (A - I* B)*(Cos[3*e]/(5*c^3) + ((I/5)*Sin[3*e])/c^3)*Sin[6*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 = 274, normalized size of antiderivative = 0.97, 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, 60, 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.
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {(i \tan (e+f x) a+a)^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {a c \left (-\frac {(A-6 i B) \int \frac {(i \tan (e+f x) a+a)^{5/2}}{(c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{5 c}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
| \(\Big \downarrow \) 57 |
| \(\displaystyle \frac {a c \left (-\frac {(A-6 i B) \left (-\frac {5 a \int \frac {(i \tan (e+f x) a+a)^{3/2}}{(c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{3 c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{5 c}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
| \(\Big \downarrow \) 57 |
| \(\displaystyle \frac {a c \left (-\frac {(A-6 i B) \left (-\frac {5 a \left (-\frac {3 a \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{c}-\frac {2 i (a+i a \tan (e+f x))^{3/2}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{3 c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{5 c}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {a c \left (-\frac {(A-6 i B) \left (-\frac {5 a \left (-\frac {3 a \left (a \int \frac {1}{\sqrt {i \tan (e+f x) a+a} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)+\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{3/2}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{3 c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{5 c}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
| \(\Big \downarrow \) 45 |
| \(\displaystyle \frac {a c \left (-\frac {(A-6 i B) \left (-\frac {5 a \left (-\frac {3 a \left (2 a \int \frac {1}{i a+\frac {i c (i \tan (e+f x) a+a)}{c-i c \tan (e+f x)}}d\frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c-i c \tan (e+f x)}}+\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{3/2}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{3 c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{5 c}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {a c \left (-\frac {(A-6 i B) \left (-\frac {5 a \left (-\frac {3 a \left (\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c}-\frac {2 i \sqrt {a} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c}}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{3/2}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{3 c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{5 c}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
(a*c*(-1/5*((I*A + B)*(a + I*a*Tan[e + f*x])^(7/2))/(a*c*(c - I*c*Tan[e + f*x])^(5/2)) - ((A - (6*I)*B)*((((-2*I)/3)*(a + I*a*Tan[e + f*x])^(5/2))/( c*(c - I*c*Tan[e + f*x])^(3/2)) - (5*a*(((-2*I)*(a + I*a*Tan[e + f*x])^(3/ 2))/(c*Sqrt[c - I*c*Tan[e + f*x]]) - (3*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]])])/Sqrt[c ] + (I*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/c))/c))/(3*c )))/(5*c)))/f
3.9.23.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[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. 832 vs. \(2 (234 ) = 468\).
Time = 0.37 (sec) , antiderivative size = 833, normalized size of antiderivative = 2.94
| method | result | size |
| 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 (246 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3}-474 i 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 \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}+60 i A \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}-94 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}+360 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}+15 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{4}+540 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 )^{2}-60 i A \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 )-90 A \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}-46 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3}+26 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}-90 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 )^{4}-360 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 )-564 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}-90 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 +15 A \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 +74 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+141 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{15 f \,c^{3} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\, \left (i+\tan \left (f x +e \right )\right )^{4}}\) | \(833\) |
| 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 (246 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3}-474 i 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 \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}+60 i A \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}-94 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}+360 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}+15 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{4}+540 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 )^{2}-60 i A \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 )-90 A \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}-46 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3}+26 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}-90 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 )^{4}-360 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 )-564 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}-90 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 +15 A \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 +74 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+141 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{15 f \,c^{3} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\, \left (i+\tan \left (f x +e \right )\right )^{4}}\) | \(833\) |
| parts | \(\text {Expression too large to display}\) | \(891\) |
int((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x,m ethod=_RETURNVERBOSE)
-1/15/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^3/c^3*(24 6*I*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^3-474*I*B*(a*c)^ (1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)+15*A*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*tan(f*x+e)^4+60*I*A *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*tan(f*x+e)^3-94*I*A*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+ e)^2+360*B*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*tan(f*x+e)^3+15*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)* tan(f*x+e)^4+540*I*B*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*tan(f*x+e)^2-60*I*A*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*tan(f*x+e)-90*A*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))*a*c*tan(f *x+e)^2-46*A*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^3+26*I*A* (a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)-90*I*B*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*tan(f*x+e)^4-360*B*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 *tan(f*x+e)-564*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^2-90 *I*B*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+15*A*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+74*A*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 515 vs. \(2 (221) = 442\).
Time = 0.27 (sec) , antiderivative size = 515, normalized size of antiderivative = 1.82 \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {15 \, c^{3} \sqrt {\frac {{\left (A^{2} - 12 i \, A B - 36 \, B^{2}\right )} a^{7}}{c^{5} f^{2}}} f \log \left (-\frac {4 \, {\left (2 \, {\left ({\left (-i \, A - 6 \, B\right )} a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-i \, A - 6 \, B\right )} 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^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{3} f\right )} \sqrt {\frac {{\left (A^{2} - 12 i \, A B - 36 \, B^{2}\right )} a^{7}}{c^{5} f^{2}}}\right )}}{{\left (i \, A + 6 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, A + 6 \, B\right )} a^{3}}\right ) - 15 \, c^{3} \sqrt {\frac {{\left (A^{2} - 12 i \, A B - 36 \, B^{2}\right )} a^{7}}{c^{5} f^{2}}} f \log \left (-\frac {4 \, {\left (2 \, {\left ({\left (-i \, A - 6 \, B\right )} a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-i \, A - 6 \, B\right )} 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^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{3} f\right )} \sqrt {\frac {{\left (A^{2} - 12 i \, A B - 36 \, B^{2}\right )} a^{7}}{c^{5} f^{2}}}\right )}}{{\left (i \, A + 6 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, A + 6 \, B\right )} a^{3}}\right ) - 4 \, {\left (3 \, {\left (i \, A + B\right )} a^{3} e^{\left (7 i \, f x + 7 i \, e\right )} + 2 \, {\left (-i \, A - 6 \, B\right )} a^{3} e^{\left (5 i \, f x + 5 i \, e\right )} + 10 \, {\left (i \, A + 6 \, B\right )} a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + 15 \, {\left (i \, A + 6 \, B\right )} 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}}}{30 \, c^{3} f} \]
integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/ 2),x, algorithm="fricas")
1/30*(15*c^3*sqrt((A^2 - 12*I*A*B - 36*B^2)*a^7/(c^5*f^2))*f*log(-4*(2*((- I*A - 6*B)*a^3*e^(3*I*f*x + 3*I*e) + (-I*A - 6*B)*a^3*e^(I*f*x + I*e))*sqr t(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) + (c^3*f* e^(2*I*f*x + 2*I*e) - c^3*f)*sqrt((A^2 - 12*I*A*B - 36*B^2)*a^7/(c^5*f^2)) )/((I*A + 6*B)*a^3*e^(2*I*f*x + 2*I*e) + (I*A + 6*B)*a^3)) - 15*c^3*sqrt(( A^2 - 12*I*A*B - 36*B^2)*a^7/(c^5*f^2))*f*log(-4*(2*((-I*A - 6*B)*a^3*e^(3 *I*f*x + 3*I*e) + (-I*A - 6*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^3*f*e^(2*I*f*x + 2*I*e) - c^3*f)*sqrt((A^2 - 12*I*A*B - 36*B^2)*a^7/(c^5*f^2)))/((I*A + 6*B)*a^3* e^(2*I*f*x + 2*I*e) + (I*A + 6*B)*a^3)) - 4*(3*(I*A + B)*a^3*e^(7*I*f*x + 7*I*e) + 2*(-I*A - 6*B)*a^3*e^(5*I*f*x + 5*I*e) + 10*(I*A + 6*B)*a^3*e^(3* I*f*x + 3*I*e) + 15*(I*A + 6*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^3*f)
Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 936 vs. \(2 (221) = 442\).
Time = 0.50 (sec) , antiderivative size = 936, normalized size of antiderivative = 3.31 \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \]
integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/ 2),x, algorithm="maxima")
15*(30*((A - 6*I*B)*a^3*cos(2*f*x + 2*e) - (-I*A - 6*B)*a^3*sin(2*f*x + 2* e) + (A - 6*I*B)*a^3)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + 30*( (A - 6*I*B)*a^3*cos(2*f*x + 2*e) - (-I*A - 6*B)*a^3*sin(2*f*x + 2*e) + (A - 6*I*B)*a^3)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) , -sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - 12*((A - I* B)*a^3*cos(2*f*x + 2*e) + (I*A + B)*a^3*sin(2*f*x + 2*e) + (A - I*B)*a^3)* cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 20*((A - 3*I*B)*a^3 *cos(2*f*x + 2*e) - (-I*A - 3*B)*a^3*sin(2*f*x + 2*e) + (A - 3*I*B)*a^3)*c os(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 60*((A - 5*I*B)*a^3* cos(2*f*x + 2*e) + (I*A + 5*B)*a^3*sin(2*f*x + 2*e) + (A - 6*I*B)*a^3)*cos (1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 15*((-I*A - 6*B)*a^3*c os(2*f*x + 2*e) + (A - 6*I*B)*a^3*sin(2*f*x + 2*e) + (-I*A - 6*B)*a^3)*log (cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2( sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*sin(1/2*arctan2(sin(2*f*x + 2*e ), cos(2*f*x + 2*e))) + 1) - 15*((I*A + 6*B)*a^3*cos(2*f*x + 2*e) - (A - 6 *I*B)*a^3*sin(2*f*x + 2*e) + (I*A + 6*B)*a^3)*log(cos(1/2*arctan2(sin(2*f* x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f *x + 2*e)))^2 - 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1 ) - 12*((I*A + B)*a^3*cos(2*f*x + 2*e) - (A - I*B)*a^3*sin(2*f*x + 2*e)...
\[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/ 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)^(5/2), x)
Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]