Integrand size = 38, antiderivative size = 230 \[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {2 (-1)^{3/4} a^{5/2} B \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {(4+4 i) a^{5/2} (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {2 a^2 (2 i A+B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d} \]
2*(-1)^(3/4)*a^(5/2)*B*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*t an(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d+(4+4*I)*a^(5/2)*(I*A +B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot(d *x+c)^(1/2)*tan(d*x+c)^(1/2)/d-2*a^2*(2*I*A+B)*cot(d*x+c)^(1/2)*(a+I*a*tan (d*x+c))^(1/2)/d-2/3*a*A*cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(3/2)/d
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(575\) vs. \(2(230)=460\).
Time = 8.30 (sec) , antiderivative size = 575, normalized size of antiderivative = 2.50 \[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (-\frac {2 A (a+i a \tan (c+d x))^{5/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (-\frac {a (5 i A+3 B) (a+i a \tan (c+d x))^{5/2}}{d \sqrt {\tan (c+d x)}}+\frac {2 \left (\frac {i a^4 (5 i A+3 B) \sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{4} (-1)^{3/4} \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right )+\frac {5}{4} \sqrt {1+i \tan (c+d x)} \sqrt {\tan (c+d x)}+\frac {1}{2} i \sqrt {1+i \tan (c+d x)} \tan ^{\frac {3}{2}}(c+d x)\right )}{d \sqrt {1+i \tan (c+d x)}}+\frac {a \left (-\frac {1}{4} i a^3 (23 A-15 i B)+a^3 (5 i A+3 B)\right ) \left (-\frac {4 i \sqrt {2} a \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\tan (c+d x)}}{\sqrt {i a \tan (c+d x)}}+\frac {4 i a^{3/2} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+i \tan (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+i \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}+\frac {i \sqrt {a} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{\sqrt {1+i \tan (c+d x)} \sqrt {i a \tan (c+d x)}}\right )}{d}\right )}{a}\right )}{3 a}\right ) \]
Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((-2*A*(a + I*a*Tan[c + d*x])^(5/2)) /(3*d*Tan[c + d*x]^(3/2)) + (2*(-((a*((5*I)*A + 3*B)*(a + I*a*Tan[c + d*x] )^(5/2))/(d*Sqrt[Tan[c + d*x]])) + (2*((I*a^4*((5*I)*A + 3*B)*Sqrt[a + I*a *Tan[c + d*x]]*((-3*(-1)^(3/4)*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]])/4 + (5*Sqrt[1 + I*Tan[c + d*x]]*Sqrt[Tan[c + d*x]])/4 + (I/2)*Sqrt[1 + I*Tan[ c + d*x]]*Tan[c + d*x]^(3/2)))/(d*Sqrt[1 + I*Tan[c + d*x]]) + (a*((-1/4*I) *a^3*(23*A - (15*I)*B) + a^3*((5*I)*A + 3*B))*(((-4*I)*Sqrt[2]*a*ArcTanh[( Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Tan[c + d *x]])/Sqrt[I*a*Tan[c + d*x]] + ((4*I)*a^(3/2)*ArcSinh[Sqrt[I*a*Tan[c + d*x ]]/Sqrt[a]]*Sqrt[1 + I*Tan[c + d*x]]*Sqrt[Tan[c + d*x]])/(Sqrt[I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + I*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan [c + d*x]] + (I*Sqrt[a]*ArcSinh[Sqrt[I*a*Tan[c + d*x]]/Sqrt[a]]*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(Sqrt[1 + I*Tan[c + d*x]]*Sqrt[I*a*Ta n[c + d*x]])))/d))/a))/(3*a))
Time = 1.39 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.92, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 4729, 3042, 4076, 27, 3042, 4076, 27, 3042, 4084, 3042, 4027, 218, 4082, 65, 216}
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.
| \(\displaystyle \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cot (c+d x)^{5/2} (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))dx\) |
| \(\Big \downarrow \) 4729 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {(i \tan (c+d x) a+a)^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {(i \tan (c+d x) a+a)^{5/2} (A+B \tan (c+d x))}{\tan (c+d x)^{5/2}}dx\) |
| \(\Big \downarrow \) 4076 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2}{3} \int \frac {3 (i \tan (c+d x) a+a)^{3/2} (a (2 i A+B)+i a B \tan (c+d x))}{2 \tan ^{\frac {3}{2}}(c+d x)}dx-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\int \frac {(i \tan (c+d x) a+a)^{3/2} (a (2 i A+B)+i a B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)}dx-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\int \frac {(i \tan (c+d x) a+a)^{3/2} (a (2 i A+B)+i a B \tan (c+d x))}{\tan (c+d x)^{3/2}}dx-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 4076 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (2 \int -\frac {\sqrt {i \tan (c+d x) a+a} \left ((4 A-3 i B) a^2+B \tan (c+d x) a^2\right )}{2 \sqrt {\tan (c+d x)}}dx-\frac {2 a^2 (B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((4 A-3 i B) a^2+B \tan (c+d x) a^2\right )}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^2 (B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((4 A-3 i B) a^2+B \tan (c+d x) a^2\right )}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^2 (B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 4084 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-4 a^2 (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-i a B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^2 (B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-4 a^2 (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-i a B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^2 (B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {8 i a^4 (A-i B) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-i a B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^2 (B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-i a B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {(4-4 i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 (B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 4082 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {i a^3 B \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {(4-4 i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 (B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 65 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 i a^3 B \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {(4-4 i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 (B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {(4-4 i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {2 (-1)^{3/4} a^{5/2} B \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 (B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )\) |
Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((2*(-1)^(3/4)*a^(5/2)*B*ArcTan[((-1 )^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - ((4 - 4*I)*a^(5/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[ a + I*a*Tan[c + d*x]]])/d - (2*a^2*((2*I)*A + B)*Sqrt[a + I*a*Tan[c + d*x] ])/(d*Sqrt[Tan[c + d*x]]) - (2*a*A*(a + I*a*Tan[c + d*x])^(3/2))/(3*d*Tan[ c + d*x]^(3/2)))
3.6.50.3.1 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[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2 Sub st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d }, x] && !GtQ[c, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f) Subst[Int[1/(a*c - b*d - 2* a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N eQ[c^2 + d^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] :> Sim p[(-a^2)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] - Simp[a/(d*(b*c + a*d)*(n + 1)) Int[ (a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m - 1) + b *d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(B/f) Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[ a^2 + b^2, 0] && EqQ[A*b + a*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] :> Sim p[(A*b + a*B)/b Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x], x] - Simp[B/b Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[ e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m Int[ActivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownTangentIntegrandQ[u, x]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 627 vs. \(2 (186 ) = 372\).
Time = 1.66 (sec) , antiderivative size = 628, normalized size of antiderivative = 2.73
| method | result | size |
| derivativedivides | \(-\frac {\left (\frac {1}{\tan \left (d x +c \right )}\right )^{\frac {5}{2}} \tan \left (d x +c \right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (-9 i B \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{2}+3 i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}+14 i A \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+12 A \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{2}+6 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{2}-3 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}+6 B \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+6 \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{2}+2 A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{3 d \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}}\) | \(628\) |
| default | \(-\frac {\left (\frac {1}{\tan \left (d x +c \right )}\right )^{\frac {5}{2}} \tan \left (d x +c \right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (-9 i B \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{2}+3 i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}+14 i A \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+12 A \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{2}+6 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{2}-3 \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}+6 B \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+6 \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{2}+2 A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{3 d \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}}\) | \(628\) |
-1/3/d*(1/tan(d*x+c))^(5/2)*tan(d*x+c)*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*(-9* I*B*(-I*a)^(1/2)*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)) )^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a*tan(d*x+c)^2+3*I*(I*a)^(1/2)*2^(1/2) *ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3* a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^2+14*I*A*(-I*a)^(1/2)*(I*a)^(1/ 2)*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+12*A*(-I*a)^(1/2)*ln(1 /2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a )/(I*a)^(1/2))*a*tan(d*x+c)^2+6*I*(-I*a)^(1/2)*ln(1/2*(2*I*a*tan(d*x+c)+2* (a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a*tan(d* x+c)^2-3*(I*a)^(1/2)*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1 +I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^2+6 *B*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/ 2)+6*(-I*a)^(1/2)*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c) ))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a*tan(d*x+c)^2+2*A*(a*tan(d*x+c)*(1+I *tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2))/(-I*a)^(1/2)/(I*a)^(1/2)/(a* tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 780 vs. \(2 (176) = 352\).
Time = 0.28 (sec) , antiderivative size = 780, normalized size of antiderivative = 3.39 \[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {24 \, \sqrt {2} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {4 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 24 \, \sqrt {2} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {4 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) + 8 \, \sqrt {2} {\left ({\left (8 i \, A + 3 \, B\right )} a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 3 \, {\left (-2 i \, A - B\right )} a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} - 3 \, \sqrt {\frac {4 i \, B^{2} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (-\frac {16 \, {\left (3 \, B a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - B a^{3} + \sqrt {2} \sqrt {\frac {4 i \, B^{2} a^{5}}{d^{2}}} {\left (i \, d e^{\left (3 i \, d x + 3 i \, c\right )} - i \, d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{B a}\right ) + 3 \, \sqrt {\frac {4 i \, B^{2} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (-\frac {16 \, {\left (3 \, B a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - B a^{3} + \sqrt {2} \sqrt {\frac {4 i \, B^{2} a^{5}}{d^{2}}} {\left (-i \, d e^{\left (3 i \, d x + 3 i \, c\right )} + i \, d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{B a}\right )}{12 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]
-1/12*(24*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2 *I*c) - d)*log(4*((A - I*B)*a^3*e^(I*d*x + I*c) + sqrt(-(I*A^2 + 2*A*B - I *B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1 ))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) - 24*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^ 2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(4*((A - I*B)*a^3*e^(I*d*x + I*c) - sqrt (-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(a/(e^( 2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I* c) - 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) + 8*sqrt(2)*((8*I*A + 3*B)*a^ 2*e^(3*I*d*x + 3*I*c) + 3*(-2*I*A - B)*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I *d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)) - 3*sqrt(4*I*B^2*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(-16*(3*B*a ^3*e^(2*I*d*x + 2*I*c) - B*a^3 + sqrt(2)*sqrt(4*I*B^2*a^5/d^2)*(I*d*e^(3*I *d*x + 3*I*c) - I*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqr t((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-2*I*d*x - 2* I*c)/(B*a)) + 3*sqrt(4*I*B^2*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(-16* (3*B*a^3*e^(2*I*d*x + 2*I*c) - B*a^3 + sqrt(2)*sqrt(4*I*B^2*a^5/d^2)*(-I*d *e^(3*I*d*x + 3*I*c) + I*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-2*I*d *x - 2*I*c)/(B*a)))/(d*e^(2*I*d*x + 2*I*c) - d)
Timed out. \[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
\[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right )^{\frac {5}{2}} \,d x } \]
\[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right )^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^{5/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \]