Integrand size = 38, antiderivative size = 201 \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {(2+2 i) a^{3/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {4 a (9 A-10 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 a (6 i A+5 B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 a A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d} \]
(-2-2*I)*a^(3/2)*(A-I*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-2/15*a*(6*I*A+5*B)*cot (d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(1/2)/d-2/5*a*A*cot(d*x+c)^(5/2)*(a+I*a*t an(d*x+c))^(1/2)/d+4/15*a*(9*A-10*I*B)*cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c)) ^(1/2)/d
Time = 6.64 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.62 \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {2 \cot ^{\frac {3}{2}}(c+d x) \left (-3 a^2 A (i+\cot (c+d x))^2 \tan (c+d x)+a^2 (3 i A+5 B) (-i+\tan (c+d x))^2+\frac {15 a (A-i B) \tan (c+d x) \left (\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) (1+i \tan (c+d x)) \sqrt {i a \tan (c+d x)}-(-1)^{3/4} a \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sqrt {\tan (c+d x)} (-i+\tan (c+d x))+\sqrt {1+i \tan (c+d x)} \left (a+i a \tan (c+d x)-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}\right )\right )}{\sqrt {1+i \tan (c+d x)}}\right )}{15 d \sqrt {a+i a \tan (c+d x)}} \]
(2*Cot[c + d*x]^(3/2)*(-3*a^2*A*(I + Cot[c + d*x])^2*Tan[c + d*x] + a^2*(( 3*I)*A + 5*B)*(-I + Tan[c + d*x])^2 + (15*a*(A - I*B)*Tan[c + d*x]*(Sqrt[a ]*ArcSinh[Sqrt[I*a*Tan[c + d*x]]/Sqrt[a]]*(1 + I*Tan[c + d*x])*Sqrt[I*a*Ta n[c + d*x]] - (-1)^(3/4)*a*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]]*Sqrt[Tan [c + d*x]]*(-I + Tan[c + d*x]) + Sqrt[1 + I*Tan[c + d*x]]*(a + I*a*Tan[c + d*x] - Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[ c + d*x]]]*Sqrt[I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])))/Sqrt[1 + I *Tan[c + d*x]]))/(15*d*Sqrt[a + I*a*Tan[c + d*x]])
Time = 1.22 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.07, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 4729, 3042, 4076, 27, 3042, 4081, 25, 3042, 4081, 27, 3042, 4027, 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.
| \(\displaystyle \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cot (c+d x)^{7/2} (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))dx\) |
| \(\Big \downarrow \) 4729 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {(i \tan (c+d x) a+a)^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {(i \tan (c+d x) a+a)^{3/2} (A+B \tan (c+d x))}{\tan (c+d x)^{7/2}}dx\) |
| \(\Big \downarrow \) 4076 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2}{5} \int \frac {\sqrt {i \tan (c+d x) a+a} (a (6 i A+5 B)-a (4 A-5 i B) \tan (c+d x))}{2 \tan ^{\frac {5}{2}}(c+d x)}dx-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{5} \int \frac {\sqrt {i \tan (c+d x) a+a} (a (6 i A+5 B)-a (4 A-5 i B) \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)}dx-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{5} \int \frac {\sqrt {i \tan (c+d x) a+a} (a (6 i A+5 B)-a (4 A-5 i B) \tan (c+d x))}{\tan (c+d x)^{5/2}}dx-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 4081 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{5} \left (\frac {2 \int -\frac {\sqrt {i \tan (c+d x) a+a} \left ((9 A-10 i B) a^2+(6 i A+5 B) \tan (c+d x) a^2\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {2 a (5 B+6 i A) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{5} \left (-\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a} \left ((9 A-10 i B) a^2+(6 i A+5 B) \tan (c+d x) a^2\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {2 a (5 B+6 i A) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{5} \left (-\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a} \left ((9 A-10 i B) a^2+(6 i A+5 B) \tan (c+d x) a^2\right )}{\tan (c+d x)^{3/2}}dx}{3 a}-\frac {2 a (5 B+6 i A) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 4081 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{5} \left (-\frac {2 \left (\frac {2 \int \frac {15 a^3 (i A+B) \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a}-\frac {2 a^2 (9 A-10 i B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )}{3 a}-\frac {2 a (5 B+6 i A) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{5} \left (-\frac {2 \left (15 a^2 (B+i A) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^2 (9 A-10 i B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )}{3 a}-\frac {2 a (5 B+6 i A) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{5} \left (-\frac {2 \left (15 a^2 (B+i A) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^2 (9 A-10 i B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )}{3 a}-\frac {2 a (5 B+6 i A) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{5} \left (-\frac {2 \left (-\frac {30 i a^4 (B+i A) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {2 a^2 (9 A-10 i B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )}{3 a}-\frac {2 a (5 B+6 i A) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{5} \left (-\frac {2 \left (\frac {(15-15 i) a^{5/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 (9 A-10 i B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\right )}{3 a}-\frac {2 a (5 B+6 i A) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a A \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )\) |
Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((-2*a*A*Sqrt[a + I*a*Tan[c + d*x]]) /(5*d*Tan[c + d*x]^(5/2)) + ((-2*a*((6*I)*A + 5*B)*Sqrt[a + I*a*Tan[c + d* x]])/(3*d*Tan[c + d*x]^(3/2)) - (2*(((15 - 15*I)*a^(5/2)*(I*A + B)*ArcTanh [((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2* a^2*(9*A - (10*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]])))/ (3*a))/5)
3.6.42.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^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[(A*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(a*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c* m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Fr eeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]
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. 714 vs. \(2 (164 ) = 328\).
Time = 0.53 (sec) , antiderivative size = 715, normalized size of antiderivative = 3.56
| method | result | size |
| derivativedivides | \(-\frac {\left (\frac {1}{\tan \left (d x +c \right )}\right )^{\frac {7}{2}} \tan \left (d x +c \right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (-72 A \sqrt {i a}\, \sqrt {-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 )}+60 i 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 )^{3}-15 i \sqrt {2}\, \sqrt {i a}\, \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 )^{3}+80 i B \sqrt {i a}\, \sqrt {-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 )}+60 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 )^{3}+30 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 )^{3}-15 \sqrt {2}\, \sqrt {i a}\, \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 )^{3}+24 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 )}-30 \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 )^{3}+20 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 )}+12 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 )}{30 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 )}}\) | \(715\) |
| default | \(-\frac {\left (\frac {1}{\tan \left (d x +c \right )}\right )^{\frac {7}{2}} \tan \left (d x +c \right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (-72 A \sqrt {i a}\, \sqrt {-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 )}+60 i 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 )^{3}-15 i \sqrt {2}\, \sqrt {i a}\, \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 )^{3}+80 i B \sqrt {i a}\, \sqrt {-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 )}+60 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 )^{3}+30 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 )^{3}-15 \sqrt {2}\, \sqrt {i a}\, \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 )^{3}+24 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 )}-30 \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 )^{3}+20 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 )}+12 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 )}{30 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 )}}\) | \(715\) |
-1/30/d*(1/tan(d*x+c))^(7/2)*tan(d*x+c)*(a*(1+I*tan(d*x+c)))^(1/2)*a*(-72* A*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1 /2)+60*I*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)^3-15*I*2^(1/2)*(I* a)^(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)^3+80*I*B*(-I*a)^(1/2)*( I*a)^(1/2)*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+60*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)^3+30*I*(-I*a)^(1/2)*ln(1/2*(2*I*a*ta n(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)^3-15*2^(1/2)*(I*a)^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*t an(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*ta n(d*x+c)^3+24*I*A*(-I*a)^(1/2)*(I*a)^(1/2)*tan(d*x+c)*(a*tan(d*x+c)*(1+I*t an(d*x+c)))^(1/2)-30*(-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)^3+20*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)+1 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. 510 vs. \(2 (153) = 306\).
Time = 0.27 (sec) , antiderivative size = 510, normalized size of antiderivative = 2.54 \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {15 \, \sqrt {2} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{3}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, 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}\right ) - 15 \, \sqrt {2} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{3}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, 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}\right ) - 2 \, \sqrt {2} {\left ({\left (27 \, A - 25 i \, B\right )} a e^{\left (5 i \, d x + 5 i \, c\right )} - 10 \, {\left (3 \, A - 4 i \, B\right )} a e^{\left (3 i \, d x + 3 i \, c\right )} + 15 \, {\left (A - i \, B\right )} a 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}}}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
-1/15*(15*sqrt(2)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^3/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log(4*((A - I*B)*a^2*e^(I*d*x + I*c) - sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^3/d^2)*(I*d*e^(2*I*d*x + 2*I*c) - I*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)) - 15*sqrt(2)*sqrt(- (-I*A^2 - 2*A*B + I*B^2)*a^3/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log(4*((A - I*B)*a^2*e^(I*d*x + I*c) - sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^3/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) + I*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*sqrt(2)*((27*A - 25*I*B)*a*e^(5*I*d*x + 5*I*c) - 10*(3*A - 4*I*B)*a*e^(3*I*d*x + 3*I*c) + 15*(A - I*B)*a*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)))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2 *I*c) + d)
Timed out. \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1457 vs. \(2 (153) = 306\).
Time = 0.74 (sec) , antiderivative size = 1457, normalized size of antiderivative = 7.25 \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
1/15*(2*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*((15*((I + 1)*A - (I - 1)*B)*a*cos(3*d*x + 3*c) + (-(16*I + 16)*A + ( 15*I - 15)*B)*a*cos(d*x + c) + 15*((I - 1)*A + (I + 1)*B)*a*sin(3*d*x + 3* c) + (-(16*I - 16)*A - (15*I + 15)*B)*a*sin(d*x + c))*cos(3/2*arctan2(sin( 2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + (15*(-(I - 1)*A - (I + 1)*B)*a*cos( 3*d*x + 3*c) + ((16*I - 16)*A + (15*I + 15)*B)*a*cos(d*x + c) + 15*((I + 1 )*A - (I - 1)*B)*a*sin(3*d*x + 3*c) + (-(16*I + 16)*A + (15*I - 15)*B)*a*s in(d*x + c))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)))*sqr t(a) + 15*(2*((-(I - 1)*A - (I + 1)*B)*a*cos(2*d*x + 2*c)^2 + (-(I - 1)*A - (I + 1)*B)*a*sin(2*d*x + 2*c)^2 + 2*((I - 1)*A + (I + 1)*B)*a*cos(2*d*x + 2*c) + (-(I - 1)*A - (I + 1)*B)*a)*arctan2(2*(cos(2*d*x + 2*c)^2 + sin(2 *d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 2*sin(d*x + c), 2*(cos(2*d*x + 2*c)^2 + sin (2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 2*cos(d*x + c)) + ((-(I + 1)*A + (I - 1)* B)*a*cos(2*d*x + 2*c)^2 + (-(I + 1)*A + (I - 1)*B)*a*sin(2*d*x + 2*c)^2 + 2*((I + 1)*A - (I - 1)*B)*a*cos(2*d*x + 2*c) + (-(I + 1)*A + (I - 1)*B)*a) *log(4*cos(d*x + c)^2 + 4*sin(d*x + c)^2 + 4*sqrt(cos(2*d*x + 2*c)^2 + sin (2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(cos(1/2*arctan2(sin(2*d*x + 2*c ), cos(2*d*x + 2*c) - 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d...
\[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cot \left (d x + c\right )^{\frac {7}{2}} \,d x } \]
Timed out. \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \]