Integrand size = 36, antiderivative size = 210 \[ \int \cot ^4(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {a} (9 i A+14 B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{8 d}-\frac {\sqrt {2} \sqrt {a} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {(7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {(i A+6 B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d} \]
1/8*(9*I*A+14*B)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))*a^(1/2)/d-(I*A+ B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^(1/2)/d +1/8*(7*A-2*I*B)*cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d-1/12*(I*A+6*B)*cot( d*x+c)^2*(a+I*a*tan(d*x+c))^(1/2)/d-1/3*A*cot(d*x+c)^3*(a+I*a*tan(d*x+c))^ (1/2)/d
Time = 4.70 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.75 \[ \int \cot ^4(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {-3 \sqrt {a} (9 i A+14 B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )+24 \sqrt {2} \sqrt {a} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+\cot (c+d x) \left (-21 A+6 i B+2 (i A+6 B) \cot (c+d x)+8 A \cot ^2(c+d x)\right ) \sqrt {a+i a \tan (c+d x)}}{24 d} \]
-1/24*(-3*Sqrt[a]*((9*I)*A + 14*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt [a]] + 24*Sqrt[2]*Sqrt[a]*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sq rt[2]*Sqrt[a])] + Cot[c + d*x]*(-21*A + (6*I)*B + 2*(I*A + 6*B)*Cot[c + d* x] + 8*A*Cot[c + d*x]^2)*Sqrt[a + I*a*Tan[c + d*x]])/d
Time = 1.44 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.12, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.472, Rules used = {3042, 4081, 27, 3042, 4081, 27, 3042, 4081, 27, 3042, 4083, 3042, 3961, 219, 4082, 73, 221}
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.
| \(\displaystyle \int \cot ^4(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x))}{\tan (c+d x)^4}dx\) |
| \(\Big \downarrow \) 4081 |
| \(\displaystyle \frac {\int \frac {1}{2} \cot ^3(c+d x) \sqrt {i \tan (c+d x) a+a} (a (i A+6 B)-5 a A \tan (c+d x))dx}{3 a}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \cot ^3(c+d x) \sqrt {i \tan (c+d x) a+a} (a (i A+6 B)-5 a A \tan (c+d x))dx}{6 a}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {i \tan (c+d x) a+a} (a (i A+6 B)-5 a A \tan (c+d x))}{\tan (c+d x)^3}dx}{6 a}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
| \(\Big \downarrow \) 4081 |
| \(\displaystyle \frac {\frac {\int -\frac {3}{2} \cot ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left ((7 A-2 i B) a^2+(i A+6 B) \tan (c+d x) a^2\right )dx}{2 a}-\frac {a (6 B+i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {3 \int \cot ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left ((7 A-2 i B) a^2+(i A+6 B) \tan (c+d x) a^2\right )dx}{4 a}-\frac {a (6 B+i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {3 \int \frac {\sqrt {i \tan (c+d x) a+a} \left ((7 A-2 i B) a^2+(i A+6 B) \tan (c+d x) a^2\right )}{\tan (c+d x)^2}dx}{4 a}-\frac {a (6 B+i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
| \(\Big \downarrow \) 4081 |
| \(\displaystyle \frac {-\frac {3 \left (\frac {\int \frac {1}{2} \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (a^3 (9 i A+14 B)-a^3 (7 A-2 i B) \tan (c+d x)\right )dx}{a}-\frac {a^2 (7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {a (6 B+i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {3 \left (\frac {\int \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (a^3 (9 i A+14 B)-a^3 (7 A-2 i B) \tan (c+d x)\right )dx}{2 a}-\frac {a^2 (7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {a (6 B+i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {3 \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^3 (9 i A+14 B)-a^3 (7 A-2 i B) \tan (c+d x)\right )}{\tan (c+d x)}dx}{2 a}-\frac {a^2 (7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {a (6 B+i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
| \(\Big \downarrow \) 4083 |
| \(\displaystyle \frac {-\frac {3 \left (\frac {a^2 (14 B+9 i A) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx-16 a^3 (A-i B) \int \sqrt {i \tan (c+d x) a+a}dx}{2 a}-\frac {a^2 (7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {a (6 B+i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {3 \left (\frac {a^2 (14 B+9 i A) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx-16 a^3 (A-i B) \int \sqrt {i \tan (c+d x) a+a}dx}{2 a}-\frac {a^2 (7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {a (6 B+i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
| \(\Big \downarrow \) 3961 |
| \(\displaystyle \frac {-\frac {3 \left (\frac {\frac {32 i a^4 (A-i B) \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}+a^2 (14 B+9 i A) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx}{2 a}-\frac {a^2 (7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {a (6 B+i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {-\frac {3 \left (\frac {a^2 (14 B+9 i A) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx+\frac {16 i \sqrt {2} a^{7/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a}-\frac {a^2 (7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {a (6 B+i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
| \(\Big \downarrow \) 4082 |
| \(\displaystyle \frac {-\frac {3 \left (\frac {\frac {a^4 (14 B+9 i A) \int \frac {\cot (c+d x)}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}+\frac {16 i \sqrt {2} a^{7/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a}-\frac {a^2 (7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {a (6 B+i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {-\frac {3 \left (\frac {\frac {16 i \sqrt {2} a^{7/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {2 i a^3 (14 B+9 i A) \int \frac {1}{i-\frac {i (i \tan (c+d x) a+a)}{a}}d\sqrt {i \tan (c+d x) a+a}}{d}}{2 a}-\frac {a^2 (7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {a (6 B+i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {-\frac {3 \left (\frac {\frac {16 i \sqrt {2} a^{7/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {2 a^{7/2} (14 B+9 i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}}{2 a}-\frac {a^2 (7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {a (6 B+i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}}{6 a}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
-1/3*(A*Cot[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d + (-1/2*(a*(I*A + 6*B )*Cot[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/d - (3*(((-2*a^(7/2)*((9*I)*A + 14*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d + ((16*I)*Sqrt[2]* a^(7/2)*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d )/(2*a) - (a^2*(7*A - (2*I)*B)*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d) )/(4*a))/(6*a)
3.1.74.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(2*a - x^2), x], x, Sqrt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a , b, c, d}, x] && EqQ[a^2 + b^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*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[((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_)]), x_Symbol] :> Simp[( A*b + a*B)/(b*c + a*d) Int[(a + b*Tan[e + f*x])^m, x], x] - Simp[(B*c - A *d)/(b*c + a*d) Int[(a + b*Tan[e + f*x])^m*((a - b*Tan[e + f*x])/(c + d*T an[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
Time = 0.29 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.81
| method | result | size |
| derivativedivides | \(\frac {2 i a^{4} \left (-\frac {\left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 a^{\frac {7}{2}}}-\frac {-\frac {i \left (\left (-\frac {i B}{8}+\frac {7 A}{16}\right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}-\frac {5 A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{6}+\left (\frac {9}{16} A \,a^{2}+\frac {1}{8} i B \,a^{2}\right ) \sqrt {a +i a \tan \left (d x +c \right )}\right )}{a^{3} \tan \left (d x +c \right )^{3}}-\frac {\left (-14 i B +9 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{16 \sqrt {a}}}{a^{3}}\right )}{d}\) | \(171\) |
| default | \(\frac {2 i a^{4} \left (-\frac {\left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 a^{\frac {7}{2}}}-\frac {-\frac {i \left (\left (-\frac {i B}{8}+\frac {7 A}{16}\right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}-\frac {5 A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{6}+\left (\frac {9}{16} A \,a^{2}+\frac {1}{8} i B \,a^{2}\right ) \sqrt {a +i a \tan \left (d x +c \right )}\right )}{a^{3} \tan \left (d x +c \right )^{3}}-\frac {\left (-14 i B +9 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{16 \sqrt {a}}}{a^{3}}\right )}{d}\) | \(171\) |
2*I/d*a^4*(-1/2*(A-I*B)/a^(7/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/ 2)*2^(1/2)/a^(1/2))-1/a^3*(-I*((-1/8*I*B+7/16*A)*(a+I*a*tan(d*x+c))^(5/2)- 5/6*A*a*(a+I*a*tan(d*x+c))^(3/2)+(9/16*A*a^2+1/8*I*B*a^2)*(a+I*a*tan(d*x+c ))^(1/2))/a^3/tan(d*x+c)^3-1/16*(-14*I*B+9*A)/a^(1/2)*arctanh((a+I*a*tan(d *x+c))^(1/2)/a^(1/2))))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 823 vs. \(2 (163) = 326\).
Time = 0.28 (sec) , antiderivative size = 823, normalized size of antiderivative = 3.92 \[ \int \cot ^4(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
1/96*(48*sqrt(2)*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^ (2*I*d*x + 2*I*c) - d)*sqrt(-(A^2 - 2*I*A*B - B^2)*a/d^2)*log(-4*((-I*A - B)*a*e^(I*d*x + I*c) + (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(A^2 - 2*I*A*B - B^2)*a/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/(I*A + B)) - 48*sqrt(2)*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2* I*d*x + 2*I*c) - d)*sqrt(-(A^2 - 2*I*A*B - B^2)*a/d^2)*log(-4*((-I*A - B)* a*e^(I*d*x + I*c) - (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(A^2 - 2*I*A*B - B^2 )*a/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/(I*A + B)) - 3*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I* c) - d)*sqrt(-(81*A^2 - 252*I*A*B - 196*B^2)*a/d^2)*log(-16*(3*(-9*I*A - 1 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (-9*I*A - 14*B)*a^2 + 2*sqrt(2)*(a*d*e^(3*I *d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(-(81*A^2 - 252*I*A*B - 196*B^2)* a/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/(9*I*A + 14 *B)) + 3*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-(81*A^2 - 252*I*A*B - 196*B^2)*a/d^2)*log(-16*(3*(-9* I*A - 14*B)*a^2*e^(2*I*d*x + 2*I*c) + (-9*I*A - 14*B)*a^2 - 2*sqrt(2)*(a*d *e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(-(81*A^2 - 252*I*A*B - 19 6*B^2)*a/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/(9*I *A + 14*B)) + 4*sqrt(2)*((31*I*A + 18*B)*e^(7*I*d*x + 7*I*c) + (5*I*A + 6* B)*e^(5*I*d*x + 5*I*c) + (I*A - 18*B)*e^(3*I*d*x + 3*I*c) - 3*(-9*I*A +...
\[ \int \cot ^4(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \left (A + B \tan {\left (c + d x \right )}\right ) \cot ^{4}{\left (c + d x \right )}\, dx \]
Time = 0.31 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.19 \[ \int \cot ^4(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {i \, a^{3} {\left (\frac {2 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} {\left (7 \, A - 2 i \, B\right )} - 40 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} A a + 3 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (9 \, A + 2 i \, B\right )} a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} - 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} + 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} - a^{5}} + \frac {24 \, \sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {5}{2}}} - \frac {3 \, {\left (9 \, A - 14 i \, B\right )} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}}}\right )}}{48 \, d} \]
1/48*I*a^3*(2*(3*(I*a*tan(d*x + c) + a)^(5/2)*(7*A - 2*I*B) - 40*(I*a*tan( d*x + c) + a)^(3/2)*A*a + 3*sqrt(I*a*tan(d*x + c) + a)*(9*A + 2*I*B)*a^2)/ ((I*a*tan(d*x + c) + a)^3*a^2 - 3*(I*a*tan(d*x + c) + a)^2*a^3 + 3*(I*a*ta n(d*x + c) + a)*a^4 - a^5) + 24*sqrt(2)*(A - I*B)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a)) )/a^(5/2) - 3*(9*A - 14*I*B)*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(s qrt(I*a*tan(d*x + c) + a) + sqrt(a)))/a^(5/2))/d
\[ \int \cot ^4(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {i \, a \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )^{4} \,d x } \]
Time = 8.45 (sec) , antiderivative size = 735, normalized size of antiderivative = 3.50 \[ \int \cot ^4(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {\frac {\left (9\,A\,a^3+B\,a^3\,2{}\mathrm {i}\right )\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{8\,d}+\frac {\left (7\,A\,a-B\,a\,2{}\mathrm {i}\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,1{}\mathrm {i}}{8\,d}-\frac {A\,a^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,5{}\mathrm {i}}{3\,d}}{3\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2-3\,a^2\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )-{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3+a^3}-\frac {\mathrm {atan}\left (\frac {47\,\sqrt {32}\,A^3\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{8\,\left (47{}\mathrm {i}\,d\,A^3\,a^5+51\,d\,A^2\,B\,a^5+64{}\mathrm {i}\,d\,A\,B^2\,a^5+68\,d\,B^3\,a^5\right )}-\frac {\sqrt {32}\,B^3\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,17{}\mathrm {i}}{2\,\left (47{}\mathrm {i}\,d\,A^3\,a^5+51\,d\,A^2\,B\,a^5+64{}\mathrm {i}\,d\,A\,B^2\,a^5+68\,d\,B^3\,a^5\right )}+\frac {8\,\sqrt {32}\,A\,B^2\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{47{}\mathrm {i}\,d\,A^3\,a^5+51\,d\,A^2\,B\,a^5+64{}\mathrm {i}\,d\,A\,B^2\,a^5+68\,d\,B^3\,a^5}-\frac {\sqrt {32}\,A^2\,B\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,51{}\mathrm {i}}{8\,\left (47{}\mathrm {i}\,d\,A^3\,a^5+51\,d\,A^2\,B\,a^5+64{}\mathrm {i}\,d\,A\,B^2\,a^5+68\,d\,B^3\,a^5\right )}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,\sqrt {\frac {a}{32}}\,8{}\mathrm {i}}{d}+\frac {\sqrt {a}\,\mathrm {atan}\left (\frac {423\,A^3\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{8\,\left (\frac {423{}\mathrm {i}\,d\,A^3\,a^5}{8}+\frac {347\,d\,A^2\,B\,a^5}{4}+\frac {139{}\mathrm {i}\,d\,A\,B^2\,a^5}{2}+119\,d\,B^3\,a^5\right )}-\frac {B^3\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,119{}\mathrm {i}}{\frac {423{}\mathrm {i}\,d\,A^3\,a^5}{8}+\frac {347\,d\,A^2\,B\,a^5}{4}+\frac {139{}\mathrm {i}\,d\,A\,B^2\,a^5}{2}+119\,d\,B^3\,a^5}+\frac {139\,A\,B^2\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\left (\frac {423{}\mathrm {i}\,d\,A^3\,a^5}{8}+\frac {347\,d\,A^2\,B\,a^5}{4}+\frac {139{}\mathrm {i}\,d\,A\,B^2\,a^5}{2}+119\,d\,B^3\,a^5\right )}-\frac {A^2\,B\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,347{}\mathrm {i}}{4\,\left (\frac {423{}\mathrm {i}\,d\,A^3\,a^5}{8}+\frac {347\,d\,A^2\,B\,a^5}{4}+\frac {139{}\mathrm {i}\,d\,A\,B^2\,a^5}{2}+119\,d\,B^3\,a^5\right )}\right )\,\left (14\,B+A\,9{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,d} \]
(a^(1/2)*atan((423*A^3*a^(9/2)*d*(a + a*tan(c + d*x)*1i)^(1/2))/(8*((A^3*a ^5*d*423i)/8 + 119*B^3*a^5*d + (A*B^2*a^5*d*139i)/2 + (347*A^2*B*a^5*d)/4) ) - (B^3*a^(9/2)*d*(a + a*tan(c + d*x)*1i)^(1/2)*119i)/((A^3*a^5*d*423i)/8 + 119*B^3*a^5*d + (A*B^2*a^5*d*139i)/2 + (347*A^2*B*a^5*d)/4) + (139*A*B^ 2*a^(9/2)*d*(a + a*tan(c + d*x)*1i)^(1/2))/(2*((A^3*a^5*d*423i)/8 + 119*B^ 3*a^5*d + (A*B^2*a^5*d*139i)/2 + (347*A^2*B*a^5*d)/4)) - (A^2*B*a^(9/2)*d* (a + a*tan(c + d*x)*1i)^(1/2)*347i)/(4*((A^3*a^5*d*423i)/8 + 119*B^3*a^5*d + (A*B^2*a^5*d*139i)/2 + (347*A^2*B*a^5*d)/4)))*(A*9i + 14*B)*1i)/(8*d) - (atan((47*32^(1/2)*A^3*a^(9/2)*d*(a + a*tan(c + d*x)*1i)^(1/2))/(8*(A^3*a ^5*d*47i + 68*B^3*a^5*d + A*B^2*a^5*d*64i + 51*A^2*B*a^5*d)) - (32^(1/2)*B ^3*a^(9/2)*d*(a + a*tan(c + d*x)*1i)^(1/2)*17i)/(2*(A^3*a^5*d*47i + 68*B^3 *a^5*d + A*B^2*a^5*d*64i + 51*A^2*B*a^5*d)) + (8*32^(1/2)*A*B^2*a^(9/2)*d* (a + a*tan(c + d*x)*1i)^(1/2))/(A^3*a^5*d*47i + 68*B^3*a^5*d + A*B^2*a^5*d *64i + 51*A^2*B*a^5*d) - (32^(1/2)*A^2*B*a^(9/2)*d*(a + a*tan(c + d*x)*1i) ^(1/2)*51i)/(8*(A^3*a^5*d*47i + 68*B^3*a^5*d + A*B^2*a^5*d*64i + 51*A^2*B* a^5*d)))*(A*1i + B)*(a/32)^(1/2)*8i)/d - (((9*A*a^3 + B*a^3*2i)*(a + a*tan (c + d*x)*1i)^(1/2)*1i)/(8*d) + ((7*A*a - B*a*2i)*(a + a*tan(c + d*x)*1i)^ (5/2)*1i)/(8*d) - (A*a^2*(a + a*tan(c + d*x)*1i)^(3/2)*5i)/(3*d))/(3*a*(a + a*tan(c + d*x)*1i)^2 - 3*a^2*(a + a*tan(c + d*x)*1i) - (a + a*tan(c + d* x)*1i)^3 + a^3)