Integrand size = 34, antiderivative size = 192 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {2 A \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 A+i B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 A+i B}{4 a^2 d \sqrt {a+i a \tan (c+d x)}} \] Output:
-2*A*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/a^(5/2)/d+1/8*(A-I*B)*arcta nh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/a^(5/2)/d+1/5*(A+ I*B)/d/(a+I*a*tan(d*x+c))^(5/2)+1/6*(3*A+I*B)/a/d/(a+I*a*tan(d*x+c))^(3/2) +1/4*(7*A+I*B)/a^2/d/(a+I*a*tan(d*x+c))^(1/2)
Time = 2.17 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.13 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {3 a^{13/2} (A+i B)+(a+i a \tan (c+d x)) \left (\frac {5}{2} a^{11/2} (3 A+i B)+\frac {15}{8} a^4 (a+i a \tan (c+d x)) \left (2 \sqrt {a} (7 A+i B)-16 A \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {a+i a \tan (c+d x)}+\sqrt {2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right ) \sqrt {a+i a \tan (c+d x)}\right )\right )}{15 a^{13/2} d (a+i a \tan (c+d x))^{5/2}} \] Input:
Integrate[(Cot[c + d*x]*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(5/2) ,x]
Output:
(3*a^(13/2)*(A + I*B) + (a + I*a*Tan[c + d*x])*((5*a^(11/2)*(3*A + I*B))/2 + (15*a^4*(a + I*a*Tan[c + d*x])*(2*Sqrt[a]*(7*A + I*B) - 16*A*ArcTanh[Sq rt[a + I*a*Tan[c + d*x]]/Sqrt[a]]*Sqrt[a + I*a*Tan[c + d*x]] + Sqrt[2]*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]*Sqrt[a + I*a* Tan[c + d*x]]))/8))/(15*a^(13/2)*d*(a + I*a*Tan[c + d*x])^(5/2))
Time = 1.37 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.10, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4079, 27, 3042, 4079, 27, 3042, 4079, 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.
| \(\displaystyle \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan (c+d x) (a+i a \tan (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\int \frac {5 \cot (c+d x) (2 a A-a (i A-B) \tan (c+d x))}{2 (i \tan (c+d x) a+a)^{3/2}}dx}{5 a^2}+\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\cot (c+d x) (2 a A-a (i A-B) \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{2 a^2}+\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 a A-a (i A-B) \tan (c+d x)}{\tan (c+d x) (i \tan (c+d x) a+a)^{3/2}}dx}{2 a^2}+\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\frac {\int \frac {3 \cot (c+d x) \left (4 a^2 A-a^2 (3 i A-B) \tan (c+d x)\right )}{2 \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}+\frac {a (3 A+i B)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\cot (c+d x) \left (4 a^2 A-a^2 (3 i A-B) \tan (c+d x)\right )}{\sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}+\frac {a (3 A+i B)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {4 a^2 A-a^2 (3 i A-B) \tan (c+d x)}{\tan (c+d x) \sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}+\frac {a (3 A+i B)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {1}{2} \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (8 a^3 A-a^3 (7 i A-B) \tan (c+d x)\right )dx}{a^2}+\frac {a^2 (7 A+i B)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a (3 A+i B)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\int \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (8 a^3 A-a^3 (7 i A-B) \tan (c+d x)\right )dx}{2 a^2}+\frac {a^2 (7 A+i B)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a (3 A+i B)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (8 a^3 A-a^3 (7 i A-B) \tan (c+d x)\right )}{\tan (c+d x)}dx}{2 a^2}+\frac {a^2 (7 A+i B)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a (3 A+i B)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4083 |
| \(\displaystyle \frac {\frac {\frac {a^3 (B+i A) \int \sqrt {i \tan (c+d x) a+a}dx+8 a^2 A \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{2 a^2}+\frac {a^2 (7 A+i B)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a (3 A+i B)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {a^3 (B+i A) \int \sqrt {i \tan (c+d x) a+a}dx+8 a^2 A \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx}{2 a^2}+\frac {a^2 (7 A+i B)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a (3 A+i B)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3961 |
| \(\displaystyle \frac {\frac {\frac {8 a^2 A \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx-\frac {2 i a^4 (B+i A) \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}}{2 a^2}+\frac {a^2 (7 A+i B)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a (3 A+i B)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {8 a^2 A \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx-\frac {i \sqrt {2} a^{7/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a^2}+\frac {a^2 (7 A+i B)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a (3 A+i B)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4082 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 a^4 A \int \frac {\cot (c+d x)}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {i \sqrt {2} a^{7/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a^2}+\frac {a^2 (7 A+i B)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a (3 A+i B)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {\frac {-\frac {16 i a^3 A \int \frac {1}{i-\frac {i (i \tan (c+d x) a+a)}{a}}d\sqrt {i \tan (c+d x) a+a}}{d}-\frac {i \sqrt {2} a^{7/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a^2}+\frac {a^2 (7 A+i B)}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}+\frac {a (3 A+i B)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\frac {a^2 (7 A+i B)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {-\frac {i \sqrt {2} a^{7/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {16 a^{7/2} A \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}}{2 a^2}}{2 a^2}+\frac {a (3 A+i B)}{3 d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}+\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
Input:
Int[(Cot[c + d*x]*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(5/2),x]
Output:
(A + I*B)/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((a*(3*A + I*B))/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + (((-16*a^(7/2)*A*ArcTanh[Sqrt[a + I*a*Tan[c + d *x]]/Sqrt[a]])/d - (I*Sqrt[2]*a^(7/2)*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d)/(2*a^2) + (a^2*(7*A + I*B))/(d*Sqrt[a + I* a*Tan[c + d*x]]))/(2*a^2))/(2*a^2)
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*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && !GtQ[n, 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[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 = 0.24 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.80
| method | result | size |
| derivativedivides | \(\frac {2 a \left (-\frac {\left (i B -A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {7}{2}}}-\frac {-i B -7 A}{8 a^{3} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {-i B -3 A}{12 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {-i B -A}{10 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {A \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{a^{\frac {7}{2}}}\right )}{d}\) | \(153\) |
| default | \(\frac {2 a \left (-\frac {\left (i B -A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {7}{2}}}-\frac {-i B -7 A}{8 a^{3} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {-i B -3 A}{12 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {-i B -A}{10 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {A \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{a^{\frac {7}{2}}}\right )}{d}\) | \(153\) |
Input:
int(cot(d*x+c)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNV ERBOSE)
Output:
2/d*a*(-1/16*(-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))-1/8/a^3*(-I*B-7*A)/(a+I*a*tan(d*x+c))^(1/2)-1/12*(-I*B-3 *A)/a^2/(a+I*a*tan(d*x+c))^(3/2)-1/10*(-I*B-A)/a/(a+I*a*tan(d*x+c))^(5/2)- 1/a^(7/2)*A*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. 644 vs. \(2 (145) = 290\).
Time = 0.12 (sec) , antiderivative size = 644, normalized size of antiderivative = 3.35 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:
integrate(cot(d*x+c)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorith m="fricas")
Output:
-1/120*(15*sqrt(1/2)*a^3*d*sqrt((A^2 - 2*I*A*B - B^2)/(a^5*d^2))*e^(5*I*d* x + 5*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(I*a^3*d*e^(2*I*d*x + 2*I*c) + I*a^3* d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((A^2 - 2*I*A*B - B^2)/(a^5*d^2)) + (-I*A - B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - 15*sqrt(1/2 )*a^3*d*sqrt((A^2 - 2*I*A*B - B^2)/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*( sqrt(2)*sqrt(1/2)*(-I*a^3*d*e^(2*I*d*x + 2*I*c) - I*a^3*d)*sqrt(a/(e^(2*I* d*x + 2*I*c) + 1))*sqrt((A^2 - 2*I*A*B - B^2)/(a^5*d^2)) + (-I*A - B)*a*e^ (I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) + 60*a^3*d*sqrt(A^2/(a^5*d^2))* e^(5*I*d*x + 5*I*c)*log(16*(3*A*a^2*e^(2*I*d*x + 2*I*c) + A*a^2 + 2*sqrt(2 )*(a^4*d*e^(3*I*d*x + 3*I*c) + a^4*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(A^2/(a^5*d^2)))*e^(-2*I*d*x - 2*I*c)/A) - 60*a^3*d*sqrt (A^2/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(16*(3*A*a^2*e^(2*I*d*x + 2*I*c) + A*a^2 - 2*sqrt(2)*(a^4*d*e^(3*I*d*x + 3*I*c) + a^4*d*e^(I*d*x + I*c))*sqrt (a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(A^2/(a^5*d^2)))*e^(-2*I*d*x - 2*I*c)/A) - sqrt(2)*((123*A + 23*I*B)*e^(6*I*d*x + 6*I*c) + 2*(72*A + 17*I*B)*e^(4* I*d*x + 4*I*c) + 2*(12*A + 7*I*B)*e^(2*I*d*x + 2*I*c) + 3*A + 3*I*B)*sqrt( a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-5*I*d*x - 5*I*c)/(a^3*d)
\[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \cot {\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(cot(d*x+c)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(5/2),x)
Output:
Integral((A + B*tan(c + d*x))*cot(c + d*x)/(I*a*(tan(c + d*x) - I))**(5/2) , x)
Time = 0.15 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.97 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\frac {15 \, \sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {5}{2}}} - \frac {240 \, A \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}}} - \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} {\left (7 \, A + i \, B\right )} + 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (3 \, A + i \, B\right )} a + 12 \, {\left (A + i \, B\right )} a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2}}}{240 \, d} \] Input:
integrate(cot(d*x+c)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorith m="maxima")
Output:
-1/240*(15*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) - 240*A*log ((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt (a)))/a^(5/2) - 4*(15*(I*a*tan(d*x + c) + a)^2*(7*A + I*B) + 10*(I*a*tan(d *x + c) + a)*(3*A + I*B)*a + 12*(A + I*B)*a^2)/((I*a*tan(d*x + c) + a)^(5/ 2)*a^2))/d
Exception generated. \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(cot(d*x+c)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorith m="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad Argument TypeDone
Time = 3.73 (sec) , antiderivative size = 528, normalized size of antiderivative = 2.75 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:
int((cot(c + d*x)*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^(5/2),x)
Output:
((A + B*1i)/(5*d) + ((3*A + B*1i)*(a + a*tan(c + d*x)*1i))/(6*a*d) + ((7*A + B*1i)*(a + a*tan(c + d*x)*1i)^2)/(4*a^2*d))/(a + a*tan(c + d*x)*1i)^(5/ 2) - (2*A*atanh((127*A^3*a*d*(1/a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/ (127*A^3*d + A*B^2*d + A^2*B*d*2i) + (A*B^2*a*d*(1/a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(127*A^3*d + A*B^2*d + A^2*B*d*2i) + (A^2*B*a*d*(1/a^3) ^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*2i)/(127*A^3*d + A*B^2*d + A^2*B*d*2i ))*(1/a^3)^(1/2))/(a*d) + (2^(1/2)*atanh((2^(1/2)*A^3*a*d*(-1/a^3)^(1/2)*( a + a*tan(c + d*x)*1i)^(1/2)*127i)/(32*((127*A^3*d)/16 - (B^3*d*1i)/16 + ( 3*A*B^2*d)/16 - (A^2*B*d*125i)/16)) + (2^(1/2)*B^3*a*d*(-1/a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(32*((127*A^3*d)/16 - (B^3*d*1i)/16 + (3*A*B^2* d)/16 - (A^2*B*d*125i)/16)) + (2^(1/2)*A*B^2*a*d*(-1/a^3)^(1/2)*(a + a*tan (c + d*x)*1i)^(1/2)*3i)/(32*((127*A^3*d)/16 - (B^3*d*1i)/16 + (3*A*B^2*d)/ 16 - (A^2*B*d*125i)/16)) + (125*2^(1/2)*A^2*B*a*d*(-1/a^3)^(1/2)*(a + a*ta n(c + d*x)*1i)^(1/2))/(32*((127*A^3*d)/16 - (B^3*d*1i)/16 + (3*A*B^2*d)/16 - (A^2*B*d*125i)/16)))*(A*1i + B)*(-1/a^3)^(1/2))/(8*a*d)
\[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {-\left (\int \frac {\cot \left (d x +c \right )}{\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right ) i +1}}d x \right ) a -\left (\int \frac {\cot \left (d x +c \right ) \tan \left (d x +c \right )}{\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right ) i +1}}d x \right ) b}{\sqrt {a}\, a^{2}} \] Input:
int(cot(d*x+c)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x)
Output:
( - (int(cot(c + d*x)/(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 2*sqrt(t an(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + d*x)*i + 1)),x)*a + int(( cot(c + d*x)*tan(c + d*x))/(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 2*s qrt(tan(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + d*x)*i + 1)),x)*b))/ (sqrt(a)*a**2)