Integrand size = 26, antiderivative size = 190 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {45 i a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{8 d}-\frac {4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {19 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d} \] Output:
45/8*I*a^(5/2)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/d-4*I*2^(1/2)*a^( 5/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d+19/8*a^2*cot( d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d-13/12*I*a^2*cot(d*x+c)^2*(a+I*a*tan(d*x+ c))^(1/2)/d-1/3*a^2*cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)/d
Time = 0.62 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.70 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {135 i a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )-96 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+a^2 \cot (c+d x) \left (57-26 i \cot (c+d x)-8 \cot ^2(c+d x)\right ) \sqrt {a+i a \tan (c+d x)}}{24 d} \] Input:
Integrate[Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^(5/2),x]
Output:
((135*I)*a^(5/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]] - (96*I)*Sqrt [2]*a^(5/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])] + a^2*Co t[c + d*x]*(57 - (26*I)*Cot[c + d*x] - 8*Cot[c + d*x]^2)*Sqrt[a + I*a*Tan[ c + d*x]])/(24*d)
Time = 1.29 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.09, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.654, Rules used = {3042, 4036, 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.
\(\displaystyle \int \cot ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan (c+d x)^4}dx\) |
\(\Big \downarrow \) 4036 |
\(\displaystyle -\frac {1}{3} \int -\frac {1}{2} \cot ^3(c+d x) \sqrt {i \tan (c+d x) a+a} \left (13 i a^2-11 a^2 \tan (c+d x)\right )dx-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} \int \cot ^3(c+d x) \sqrt {i \tan (c+d x) a+a} \left (13 i a^2-11 a^2 \tan (c+d x)\right )dx-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{6} \int \frac {\sqrt {i \tan (c+d x) a+a} \left (13 i a^2-11 a^2 \tan (c+d x)\right )}{\tan (c+d x)^3}dx-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 4081 |
\(\displaystyle \frac {1}{6} \left (\frac {\int -\frac {3}{2} \cot ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left (13 i \tan (c+d x) a^3+19 a^3\right )dx}{2 a}-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} \left (-\frac {3 \int \cot ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left (13 i \tan (c+d x) a^3+19 a^3\right )dx}{4 a}-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{6} \left (-\frac {3 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (13 i \tan (c+d x) a^3+19 a^3\right )}{\tan (c+d x)^2}dx}{4 a}-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 4081 |
\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (\frac {\int \frac {1}{2} \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (45 i a^4-19 a^4 \tan (c+d x)\right )dx}{a}-\frac {19 a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (\frac {\int \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (45 i a^4-19 a^4 \tan (c+d x)\right )dx}{2 a}-\frac {19 a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (45 i a^4-19 a^4 \tan (c+d x)\right )}{\tan (c+d x)}dx}{2 a}-\frac {19 a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 4083 |
\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (\frac {45 i a^3 \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx-64 a^4 \int \sqrt {i \tan (c+d x) a+a}dx}{2 a}-\frac {19 a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (\frac {45 i a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx-64 a^4 \int \sqrt {i \tan (c+d x) a+a}dx}{2 a}-\frac {19 a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 3961 |
\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (\frac {\frac {128 i a^5 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}+45 i a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx}{2 a}-\frac {19 a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (\frac {45 i a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx+\frac {64 i \sqrt {2} a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a}-\frac {19 a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 4082 |
\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (\frac {\frac {45 i a^5 \int \frac {\cot (c+d x)}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}+\frac {64 i \sqrt {2} a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a}-\frac {19 a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{6} \left (-\frac {3 \left (\frac {\frac {90 a^4 \int \frac {1}{i-\frac {i (i \tan (c+d x) a+a)}{a}}d\sqrt {i \tan (c+d x) a+a}}{d}+\frac {64 i \sqrt {2} a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a}-\frac {19 a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{6} \left (-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {3 \left (\frac {\frac {64 i \sqrt {2} a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {90 i a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}}{2 a}-\frac {19 a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}\right )-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\) |
Input:
Int[Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^(5/2),x]
Output:
-1/3*(a^2*Cot[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d + ((((-13*I)/2)*a^2 *Cot[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/d - (3*((((-90*I)*a^(9/2)*ArcT anh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d + ((64*I)*Sqrt[2]*a^(9/2)*ArcTa nh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d)/(2*a) - (19*a^3*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d))/(4*a))/6
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_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-a^2)*(b*c - a*d)*(a + b*Tan[e + f*x] )^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] + Si mp[a/(d*(b*c + a*d)*(n + 1)) Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[ e + f*x])^(n + 1)*Simp[b*(b*c*(m - 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
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 = 1.56 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.73
method | result | size |
derivativedivides | \(\frac {2 i a^{5} \left (-\frac {2 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{a^{\frac {5}{2}}}+\frac {-\frac {i \left (-\frac {19 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{16}+\frac {11 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{6}-\frac {13 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}}{16}\right )}{a^{3} \tan \left (d x +c \right )^{3}}+\frac {45 \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{16 \sqrt {a}}}{a^{2}}\right )}{d}\) | \(138\) |
default | \(\frac {2 i a^{5} \left (-\frac {2 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{a^{\frac {5}{2}}}+\frac {-\frac {i \left (-\frac {19 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{16}+\frac {11 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{6}-\frac {13 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}}{16}\right )}{a^{3} \tan \left (d x +c \right )^{3}}+\frac {45 \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{16 \sqrt {a}}}{a^{2}}\right )}{d}\) | \(138\) |
Input:
int(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
2*I/d*a^5*(-2/a^(5/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2) /a^(1/2))+1/a^2*(-I*(-19/16*(a+I*a*tan(d*x+c))^(5/2)+11/6*a*(a+I*a*tan(d*x +c))^(3/2)-13/16*a^2*(a+I*a*tan(d*x+c))^(1/2))/a^3/tan(d*x+c)^3+45/16/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. 659 vs. \(2 (145) = 290\).
Time = 0.10 (sec) , antiderivative size = 659, normalized size of antiderivative = 3.47 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx =\text {Too large to display} \] Input:
integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
Output:
1/96*(192*sqrt(2)*sqrt(-a^5/d^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)*log(4*(a^3*e^(I*d*x + I*c) + sqrt(- a^5/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)) )*e^(-I*d*x - I*c)/a^2) - 192*sqrt(2)*sqrt(-a^5/d^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)*log(4*(a^3*e^(I *d*x + I*c) + sqrt(-a^5/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)))*e^(-I*d*x - I*c)/a^2) + 135*sqrt(-a^5/d^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)*lo g(16*(3*a^3*e^(2*I*d*x + 2*I*c) + a^3 - 2*sqrt(2)*sqrt(-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)))* e^(-2*I*d*x - 2*I*c)/a) - 135*sqrt(-a^5/d^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)*log(16*(3*a^3*e^(2*I*d* x + 2*I*c) + a^3 - 2*sqrt(2)*sqrt(-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)))*e^(-2*I*d*x - 2*I*c) /a) + 4*sqrt(2)*(91*I*a^2*e^(7*I*d*x + 7*I*c) - 7*I*a^2*e^(5*I*d*x + 5*I*c ) - 59*I*a^2*e^(3*I*d*x + 3*I*c) + 39*I*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2* I*d*x + 2*I*c) + 1)))/(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)
Timed out. \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:
integrate(cot(d*x+c)**4*(a+I*a*tan(d*x+c))**(5/2),x)
Output:
Timed out
Time = 0.13 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.14 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {i \, a^{3} {\left (\frac {96 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{\sqrt {a}} - \frac {135 \, \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{\sqrt {a}} + \frac {2 \, {\left (57 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} - 88 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a + 39 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} - 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a + 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} - a^{3}}\right )}}{48 \, d} \] Input:
integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
Output:
1/48*I*a^3*(96*sqrt(2)*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)))/sqrt(a) - 135*log((sqrt(I *a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a)))/sq rt(a) + 2*(57*(I*a*tan(d*x + c) + a)^(5/2) - 88*(I*a*tan(d*x + c) + a)^(3/ 2)*a + 39*sqrt(I*a*tan(d*x + c) + a)*a^2)/((I*a*tan(d*x + c) + a)^3 - 3*(I *a*tan(d*x + c) + a)^2*a + 3*(I*a*tan(d*x + c) + a)*a^2 - a^3))/d
Exception generated. \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="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 = 0.24 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.90 \[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {19\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{8\,d\,{\mathrm {tan}\left (c+d\,x\right )}^3}-\frac {\mathrm {atan}\left (\frac {\sqrt {-a^5}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^3}\right )\,\sqrt {-a^5}\,45{}\mathrm {i}}{8\,d}-\frac {13\,a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{8\,d\,{\mathrm {tan}\left (c+d\,x\right )}^3}+\frac {11\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,d\,{\mathrm {tan}\left (c+d\,x\right )}^3}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {-a^5}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,a^3}\right )\,\sqrt {-a^5}\,4{}\mathrm {i}}{d} \] Input:
int(cot(c + d*x)^4*(a + a*tan(c + d*x)*1i)^(5/2),x)
Output:
(11*a*(a + a*tan(c + d*x)*1i)^(3/2))/(3*d*tan(c + d*x)^3) - (atan(((-a^5)^ (1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/a^3)*(-a^5)^(1/2)*45i)/(8*d) - (13*a^ 2*(a + a*tan(c + d*x)*1i)^(1/2))/(8*d*tan(c + d*x)^3) - (19*(a + a*tan(c + d*x)*1i)^(5/2))/(8*d*tan(c + d*x)^3) + (2^(1/2)*atan((2^(1/2)*(-a^5)^(1/2 )*(a + a*tan(c + d*x)*1i)^(1/2))/(2*a^3))*(-a^5)^(1/2)*4i)/d
\[ \int \cot ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (-\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cot \left (d x +c \right )^{4} \tan \left (d x +c \right )^{2}d x \right )+2 \left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cot \left (d x +c \right )^{4} \tan \left (d x +c \right )d x \right ) i +\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cot \left (d x +c \right )^{4}d x \right ) \] Input:
int(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x)
Output:
sqrt(a)*a**2*( - int(sqrt(tan(c + d*x)*i + 1)*cot(c + d*x)**4*tan(c + d*x) **2,x) + 2*int(sqrt(tan(c + d*x)*i + 1)*cot(c + d*x)**4*tan(c + d*x),x)*i + int(sqrt(tan(c + d*x)*i + 1)*cot(c + d*x)**4,x))