Integrand size = 28, antiderivative size = 222 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\sqrt {\cot (c+d x)}} \, dx=-\frac {23 (-1)^{3/4} a^{5/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{4 d}-\frac {(4+4 i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {a^2 \sqrt {a+i a \tan (c+d x)}}{2 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {9 i a^2 \sqrt {a+i a \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}} \] Output:
-23/4*(-1)^(3/4)*a^(5/2)*arctan((-1)^(3/4)*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-(4+4*I)*a^(5/2)*ar ctanh((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-1/2*a^2*(a+I*a*tan(d*x+c))^(1/2)/d/cot(d*x+c)^(3/ 2)+9/4*I*a^2*(a+I*a*tan(d*x+c))^(1/2)/d/cot(d*x+c)^(1/2)
Time = 4.71 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.09 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\sqrt {\cot (c+d x)}} \, dx=-\frac {a^4 \left (23 \sqrt {a} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \cot (c+d x) (i+\cot (c+d x))+\sqrt {1+i \tan (c+d x)} \left (\left (-2+11 i \cot (c+d x)+9 \cot ^2(c+d x)\right ) \sqrt {i a \tan (c+d x)}-16 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}\right )\right )}{4 d \cot ^{\frac {7}{2}}(c+d x) \sqrt {1+i \tan (c+d x)} (i a \tan (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \] Input:
Integrate[(a + I*a*Tan[c + d*x])^(5/2)/Sqrt[Cot[c + d*x]],x]
Output:
-1/4*(a^4*(23*Sqrt[a]*ArcSinh[Sqrt[I*a*Tan[c + d*x]]/Sqrt[a]]*Cot[c + d*x] *(I + Cot[c + d*x]) + Sqrt[1 + I*Tan[c + d*x]]*((-2 + (11*I)*Cot[c + d*x] + 9*Cot[c + d*x]^2)*Sqrt[I*a*Tan[c + d*x]] - 16*Sqrt[2]*ArcTanh[(Sqrt[2]*S qrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Cot[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])))/(d*Cot[c + d*x]^(7/2)*Sqrt[1 + I*Tan[c + d*x]]*(I*a* Tan[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])
Time = 1.23 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.95, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4729, 3042, 4039, 27, 3042, 4080, 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.
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{5/2}}{\sqrt {\cot (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{5/2}}{\sqrt {\cot (c+d x)}}dx\) |
\(\Big \downarrow \) 4729 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^{5/2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^{5/2}dx\) |
\(\Big \downarrow \) 4039 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{2} a \int \frac {1}{2} \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} (9 i \tan (c+d x) a+7 a)dx-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} a \int \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} (9 i \tan (c+d x) a+7 a)dx-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} a \int \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} (9 i \tan (c+d x) a+7 a)dx-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )\) |
\(\Big \downarrow \) 4080 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} a \left (\frac {\int -\frac {\sqrt {i \tan (c+d x) a+a} \left (9 i a^2-23 a^2 \tan (c+d x)\right )}{2 \sqrt {\tan (c+d x)}}dx}{a}+\frac {9 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} a \left (\frac {9 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (9 i a^2-23 a^2 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} a \left (\frac {9 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (9 i a^2-23 a^2 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )\) |
\(\Big \downarrow \) 4084 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} a \left (\frac {9 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {32 i a^2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-23 i a \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} a \left (\frac {9 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {32 i a^2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-23 i a \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )\) |
\(\Big \downarrow \) 4027 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} a \left (\frac {9 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {64 a^4 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-23 i a \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} a \left (\frac {9 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {(32+32 i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-23 i a \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )\) |
\(\Big \downarrow \) 4082 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} a \left (\frac {9 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {(32+32 i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {23 i a^3 \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}}{2 a}\right )-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )\) |
\(\Big \downarrow \) 65 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} a \left (\frac {9 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {(32+32 i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {46 i a^3 \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{2 a}\right )-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{4} a \left (\frac {9 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {46 (-1)^{3/4} a^{5/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(32+32 i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a}\right )-\frac {a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\right )\) |
Input:
Int[(a + I*a*Tan[c + d*x])^(5/2)/Sqrt[Cot[c + d*x]],x]
Output:
Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-1/2*(a^2*Tan[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/d + (a*(-1/2*((46*(-1)^(3/4)*a^(5/2)*ArcTan[((-1)^(3 /4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d + ((32 + 32 *I)*a^(5/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[ c + d*x]]])/d)/a + ((9*I)*a*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) /d))/4)
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_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[a/(d*(m + n - 1)) Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) + b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x ] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (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[B*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] + Simp[ 1/(a*(m + n)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Sim p[a*A*c*(m + n) - B*(b*c*m + a*d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*T an[e + f*x], 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] && 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_)])^(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. 354 vs. \(2 (175 ) = 350\).
Time = 0.40 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.60
method | result | size |
derivativedivides | \(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (18 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+23 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \sqrt {i a}\, \tan \left (d x +c \right )+16 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +32 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \right )}{8 d \sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \left (1+i \tan \left (d x +c \right )\right ) \tan \left (d x +c \right ) \sqrt {i a}\, \sqrt {-i a}}\) | \(355\) |
default | \(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (18 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+23 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \sqrt {i a}\, \tan \left (d x +c \right )+16 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +32 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \right )}{8 d \sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \left (1+i \tan \left (d x +c \right )\right ) \tan \left (d x +c \right ) \sqrt {i a}\, \sqrt {-i a}}\) | \(355\) |
Input:
int((a+I*a*tan(d*x+c))^(5/2)/cot(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/8/d*(a*(1+I*tan(d*x+c)))^(1/2)*a*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*( 18*I*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2)+23*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-4*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*( -I*a)^(1/2)*(I*a)^(1/2)*tan(d*x+c)+16*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)) *(I*a)^(1/2)*a+32*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))*(-I*a)^(1/2)*a)/(1/tan(d*x+c))^(1/2)/ (1+I*tan(d*x+c))/tan(d*x+c)/(I*a)^(1/2)/(-I*a)^(1/2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 722 vs. \(2 (166) = 332\).
Time = 0.12 (sec) , antiderivative size = 722, normalized size of antiderivative = 3.25 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\sqrt {\cot (c+d x)}} \, dx =\text {Too large to display} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/2)/cot(d*x+c)^(1/2),x, algorithm="fricas")
Output:
1/4*(sqrt(2)*(11*a^2*e^(5*I*d*x + 5*I*c) - 4*a^2*e^(3*I*d*x + 3*I*c) - 7*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)) - sqrt(529/16*I*a^5/d^2)*(d*e^(4*I *d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(-16/23*(69*a^3*e^(2*I*d*x + 2*I*c) - 23*a^3 + 8*sqrt(2)*sqrt(529/16*I*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)/a) + sqrt(529/16*I*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(-16/23*(69*a^3*e^(2*I*d*x + 2*I*c) - 23*a^3 + 8*sqrt(2)*sqrt(529/1 6*I*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)/a) - sqrt(128*I*a^5/d^2)*(d*e^(4*I*d*x + 4*I *c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(1/4*(16*I*a^3*e^(I*d*x + I*c) + sqr t(2)*sqrt(128*I*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)/a^2) + sqrt(128*I*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e ^(2*I*d*x + 2*I*c) + d)*log(1/4*(16*I*a^3*e^(I*d*x + I*c) - sqrt(2)*sqrt(1 28*I*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)/a^2))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\sqrt {\cot (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate((a+I*a*tan(d*x+c))**(5/2)/cot(d*x+c)**(1/2),x)
Output:
Timed out
\[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\sqrt {\cot (c+d x)}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sqrt {\cot \left (d x + c\right )}} \,d x } \] Input:
integrate((a+I*a*tan(d*x+c))^(5/2)/cot(d*x+c)^(1/2),x, algorithm="maxima")
Output:
integrate((I*a*tan(d*x + c) + a)^(5/2)/sqrt(cot(d*x + c)), x)
Exception generated. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\sqrt {\cot (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/2)/cot(d*x+c)^(1/2),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
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\sqrt {\cot (c+d x)}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}} \,d x \] Input:
int((a + a*tan(c + d*x)*1i)^(5/2)/cot(c + d*x)^(1/2),x)
Output:
int((a + a*tan(c + d*x)*1i)^(5/2)/cot(c + d*x)^(1/2), x)
\[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\sqrt {\cot (c+d x)}} \, dx=\sqrt {a}\, a^{2} \left (-\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sqrt {\cot \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}}{\cot \left (d x +c \right )}d x \right )+2 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sqrt {\cot \left (d x +c \right )}\, \tan \left (d x +c \right )}{\cot \left (d x +c \right )}d x \right ) i +\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \sqrt {\cot \left (d x +c \right )}}{\cot \left (d x +c \right )}d x \right ) \] Input:
int((a+I*a*tan(d*x+c))^(5/2)/cot(d*x+c)^(1/2),x)
Output:
sqrt(a)*a**2*( - int((sqrt(tan(c + d*x)*i + 1)*sqrt(cot(c + d*x))*tan(c + d*x)**2)/cot(c + d*x),x) + 2*int((sqrt(tan(c + d*x)*i + 1)*sqrt(cot(c + d* x))*tan(c + d*x))/cot(c + d*x),x)*i + int((sqrt(tan(c + d*x)*i + 1)*sqrt(c ot(c + d*x)))/cot(c + d*x),x))