Integrand size = 28, antiderivative size = 235 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\frac {(2-2 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 i a^2}{7 d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {16 i a \sqrt {a+i a \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {76 a \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {268 i a \sqrt {a+i a \tan (c+d x)}}{105 d \sqrt {\tan (c+d x)}} \] Output:
(2-2*I)*a^(3/2)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^ (1/2))/d-2/7*a^2/d/tan(d*x+c)^(7/2)/(a+I*a*tan(d*x+c))^(1/2)-2/7*I*a^2/d/t an(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(1/2)-16/35*I*a*(a+I*a*tan(d*x+c))^(1/2 )/d/tan(d*x+c)^(5/2)+76/105*a*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(3/2)+ 268/105*I*a*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)
Time = 6.53 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.60 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {2 i \sqrt {2} a \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {2 i a^{3/2} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+i \tan (c+d x)} \sqrt {i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {2 (-1)^{3/4} a \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {1+i \tan (c+d x)}}-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {16 i a \sqrt {a+i a \tan (c+d x)}}{35 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {76 a \sqrt {a+i a \tan (c+d x)}}{105 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {268 i a \sqrt {a+i a \tan (c+d x)}}{105 d \sqrt {\tan (c+d x)}} \] Input:
Integrate[(a + I*a*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(9/2),x]
Output:
((-2*I)*Sqrt[2]*a*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Ta n[c + d*x]]]*Sqrt[I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]) + ((2*I)*a^(3/ 2)*ArcSinh[Sqrt[I*a*Tan[c + d*x]]/Sqrt[a]]*Sqrt[1 + I*Tan[c + d*x]]*Sqrt[I *a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (2*( -1)^(3/4)*a*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]]*Sqrt[a + I*a*Tan[c + d* x]])/(d*Sqrt[1 + I*Tan[c + d*x]]) - (2*a*Sqrt[a + I*a*Tan[c + d*x]])/(7*d* Tan[c + d*x]^(7/2)) - (((16*I)/35)*a*Sqrt[a + I*a*Tan[c + d*x]])/(d*Tan[c + d*x]^(5/2)) + (76*a*Sqrt[a + I*a*Tan[c + d*x]])/(105*d*Tan[c + d*x]^(3/2 )) + (((268*I)/105)*a*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]])
Time = 1.48 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.12, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {3042, 4036, 27, 3042, 4079, 27, 3042, 4081, 27, 3042, 4081, 27, 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.
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan (c+d x)^{9/2}}dx\) |
\(\Big \downarrow \) 4036 |
\(\displaystyle -\frac {2}{7} \int -\frac {13 i a^2-15 a^2 \tan (c+d x)}{2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {i \tan (c+d x) a+a}}dx-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \int \frac {13 i a^2-15 a^2 \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {i \tan (c+d x) a+a}}dx-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \int \frac {13 i a^2-15 a^2 \tan (c+d x)}{\tan (c+d x)^{7/2} \sqrt {i \tan (c+d x) a+a}}dx-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 4079 |
\(\displaystyle \frac {1}{7} \left (\frac {\int \frac {2 \sqrt {i \tan (c+d x) a+a} \left (4 i a^3-3 a^3 \tan (c+d x)\right )}{\tan ^{\frac {7}{2}}(c+d x)}dx}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (4 i a^3-3 a^3 \tan (c+d x)\right )}{\tan ^{\frac {7}{2}}(c+d x)}dx}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (4 i a^3-3 a^3 \tan (c+d x)\right )}{\tan (c+d x)^{7/2}}dx}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 4081 |
\(\displaystyle \frac {1}{7} \left (\frac {2 \left (\frac {2 \int -\frac {\sqrt {i \tan (c+d x) a+a} \left (16 i \tan (c+d x) a^4+19 a^4\right )}{2 \tan ^{\frac {5}{2}}(c+d x)}dx}{5 a}-\frac {8 i a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \left (\frac {2 \left (-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (16 i \tan (c+d x) a^4+19 a^4\right )}{\tan ^{\frac {5}{2}}(c+d x)}dx}{5 a}-\frac {8 i a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (\frac {2 \left (-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (16 i \tan (c+d x) a^4+19 a^4\right )}{\tan (c+d x)^{5/2}}dx}{5 a}-\frac {8 i a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 4081 |
\(\displaystyle \frac {1}{7} \left (\frac {2 \left (-\frac {\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (67 i a^5-38 a^5 \tan (c+d x)\right )}{2 \tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {38 a^4 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {8 i a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \left (\frac {2 \left (-\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (67 i a^5-38 a^5 \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {38 a^4 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {8 i a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (\frac {2 \left (-\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (67 i a^5-38 a^5 \tan (c+d x)\right )}{\tan (c+d x)^{3/2}}dx}{3 a}-\frac {38 a^4 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {8 i a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 4081 |
\(\displaystyle \frac {1}{7} \left (\frac {2 \left (-\frac {\frac {\frac {2 \int -\frac {105 a^6 \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a}-\frac {134 i a^5 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {38 a^4 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {8 i a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \left (\frac {2 \left (-\frac {\frac {-105 a^5 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {134 i a^5 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {38 a^4 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {8 i a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (\frac {2 \left (-\frac {\frac {-105 a^5 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {134 i a^5 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {38 a^4 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {8 i a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 4027 |
\(\displaystyle \frac {1}{7} \left (\frac {2 \left (-\frac {\frac {\frac {210 i a^7 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {134 i a^5 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {38 a^4 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {8 i a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\right )}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {1}{7} \left (\frac {2 \left (-\frac {8 i a^3 \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {\frac {-\frac {(105-105 i) a^{11/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {134 i a^5 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {38 a^4 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}\right )}{a^2}-\frac {2 i a^2}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\right )-\frac {2 a^2}{7 d \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
Input:
Int[(a + I*a*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(9/2),x]
Output:
(-2*a^2)/(7*d*Tan[c + d*x]^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]) + (((-2*I)*a^ 2)/(d*Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]) + (2*((((-8*I)/5)*a^3 *Sqrt[a + I*a*Tan[c + d*x]])/(d*Tan[c + d*x]^(5/2)) - ((-38*a^4*Sqrt[a + I *a*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2)) + (((-105 + 105*I)*a^(11/2)*Arc Tanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - ((134*I)*a^5*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))/(3*a))/( 5*a)))/a^2)/7
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_)*((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*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[(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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 454 vs. \(2 (188 ) = 376\).
Time = 1.64 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.94
method | result | size |
derivativedivides | \(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (105 i \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{4}+420 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{4}-105 \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{4}+152 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )^{2}+536 i \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{3} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-96 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-60 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{210 d \tan \left (d x +c \right )^{\frac {7}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\) | \(455\) |
default | \(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (105 i \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{4}+420 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{4}-105 \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{4}+152 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )^{2}+536 i \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{3} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-96 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-60 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{210 d \tan \left (d x +c \right )^{\frac {7}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\) | \(455\) |
Input:
int((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(9/2),x,method=_RETURNVERBOSE)
Output:
1/210/d*(a*(1+I*tan(d*x+c)))^(1/2)*a/tan(d*x+c)^(7/2)*(105*I*(I*a)^(1/2)*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))*a*tan(d*x+c)^4+420*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*tan(d*x+c)^4-105*(I*a)^(1/2)*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))*a*tan(d*x+c)^4+152*(-I*a)^(1/2)*(I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d* x+c)))^(1/2)*tan(d*x+c)^2+536*I*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3*(a*t an(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-96*I*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d *x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)-60*(-I*a)^(1/2)*(I*a)^(1/2)*(a*tan( d*x+c)*(1+I*tan(d*x+c)))^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(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. 491 vs. \(2 (175) = 350\).
Time = 0.12 (sec) , antiderivative size = 491, normalized size of antiderivative = 2.09 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {4 \, \sqrt {2} {\left (211 \, a e^{\left (9 i \, d x + 9 i \, c\right )} - 160 \, a e^{\left (7 i \, d x + 7 i \, c\right )} + 14 \, a e^{\left (5 i \, d x + 5 i \, c\right )} + 280 \, a e^{\left (3 i \, d x + 3 i \, c\right )} - 105 \, 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}} - 105 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {8 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (2 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\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}} + \sqrt {-\frac {8 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right ) + 105 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {8 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (2 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\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}} - \sqrt {-\frac {8 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right )}{210 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:
integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(9/2),x, algorithm="fricas")
Output:
-1/210*(4*sqrt(2)*(211*a*e^(9*I*d*x + 9*I*c) - 160*a*e^(7*I*d*x + 7*I*c) + 14*a*e^(5*I*d*x + 5*I*c) + 280*a*e^(3*I*d*x + 3*I*c) - 105*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)) - 105*(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) - 4*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-8*I* a^3/d^2)*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*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(-8*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a) + 105*(d*e^ (8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) - 4* d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-8*I*a^3/d^2)*log(1/2*(2*sqrt(2)*(a*e^(2*I *d*x + 2*I*c) + a)*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(-8*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a))/(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c ) + 6*d*e^(4*I*d*x + 4*I*c) - 4*d*e^(2*I*d*x + 2*I*c) + d)
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+I*a*tan(d*x+c))**(3/2)/tan(d*x+c)**(9/2),x)
Output:
Timed out
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2883 vs. \(2 (175) = 350\).
Time = 0.44 (sec) , antiderivative size = 2883, normalized size of antiderivative = 12.27 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(9/2),x, algorithm="maxima")
Output:
-1/420*(3*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c ) + 1)*(((280*I + 280)*a*cos(7*d*x + 7*c) - (140*I + 140)*a*cos(5*d*x + 5* c) + (133*I + 133)*a*cos(3*d*x + 3*c) + (47*I + 47)*a*cos(d*x + c) + (280* I - 280)*a*sin(7*d*x + 7*c) - (140*I - 140)*a*sin(5*d*x + 5*c) + (133*I - 133)*a*sin(3*d*x + 3*c) + (47*I - 47)*a*sin(d*x + c))*cos(7/2*arctan2(sin( 2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + 4*(47*(-(I + 1)*a*cos(d*x + c) - ( I - 1)*a*sin(d*x + c))*cos(2*d*x + 2*c)^2 + 47*(-(I + 1)*a*cos(d*x + c) - (I - 1)*a*sin(d*x + c))*sin(2*d*x + 2*c)^2 + 70*((I + 1)*a*cos(2*d*x + 2*c )^2 + (I + 1)*a*sin(2*d*x + 2*c)^2 - (2*I + 2)*a*cos(2*d*x + 2*c) + (I + 1 )*a)*cos(3*d*x + 3*c) + 94*((I + 1)*a*cos(d*x + c) + (I - 1)*a*sin(d*x + c ))*cos(2*d*x + 2*c) - (47*I + 47)*a*cos(d*x + c) + 70*((I - 1)*a*cos(2*d*x + 2*c)^2 + (I - 1)*a*sin(2*d*x + 2*c)^2 - (2*I - 2)*a*cos(2*d*x + 2*c) + (I - 1)*a)*sin(3*d*x + 3*c) - (47*I - 47)*a*sin(d*x + c))*cos(3/2*arctan2( sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + ((280*I - 280)*a*cos(7*d*x + 7 *c) - (140*I - 140)*a*cos(5*d*x + 5*c) + (133*I - 133)*a*cos(3*d*x + 3*c) + (47*I - 47)*a*cos(d*x + c) - (280*I + 280)*a*sin(7*d*x + 7*c) + (140*I + 140)*a*sin(5*d*x + 5*c) - (133*I + 133)*a*sin(3*d*x + 3*c) - (47*I + 47)* a*sin(d*x + c))*sin(7/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + 4*(47*(-(I - 1)*a*cos(d*x + c) + (I + 1)*a*sin(d*x + c))*cos(2*d*x + 2*c )^2 + 47*(-(I - 1)*a*cos(d*x + c) + (I + 1)*a*sin(d*x + c))*sin(2*d*x +...
Exception generated. \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(9/2),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. \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{9/2}} \,d x \] Input:
int((a + a*tan(c + d*x)*1i)^(3/2)/tan(c + d*x)^(9/2),x)
Output:
int((a + a*tan(c + d*x)*1i)^(3/2)/tan(c + d*x)^(9/2), x)
\[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 \sqrt {a}\, a \left (-76 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{3} i +38 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-24 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -15 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}-105 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{2}}d x \right ) \tan \left (d x +c \right )^{4} d i \right )}{105 \tan \left (d x +c \right )^{4} d} \] Input:
int((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(9/2),x)
Output:
(2*sqrt(a)*a*( - 76*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d* x)**3*i + 38*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 24*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*i - 15*sqrt(t an(c + d*x))*sqrt(tan(c + d*x)*i + 1) - 105*int((sqrt(tan(c + d*x))*sqrt(t an(c + d*x)*i + 1))/tan(c + d*x)**2,x)*tan(c + d*x)**4*d*i))/(105*tan(c + d*x)**4*d)