Integrand size = 28, antiderivative size = 202 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\frac {(4-4 i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {32 a^2 \sqrt {a+i a \tan (c+d x)}}{21 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {104 i a^2 \sqrt {a+i a \tan (c+d x)}}{21 d \sqrt {\tan (c+d x)}} \] Output:
(4-4*I)*a^(5/2)*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*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(7/2)-6/7*I*a^2*(a+ I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(5/2)+32/21*a^2*(a+I*a*tan(d*x+c))^(1/2 )/d/tan(d*x+c)^(3/2)+104/21*I*a^2*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1 /2)
Time = 3.72 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.93 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\frac {4 i \sqrt {2} a^2 e^{-i (c+d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (e^{i (c+d x)} \left (-21+70 e^{2 i (c+d x)}-77 e^{4 i (c+d x)}+40 e^{6 i (c+d x)}\right )-21 \left (-1+e^{2 i (c+d x)}\right )^{7/2} \text {arctanh}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right )}{21 d \left (-1+e^{2 i (c+d x)}\right )^3 \sqrt {\tan (c+d x)}} \] Input:
Integrate[(a + I*a*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(9/2),x]
Output:
(((4*I)/21)*Sqrt[2]*a^2*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d* x)))]*(E^(I*(c + d*x))*(-21 + 70*E^((2*I)*(c + d*x)) - 77*E^((4*I)*(c + d* x)) + 40*E^((6*I)*(c + d*x))) - 21*(-1 + E^((2*I)*(c + d*x)))^(7/2)*ArcTan h[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]]))/(d*E^(I*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))^3*Sqrt[Tan[c + d*x]])
Time = 1.19 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.07, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.536, Rules used = {3042, 4036, 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.
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan (c+d x)^{9/2}}dx\) |
\(\Big \downarrow \) 4036 |
\(\displaystyle -\frac {2}{7} \int -\frac {\sqrt {i \tan (c+d x) a+a} \left (15 i a^2-13 a^2 \tan (c+d x)\right )}{2 \tan ^{\frac {7}{2}}(c+d x)}dx-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \int \frac {\sqrt {i \tan (c+d x) a+a} \left (15 i a^2-13 a^2 \tan (c+d x)\right )}{\tan ^{\frac {7}{2}}(c+d x)}dx-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \int \frac {\sqrt {i \tan (c+d x) a+a} \left (15 i a^2-13 a^2 \tan (c+d x)\right )}{\tan (c+d x)^{7/2}}dx-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4081 |
\(\displaystyle \frac {1}{7} \left (\frac {2 \int -\frac {10 \sqrt {i \tan (c+d x) a+a} \left (3 i \tan (c+d x) a^3+4 a^3\right )}{\tan ^{\frac {5}{2}}(c+d x)}dx}{5 a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \left (-\frac {4 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (3 i \tan (c+d x) a^3+4 a^3\right )}{\tan ^{\frac {5}{2}}(c+d x)}dx}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (-\frac {4 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (3 i \tan (c+d x) a^3+4 a^3\right )}{\tan (c+d x)^{5/2}}dx}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4081 |
\(\displaystyle \frac {1}{7} \left (-\frac {4 \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (13 i a^4-8 a^4 \tan (c+d x)\right )}{2 \tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \left (-\frac {4 \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (13 i a^4-8 a^4 \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (-\frac {4 \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (13 i a^4-8 a^4 \tan (c+d x)\right )}{\tan (c+d x)^{3/2}}dx}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4081 |
\(\displaystyle \frac {1}{7} \left (-\frac {4 \left (\frac {\frac {2 \int -\frac {21 a^5 \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a}-\frac {26 i a^4 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \left (-\frac {4 \left (\frac {-21 a^4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {26 i a^4 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (-\frac {4 \left (\frac {-21 a^4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {26 i a^4 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4027 |
\(\displaystyle \frac {1}{7} \left (-\frac {4 \left (\frac {\frac {42 i a^6 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {26 i a^4 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {1}{7} \left (-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 \left (\frac {-\frac {(21-21 i) a^{9/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {26 i a^4 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\) |
Input:
Int[(a + I*a*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(9/2),x]
Output:
(-2*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(7*d*Tan[c + d*x]^(7/2)) + (((-6*I)*a^ 2*Sqrt[a + I*a*Tan[c + d*x]])/(d*Tan[c + d*x]^(5/2)) - (4*((-8*a^3*Sqrt[a + I*a*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2)) + (((-21 + 21*I)*a^(9/2)*Arc Tanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - ((26*I)*a^4*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))/(3*a)))/a )/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*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. 456 vs. \(2 (163 ) = 326\).
Time = 1.56 (sec) , antiderivative size = 457, normalized size of antiderivative = 2.26
method | result | size |
derivativedivides | \(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (21 i \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{4}-21 \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{4}+84 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{4}+32 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )^{2}+104 i \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{3} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-18 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-6 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{21 d \tan \left (d x +c \right )^{\frac {7}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\) | \(457\) |
default | \(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (21 i \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{4}-21 \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{4}+84 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{4}+32 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )^{2}+104 i \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{3} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-18 i \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-6 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{21 d \tan \left (d x +c \right )^{\frac {7}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\) | \(457\) |
Input:
int((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(9/2),x,method=_RETURNVERBOSE)
Output:
1/21/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2/tan(d*x+c)^(7/2)*(21*I*2^(1/2)*(I*a) ^(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)^4-21*(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+84*I*(-I*a)^(1/2)*ln(1/2*(2*I*a*ta n(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+32*(-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+104*I*(-I*a)^(1/2)*(I*a)^(1/2)*tan(d*x+c)^3*(a*tan( d*x+c)*(1+I*tan(d*x+c)))^(1/2)-18*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)-6*(-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. 509 vs. \(2 (152) = 304\).
Time = 0.09 (sec) , antiderivative size = 509, normalized size of antiderivative = 2.52 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {8 \, \sqrt {2} {\left (40 \, a^{2} e^{\left (9 i \, d x + 9 i \, c\right )} - 37 \, a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} - 7 \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + 49 \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} - 21 \, a^{2} 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}} - 21 \, \sqrt {-\frac {32 i \, a^{5}}{d^{2}}} {\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 )} \log \left (\frac {{\left (4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\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 {32 i \, a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right ) + 21 \, \sqrt {-\frac {32 i \, a^{5}}{d^{2}}} {\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 )} \log \left (\frac {{\left (4 \, \sqrt {2} {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\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 {32 i \, a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right )}{42 \, {\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))^(5/2)/tan(d*x+c)^(9/2),x, algorithm="fricas")
Output:
-1/42*(8*sqrt(2)*(40*a^2*e^(9*I*d*x + 9*I*c) - 37*a^2*e^(7*I*d*x + 7*I*c) - 7*a^2*e^(5*I*d*x + 5*I*c) + 49*a^2*e^(3*I*d*x + 3*I*c) - 21*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)) - 21*sqrt(-32*I*a^5/d^2)*(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)*log(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*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(-32*I*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a^2) + 21*sqrt(-32*I*a^5/d^2)*(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)*log(1/4*(4*sqrt(2 )*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*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(-32*I*a^5/d^ 2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a^2))/(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. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+I*a*tan(d*x+c))**(5/2)/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. 3186 vs. \(2 (152) = 304\).
Time = 0.73 (sec) , antiderivative size = 3186, normalized size of antiderivative = 15.77 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(9/2),x, algorithm="maxima")
Output:
-2/105*(2*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c ) + 1)*(3*((35*I + 35)*a^2*cos(7*d*x + 7*c) - (35*I + 35)*a^2*cos(5*d*x + 5*c) + (21*I + 21)*a^2*cos(3*d*x + 3*c) - (I + 1)*a^2*cos(d*x + c) + (35*I - 35)*a^2*sin(7*d*x + 7*c) - (35*I - 35)*a^2*sin(5*d*x + 5*c) + (21*I - 2 1)*a^2*sin(3*d*x + 3*c) - (I - 1)*a^2*sin(d*x + c))*cos(7/2*arctan2(sin(2* d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + 5*(13*(-(I + 1)*a^2*cos(d*x + c) - ( I - 1)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 - (13*I + 13)*a^2*cos(d*x + c) + 13*(-(I + 1)*a^2*cos(d*x + c) - (I - 1)*a^2*sin(d*x + c))*sin(2*d*x + 2 *c)^2 - (13*I - 13)*a^2*sin(d*x + c) + 21*((I + 1)*a^2*cos(2*d*x + 2*c)^2 + (I + 1)*a^2*sin(2*d*x + 2*c)^2 - (2*I + 2)*a^2*cos(2*d*x + 2*c) + (I + 1 )*a^2)*cos(3*d*x + 3*c) + 26*((I + 1)*a^2*cos(d*x + c) + (I - 1)*a^2*sin(d *x + c))*cos(2*d*x + 2*c) + 21*((I - 1)*a^2*cos(2*d*x + 2*c)^2 + (I - 1)*a ^2*sin(2*d*x + 2*c)^2 - (2*I - 2)*a^2*cos(2*d*x + 2*c) + (I - 1)*a^2)*sin( 3*d*x + 3*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + 3*((35*I - 35)*a^2*cos(7*d*x + 7*c) - (35*I - 35)*a^2*cos(5*d*x + 5*c) + ( 21*I - 21)*a^2*cos(3*d*x + 3*c) - (I - 1)*a^2*cos(d*x + c) - (35*I + 35)*a ^2*sin(7*d*x + 7*c) + (35*I + 35)*a^2*sin(5*d*x + 5*c) - (21*I + 21)*a^2*s in(3*d*x + 3*c) + (I + 1)*a^2*sin(d*x + c))*sin(7/2*arctan2(sin(2*d*x + 2* c), -cos(2*d*x + 2*c) + 1)) + 5*(13*(-(I - 1)*a^2*cos(d*x + c) + (I + 1)*a ^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 - (13*I - 13)*a^2*cos(d*x + c) + 13...
Exception generated. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/2)/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. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{9/2}} \,d x \] Input:
int((a + a*tan(c + d*x)*1i)^(5/2)/tan(c + d*x)^(9/2),x)
Output:
int((a + a*tan(c + d*x)*1i)^(5/2)/tan(c + d*x)^(9/2), x)
\[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 \sqrt {a}\, a^{2} \left (-32 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{3} i +16 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-9 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -3 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}-42 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{2}}d x \right ) \tan \left (d x +c \right )^{4} d i \right )}{21 \tan \left (d x +c \right )^{4} d} \] Input:
int((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(9/2),x)
Output:
(2*sqrt(a)*a**2*( - 32*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**3*i + 16*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)** 2 - 9*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*i - 3*sqrt( tan(c + d*x))*sqrt(tan(c + d*x)*i + 1) - 42*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))/(21*tan(c + d *x)**4*d)