Integrand size = 38, antiderivative size = 237 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(7 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {(23 i A-25 B) \sqrt {a+i a \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(61 A+35 i B) \sqrt {a+i a \tan (c+d x)}}{15 a d \sqrt {\tan (c+d x)}} \]
(-1/2-1/2*I)*(A-I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x +c))^(1/2))/d/a^(1/2)+1/15*(61*A+35*I*B)*(a+I*a*tan(d*x+c))^(1/2)/a/d/tan( d*x+c)^(1/2)+(A+I*B)/d/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2)-1/5*(7*A+ 5*I*B)*(a+I*a*tan(d*x+c))^(1/2)/a/d/tan(d*x+c)^(5/2)+1/15*(23*I*A-25*B)*(a +I*a*tan(d*x+c))^(1/2)/a/d/tan(d*x+c)^(3/2)
Time = 3.82 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.73 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\frac {15 \sqrt {2} a (A-i B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \tan ^4(c+d x)}{(i a \tan (c+d x))^{3/2}}+\frac {2 \left (-6 A+2 i (A+5 i B) \tan (c+d x)+2 (19 A+5 i B) \tan ^2(c+d x)+(61 i A-35 B) \tan ^3(c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}}}{30 d \tan ^{\frac {5}{2}}(c+d x)} \]
((15*Sqrt[2]*a*(A - I*B)*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Tan[c + d*x]^4)/(I*a*Tan[c + d*x])^(3/2) + (2*(-6*A + (2*I)*(A + (5*I)*B)*Tan[c + d*x] + 2*(19*A + (5*I)*B)*Tan[c + d*x]^2 + ((6 1*I)*A - 35*B)*Tan[c + d*x]^3))/Sqrt[a + I*a*Tan[c + d*x]])/(30*d*Tan[c + d*x]^(5/2))
Time = 1.35 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.07, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {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.
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan (c+d x)^{7/2} \sqrt {a+i a \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\int \frac {\sqrt {i \tan (c+d x) a+a} (a (7 A+5 i B)-6 a (i A-B) \tan (c+d x))}{2 \tan ^{\frac {7}{2}}(c+d x)}dx}{a^2}+\frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sqrt {i \tan (c+d x) a+a} (a (7 A+5 i B)-6 a (i A-B) \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)}dx}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {i \tan (c+d x) a+a} (a (7 A+5 i B)-6 a (i A-B) \tan (c+d x))}{\tan (c+d x)^{7/2}}dx}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 4081 |
| \(\displaystyle \frac {\frac {2 \int -\frac {\sqrt {i \tan (c+d x) a+a} \left ((23 i A-25 B) a^2+4 (7 A+5 i B) \tan (c+d x) a^2\right )}{2 \tan ^{\frac {5}{2}}(c+d x)}dx}{5 a}-\frac {2 a (7 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((23 i A-25 B) a^2+4 (7 A+5 i B) \tan (c+d x) a^2\right )}{\tan ^{\frac {5}{2}}(c+d x)}dx}{5 a}-\frac {2 a (7 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((23 i A-25 B) a^2+4 (7 A+5 i B) \tan (c+d x) a^2\right )}{\tan (c+d x)^{5/2}}dx}{5 a}-\frac {2 a (7 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 4081 |
| \(\displaystyle \frac {-\frac {\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^3 (61 A+35 i B)-2 a^3 (23 i A-25 B) \tan (c+d x)\right )}{2 \tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {2 a^2 (-25 B+23 i A) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 a (7 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^3 (61 A+35 i B)-2 a^3 (23 i A-25 B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {2 a^2 (-25 B+23 i A) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 a (7 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (a^3 (61 A+35 i B)-2 a^3 (23 i A-25 B) \tan (c+d x)\right )}{\tan (c+d x)^{3/2}}dx}{3 a}-\frac {2 a^2 (-25 B+23 i A) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 a (7 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 4081 |
| \(\displaystyle \frac {-\frac {\frac {\frac {2 \int \frac {15 a^4 (i A+B) \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a}-\frac {2 a^3 (61 A+35 i B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 a^2 (-25 B+23 i A) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 a (7 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\frac {15 a^3 (B+i A) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^3 (61 A+35 i B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 a^2 (-25 B+23 i A) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 a (7 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {15 a^3 (B+i A) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^3 (61 A+35 i B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 a^2 (-25 B+23 i A) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 a (7 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \frac {-\frac {\frac {-\frac {30 i a^5 (B+i A) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {2 a^3 (61 A+35 i B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 a^2 (-25 B+23 i A) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 a (7 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {-\frac {\frac {\frac {(15-15 i) a^{7/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^3 (61 A+35 i B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 a^2 (-25 B+23 i A) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 a (7 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}}{2 a^2}+\frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}\) |
(A + I*B)/(d*Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]) + ((-2*a*(7*A + (5*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(5*d*Tan[c + d*x]^(5/2)) - ((-2*a^2 *((23*I)*A - 25*B)*Sqrt[a + I*a*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2)) + (((15 - 15*I)*a^(7/2)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x] ])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*a^3*(61*A + (35*I)*B)*Sqrt[a + I*a* Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))/(3*a))/(5*a))/(2*a^2)
3.2.84.3.1 Defintions of rubi rules used
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_)*((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. 814 vs. \(2 (193 ) = 386\).
Time = 0.18 (sec) , antiderivative size = 815, normalized size of antiderivative = 3.44
| method | result | size |
| derivativedivides | \(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (-396 i A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{3}\left (d x +c \right )\right )+140 i B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{4}\left (d x +c \right )\right )-15 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 \left (\tan ^{5}\left (d x +c \right )\right )+244 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{4}\left (d x +c \right )\right )-15 i B \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 {2}\, a \left (\tan ^{3}\left (d x +c \right )\right )+30 B \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 \left (\tan ^{4}\left (d x +c \right )\right )+180 B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{3}\left (d x +c \right )\right )+16 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 )+15 A \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 {2}\, a \left (\tan ^{3}\left (d x +c \right )\right )-144 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )+30 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 \left (\tan ^{4}\left (d x +c \right )\right )+15 i B \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 \left (\tan ^{5}\left (d x +c \right )\right )+40 B \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 )+24 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{60 d a \tan \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )^{2}}\) | \(815\) |
| default | \(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (-396 i A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{3}\left (d x +c \right )\right )+140 i B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{4}\left (d x +c \right )\right )-15 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 \left (\tan ^{5}\left (d x +c \right )\right )+244 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{4}\left (d x +c \right )\right )-15 i B \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 {2}\, a \left (\tan ^{3}\left (d x +c \right )\right )+30 B \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 \left (\tan ^{4}\left (d x +c \right )\right )+180 B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{3}\left (d x +c \right )\right )+16 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 )+15 A \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 {2}\, a \left (\tan ^{3}\left (d x +c \right )\right )-144 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )+30 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 \left (\tan ^{4}\left (d x +c \right )\right )+15 i B \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 \left (\tan ^{5}\left (d x +c \right )\right )+40 B \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 )+24 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{60 d a \tan \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )^{2}}\) | \(815\) |
| parts | \(\text {Expression too large to display}\) | \(870\) |
1/60/d*(a*(1+I*tan(d*x+c)))^(1/2)/a/tan(d*x+c)^(5/2)*(-396*I*A*(-I*a)^(1/2 )*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3+140*I*B*(-I*a)^(1/2)* (a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^4-15*A*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)^5+244*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan( d*x+c)))^(1/2)*tan(d*x+c)^4-15*I*B*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*t an(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*ta n(d*x+c)^3+30*B*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+180*B*(- I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3+16*I*A*(-I*a )^(1/2)*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+15*A*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))*2^(1/2)*a*tan(d*x+c)^3-144*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+ I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2+30*I*A*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+15*I*B*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)^5+ 40*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)+24*A*(- 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)/(-tan(d*x+c)+I)^2
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 596 vs. \(2 (181) = 362\).
Time = 0.28 (sec) , antiderivative size = 596, normalized size of antiderivative = 2.51 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {15 \, \sqrt {2} {\left (a d e^{\left (7 i \, d x + 7 i \, c\right )} - 3 \, a d e^{\left (5 i \, d x + 5 i \, c\right )} + 3 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} - a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a d^{2}}} \log \left (\frac {\sqrt {2} a d \sqrt {-\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\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}}}{4 i \, A + 4 \, B}\right ) - 15 \, \sqrt {2} {\left (a d e^{\left (7 i \, d x + 7 i \, c\right )} - 3 \, a d e^{\left (5 i \, d x + 5 i \, c\right )} + 3 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} - a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a d^{2}}} \log \left (-\frac {\sqrt {2} a d \sqrt {-\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\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}}}{4 i \, A + 4 \, B}\right ) + 2 \, \sqrt {2} {\left ({\left (-103 i \, A + 35 \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + 6 \, {\left (17 i \, A - 15 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 20 \, {\left (2 i \, A - B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 30 \, {\left (-5 i \, A + 3 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 15 i \, A - 15 \, B\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}}}{60 \, {\left (a d e^{\left (7 i \, d x + 7 i \, c\right )} - 3 \, a d e^{\left (5 i \, d x + 5 i \, c\right )} + 3 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} - a d e^{\left (i \, d x + i \, c\right )}\right )}} \]
-1/60*(15*sqrt(2)*(a*d*e^(7*I*d*x + 7*I*c) - 3*a*d*e^(5*I*d*x + 5*I*c) + 3 *a*d*e^(3*I*d*x + 3*I*c) - a*d*e^(I*d*x + I*c))*sqrt(-(-I*A^2 - 2*A*B + I* B^2)/(a*d^2))*log((sqrt(2)*a*d*sqrt(-(-I*A^2 - 2*A*B + I*B^2)/(a*d^2))*e^( I*d*x + I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*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)))/(4*I*A + 4*B)) - 15*sqrt(2)*(a*d*e^(7*I*d*x + 7*I*c) - 3*a*d* e^(5*I*d*x + 5*I*c) + 3*a*d*e^(3*I*d*x + 3*I*c) - a*d*e^(I*d*x + I*c))*sqr t(-(-I*A^2 - 2*A*B + I*B^2)/(a*d^2))*log(-(sqrt(2)*a*d*sqrt(-(-I*A^2 - 2*A *B + I*B^2)/(a*d^2))*e^(I*d*x + I*c) - sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I *c) + I*A + B)*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)))/(4*I*A + 4*B)) + 2*sqrt(2)*((-103*I*A + 35*B)*e^(8*I*d*x + 8*I*c) + 6*(17*I*A - 15*B)*e^(6*I*d*x + 6*I*c) + 20* (2*I*A - B)*e^(4*I*d*x + 4*I*c) + 30*(-5*I*A + 3*B)*e^(2*I*d*x + 2*I*c) + 15*I*A - 15*B)*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)))/(a*d*e^(7*I*d*x + 7*I*c) - 3*a*d*e^(5 *I*d*x + 5*I*c) + 3*a*d*e^(3*I*d*x + 3*I*c) - a*d*e^(I*d*x + I*c))
\[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {A + B \tan {\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \tan ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: RuntimeError} \]
Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: NotImplementedError} \]
Exception raised: NotImplementedError >> unable to parse Giac output: 2*(( -(345*i*sageVARa^3*sageVARA*sageVARd^2-150*sageVARa^3*sageVARd^2*sageVARB) /(225*i)/sageVARa^6/sageVARd^3*sqrt(i*sageVARa*tan(sageVARc+sageVARd*sageV ARx)+sageVARa)*sqrt
Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]