Integrand size = 36, antiderivative size = 267 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {((5+3 i) A-(3-i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}+\frac {((-5-3 i) A+(3-i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) ((4+i) A+(1+2 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a d}+\frac {((5-3 i) A+(3+i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}-\frac {5 A+i B}{2 a d \sqrt {\tan (c+d x)}}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))} \]
-1/8*((5+3*I)*A+(-3+I)*B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/a/d*2^(1/2)+ 1/8*((-5-3*I)*A+(3-I)*B)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/a/d*2^(1/2)+(- 1/16+1/16*I)*((4+I)*A+(1+2*I)*B)*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)) /a/d*2^(1/2)+1/16*((5-3*I)*A+(3+I)*B)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d* x+c))/a/d*2^(1/2)+1/2*(-5*A-I*B)/a/d/tan(d*x+c)^(1/2)+1/2*(A+I*B)/d/tan(d* x+c)^(1/2)/(a+I*a*tan(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 1.38 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.43 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {-i A+B-2 (2 A+i B) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-i \tan (c+d x)\right ) (-i+\tan (c+d x))-(A-i B) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},i \tan (c+d x)\right ) (-i+\tan (c+d x))}{2 a d \sqrt {\tan (c+d x)} (-i+\tan (c+d x))} \]
((-I)*A + B - 2*(2*A + I*B)*Hypergeometric2F1[-1/2, 1, 1/2, (-I)*Tan[c + d *x]]*(-I + Tan[c + d*x]) - (A - I*B)*Hypergeometric2F1[-1/2, 1, 1/2, I*Tan [c + d*x]]*(-I + Tan[c + d*x]))/(2*a*d*Sqrt[Tan[c + d*x]]*(-I + Tan[c + d* x]))
Time = 0.73 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.89, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.472, Rules used = {3042, 4079, 27, 3042, 4012, 25, 3042, 4017, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan (c+d x)^{3/2} (a+i a \tan (c+d x))}dx\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\int \frac {a (5 A+i B)-3 a (i A-B) \tan (c+d x)}{2 \tan ^{\frac {3}{2}}(c+d x)}dx}{2 a^2}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (5 A+i B)-3 a (i A-B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)}dx}{4 a^2}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (5 A+i B)-3 a (i A-B) \tan (c+d x)}{\tan (c+d x)^{3/2}}dx}{4 a^2}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {\int -\frac {3 a (i A-B)+a (5 A+i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx-\frac {2 a (5 A+i B)}{d \sqrt {\tan (c+d x)}}}{4 a^2}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\int \frac {3 a (i A-B)+a (5 A+i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx-\frac {2 a (5 A+i B)}{d \sqrt {\tan (c+d x)}}}{4 a^2}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\int \frac {3 a (i A-B)+a (5 A+i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx-\frac {2 a (5 A+i B)}{d \sqrt {\tan (c+d x)}}}{4 a^2}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 4017 |
| \(\displaystyle \frac {-\frac {2 \int \frac {a (3 (i A-B)+(5 A+i B) \tan (c+d x))}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}-\frac {2 a (5 A+i B)}{d \sqrt {\tan (c+d x)}}}{4 a^2}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {2 a \int \frac {3 (i A-B)+(5 A+i B) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}-\frac {2 a (5 A+i B)}{d \sqrt {\tan (c+d x)}}}{4 a^2}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 1482 |
| \(\displaystyle \frac {-\frac {2 a \left (\frac {1}{2} ((5+3 i) A-(3-i) B) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\left (\frac {1}{2}-\frac {i}{2}\right ) ((4+i) A+(1+2 i) B) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a (5 A+i B)}{d \sqrt {\tan (c+d x)}}}{4 a^2}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {-\frac {2 a \left (\frac {1}{2} ((5+3 i) A-(3-i) B) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )-\left (\frac {1}{2}-\frac {i}{2}\right ) ((4+i) A+(1+2 i) B) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a (5 A+i B)}{d \sqrt {\tan (c+d x)}}}{4 a^2}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {-\frac {2 a \left (\frac {1}{2} ((5+3 i) A-(3-i) B) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )-\left (\frac {1}{2}-\frac {i}{2}\right ) ((4+i) A+(1+2 i) B) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a (5 A+i B)}{d \sqrt {\tan (c+d x)}}}{4 a^2}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {-\frac {2 a \left (\frac {1}{2} ((5+3 i) A-(3-i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\left (\frac {1}{2}-\frac {i}{2}\right ) ((4+i) A+(1+2 i) B) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a (5 A+i B)}{d \sqrt {\tan (c+d x)}}}{4 a^2}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {-\frac {2 a \left (\frac {1}{2} ((5+3 i) A-(3-i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\left (\frac {1}{2}-\frac {i}{2}\right ) ((4+i) A+(1+2 i) B) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )\right )}{d}-\frac {2 a (5 A+i B)}{d \sqrt {\tan (c+d x)}}}{4 a^2}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {2 a \left (\frac {1}{2} ((5+3 i) A-(3-i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\left (\frac {1}{2}-\frac {i}{2}\right ) ((4+i) A+(1+2 i) B) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )\right )}{d}-\frac {2 a (5 A+i B)}{d \sqrt {\tan (c+d x)}}}{4 a^2}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {2 a \left (\frac {1}{2} ((5+3 i) A-(3-i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\left (\frac {1}{2}-\frac {i}{2}\right ) ((4+i) A+(1+2 i) B) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d}-\frac {2 a (5 A+i B)}{d \sqrt {\tan (c+d x)}}}{4 a^2}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {-\frac {2 a \left (\frac {1}{2} ((5+3 i) A-(3-i) B) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\left (\frac {1}{2}-\frac {i}{2}\right ) ((4+i) A+(1+2 i) B) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d}-\frac {2 a (5 A+i B)}{d \sqrt {\tan (c+d x)}}}{4 a^2}+\frac {A+i B}{2 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))}\) |
((-2*a*((((5 + 3*I)*A - (3 - I)*B)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x] ]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]))/2 - (1/2 - I/2)*((4 + I)*A + (1 + 2*I)*B)*(-1/2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/Sqrt[2] + Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] /(2*Sqrt[2]))))/d - (2*a*(5*A + I*B))/(d*Sqrt[Tan[c + d*x]]))/(4*a^2) + (A + I*B)/(2*d*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x]))
3.2.38.3.1 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, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ a*c, 2]}, Simp[(d*q + a*e)/(2*a*c) Int[(q + c*x^2)/(a + c*x^4), x], x] + Simp[(d*q - a*e)/(2*a*c) Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a , c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- a)*c]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ (f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x] )^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 ]
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ )]], x_Symbol] :> Simp[2/f Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & & NeQ[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]
Time = 0.05 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.52
| method | result | size |
| derivativedivides | \(\frac {\frac {i \left (i A -B \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{2 \tan \left (d x +c \right )-2 i}-\frac {2 \left (i B +2 A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}-\frac {2 A}{\sqrt {\tan \left (d x +c \right )}}+\frac {4 \left (-\frac {A}{4}+\frac {i B}{4}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d a}\) | \(140\) |
| default | \(\frac {\frac {i \left (i A -B \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{2 \tan \left (d x +c \right )-2 i}-\frac {2 \left (i B +2 A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}-\frac {2 A}{\sqrt {\tan \left (d x +c \right )}}+\frac {4 \left (-\frac {A}{4}+\frac {i B}{4}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d a}\) | \(140\) |
1/d/a*(1/2*I*(I*A-B)*tan(d*x+c)^(1/2)/(tan(d*x+c)-I)-2*(2*A+I*B)/(2^(1/2)- I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)-I*2^(1/2)))-2*A/tan(d*x+c)^( 1/2)+4*(-1/4*A+1/4*I*B)/(2^(1/2)+I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^( 1/2)+I*2^(1/2))))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 703 vs. \(2 (200) = 400\).
Time = 0.27 (sec) , antiderivative size = 703, normalized size of antiderivative = 2.63 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} \log \left (\frac {2 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \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 {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} + {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) - {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} \log \left (-\frac {2 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \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 {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} - {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) + 2 \, {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {-4 i \, A^{2} + 4 \, A B + i \, B^{2}}{a^{2} d^{2}}} \log \left (\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \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 {-4 i \, A^{2} + 4 \, A B + i \, B^{2}}{a^{2} d^{2}}} + 2 \, A + i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) - 2 \, {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {-4 i \, A^{2} + 4 \, A B + i \, B^{2}}{a^{2} d^{2}}} \log \left (-\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \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 {-4 i \, A^{2} + 4 \, A B + i \, B^{2}}{a^{2} d^{2}}} - 2 \, A - i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) + 2 \, {\left ({\left (-9 i \, A + B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 8 i \, A e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A - B\right )} \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}}}{8 \, {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]
1/8*((a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))*sqrt((I*A^2 + 2*A *B - I*B^2)/(a^2*d^2))*log(2*((a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt((-I*e^( 2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*A^2 + 2*A*B - I*B ^2)/(a^2*d^2)) + (A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) - (a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^2*d^2))*log(-2*((a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt((-I *e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^2*d^2)) - (A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/( I*A + B)) + 2*(a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))*sqrt((-4 *I*A^2 + 4*A*B + I*B^2)/(a^2*d^2))*log(((a*d*e^(2*I*d*x + 2*I*c) + a*d)*sq rt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-4*I*A^2 + 4*A*B + I*B^2)/(a^2*d^2)) + 2*A + I*B)*e^(-2*I*d*x - 2*I*c)/(a*d)) - 2*( a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))*sqrt((-4*I*A^2 + 4*A*B + I*B^2)/(a^2*d^2))*log(-((a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt((-I*e^(2*I* d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-4*I*A^2 + 4*A*B + I*B^ 2)/(a^2*d^2)) - 2*A - I*B)*e^(-2*I*d*x - 2*I*c)/(a*d)) + 2*((-9*I*A + B)*e ^(4*I*d*x + 4*I*c) - 8*I*A*e^(2*I*d*x + 2*I*c) + I*A - B)*sqrt((-I*e^(2*I* d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/(a*d*e^(4*I*d*x + 4*I*c) - a *d*e^(2*I*d*x + 2*I*c))
Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\text {Exception raised: TypeError} \]
Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.77 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.42 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=-\frac {\left (i + 1\right ) \, \sqrt {2} {\left (2 \, A + i \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{2 \, a d} + \frac {\left (i - 1\right ) \, \sqrt {2} {\left (A - i \, B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{4 \, a d} + \frac {-5 i \, A \tan \left (d x + c\right ) + B \tan \left (d x + c\right ) - 4 \, A}{2 \, {\left (i \, \tan \left (d x + c\right )^{\frac {3}{2}} + \sqrt {\tan \left (d x + c\right )}\right )} a d} \]
-(1/2*I + 1/2)*sqrt(2)*(2*A + I*B)*arctan((1/2*I + 1/2)*sqrt(2)*sqrt(tan(d *x + c)))/(a*d) + (1/4*I - 1/4)*sqrt(2)*(A - I*B)*arctan(-(1/2*I - 1/2)*sq rt(2)*sqrt(tan(d*x + c)))/(a*d) + 1/2*(-5*I*A*tan(d*x + c) + B*tan(d*x + c ) - 4*A)/((I*tan(d*x + c)^(3/2) + sqrt(tan(d*x + c)))*a*d)
Time = 9.16 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.00 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=2\,\mathrm {atanh}\left (\frac {a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{a^2\,d^2}}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{a^2\,d^2}}+2\,\mathrm {atanh}\left (\frac {4\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}-\mathrm {atan}\left (\frac {2\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{4\,a^2\,d^2}}}{B}\right )\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{4\,a^2\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {4\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}\,2{}\mathrm {i}-\frac {\frac {2\,A}{a\,d}+\frac {A\,\mathrm {tan}\left (c+d\,x\right )\,5{}\mathrm {i}}{2\,a\,d}}{\sqrt {\mathrm {tan}\left (c+d\,x\right )}+{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,1{}\mathrm {i}}+\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \]
2*atanh((a*d*tan(c + d*x)^(1/2)*(-(A^2*1i)/(a^2*d^2))^(1/2))/A)*(-(A^2*1i) /(a^2*d^2))^(1/2) + 2*atanh((4*a*d*tan(c + d*x)^(1/2)*((A^2*1i)/(16*a^2*d^ 2))^(1/2))/A)*((A^2*1i)/(16*a^2*d^2))^(1/2) - atan((2*a*d*tan(c + d*x)^(1/ 2)*((B^2*1i)/(4*a^2*d^2))^(1/2))/B)*((B^2*1i)/(4*a^2*d^2))^(1/2)*2i + atan ((4*a*d*tan(c + d*x)^(1/2)*(-(B^2*1i)/(16*a^2*d^2))^(1/2))/B)*(-(B^2*1i)/( 16*a^2*d^2))^(1/2)*2i - ((2*A)/(a*d) + (A*tan(c + d*x)*5i)/(2*a*d))/(tan(c + d*x)^(1/2) + tan(c + d*x)^(3/2)*1i) + (B*tan(c + d*x)^(1/2))/(2*a*d*(ta n(c + d*x)*1i + 1))