Integrand size = 31, antiderivative size = 205 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(a (A-B)-b (A+B)) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a (A-B)-b (A+B)) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(b (A-B)+a (A+B)) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {(b (A-B)+a (A+B)) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 a A}{d \sqrt {\tan (c+d x)}} \]
-1/2*(a*(A-B)-b*(A+B))*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)-1/2*( a*(A-B)-b*(A+B))*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)-1/4*(b*(A-B) +a*(A+B))*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)+1/4*(b*(A-B) +a*(A+B))*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)-2*a*A/d/tan( d*x+c)^(1/2)
Time = 0.68 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.77 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {-2 \sqrt {2} (a (A-B)-b (A+B)) \left (\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )\right )+\sqrt {2} (b (A-B)+a (A+B)) \left (\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )+\frac {8 a A}{\sqrt {\tan (c+d x)}}}{4 d} \]
-1/4*(-2*Sqrt[2]*(a*(A - B) - b*(A + B))*(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]) + Sqrt[2]*(b*(A - B) + a* (A + B))*(Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - Log[1 + Sqr t[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]) + (8*a*A)/Sqrt[Tan[c + d*x]])/d
Time = 0.53 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.89, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 4074, 3042, 4017, 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.
| \(\displaystyle \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan (c+d x)^{3/2}}dx\) |
| \(\Big \downarrow \) 4074 |
| \(\displaystyle \int \frac {A b+a B-(a A-b B) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx-\frac {2 a A}{d \sqrt {\tan (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A b+a B-(a A-b B) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx-\frac {2 a A}{d \sqrt {\tan (c+d x)}}\) |
| \(\Big \downarrow \) 4017 |
| \(\displaystyle \frac {2 \int \frac {A b+a B-(a A-b B) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}-\frac {2 a A}{d \sqrt {\tan (c+d x)}}\) |
| \(\Big \downarrow \) 1482 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a (A+B)+b (A-B)) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (a (A-B)-b (A+B)) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a A}{d \sqrt {\tan (c+d x)}}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a (A+B)+b (A-B)) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (a (A-B)-b (A+B)) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d}-\frac {2 a A}{d \sqrt {\tan (c+d x)}}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a (A+B)+b (A-B)) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (a (A-B)-b (A+B)) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d}-\frac {2 a A}{d \sqrt {\tan (c+d x)}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a (A+B)+b (A-B)) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (a (A-B)-b (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}-\frac {2 a A}{d \sqrt {\tan (c+d x)}}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a (A+B)+b (A-B)) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )-\frac {1}{2} (a (A-B)-b (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}-\frac {2 a A}{d \sqrt {\tan (c+d x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a (A+B)+b (A-B)) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )-\frac {1}{2} (a (A-B)-b (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}-\frac {2 a A}{d \sqrt {\tan (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a (A+B)+b (A-B)) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )-\frac {1}{2} (a (A-B)-b (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}-\frac {2 a A}{d \sqrt {\tan (c+d x)}}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a (A+B)+b (A-B)) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )-\frac {1}{2} (a (A-B)-b (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}-\frac {2 a A}{d \sqrt {\tan (c+d x)}}\) |
(2*(-1/2*((a*(A - B) - b*(A + B))*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] ]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2])) + ((b*(A - B ) + a*(A + B))*(-1/2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/Sq rt[2] + Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/(2*Sqrt[2])))/2 ))/d - (2*a*A)/(d*Sqrt[Tan[c + d*x]])
3.4.82.3.1 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[((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_)]), x_Symbol] :> Simp[(b *c - a*d)*(A*b - a*B)*((a + b*Tan[e + f*x])^(m + 1)/(b*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*A*c + b*B*c + A*b*d - a*B*d - (A*b*c - a*B*c - a*A*d - b*B*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && LtQ[m , -1] && NeQ[a^2 + b^2, 0]
Time = 0.04 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.99
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 a A}{\sqrt {\tan \left (d x +c \right )}}+\frac {\left (A b +B a \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-a A +B b \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\) | \(203\) |
| default | \(\frac {-\frac {2 a A}{\sqrt {\tan \left (d x +c \right )}}+\frac {\left (A b +B a \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-a A +B b \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\) | \(203\) |
| parts | \(\frac {\left (A b +B a \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}+\frac {a A \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}+\frac {B b \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}\) | \(289\) |
1/d*(-2*a*A/tan(d*x+c)^(1/2)+1/4*(A*b+B*a)*2^(1/2)*(ln((1+2^(1/2)*tan(d*x+ c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2 ^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/4*(-A*a+ B*b)*2^(1/2)*(ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d* x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2 ^(1/2)*tan(d*x+c)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 2244 vs. \(2 (175) = 350\).
Time = 0.39 (sec) , antiderivative size = 2244, normalized size of antiderivative = 10.95 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \]
-1/2*(d*sqrt((2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2)*a*b + d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B ^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4)) /d^2)*log(((A*a - B*b)*d^3*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b ^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4) - ((A^2*B - B^3)*a^3 + (A^3 - 5*A*B ^2)*a^2*b - (5*A^2*B - B^3)*a*b^2 - (A^3 - A*B^2)*b^3)*d)*sqrt((2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2)*a*b + d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4))/d^2) + ((A^4 - B^4)*a^4 - 4*(A^3*B + A*B^3)*a^3*b - 4*(A^3*B + A*B^3)*a*b^3 - (A^4 - B^4)*b^4)*sq rt(tan(d*x + c)))*tan(d*x + c) - d*sqrt((2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2)*a*b + d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3* b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4))/d^2)*log(-((A*a - B*b)*d^3*sqrt(-((A^4 - 2*A^2 *B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2 *b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4) - ((A^2 *B - B^3)*a^3 + (A^3 - 5*A*B^2)*a^2*b - (5*A^2*B - B^3)*a*b^2 - (A^3 - A*B ^2)*b^3)*d)*sqrt((2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2)*a*b + d^2*sqrt(-(( A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*...
\[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
Time = 0.31 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.85 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 \, \sqrt {2} {\left ({\left (A - B\right )} a - {\left (A + B\right )} b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left ({\left (A - B\right )} a - {\left (A + B\right )} b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left ({\left (A + B\right )} a + {\left (A - B\right )} b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left ({\left (A + B\right )} a + {\left (A - B\right )} b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \frac {8 \, A a}{\sqrt {\tan \left (d x + c\right )}}}{4 \, d} \]
-1/4*(2*sqrt(2)*((A - B)*a - (A + B)*b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sq rt(tan(d*x + c)))) + 2*sqrt(2)*((A - B)*a - (A + B)*b)*arctan(-1/2*sqrt(2) *(sqrt(2) - 2*sqrt(tan(d*x + c)))) - sqrt(2)*((A + B)*a + (A - B)*b)*log(s qrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + sqrt(2)*((A + B)*a + (A - B)*b)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 8*A*a/sqrt(tan (d*x + c)))/d
Timed out. \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]
Time = 10.79 (sec) , antiderivative size = 1420, normalized size of antiderivative = 6.93 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=2\,\mathrm {atanh}\left (\frac {32\,A^2\,a^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,a\,b}{2\,d^2}-\frac {\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{4\,d^4}}}{16\,A\,b\,\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}+16\,A^3\,a^3\,d^2-16\,A^3\,a\,b^2\,d^2}-\frac {32\,A^2\,b^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,a\,b}{2\,d^2}-\frac {\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{4\,d^4}}}{16\,A\,b\,\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}+16\,A^3\,a^3\,d^2-16\,A^3\,a\,b^2\,d^2}\right )\,\sqrt {\frac {A^2\,a\,b}{2\,d^2}-\frac {\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{4\,d^4}}-2\,\mathrm {atanh}\left (\frac {32\,A^2\,a^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{4\,d^4}+\frac {A^2\,a\,b}{2\,d^2}}}{16\,A\,b\,\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}-16\,A^3\,a^3\,d^2+16\,A^3\,a\,b^2\,d^2}-\frac {32\,A^2\,b^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{4\,d^4}+\frac {A^2\,a\,b}{2\,d^2}}}{16\,A\,b\,\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}-16\,A^3\,a^3\,d^2+16\,A^3\,a\,b^2\,d^2}\right )\,\sqrt {\frac {\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{4\,d^4}+\frac {A^2\,a\,b}{2\,d^2}}-2\,\mathrm {atanh}\left (\frac {32\,B^2\,a^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{4\,d^4}-\frac {B^2\,a\,b}{2\,d^2}}}{\frac {16\,B\,a\,\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{d^3}-\frac {16\,B^3\,b^3}{d}+\frac {16\,B^3\,a^2\,b}{d}}-\frac {32\,B^2\,b^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{4\,d^4}-\frac {B^2\,a\,b}{2\,d^2}}}{\frac {16\,B\,a\,\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{d^3}-\frac {16\,B^3\,b^3}{d}+\frac {16\,B^3\,a^2\,b}{d}}\right )\,\sqrt {-\frac {\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{4\,d^4}-\frac {B^2\,a\,b}{2\,d^2}}+2\,\mathrm {atanh}\left (\frac {32\,B^2\,a^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{4\,d^4}-\frac {B^2\,a\,b}{2\,d^2}}}{\frac {16\,B^3\,b^3}{d}+\frac {16\,B\,a\,\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{d^3}-\frac {16\,B^3\,a^2\,b}{d}}-\frac {32\,B^2\,b^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{4\,d^4}-\frac {B^2\,a\,b}{2\,d^2}}}{\frac {16\,B^3\,b^3}{d}+\frac {16\,B\,a\,\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{d^3}-\frac {16\,B^3\,a^2\,b}{d}}\right )\,\sqrt {\frac {\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{4\,d^4}-\frac {B^2\,a\,b}{2\,d^2}}-\frac {2\,A\,a}{d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}} \]
2*atanh((32*A^2*a^2*d^3*tan(c + d*x)^(1/2)*((A^2*a*b)/(2*d^2) - (2*A^4*a^2 *b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2)/(4*d^4))^(1/2))/(16*A*b*(2*A^4 *a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2) + 16*A^3*a^3*d^2 - 16*A^3* a*b^2*d^2) - (32*A^2*b^2*d^3*tan(c + d*x)^(1/2)*((A^2*a*b)/(2*d^2) - (2*A^ 4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2)/(4*d^4))^(1/2))/(16*A*b*( 2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2) + 16*A^3*a^3*d^2 - 16 *A^3*a*b^2*d^2))*((A^2*a*b)/(2*d^2) - (2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A ^4*a^4*d^4)^(1/2)/(4*d^4))^(1/2) - 2*atanh((32*A^2*a^2*d^3*tan(c + d*x)^(1 /2)*((2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2)/(4*d^4) + (A^2* a*b)/(2*d^2))^(1/2))/(16*A*b*(2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^ 4)^(1/2) - 16*A^3*a^3*d^2 + 16*A^3*a*b^2*d^2) - (32*A^2*b^2*d^3*tan(c + d* x)^(1/2)*((2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2)/(4*d^4) + (A^2*a*b)/(2*d^2))^(1/2))/(16*A*b*(2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a ^4*d^4)^(1/2) - 16*A^3*a^3*d^2 + 16*A^3*a*b^2*d^2))*((2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2)/(4*d^4) + (A^2*a*b)/(2*d^2))^(1/2) - 2*at anh((32*B^2*a^2*tan(c + d*x)^(1/2)*(- (2*B^4*a^2*b^2*d^4 - B^4*b^4*d^4 - B ^4*a^4*d^4)^(1/2)/(4*d^4) - (B^2*a*b)/(2*d^2))^(1/2))/((16*B*a*(2*B^4*a^2* b^2*d^4 - B^4*b^4*d^4 - B^4*a^4*d^4)^(1/2))/d^3 - (16*B^3*b^3)/d + (16*B^3 *a^2*b)/d) - (32*B^2*b^2*tan(c + d*x)^(1/2)*(- (2*B^4*a^2*b^2*d^4 - B^4*b^ 4*d^4 - B^4*a^4*d^4)^(1/2)/(4*d^4) - (B^2*a*b)/(2*d^2))^(1/2))/((16*B*a...