Integrand size = 33, antiderivative size = 278 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx=\frac {(b (A-B)-a (A+B)) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}-\frac {(b (A-B)-a (A+B)) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {2 \sqrt {b} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} \left (a^2+b^2\right ) d}-\frac {(a (A-B)+b (A+B)) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a (A-B)+b (A+B)) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d} \]
-1/2*(b*(A-B)-a*(A+B))*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b^2)/d*2^( 1/2)-1/2*(b*(A-B)-a*(A+B))*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b^2)/d* 2^(1/2)-1/4*(a*(A-B)+b*(A+B))*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(a ^2+b^2)/d*2^(1/2)+1/4*(a*(A-B)+b*(A+B))*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan( d*x+c))/(a^2+b^2)/d*2^(1/2)+2*(A*b-B*a)*arctan(b^(1/2)*tan(d*x+c)^(1/2)/a^ (1/2))*b^(1/2)/(a^2+b^2)/d/a^(1/2)
Time = 0.37 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.70 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx=-\frac {2 \sqrt {2} (b (-A+B)+a (A+B)) \left (\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )\right )+\frac {8 \sqrt {b} (-A b+a B) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}+\sqrt {2} (a (A-B)+b (A+B)) \left (\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )}{4 \left (a^2+b^2\right ) d} \]
-1/4*(2*Sqrt[2]*(b*(-A + B) + a*(A + B))*(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]) + (8*Sqrt[b]*(-(A*b) + a* B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/Sqrt[a] + Sqrt[2]*(a*(A - B) + b*(A + B))*(Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - Log [1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]))/((a^2 + b^2)*d)
Time = 0.81 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.82, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4096, 3042, 4017, 1482, 1476, 1082, 217, 1479, 25, 27, 1103, 4117, 73, 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.
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx\) |
| \(\Big \downarrow \) 4096 |
| \(\displaystyle \frac {b (A b-a B) \int \frac {\tan ^2(c+d x)+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}+\frac {\int \frac {a A+b B-(A b-a B) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a A+b B-(A b-a B) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}+\frac {b (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 4017 |
| \(\displaystyle \frac {2 \int \frac {a A+b B-(A b-a B) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}+\frac {b (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 1482 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (b (A-B)-a (A+B)) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}+\frac {b (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d \left (a^2+b^2\right )}+\frac {b (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {b (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {b (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {b (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {b (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {b (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {b (A b-a B) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}+\frac {2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 4117 |
| \(\displaystyle \frac {b (A b-a B) \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}d\tan (c+d x)}{d \left (a^2+b^2\right )}+\frac {2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {2 b (A b-a B) \int \frac {1}{a+b \tan (c+d x)}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}+\frac {2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {2 \sqrt {b} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d \left (a^2+b^2\right )}+\frac {2 \left (\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\) |
(2*Sqrt[b]*(A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(Sqrt [a]*(a^2 + b^2)*d) + (2*(-1/2*((b*(A - B) - a*(A + B))*(-(ArcTan[1 - Sqrt[ 2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/S qrt[2])) + ((a*(A - B) + b*(A + 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])))/2))/((a^2 + b^2)*d)
3.5.1.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, 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_)^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[((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_)])*((c_.) + (d_.)*tan[(e_.) + (f_ .)*(x_)])^(n_))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ 1/(a^2 + b^2) Int[(c + d*Tan[e + f*x])^n*Simp[a*A + b*B - (A*b - a*B)*Tan [e + f*x], x], x], x] + Simp[b*((A*b - a*B)/(a^2 + b^2)) Int[(c + d*Tan[e + f*x])^n*((1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[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_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A/f Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
Time = 0.06 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.88
| method | result | size |
| derivativedivides | \(\frac {\frac {2 \left (A b -B a \right ) b \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right ) \sqrt {a b}}+\frac {\frac {\left (a A +B b \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-A b +B a \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{a^{2}+b^{2}}}{d}\) | \(244\) |
| default | \(\frac {\frac {2 \left (A b -B a \right ) b \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right ) \sqrt {a b}}+\frac {\frac {\left (a A +B b \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-A b +B a \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{a^{2}+b^{2}}}{d}\) | \(244\) |
1/d*(2*(A*b-B*a)*b/(a^2+b^2)/(a*b)^(1/2)*arctan(b*tan(d*x+c)^(1/2)/(a*b)^( 1/2))+2/(a^2+b^2)*(1/8*(A*a+B*b)*2^(1/2)*(ln((1+2^(1/2)*tan(d*x+c)^(1/2)+t an(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)))+1/8*(-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)))))
Leaf count of result is larger than twice the leaf count of optimal. 2999 vs. \(2 (240) = 480\).
Time = 4.61 (sec) , antiderivative size = 6024, normalized size of antiderivative = 21.67 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx=\text {Too large to display} \]
\[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx=\int \frac {A + B \tan {\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right ) \sqrt {\tan {\left (c + d x \right )}}}\, dx \]
Time = 0.33 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.78 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx=-\frac {\frac {8 \, {\left (B a b - A b^{2}\right )} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{2} + b^{2}\right )} \sqrt {a b}} - \frac {2 \, \sqrt {2} {\left ({\left (A + B\right )} a - {\left (A - B\right )} b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left ({\left (A + B\right )} a - {\left (A - B\right )} b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left ({\left (A - B\right )} a + {\left (A + B\right )} b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left ({\left (A - B\right )} a + {\left (A + B\right )} b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{a^{2} + b^{2}}}{4 \, d} \]
-1/4*(8*(B*a*b - A*b^2)*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^2 + b^2 )*sqrt(a*b)) - (2*sqrt(2)*((A + B)*a - (A - B)*b)*arctan(1/2*sqrt(2)*(sqrt (2) + 2*sqrt(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(sqrt(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))/(a^2 + b^2))/d
Timed out. \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx=\text {Timed out} \]
Time = 13.41 (sec) , antiderivative size = 14816, normalized size of antiderivative = 53.29 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx=\text {Too large to display} \]
atan(((((32*(13*B^3*a^2*b^4*d^2 + B^3*a^4*b^2*d^2))/d^5 + (((32*(12*B*a*b^ 7*d^4 + 24*B*a^3*b^5*d^4 + 12*B*a^5*b^3*d^4))/d^5 - (32*tan(c + d*x)^(1/2) *(((64*B^4*a^2*b^2*d^4 - B^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^( 1/2) - 8*B^2*a*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2)*(16* b^9*d^4 + 16*a^2*b^7*d^4 - 16*a^4*b^5*d^4 - 16*a^6*b^3*d^4))/d^4)*(((64*B^ 4*a^2*b^2*d^4 - B^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8* B^2*a*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2) + (32*tan(c + d*x)^(1/2)*(20*B^2*a^3*b^4*d^2 + 2*B^2*a^5*b^2*d^2 - 14*B^2*a*b^6*d^2))/d ^4)*(((64*B^4*a^2*b^2*d^4 - B^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4) )^(1/2) - 8*B^2*a*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2))* (((64*B^4*a^2*b^2*d^4 - B^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1 /2) - 8*B^2*a*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2) - (32 *tan(c + d*x)^(1/2)*(B^4*b^5 - 2*B^4*a^2*b^3))/d^4)*(((64*B^4*a^2*b^2*d^4 - B^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a*b*d^2)/( 16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2)*1i - (((32*(13*B^3*a^2*b^4* d^2 + B^3*a^4*b^2*d^2))/d^5 + (((32*(12*B*a*b^7*d^4 + 24*B*a^3*b^5*d^4 + 1 2*B*a^5*b^3*d^4))/d^5 + (32*tan(c + d*x)^(1/2)*(((64*B^4*a^2*b^2*d^4 - B^4 *(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a*b*d^2)/(16*(a ^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2)*(16*b^9*d^4 + 16*a^2*b^7*d^4 - 1 6*a^4*b^5*d^4 - 16*a^6*b^3*d^4))/d^4)*(((64*B^4*a^2*b^2*d^4 - B^4*(16*a...