Integrand size = 23, antiderivative size = 232 \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx=-\frac {(a-b) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a-b) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {2 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} \left (a^2+b^2\right ) d}-\frac {(a+b) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a+b) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d} \]
1/2*(a-b)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b^2)/d*2^(1/2)+1/2*(a-b )*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b^2)/d*2^(1/2)-1/4*(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+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*b^(3/2)*arctan(b^(1 /2)*tan(d*x+c)^(1/2)/a^(1/2))/(a^2+b^2)/d/a^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.19 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx=\frac {2 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} \left (a^2+b^2\right ) d}-\frac {a \left (2 \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-2 \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )+\sqrt {2} \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\sqrt {2} \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )}{4 \left (a^2+b^2\right ) d}-\frac {2 b \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},-\tan ^2(c+d x)\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 \left (a^2+b^2\right ) d} \]
(2*b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(Sqrt[a]*(a^2 + b ^2)*d) - (a*(2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - 2*Sqrt[2]* ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] + Sqrt[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]] + T an[c + d*x]]))/(4*(a^2 + b^2)*d) - (2*b*Hypergeometric2F1[3/4, 1, 7/4, -Ta n[c + d*x]^2]*Tan[c + d*x]^(3/2))/(3*(a^2 + b^2)*d)
Time = 0.73 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.87, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {3042, 4057, 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.
| \(\displaystyle \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx\) |
| \(\Big \downarrow \) 4057 |
| \(\displaystyle \frac {b^2 \int \frac {\tan ^2(c+d x)+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}+\frac {\int \frac {a-b \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b^2 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}+\frac {\int \frac {a-b \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 4017 |
| \(\displaystyle \frac {2 \int \frac {a-b \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 1482 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a+b) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a-b) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a+b) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a-b) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d \left (a^2+b^2\right )}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a+b) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a-b) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a+b) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a+b) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a+b) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a+b) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )+\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {b^2 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {b^2 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}+\frac {2 \left (\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a+b) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 4117 |
| \(\displaystyle \frac {b^2 \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}d\tan (c+d x)}{d \left (a^2+b^2\right )}+\frac {2 \left (\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a+b) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {2 b^2 \int \frac {1}{a+b \tan (c+d x)}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}+\frac {2 \left (\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a+b) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a+b) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}+\frac {2 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d \left (a^2+b^2\right )}\) |
(2*b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(Sqrt[a]*(a^2 + b ^2)*d) + (2*(((a - b)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]))/2 + ((a + b)*(-1/2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/Sqrt[2] + Log[1 + Sqrt[2]*Sq rt[Tan[c + d*x]] + Tan[c + d*x]]/(2*Sqrt[2])))/2))/((a^2 + b^2)*d)
3.6.87.3.1 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_)])^(m_)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/(c^2 + d^2) Int[(a + b*Tan[e + f*x])^m *(c - d*Tan[e + f*x]), x], x] + Simp[d^2/(c^2 + d^2) Int[(a + b*Tan[e + f *x])^m*((1 + Tan[e + f*x]^2)/(c + d*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d ^2, 0] && !IntegerQ[m]
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 = 0.09 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.97
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {a \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {b \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{a^{2}+b^{2}}+\frac {2 b^{2} \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right ) \sqrt {a b}}}{d}\) | \(225\) |
| default | \(\frac {\frac {\frac {a \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {b \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{a^{2}+b^{2}}+\frac {2 b^{2} \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right ) \sqrt {a b}}}{d}\) | \(225\) |
1/d*(2/(a^2+b^2)*(1/8*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)))-1/8*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*ar ctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))))+2 *b^2/(a^2+b^2)/(a*b)^(1/2)*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 1352 vs. \(2 (194) = 388\).
Time = 0.32 (sec) , antiderivative size = 2730, normalized size of antiderivative = 11.77 \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx=\text {Too large to display} \]
[-1/2*((a^2 + b^2)*d*sqrt(((a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(-(a^4 - 2*a^2* b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) + 2*a*b) /((a^4 + 2*a^2*b^2 + b^4)*d^2))*log(((a^4*b + 2*a^2*b^3 + b^5)*d^3*sqrt(-( a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^ 4)) + (a^3 - a*b^2)*d)*sqrt(((a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(-(a^4 - 2*a^ 2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) + 2*a* b)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - (a^2 - b^2)*sqrt(tan(d*x + c))) - (a^2 + b^2)*d*sqrt(((a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/ ((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) + 2*a*b)/((a^4 + 2* a^2*b^2 + b^4)*d^2))*log(-((a^4*b + 2*a^2*b^3 + b^5)*d^3*sqrt(-(a^4 - 2*a^ 2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) + (a^3 - a*b^2)*d)*sqrt(((a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^ 4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) + 2*a*b)/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - (a^2 - b^2)*sqrt(tan(d*x + c))) - (a^2 + b^2)*d* sqrt(-((a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4 *a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) - 2*a*b)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*log(((a^4*b + 2*a^2*b^3 + b^5)*d^3*sqrt(-(a^4 - 2*a^2*b^2 + b^ 4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) - (a^3 - a*b^2)* d)*sqrt(-((a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) - 2*a*b)/((a^4 + 2*a^2...
\[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx=\int \frac {1}{\left (a + b \tan {\left (c + d x \right )}\right ) \sqrt {\tan {\left (c + d x \right )}}}\, dx \]
Time = 0.40 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx=\frac {\frac {8 \, b^{2} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{2} + b^{2}\right )} \sqrt {a b}} + \frac {2 \, \sqrt {2} {\left (a - b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (a - b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (a + b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (a + b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{a^{2} + b^{2}}}{4 \, d} \]
1/4*(8*b^2*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^2 + b^2)*sqrt(a*b)) + (2*sqrt(2)*(a - b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*(a - b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) + sqrt(2)*(a + b)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - sqr t(2)*(a + b)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))/(a^2 + b ^2))/d
Timed out. \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx=\text {Timed out} \]
Time = 6.39 (sec) , antiderivative size = 4102, normalized size of antiderivative = 17.68 \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))} \, dx=\text {Too large to display} \]
- atan(((((1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2)*(((32*(16*b^8*d^ 2 + 28*a^2*b^6*d^2 + 8*a^4*b^4*d^2 - 4*a^6*b^2*d^2))/d^3 - (32*tan(c + d*x )^(1/2)*(1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(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)*(1i/(4*(b^2*d^2 - a^2* d^2 + a*b*d^2*2i)))^(1/2) - (32*tan(c + d*x)^(1/2)*(4*a^3*b^4*d^2 - 30*a*b ^6*d^2 + 2*a^5*b^2*d^2))/d^4) - (32*(5*a*b^5 + a^3*b^3))/d^3)*(1i/(4*(b^2* d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2) - (96*b^5*tan(c + d*x)^(1/2))/d^4)*(1i /(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2)*1i - (((1i/(4*(b^2*d^2 - a^2* d^2 + a*b*d^2*2i)))^(1/2)*(((32*(16*b^8*d^2 + 28*a^2*b^6*d^2 + 8*a^4*b^4*d ^2 - 4*a^6*b^2*d^2))/d^3 + (32*tan(c + d*x)^(1/2)*(1i/(4*(b^2*d^2 - a^2*d^ 2 + a*b*d^2*2i)))^(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)*(1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2) + (32* tan(c + d*x)^(1/2)*(4*a^3*b^4*d^2 - 30*a*b^6*d^2 + 2*a^5*b^2*d^2))/d^4) - (32*(5*a*b^5 + a^3*b^3))/d^3)*(1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1 /2) + (96*b^5*tan(c + d*x)^(1/2))/d^4)*(1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2 *2i)))^(1/2)*1i)/((((1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2)*(((32* (16*b^8*d^2 + 28*a^2*b^6*d^2 + 8*a^4*b^4*d^2 - 4*a^6*b^2*d^2))/d^3 - (32*t an(c + d*x)^(1/2)*(1i/(4*(b^2*d^2 - a^2*d^2 + a*b*d^2*2i)))^(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)*(1i/(4*(b^2* d^2 - a^2*d^2 + a*b*d^2*2i)))^(1/2) - (32*tan(c + d*x)^(1/2)*(4*a^3*b^4...