Integrand size = 12, antiderivative size = 192 \[ \int \sqrt {b \tan (c+d x)} \, dx=-\frac {\sqrt {b} \arctan \left (1-\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}\right )}{\sqrt {2} d}+\frac {\sqrt {b} \arctan \left (1+\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}\right )}{\sqrt {2} d}+\frac {\sqrt {b} \log \left (\sqrt {b}+\sqrt {b} \tan (c+d x)-\sqrt {2} \sqrt {b \tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {\sqrt {b} \log \left (\sqrt {b}+\sqrt {b} \tan (c+d x)+\sqrt {2} \sqrt {b \tan (c+d x)}\right )}{2 \sqrt {2} d} \]
-1/2*arctan(1-2^(1/2)*(b*tan(d*x+c))^(1/2)/b^(1/2))*b^(1/2)/d*2^(1/2)+1/2* arctan(1+2^(1/2)*(b*tan(d*x+c))^(1/2)/b^(1/2))*b^(1/2)/d*2^(1/2)+1/4*ln(b^ (1/2)-2^(1/2)*(b*tan(d*x+c))^(1/2)+b^(1/2)*tan(d*x+c))*b^(1/2)/d*2^(1/2)-1 /4*ln(b^(1/2)+2^(1/2)*(b*tan(d*x+c))^(1/2)+b^(1/2)*tan(d*x+c))*b^(1/2)/d*2 ^(1/2)
Time = 0.06 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.37 \[ \int \sqrt {b \tan (c+d x)} \, dx=\frac {\left (\arctan \left (\sqrt [4]{-\tan ^2(c+d x)}\right )-\text {arctanh}\left (\sqrt [4]{-\tan ^2(c+d x)}\right )\right ) \sqrt [4]{-\tan (c+d x)} \sqrt {b \tan (c+d x)}}{d \tan ^{\frac {3}{4}}(c+d x)} \]
((ArcTan[(-Tan[c + d*x]^2)^(1/4)] - ArcTanh[(-Tan[c + d*x]^2)^(1/4)])*(-Ta n[c + d*x])^(1/4)*Sqrt[b*Tan[c + d*x]])/(d*Tan[c + d*x]^(3/4))
Time = 0.39 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.91, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {3042, 3957, 266, 826, 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.
\(\displaystyle \int \sqrt {b \tan (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {b \tan (c+d x)}dx\) |
\(\Big \downarrow \) 3957 |
\(\displaystyle \frac {b \int \frac {\sqrt {b \tan (c+d x)}}{\tan ^2(c+d x) b^2+b^2}d(b \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 b \int \frac {b^2 \tan ^2(c+d x)}{b^4 \tan ^4(c+d x)+b^2}d\sqrt {b \tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 826 |
\(\displaystyle \frac {2 b \left (\frac {1}{2} \int \frac {b^2 \tan ^2(c+d x)+b}{b^4 \tan ^4(c+d x)+b^2}d\sqrt {b \tan (c+d x)}-\frac {1}{2} \int \frac {b-b^2 \tan ^2(c+d x)}{b^4 \tan ^4(c+d x)+b^2}d\sqrt {b \tan (c+d x)}\right )}{d}\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle \frac {2 b \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{b^2 \tan ^2(c+d x)-\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}+\frac {1}{2} \int \frac {1}{b^2 \tan ^2(c+d x)+\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}\right )-\frac {1}{2} \int \frac {b-b^2 \tan ^2(c+d x)}{b^4 \tan ^4(c+d x)+b^2}d\sqrt {b \tan (c+d x)}\right )}{d}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {2 b \left (\frac {1}{2} \left (\frac {\int \frac {1}{-b^2 \tan ^2(c+d x)-1}d\left (1-\sqrt {2} \sqrt {b} \tan (c+d x)\right )}{\sqrt {2} \sqrt {b}}-\frac {\int \frac {1}{-b^2 \tan ^2(c+d x)-1}d\left (\sqrt {2} \sqrt {b} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {b}}\right )-\frac {1}{2} \int \frac {b-b^2 \tan ^2(c+d x)}{b^4 \tan ^4(c+d x)+b^2}d\sqrt {b \tan (c+d x)}\right )}{d}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {2 b \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {b} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {b}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {b} \tan (c+d x)\right )}{\sqrt {2} \sqrt {b}}\right )-\frac {1}{2} \int \frac {b-b^2 \tan ^2(c+d x)}{b^4 \tan ^4(c+d x)+b^2}d\sqrt {b \tan (c+d x)}\right )}{d}\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle \frac {2 b \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2} \sqrt {b}-2 \sqrt {b \tan (c+d x)}}{b^2 \tan ^2(c+d x)-\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}}{2 \sqrt {2} \sqrt {b}}+\frac {\int -\frac {\sqrt {2} \left (\sqrt {b}+\sqrt {2} \sqrt {b \tan (c+d x)}\right )}{b^2 \tan ^2(c+d x)+\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}}{2 \sqrt {2} \sqrt {b}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {b} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {b}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {b} \tan (c+d x)\right )}{\sqrt {2} \sqrt {b}}\right )\right )}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 b \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {b}-2 \sqrt {b \tan (c+d x)}}{b^2 \tan ^2(c+d x)-\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}}{2 \sqrt {2} \sqrt {b}}-\frac {\int \frac {\sqrt {2} \left (\sqrt {b}+\sqrt {2} \sqrt {b \tan (c+d x)}\right )}{b^2 \tan ^2(c+d x)+\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}}{2 \sqrt {2} \sqrt {b}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {b} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {b}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {b} \tan (c+d x)\right )}{\sqrt {2} \sqrt {b}}\right )\right )}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 b \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {b}-2 \sqrt {b \tan (c+d x)}}{b^2 \tan ^2(c+d x)-\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}}{2 \sqrt {2} \sqrt {b}}-\frac {\int \frac {\sqrt {b}+\sqrt {2} \sqrt {b \tan (c+d x)}}{b^2 \tan ^2(c+d x)+\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}}{2 \sqrt {b}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {b} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {b}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {b} \tan (c+d x)\right )}{\sqrt {2} \sqrt {b}}\right )\right )}{d}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {2 b \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {b} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {b}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {b} \tan (c+d x)\right )}{\sqrt {2} \sqrt {b}}\right )+\frac {1}{2} \left (\frac {\log \left (-\sqrt {2} b^{3/2} \tan (c+d x)+b^2 \tan ^2(c+d x)+b\right )}{2 \sqrt {2} \sqrt {b}}-\frac {\log \left (\sqrt {2} b^{3/2} \tan (c+d x)+b^2 \tan ^2(c+d x)+b\right )}{2 \sqrt {2} \sqrt {b}}\right )\right )}{d}\) |
(2*b*((-(ArcTan[1 - Sqrt[2]*Sqrt[b]*Tan[c + d*x]]/(Sqrt[2]*Sqrt[b])) + Arc Tan[1 + Sqrt[2]*Sqrt[b]*Tan[c + d*x]]/(Sqrt[2]*Sqrt[b]))/2 + (Log[b - Sqrt [2]*b^(3/2)*Tan[c + d*x] + b^2*Tan[c + d*x]^2]/(2*Sqrt[2]*Sqrt[b]) - Log[b + Sqrt[2]*b^(3/2)*Tan[c + d*x] + b^2*Tan[c + d*x]^2]/(2*Sqrt[2]*Sqrt[b])) /2))/d
3.1.12.3.1 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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s) Int[(r + s*x^2)/(a + b*x^ 4), x], x] - Simp[1/(2*s) Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, 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[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d Subst[Int [x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
Time = 0.17 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {b \sqrt {2}\, \left (\ln \left (\frac {b \tan \left (d x +c \right )-\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}{b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d \left (b^{2}\right )^{\frac {1}{4}}}\) | \(136\) |
default | \(\frac {b \sqrt {2}\, \left (\ln \left (\frac {b \tan \left (d x +c \right )-\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}{b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d \left (b^{2}\right )^{\frac {1}{4}}}\) | \(136\) |
1/4/d*b/(b^2)^(1/4)*2^(1/2)*(ln((b*tan(d*x+c)-(b^2)^(1/4)*(b*tan(d*x+c))^( 1/2)*2^(1/2)+(b^2)^(1/2))/(b*tan(d*x+c)+(b^2)^(1/4)*(b*tan(d*x+c))^(1/2)*2 ^(1/2)+(b^2)^(1/2)))+2*arctan(2^(1/2)/(b^2)^(1/4)*(b*tan(d*x+c))^(1/2)+1)- 2*arctan(-2^(1/2)/(b^2)^(1/4)*(b*tan(d*x+c))^(1/2)+1))
Result contains complex when optimal does not.
Time = 0.23 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.85 \[ \int \sqrt {b \tan (c+d x)} \, dx=\frac {1}{2} \, \left (-\frac {b^{2}}{d^{4}}\right )^{\frac {1}{4}} \log \left (d^{3} \left (-\frac {b^{2}}{d^{4}}\right )^{\frac {3}{4}} + \sqrt {b \tan \left (d x + c\right )} b\right ) - \frac {1}{2} i \, \left (-\frac {b^{2}}{d^{4}}\right )^{\frac {1}{4}} \log \left (i \, d^{3} \left (-\frac {b^{2}}{d^{4}}\right )^{\frac {3}{4}} + \sqrt {b \tan \left (d x + c\right )} b\right ) + \frac {1}{2} i \, \left (-\frac {b^{2}}{d^{4}}\right )^{\frac {1}{4}} \log \left (-i \, d^{3} \left (-\frac {b^{2}}{d^{4}}\right )^{\frac {3}{4}} + \sqrt {b \tan \left (d x + c\right )} b\right ) - \frac {1}{2} \, \left (-\frac {b^{2}}{d^{4}}\right )^{\frac {1}{4}} \log \left (-d^{3} \left (-\frac {b^{2}}{d^{4}}\right )^{\frac {3}{4}} + \sqrt {b \tan \left (d x + c\right )} b\right ) \]
1/2*(-b^2/d^4)^(1/4)*log(d^3*(-b^2/d^4)^(3/4) + sqrt(b*tan(d*x + c))*b) - 1/2*I*(-b^2/d^4)^(1/4)*log(I*d^3*(-b^2/d^4)^(3/4) + sqrt(b*tan(d*x + c))*b ) + 1/2*I*(-b^2/d^4)^(1/4)*log(-I*d^3*(-b^2/d^4)^(3/4) + sqrt(b*tan(d*x + c))*b) - 1/2*(-b^2/d^4)^(1/4)*log(-d^3*(-b^2/d^4)^(3/4) + sqrt(b*tan(d*x + c))*b)
\[ \int \sqrt {b \tan (c+d x)} \, dx=\int \sqrt {b \tan {\left (c + d x \right )}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.80 \[ \int \sqrt {b \tan (c+d x)} \, dx=\frac {b {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {b} + 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {b}}\right )}{\sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {b} - 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {b}}\right )}{\sqrt {b}} - \frac {\sqrt {2} \log \left (b \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {b} + b\right )}{\sqrt {b}} + \frac {\sqrt {2} \log \left (b \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {b} + b\right )}{\sqrt {b}}\right )}}{4 \, d} \]
1/4*b*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(b) + 2*sqrt(b*tan(d*x + c)))/sqrt(b))/sqrt(b) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(b) - 2 *sqrt(b*tan(d*x + c)))/sqrt(b))/sqrt(b) - sqrt(2)*log(b*tan(d*x + c) + sqr t(2)*sqrt(b*tan(d*x + c))*sqrt(b) + b)/sqrt(b) + sqrt(2)*log(b*tan(d*x + c ) - sqrt(2)*sqrt(b*tan(d*x + c))*sqrt(b) + b)/sqrt(b))/d
Time = 0.37 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.92 \[ \int \sqrt {b \tan (c+d x)} \, dx=\frac {\frac {2 \, \sqrt {2} {\left | b \right |}^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | b \right |}} + 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {{\left | b \right |}}}\right )}{d} + \frac {2 \, \sqrt {2} {\left | b \right |}^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | b \right |}} - 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {{\left | b \right |}}}\right )}{d} - \frac {\sqrt {2} {\left | b \right |}^{\frac {3}{2}} \log \left (b \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {{\left | b \right |}} + {\left | b \right |}\right )}{d} + \frac {\sqrt {2} {\left | b \right |}^{\frac {3}{2}} \log \left (b \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {{\left | b \right |}} + {\left | b \right |}\right )}{d}}{4 \, b} \]
1/4*(2*sqrt(2)*abs(b)^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(b)) + 2*s qrt(b*tan(d*x + c)))/sqrt(abs(b)))/d + 2*sqrt(2)*abs(b)^(3/2)*arctan(-1/2* sqrt(2)*(sqrt(2)*sqrt(abs(b)) - 2*sqrt(b*tan(d*x + c)))/sqrt(abs(b)))/d - sqrt(2)*abs(b)^(3/2)*log(b*tan(d*x + c) + sqrt(2)*sqrt(b*tan(d*x + c))*sqr t(abs(b)) + abs(b))/d + sqrt(2)*abs(b)^(3/2)*log(b*tan(d*x + c) - sqrt(2)* sqrt(b*tan(d*x + c))*sqrt(abs(b)) + abs(b))/d)/b
Time = 2.77 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.26 \[ \int \sqrt {b \tan (c+d x)} \, dx=\frac {{\left (-1\right )}^{1/4}\,\sqrt {b}\,\left (\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {b}}\right )-\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {b}}\right )\right )}{d} \]