Integrand size = 22, antiderivative size = 66 \[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )} \, dx=\frac {2 \sqrt {x}}{a}+\frac {4 b \text {arctanh}\left (\frac {a+b \tan \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d} \]
Time = 0.43 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )} \, dx=\frac {2 \left (\frac {c}{d}+\sqrt {x}-\frac {2 b \arctan \left (\frac {a+b \tan \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2} d}\right )}{a} \]
(2*(c/d + Sqrt[x] - (2*b*ArcTan[(a + b*Tan[(c + d*Sqrt[x])/2])/Sqrt[-a^2 + b^2]])/(Sqrt[-a^2 + b^2]*d)))/a
Time = 0.33 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {4693, 3042, 4270, 3042, 3139, 1083, 219}
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.
\(\displaystyle \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )} \, dx\) |
\(\Big \downarrow \) 4693 |
\(\displaystyle 2 \int \frac {1}{a+b \csc \left (c+d \sqrt {x}\right )}d\sqrt {x}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 \int \frac {1}{a+b \csc \left (c+d \sqrt {x}\right )}d\sqrt {x}\) |
\(\Big \downarrow \) 4270 |
\(\displaystyle 2 \left (\frac {\sqrt {x}}{a}-\frac {\int \frac {1}{\frac {a \sin \left (c+d \sqrt {x}\right )}{b}+1}d\sqrt {x}}{a}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 \left (\frac {\sqrt {x}}{a}-\frac {\int \frac {1}{\frac {a \sin \left (c+d \sqrt {x}\right )}{b}+1}d\sqrt {x}}{a}\right )\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle 2 \left (\frac {\sqrt {x}}{a}-\frac {2 \int \frac {1}{x+\frac {2 a \tan \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{b}+1}d\tan \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{a d}\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle 2 \left (\frac {4 \int \frac {1}{-4 \left (1-\frac {a^2}{b^2}\right )-x}d\left (\frac {2 a}{b}+2 \tan \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )\right )}{a d}+\frac {\sqrt {x}}{a}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 2 \left (\frac {2 b \text {arctanh}\left (\frac {b \left (\frac {2 a}{b}+2 \tan \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )\right )}{2 \sqrt {a^2-b^2}}\right )}{a d \sqrt {a^2-b^2}}+\frac {\sqrt {x}}{a}\right )\) |
2*(Sqrt[x]/a + (2*b*ArcTanh[(b*((2*a)/b + 2*Tan[(c + d*Sqrt[x])/2]))/(2*Sq rt[a^2 - b^2])])/(a*Sqrt[a^2 - b^2]*d))
3.1.64.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ [a^2 - b^2, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Simp[1/a Int[1/(1 + (a/b)*Sin[c + d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[((a_.) + Csc[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Csc[c + d*x])^ p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m + 1)/n], 0] && IntegerQ[p]
Time = 0.35 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.11
method | result | size |
derivativedivides | \(\frac {\frac {4 \arctan \left (\tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )\right )}{a}-\frac {4 b \arctan \left (\frac {2 b \tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a \sqrt {-a^{2}+b^{2}}}}{d}\) | \(73\) |
default | \(\frac {\frac {4 \arctan \left (\tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )\right )}{a}-\frac {4 b \arctan \left (\frac {2 b \tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a \sqrt {-a^{2}+b^{2}}}}{d}\) | \(73\) |
2/d*(2/a*arctan(tan(1/2*c+1/2*d*x^(1/2)))-2*b/a/(-a^2+b^2)^(1/2)*arctan(1/ 2*(2*b*tan(1/2*c+1/2*d*x^(1/2))+2*a)/(-a^2+b^2)^(1/2)))
Time = 0.28 (sec) , antiderivative size = 275, normalized size of antiderivative = 4.17 \[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )} \, dx=\left [\frac {2 \, {\left (a^{2} - b^{2}\right )} d \sqrt {x} + \sqrt {a^{2} - b^{2}} b \log \left (\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d \sqrt {x} + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} a \cos \left (d \sqrt {x} + c\right ) + a^{2} + b^{2} + 2 \, {\left (\sqrt {a^{2} - b^{2}} b \cos \left (d \sqrt {x} + c\right ) + a b\right )} \sin \left (d \sqrt {x} + c\right )}{a^{2} \cos \left (d \sqrt {x} + c\right )^{2} - 2 \, a b \sin \left (d \sqrt {x} + c\right ) - a^{2} - b^{2}}\right )}{{\left (a^{3} - a b^{2}\right )} d}, \frac {2 \, {\left ({\left (a^{2} - b^{2}\right )} d \sqrt {x} + \sqrt {-a^{2} + b^{2}} b \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} b \sin \left (d \sqrt {x} + c\right ) + \sqrt {-a^{2} + b^{2}} a}{{\left (a^{2} - b^{2}\right )} \cos \left (d \sqrt {x} + c\right )}\right )\right )}}{{\left (a^{3} - a b^{2}\right )} d}\right ] \]
[(2*(a^2 - b^2)*d*sqrt(x) + sqrt(a^2 - b^2)*b*log(((a^2 - 2*b^2)*cos(d*sqr t(x) + c)^2 + 2*sqrt(a^2 - b^2)*a*cos(d*sqrt(x) + c) + a^2 + b^2 + 2*(sqrt (a^2 - b^2)*b*cos(d*sqrt(x) + c) + a*b)*sin(d*sqrt(x) + c))/(a^2*cos(d*sqr t(x) + c)^2 - 2*a*b*sin(d*sqrt(x) + c) - a^2 - b^2)))/((a^3 - a*b^2)*d), 2 *((a^2 - b^2)*d*sqrt(x) + sqrt(-a^2 + b^2)*b*arctan(-(sqrt(-a^2 + b^2)*b*s in(d*sqrt(x) + c) + sqrt(-a^2 + b^2)*a)/((a^2 - b^2)*cos(d*sqrt(x) + c)))) /((a^3 - a*b^2)*d)]
\[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )} \, dx=\int \frac {1}{\sqrt {x} \left (a + b \csc {\left (c + d \sqrt {x} \right )}\right )}\, dx \]
Exception generated. \[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Time = 0.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.27 \[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )} \, dx=-\frac {4 \, {\left (\pi \left \lfloor \frac {d \sqrt {x} + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, d \sqrt {x} + \frac {1}{2} \, c\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )} b}{\sqrt {-a^{2} + b^{2}} a d} + \frac {2 \, {\left (d \sqrt {x} + c\right )}}{a d} \]
-4*(pi*floor(1/2*(d*sqrt(x) + c)/pi + 1/2)*sgn(b) + arctan((b*tan(1/2*d*sq rt(x) + 1/2*c) + a)/sqrt(-a^2 + b^2)))*b/(sqrt(-a^2 + b^2)*a*d) + 2*(d*sqr t(x) + c)/(a*d)
Time = 19.52 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.41 \[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )} \, dx=\frac {2\,\sqrt {x}}{a}-\frac {2\,b\,\ln \left (b\,{\mathrm {e}}^{d\,\sqrt {x}\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,2{}\mathrm {i}-\frac {2\,b\,\left (a\,1{}\mathrm {i}+b\,{\mathrm {e}}^{d\,\sqrt {x}\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\right )}{\sqrt {a+b}\,\sqrt {a-b}}\right )}{a\,d\,\sqrt {a+b}\,\sqrt {a-b}}+\frac {2\,b\,\ln \left (b\,{\mathrm {e}}^{d\,\sqrt {x}\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,2{}\mathrm {i}+\frac {2\,b\,\left (a\,1{}\mathrm {i}+b\,{\mathrm {e}}^{d\,\sqrt {x}\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\right )}{\sqrt {a+b}\,\sqrt {a-b}}\right )}{a\,d\,\sqrt {a+b}\,\sqrt {a-b}} \]
(2*x^(1/2))/a - (2*b*log(b*exp(d*x^(1/2)*1i)*exp(c*1i)*2i - (2*b*(a*1i + b *exp(d*x^(1/2)*1i)*exp(c*1i)))/((a + b)^(1/2)*(a - b)^(1/2))))/(a*d*(a + b )^(1/2)*(a - b)^(1/2)) + (2*b*log(b*exp(d*x^(1/2)*1i)*exp(c*1i)*2i + (2*b* (a*1i + b*exp(d*x^(1/2)*1i)*exp(c*1i)))/((a + b)^(1/2)*(a - b)^(1/2))))/(a *d*(a + b)^(1/2)*(a - b)^(1/2))