\(\int \frac {1}{\sqrt {x} (a+b \csc (c+d \sqrt {x}))} \, dx\) [64]

Optimal result

Integrand size = 22, antiderivative size = 66 \[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )} \, dx=\frac {2 \sqrt {x}}{a}+\frac {4 b \text {arctanh}\left (\frac {a+b \tan \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d} \]

[Out]

4*b*arctanh((a+b*tan(1/2*c+1/2*d*x^(1/2)))/(a^2-b^2)^(1/2))/a/d/(a^2-b^2)^(1/2)+2*x^(1/2)/a

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {4290, 3868, 2739, 632, 212} \[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )} \, dx=\frac {4 b \text {arctanh}\left (\frac {a+b \tan \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{\sqrt {a^2-b^2}}\right )}{a d \sqrt {a^2-b^2}}+\frac {2 \sqrt {x}}{a} \]

[In]

Int[1/(Sqrt[x]*(a + b*Csc[c + d*Sqrt[x]])),x]

[Out]

(2*Sqrt[x])/a + (4*b*ArcTanh[(a + b*Tan[(c + d*Sqrt[x])/2])/Sqrt[a^2 - b^2]])/(a*Sqrt[a^2 - b^2]*d)

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[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]

Rule 3868

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Dist[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]

Rule 4290

Int[((a_.) + Csc[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(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]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {1}{a+b \csc (c+d x)} \, dx,x,\sqrt {x}\right ) \\ & = \frac {2 \sqrt {x}}{a}-\frac {2 \text {Subst}\left (\int \frac {1}{1+\frac {a \sin (c+d x)}{b}} \, dx,x,\sqrt {x}\right )}{a} \\ & = \frac {2 \sqrt {x}}{a}-\frac {4 \text {Subst}\left (\int \frac {1}{1+\frac {2 a x}{b}+x^2} \, dx,x,\tan \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )\right )}{a d} \\ & = \frac {2 \sqrt {x}}{a}+\frac {8 \text {Subst}\left (\int \frac {1}{-4 \left (1-\frac {a^2}{b^2}\right )-x^2} \, dx,x,\frac {2 a}{b}+2 \tan \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )\right )}{a d} \\ & = \frac {2 \sqrt {x}}{a}+\frac {4 b \text {arctanh}\left (\frac {b \left (\frac {a}{b}+\tan \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )\right )}{\sqrt {a^2-b^2}}\right )}{a \sqrt {a^2-b^2} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )} \, dx=\frac {2 \left (\frac {c}{d}+\sqrt {x}-\frac {2 b \arctan \left (\frac {a+b \tan \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2} d}\right )}{a} \]

[In]

Integrate[1/(Sqrt[x]*(a + b*Csc[c + d*Sqrt[x]])),x]

[Out]

(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

Maple [A] (verified)

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	\(\frac {\frac {4 \arctan \left (\tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )\right )}{a}-\frac {4 b \arctan \left (\frac {2 b \tan \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a \sqrt {-a^{2}+b^{2}}}}{d}\)	\(73\)

[In]

int(1/(a+b*csc(c+d*x^(1/2)))/x^(1/2),x,method=_RETURNVERBOSE)

[Out]

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)))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 275, normalized size of antiderivative = 4.17 \[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )} \, dx=\left [\frac {2 \, {\left (a^{2} - b^{2}\right )} d \sqrt {x} + \sqrt {a^{2} - b^{2}} b \log \left (\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d \sqrt {x} + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} a \cos \left (d \sqrt {x} + c\right ) + a^{2} + b^{2} + 2 \, {\left (\sqrt {a^{2} - b^{2}} b \cos \left (d \sqrt {x} + c\right ) + a b\right )} \sin \left (d \sqrt {x} + c\right )}{a^{2} \cos \left (d \sqrt {x} + c\right )^{2} - 2 \, a b \sin \left (d \sqrt {x} + c\right ) - a^{2} - b^{2}}\right )}{{\left (a^{3} - a b^{2}\right )} d}, \frac {2 \, {\left ({\left (a^{2} - b^{2}\right )} d \sqrt {x} + \sqrt {-a^{2} + b^{2}} b \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} b \sin \left (d \sqrt {x} + c\right ) + \sqrt {-a^{2} + b^{2}} a}{{\left (a^{2} - b^{2}\right )} \cos \left (d \sqrt {x} + c\right )}\right )\right )}}{{\left (a^{3} - a b^{2}\right )} d}\right ] \]

[In]

integrate(1/(a+b*csc(c+d*x^(1/2)))/x^(1/2),x, algorithm="fricas")

[Out]

[(2*(a^2 - b^2)*d*sqrt(x) + sqrt(a^2 - b^2)*b*log(((a^2 - 2*b^2)*cos(d*sqrt(x) + c)^2 + 2*sqrt(a^2 - b^2)*a*co
s(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*s
qrt(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*sin(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)]

Sympy [F]

\[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )} \, dx=\int \frac {1}{\sqrt {x} \left (a + b \csc {\left (c + d \sqrt {x} \right )}\right )}\, dx \]

[In]

integrate(1/(a+b*csc(c+d*x**(1/2)))/x**(1/2),x)

[Out]

Integral(1/(sqrt(x)*(a + b*csc(c + d*sqrt(x)))), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(a+b*csc(c+d*x^(1/2)))/x^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.27 \[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )} \, dx=-\frac {4 \, {\left (\pi \left \lfloor \frac {d \sqrt {x} + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, d \sqrt {x} + \frac {1}{2} \, c\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )} b}{\sqrt {-a^{2} + b^{2}} a d} + \frac {2 \, {\left (d \sqrt {x} + c\right )}}{a d} \]

[In]

integrate(1/(a+b*csc(c+d*x^(1/2)))/x^(1/2),x, algorithm="giac")

[Out]

-4*(pi*floor(1/2*(d*sqrt(x) + c)/pi + 1/2)*sgn(b) + arctan((b*tan(1/2*d*sqrt(x) + 1/2*c) + a)/sqrt(-a^2 + b^2)
))*b/(sqrt(-a^2 + b^2)*a*d) + 2*(d*sqrt(x) + c)/(a*d)

Mupad [B] (verification not implemented)

Time = 19.52 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.41 \[ \int \frac {1}{\sqrt {x} \left (a+b \csc \left (c+d \sqrt {x}\right )\right )} \, dx=\frac {2\,\sqrt {x}}{a}-\frac {2\,b\,\ln \left (b\,{\mathrm {e}}^{d\,\sqrt {x}\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,2{}\mathrm {i}-\frac {2\,b\,\left (a\,1{}\mathrm {i}+b\,{\mathrm {e}}^{d\,\sqrt {x}\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\right )}{\sqrt {a+b}\,\sqrt {a-b}}\right )}{a\,d\,\sqrt {a+b}\,\sqrt {a-b}}+\frac {2\,b\,\ln \left (b\,{\mathrm {e}}^{d\,\sqrt {x}\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,2{}\mathrm {i}+\frac {2\,b\,\left (a\,1{}\mathrm {i}+b\,{\mathrm {e}}^{d\,\sqrt {x}\,1{}\mathrm {i}}\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\right )}{\sqrt {a+b}\,\sqrt {a-b}}\right )}{a\,d\,\sqrt {a+b}\,\sqrt {a-b}} \]

[In]

int(1/(x^(1/2)*(a + b/sin(c + d*x^(1/2)))),x)

[Out]

(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))