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

Optimal. Leaf size=66 \[ \frac{4 b \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )}{\sqrt{a^2-b^2}}\right )}{a d \sqrt{a^2-b^2}}+\frac{2 \sqrt{x}}{a} \]

[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)

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Rubi [A]  time = 0.104868, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {4205, 3783, 2660, 618, 206} \[ \frac{4 b \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )}{\sqrt{a^2-b^2}}\right )}{a d \sqrt{a^2-b^2}}+\frac{2 \sqrt{x}}{a} \]

Antiderivative was successfully verified.

[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 4205

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]

Rule 3783

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

Rule 2660

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 618

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{x} \left (a+b \csc \left (c+d \sqrt{x}\right )\right )} \, dx &=2 \operatorname{Subst}\left (\int \frac{1}{a+b \csc (c+d x)} \, dx,x,\sqrt{x}\right )\\ &=\frac{2 \sqrt{x}}{a}-\frac{2 \operatorname{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 \operatorname{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 \operatorname{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 \tanh ^{-1}\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]  time = 0.218059, size = 68, normalized size = 1.03 \[ \frac{2 \left (-\frac{2 b \tan ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} \left (c+d \sqrt{x}\right )\right )}{\sqrt{b^2-a^2}}\right )}{d \sqrt{b^2-a^2}}+\frac{c}{d}+\sqrt{x}\right )}{a} \]

Antiderivative was successfully verified.

[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

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Maple [A]  time = 0.086, size = 74, normalized size = 1.1 \begin{align*} 4\,{\frac{\arctan \left ( \tan \left ( c/2+1/2\,d\sqrt{x} \right ) \right ) }{ad}}-4\,{\frac{b}{ad\sqrt{-{a}^{2}+{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,b\tan \left ( c/2+1/2\,d\sqrt{x} \right ) +2\,a}{\sqrt{-{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.544666, size = 630, normalized size = 9.55 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x} \left (a + b \csc{\left (c + d \sqrt{x} \right )}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.37713, size = 113, normalized size = 1.71 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)