3.27 \(\int \frac{x}{(a+b \csc (c+d x^2))^2} \, dx\)

Optimal. Leaf size=120 \[ \frac{b \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} \left (c+d x^2\right )\right )}{\sqrt{a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{3/2}}-\frac{b^2 \cot \left (c+d x^2\right )}{2 a d \left (a^2-b^2\right ) \left (a+b \csc \left (c+d x^2\right )\right )}+\frac{x^2}{2 a^2} \]

[Out]

x^2/(2*a^2) + (b*(2*a^2 - b^2)*ArcTanh[(a + b*Tan[(c + d*x^2)/2])/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2)^(3/2)*d)
- (b^2*Cot[c + d*x^2])/(2*a*(a^2 - b^2)*d*(a + b*Csc[c + d*x^2]))

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Rubi [A]  time = 0.241583, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {4205, 3785, 3919, 3831, 2660, 618, 206} \[ \frac{b \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} \left (c+d x^2\right )\right )}{\sqrt{a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{3/2}}-\frac{b^2 \cot \left (c+d x^2\right )}{2 a d \left (a^2-b^2\right ) \left (a+b \csc \left (c+d x^2\right )\right )}+\frac{x^2}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*Csc[c + d*x^2])^2,x]

[Out]

x^2/(2*a^2) + (b*(2*a^2 - b^2)*ArcTanh[(a + b*Tan[(c + d*x^2)/2])/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2)^(3/2)*d)
- (b^2*Cot[c + d*x^2])/(2*a*(a^2 - b^2)*d*(a + b*Csc[c + d*x^2]))

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 3785

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

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, 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{x}{\left (a+b \csc \left (c+d x^2\right )\right )^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{(a+b \csc (c+d x))^2} \, dx,x,x^2\right )\\ &=-\frac{b^2 \cot \left (c+d x^2\right )}{2 a \left (a^2-b^2\right ) d \left (a+b \csc \left (c+d x^2\right )\right )}-\frac{\operatorname{Subst}\left (\int \frac{-a^2+b^2+a b \csc (c+d x)}{a+b \csc (c+d x)} \, dx,x,x^2\right )}{2 a \left (a^2-b^2\right )}\\ &=\frac{x^2}{2 a^2}-\frac{b^2 \cot \left (c+d x^2\right )}{2 a \left (a^2-b^2\right ) d \left (a+b \csc \left (c+d x^2\right )\right )}-\frac{\left (b \left (2 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{\csc (c+d x)}{a+b \csc (c+d x)} \, dx,x,x^2\right )}{2 a^2 \left (a^2-b^2\right )}\\ &=\frac{x^2}{2 a^2}-\frac{b^2 \cot \left (c+d x^2\right )}{2 a \left (a^2-b^2\right ) d \left (a+b \csc \left (c+d x^2\right )\right )}-\frac{\left (2 a^2-b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a \sin (c+d x)}{b}} \, dx,x,x^2\right )}{2 a^2 \left (a^2-b^2\right )}\\ &=\frac{x^2}{2 a^2}-\frac{b^2 \cot \left (c+d x^2\right )}{2 a \left (a^2-b^2\right ) d \left (a+b \csc \left (c+d x^2\right )\right )}-\frac{\left (2 a^2-b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{2 a x}{b}+x^2} \, dx,x,\tan \left (\frac{1}{2} \left (c+d x^2\right )\right )\right )}{a^2 \left (a^2-b^2\right ) d}\\ &=\frac{x^2}{2 a^2}-\frac{b^2 \cot \left (c+d x^2\right )}{2 a \left (a^2-b^2\right ) d \left (a+b \csc \left (c+d x^2\right )\right )}+\frac{\left (2 \left (2 a^2-b^2\right )\right ) \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 x^2\right )\right )\right )}{a^2 \left (a^2-b^2\right ) d}\\ &=\frac{x^2}{2 a^2}+\frac{b \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{b \left (\frac{a}{b}+\tan \left (\frac{1}{2} \left (c+d x^2\right )\right )\right )}{\sqrt{a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{3/2} d}-\frac{b^2 \cot \left (c+d x^2\right )}{2 a \left (a^2-b^2\right ) d \left (a+b \csc \left (c+d x^2\right )\right )}\\ \end{align*}

Mathematica [A]  time = 0.666935, size = 158, normalized size = 1.32 \[ \frac{\csc \left (c+d x^2\right ) \left (a \sin \left (c+d x^2\right )+b\right ) \left (-\frac{2 b \left (b^2-2 a^2\right ) \tan ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} \left (c+d x^2\right )\right )}{\sqrt{b^2-a^2}}\right ) \left (a+b \csc \left (c+d x^2\right )\right )}{\left (b^2-a^2\right )^{3/2}}+\frac{a b^2 \cot \left (c+d x^2\right )}{(b-a) (a+b)}+\left (c+d x^2\right ) \left (a+b \csc \left (c+d x^2\right )\right )\right )}{2 a^2 d \left (a+b \csc \left (c+d x^2\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*Csc[c + d*x^2])^2,x]

[Out]

(Csc[c + d*x^2]*((a*b^2*Cot[c + d*x^2])/((-a + b)*(a + b)) + (c + d*x^2)*(a + b*Csc[c + d*x^2]) - (2*b*(-2*a^2
+ b^2)*ArcTan[(a + b*Tan[(c + d*x^2)/2])/Sqrt[-a^2 + b^2]]*(a + b*Csc[c + d*x^2]))/(-a^2 + b^2)^(3/2))*(b + a
*Sin[c + d*x^2]))/(2*a^2*d*(a + b*Csc[c + d*x^2])^2)

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Maple [B]  time = 0.086, size = 261, normalized size = 2.2 \begin{align*} -{\frac{b}{d \left ({a}^{2}-{b}^{2} \right ) }\tan \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tan \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}b+2\,a\tan \left ( 1/2\,d{x}^{2}+c/2 \right ) +b \right ) ^{-1}}-{\frac{{b}^{2}}{ad \left ({a}^{2}-{b}^{2} \right ) } \left ( \left ( \tan \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}b+2\,a\tan \left ( 1/2\,d{x}^{2}+c/2 \right ) +b \right ) ^{-1}}-2\,{\frac{b}{d \left ({a}^{2}-{b}^{2} \right ) \sqrt{-{a}^{2}+{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,b\tan \left ( 1/2\,d{x}^{2}+c/2 \right ) +2\,a}{\sqrt{-{a}^{2}+{b}^{2}}}} \right ) }+{\frac{{b}^{3}}{d{a}^{2} \left ({a}^{2}-{b}^{2} \right ) }\arctan \left ({\frac{1}{2} \left ( 2\,b\tan \left ( 1/2\,d{x}^{2}+c/2 \right ) +2\,a \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}} \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}}+{\frac{1}{d{a}^{2}}\arctan \left ( \tan \left ({\frac{d{x}^{2}}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*b/(tan(1/2*d*x^2+1/2*c)^2*b+2*a*tan(1/2*d*x^2+1/2*c)+b)/(a^2-b^2)*tan(1/2*d*x^2+1/2*c)-1/d*b^2/a/(tan(1/2
*d*x^2+1/2*c)^2*b+2*a*tan(1/2*d*x^2+1/2*c)+b)/(a^2-b^2)-2/d*b/(a^2-b^2)/(-a^2+b^2)^(1/2)*arctan(1/2*(2*b*tan(1
/2*d*x^2+1/2*c)+2*a)/(-a^2+b^2)^(1/2))+1/d*b^3/a^2/(a^2-b^2)/(-a^2+b^2)^(1/2)*arctan(1/2*(2*b*tan(1/2*d*x^2+1/
2*c)+2*a)/(-a^2+b^2)^(1/2))+1/d/a^2*arctan(tan(1/2*d*x^2+1/2*c))

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [B]  time = 0.57199, size = 1138, normalized size = 9.48 \begin{align*} \left [\frac{2 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d x^{2} \sin \left (d x^{2} + c\right ) + 2 \,{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d x^{2} +{\left (2 \, a^{2} b^{2} - b^{4} +{\left (2 \, a^{3} b - a b^{3}\right )} \sin \left (d x^{2} + c\right )\right )} \sqrt{a^{2} - b^{2}} \log \left (\frac{{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x^{2} + c\right )^{2} + 2 \, a b \sin \left (d x^{2} + c\right ) + a^{2} + b^{2} + 2 \,{\left (b \cos \left (d x^{2} + c\right ) \sin \left (d x^{2} + c\right ) + a \cos \left (d x^{2} + c\right )\right )} \sqrt{a^{2} - b^{2}}}{a^{2} \cos \left (d x^{2} + c\right )^{2} - 2 \, a b \sin \left (d x^{2} + c\right ) - a^{2} - b^{2}}\right ) - 2 \,{\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x^{2} + c\right )}{4 \,{\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \sin \left (d x^{2} + c\right ) +{\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d\right )}}, \frac{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d x^{2} \sin \left (d x^{2} + c\right ) +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d x^{2} +{\left (2 \, a^{2} b^{2} - b^{4} +{\left (2 \, a^{3} b - a b^{3}\right )} \sin \left (d x^{2} + c\right )\right )} \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \sin \left (d x^{2} + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (d x^{2} + c\right )}\right ) -{\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x^{2} + c\right )}{2 \,{\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \sin \left (d x^{2} + c\right ) +{\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*(a^5 - 2*a^3*b^2 + a*b^4)*d*x^2*sin(d*x^2 + c) + 2*(a^4*b - 2*a^2*b^3 + b^5)*d*x^2 + (2*a^2*b^2 - b^4
+ (2*a^3*b - a*b^3)*sin(d*x^2 + c))*sqrt(a^2 - b^2)*log(((a^2 - 2*b^2)*cos(d*x^2 + c)^2 + 2*a*b*sin(d*x^2 + c)
+ a^2 + b^2 + 2*(b*cos(d*x^2 + c)*sin(d*x^2 + c) + a*cos(d*x^2 + c))*sqrt(a^2 - b^2))/(a^2*cos(d*x^2 + c)^2 -
2*a*b*sin(d*x^2 + c) - a^2 - b^2)) - 2*(a^3*b^2 - a*b^4)*cos(d*x^2 + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*sin(d
*x^2 + c) + (a^6*b - 2*a^4*b^3 + a^2*b^5)*d), 1/2*((a^5 - 2*a^3*b^2 + a*b^4)*d*x^2*sin(d*x^2 + c) + (a^4*b - 2
*a^2*b^3 + b^5)*d*x^2 + (2*a^2*b^2 - b^4 + (2*a^3*b - a*b^3)*sin(d*x^2 + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^
2 + b^2)*(b*sin(d*x^2 + c) + a)/((a^2 - b^2)*cos(d*x^2 + c))) - (a^3*b^2 - a*b^4)*cos(d*x^2 + c))/((a^7 - 2*a^
5*b^2 + a^3*b^4)*d*sin(d*x^2 + c) + (a^6*b - 2*a^4*b^3 + a^2*b^5)*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\left (a + b \csc{\left (c + d x^{2} \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.3411, size = 235, normalized size = 1.96 \begin{align*} -\frac{{\left (2 \, a^{2} b - b^{3}\right )}{\left (\pi \left \lfloor \frac{d x^{2} + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (\frac{1}{2} \, d x^{2} + \frac{1}{2} \, c\right ) + a}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} d - a^{2} b^{2} d\right )} \sqrt{-a^{2} + b^{2}}} - \frac{a b \tan \left (\frac{1}{2} \, d x^{2} + \frac{1}{2} \, c\right ) + b^{2}}{{\left (a^{3} d - a b^{2} d\right )}{\left (b \tan \left (\frac{1}{2} \, d x^{2} + \frac{1}{2} \, c\right )^{2} + 2 \, a \tan \left (\frac{1}{2} \, d x^{2} + \frac{1}{2} \, c\right ) + b\right )}} + \frac{d x^{2} + c}{2 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a^2*b - b^3)*(pi*floor(1/2*(d*x^2 + c)/pi + 1/2)*sgn(b) + arctan((b*tan(1/2*d*x^2 + 1/2*c) + a)/sqrt(-a^2
+ b^2)))/((a^4*d - a^2*b^2*d)*sqrt(-a^2 + b^2)) - (a*b*tan(1/2*d*x^2 + 1/2*c) + b^2)/((a^3*d - a*b^2*d)*(b*tan
(1/2*d*x^2 + 1/2*c)^2 + 2*a*tan(1/2*d*x^2 + 1/2*c) + b)) + 1/2*(d*x^2 + c)/(a^2*d)