3.12 \(\int x (a+b \csc (c+d x^2))^2 \, dx\)

Optimal. Leaf size=45 \[ \frac{a^2 x^2}{2}-\frac{a b \tanh ^{-1}\left (\cos \left (c+d x^2\right )\right )}{d}-\frac{b^2 \cot \left (c+d x^2\right )}{2 d} \]

[Out]

(a^2*x^2)/2 - (a*b*ArcTanh[Cos[c + d*x^2]])/d - (b^2*Cot[c + d*x^2])/(2*d)

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Rubi [A]  time = 0.0508865, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {4205, 3773, 3770, 3767, 8} \[ \frac{a^2 x^2}{2}-\frac{a b \tanh ^{-1}\left (\cos \left (c+d x^2\right )\right )}{d}-\frac{b^2 \cot \left (c+d x^2\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*Csc[c + d*x^2])^2,x]

[Out]

(a^2*x^2)/2 - (a*b*ArcTanh[Cos[c + d*x^2]])/d - (b^2*Cot[c + d*x^2])/(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 3773

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int x \left (a+b \csc \left (c+d x^2\right )\right )^2 \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int (a+b \csc (c+d x))^2 \, dx,x,x^2\right )\\ &=\frac{a^2 x^2}{2}+(a b) \operatorname{Subst}\left (\int \csc (c+d x) \, dx,x,x^2\right )+\frac{1}{2} b^2 \operatorname{Subst}\left (\int \csc ^2(c+d x) \, dx,x,x^2\right )\\ &=\frac{a^2 x^2}{2}-\frac{a b \tanh ^{-1}\left (\cos \left (c+d x^2\right )\right )}{d}-\frac{b^2 \operatorname{Subst}\left (\int 1 \, dx,x,\cot \left (c+d x^2\right )\right )}{2 d}\\ &=\frac{a^2 x^2}{2}-\frac{a b \tanh ^{-1}\left (\cos \left (c+d x^2\right )\right )}{d}-\frac{b^2 \cot \left (c+d x^2\right )}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.396546, size = 86, normalized size = 1.91 \[ \frac{2 a \left (a c+a d x^2+2 b \log \left (\sin \left (\frac{1}{2} \left (c+d x^2\right )\right )\right )-2 b \log \left (\cos \left (\frac{1}{2} \left (c+d x^2\right )\right )\right )\right )+b^2 \tan \left (\frac{1}{2} \left (c+d x^2\right )\right )+b^2 \left (-\cot \left (\frac{1}{2} \left (c+d x^2\right )\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b^2*Cot[(c + d*x^2)/2]) + 2*a*(a*c + a*d*x^2 - 2*b*Log[Cos[(c + d*x^2)/2]] + 2*b*Log[Sin[(c + d*x^2)/2]]) +
b^2*Tan[(c + d*x^2)/2])/(4*d)

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Maple [A]  time = 0.03, size = 61, normalized size = 1.4 \begin{align*}{\frac{{a}^{2}{x}^{2}}{2}}-{\frac{{b}^{2}\cot \left ( d{x}^{2}+c \right ) }{2\,d}}+{\frac{ab\ln \left ( \csc \left ( d{x}^{2}+c \right ) -\cot \left ( d{x}^{2}+c \right ) \right ) }{d}}+{\frac{{a}^{2}c}{2\,d}} \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]

1/2*a^2*x^2-1/2*b^2*cot(d*x^2+c)/d+1/d*a*b*ln(csc(d*x^2+c)-cot(d*x^2+c))+1/2/d*a^2*c

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Maxima [B]  time = 0.95805, size = 132, normalized size = 2.93 \begin{align*} \frac{1}{2} \, a^{2} x^{2} - \frac{a b \log \left (\cot \left (d x^{2} + c\right ) + \csc \left (d x^{2} + c\right )\right )}{d} - \frac{b^{2} \sin \left (2 \, d x^{2} + 2 \, c\right )}{d \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + d \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} - 2 \, d \cos \left (2 \, d x^{2} + 2 \, c\right ) + 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="maxima")

[Out]

1/2*a^2*x^2 - a*b*log(cot(d*x^2 + c) + csc(d*x^2 + c))/d - b^2*sin(2*d*x^2 + 2*c)/(d*cos(2*d*x^2 + 2*c)^2 + d*
sin(2*d*x^2 + 2*c)^2 - 2*d*cos(2*d*x^2 + 2*c) + d)

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Fricas [B]  time = 0.504679, size = 236, normalized size = 5.24 \begin{align*} \frac{a^{2} d x^{2} \sin \left (d x^{2} + c\right ) - a b \log \left (\frac{1}{2} \, \cos \left (d x^{2} + c\right ) + \frac{1}{2}\right ) \sin \left (d x^{2} + c\right ) + a b \log \left (-\frac{1}{2} \, \cos \left (d x^{2} + c\right ) + \frac{1}{2}\right ) \sin \left (d x^{2} + c\right ) - b^{2} \cos \left (d x^{2} + c\right )}{2 \, d \sin \left (d x^{2} + c\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]

1/2*(a^2*d*x^2*sin(d*x^2 + c) - a*b*log(1/2*cos(d*x^2 + c) + 1/2)*sin(d*x^2 + c) + a*b*log(-1/2*cos(d*x^2 + c)
+ 1/2)*sin(d*x^2 + c) - b^2*cos(d*x^2 + c))/(d*sin(d*x^2 + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int 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 [B]  time = 1.18247, size = 113, normalized size = 2.51 \begin{align*} \frac{2 \,{\left (d x^{2} + c\right )} a^{2} + 4 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, d x^{2} + \frac{1}{2} \, c\right ) \right |}\right ) + b^{2} \tan \left (\frac{1}{2} \, d x^{2} + \frac{1}{2} \, c\right ) - \frac{4 \, a b \tan \left (\frac{1}{2} \, d x^{2} + \frac{1}{2} \, c\right ) + b^{2}}{\tan \left (\frac{1}{2} \, d x^{2} + \frac{1}{2} \, c\right )}}{4 \, 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]

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