3.77 \(\int x^2 (d+e x^2) (a+b \csc ^{-1}(c x)) \, dx\)

Optimal. Leaf size=161 \[ \frac{1}{3} d x^3 \left (a+b \csc ^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \csc ^{-1}(c x)\right )+\frac{b x^2 \sqrt{c^2 x^2-1} \left (20 c^2 d+9 e\right )}{120 c^3 \sqrt{c^2 x^2}}+\frac{b x \left (20 c^2 d+9 e\right ) \tanh ^{-1}\left (\frac{c x}{\sqrt{c^2 x^2-1}}\right )}{120 c^4 \sqrt{c^2 x^2}}+\frac{b e x^4 \sqrt{c^2 x^2-1}}{20 c \sqrt{c^2 x^2}} \]

[Out]

(b*(20*c^2*d + 9*e)*x^2*Sqrt[-1 + c^2*x^2])/(120*c^3*Sqrt[c^2*x^2]) + (b*e*x^4*Sqrt[-1 + c^2*x^2])/(20*c*Sqrt[
c^2*x^2]) + (d*x^3*(a + b*ArcCsc[c*x]))/3 + (e*x^5*(a + b*ArcCsc[c*x]))/5 + (b*(20*c^2*d + 9*e)*x*ArcTanh[(c*x
)/Sqrt[-1 + c^2*x^2]])/(120*c^4*Sqrt[c^2*x^2])

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Rubi [A]  time = 0.107796, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {14, 5239, 12, 459, 321, 217, 206} \[ \frac{1}{3} d x^3 \left (a+b \csc ^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \csc ^{-1}(c x)\right )+\frac{b x^2 \sqrt{c^2 x^2-1} \left (20 c^2 d+9 e\right )}{120 c^3 \sqrt{c^2 x^2}}+\frac{b x \left (20 c^2 d+9 e\right ) \tanh ^{-1}\left (\frac{c x}{\sqrt{c^2 x^2-1}}\right )}{120 c^4 \sqrt{c^2 x^2}}+\frac{b e x^4 \sqrt{c^2 x^2-1}}{20 c \sqrt{c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(d + e*x^2)*(a + b*ArcCsc[c*x]),x]

[Out]

(b*(20*c^2*d + 9*e)*x^2*Sqrt[-1 + c^2*x^2])/(120*c^3*Sqrt[c^2*x^2]) + (b*e*x^4*Sqrt[-1 + c^2*x^2])/(20*c*Sqrt[
c^2*x^2]) + (d*x^3*(a + b*ArcCsc[c*x]))/3 + (e*x^5*(a + b*ArcCsc[c*x]))/5 + (b*(20*c^2*d + 9*e)*x*ArcTanh[(c*x
)/Sqrt[-1 + c^2*x^2]])/(120*c^4*Sqrt[c^2*x^2])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 5239

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcCsc[c*x], u, x] + Dist[(b*c*x)/Sqrt[c^2*x^2], Int[SimplifyI
ntegrand[u/(x*Sqrt[c^2*x^2 - 1]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] &&  !(ILtQ
[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[p, 0] && GtQ[m + 2*p + 3, 0])) || (I
LtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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 x^2 \left (d+e x^2\right ) \left (a+b \csc ^{-1}(c x)\right ) \, dx &=\frac{1}{3} d x^3 \left (a+b \csc ^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \csc ^{-1}(c x)\right )+\frac{(b c x) \int \frac{x^2 \left (5 d+3 e x^2\right )}{15 \sqrt{-1+c^2 x^2}} \, dx}{\sqrt{c^2 x^2}}\\ &=\frac{1}{3} d x^3 \left (a+b \csc ^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \csc ^{-1}(c x)\right )+\frac{(b c x) \int \frac{x^2 \left (5 d+3 e x^2\right )}{\sqrt{-1+c^2 x^2}} \, dx}{15 \sqrt{c^2 x^2}}\\ &=\frac{b e x^4 \sqrt{-1+c^2 x^2}}{20 c \sqrt{c^2 x^2}}+\frac{1}{3} d x^3 \left (a+b \csc ^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \csc ^{-1}(c x)\right )-\frac{\left (b c \left (-20 d-\frac{9 e}{c^2}\right ) x\right ) \int \frac{x^2}{\sqrt{-1+c^2 x^2}} \, dx}{60 \sqrt{c^2 x^2}}\\ &=\frac{b \left (20 c^2 d+9 e\right ) x^2 \sqrt{-1+c^2 x^2}}{120 c^3 \sqrt{c^2 x^2}}+\frac{b e x^4 \sqrt{-1+c^2 x^2}}{20 c \sqrt{c^2 x^2}}+\frac{1}{3} d x^3 \left (a+b \csc ^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \csc ^{-1}(c x)\right )-\frac{\left (b \left (-20 d-\frac{9 e}{c^2}\right ) x\right ) \int \frac{1}{\sqrt{-1+c^2 x^2}} \, dx}{120 c \sqrt{c^2 x^2}}\\ &=\frac{b \left (20 c^2 d+9 e\right ) x^2 \sqrt{-1+c^2 x^2}}{120 c^3 \sqrt{c^2 x^2}}+\frac{b e x^4 \sqrt{-1+c^2 x^2}}{20 c \sqrt{c^2 x^2}}+\frac{1}{3} d x^3 \left (a+b \csc ^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \csc ^{-1}(c x)\right )-\frac{\left (b \left (-20 d-\frac{9 e}{c^2}\right ) x\right ) \operatorname{Subst}\left (\int \frac{1}{1-c^2 x^2} \, dx,x,\frac{x}{\sqrt{-1+c^2 x^2}}\right )}{120 c \sqrt{c^2 x^2}}\\ &=\frac{b \left (20 c^2 d+9 e\right ) x^2 \sqrt{-1+c^2 x^2}}{120 c^3 \sqrt{c^2 x^2}}+\frac{b e x^4 \sqrt{-1+c^2 x^2}}{20 c \sqrt{c^2 x^2}}+\frac{1}{3} d x^3 \left (a+b \csc ^{-1}(c x)\right )+\frac{1}{5} e x^5 \left (a+b \csc ^{-1}(c x)\right )+\frac{b \left (20 c^2 d+9 e\right ) x \tanh ^{-1}\left (\frac{c x}{\sqrt{-1+c^2 x^2}}\right )}{120 c^4 \sqrt{c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.166245, size = 121, normalized size = 0.75 \[ \frac{c^2 x^2 \left (8 a c^3 x \left (5 d+3 e x^2\right )+b \sqrt{1-\frac{1}{c^2 x^2}} \left (c^2 \left (20 d+6 e x^2\right )+9 e\right )\right )+b \left (20 c^2 d+9 e\right ) \log \left (x \left (\sqrt{1-\frac{1}{c^2 x^2}}+1\right )\right )+8 b c^5 x^3 \csc ^{-1}(c x) \left (5 d+3 e x^2\right )}{120 c^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*x^2*(8*a*c^3*x*(5*d + 3*e*x^2) + b*Sqrt[1 - 1/(c^2*x^2)]*(9*e + c^2*(20*d + 6*e*x^2))) + 8*b*c^5*x^3*(5*d
+ 3*e*x^2)*ArcCsc[c*x] + b*(20*c^2*d + 9*e)*Log[(1 + Sqrt[1 - 1/(c^2*x^2)])*x])/(120*c^5)

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Maple [B]  time = 0.184, size = 282, normalized size = 1.8 \begin{align*}{\frac{ae{x}^{5}}{5}}+{\frac{ad{x}^{3}}{3}}+{\frac{b{\rm arccsc} \left (cx\right )e{x}^{5}}{5}}+{\frac{b{\rm arccsc} \left (cx\right )d{x}^{3}}{3}}+{\frac{b{x}^{4}e}{20\,c}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{be{x}^{2}}{40\,{c}^{3}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{bd{x}^{2}}{6\,c}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{bd}{6\,{c}^{3}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{bd}{6\,{c}^{4}x}\sqrt{{c}^{2}{x}^{2}-1}\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}-1} \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{3\,be}{40\,{c}^{5}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{3\,be}{40\,{c}^{6}x}\sqrt{{c}^{2}{x}^{2}-1}\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}-1} \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(e*x^2+d)*(a+b*arccsc(c*x)),x)

[Out]

1/5*a*e*x^5+1/3*a*d*x^3+1/5*b*arccsc(c*x)*e*x^5+1/3*b*arccsc(c*x)*d*x^3+1/20/c*b/((c^2*x^2-1)/c^2/x^2)^(1/2)*x
^4*e+1/40/c^3*b/((c^2*x^2-1)/c^2/x^2)^(1/2)*x^2*e+1/6/c*b/((c^2*x^2-1)/c^2/x^2)^(1/2)*d*x^2-1/6/c^3*b/((c^2*x^
2-1)/c^2/x^2)^(1/2)*d+1/6/c^4*b*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x*d*ln(c*x+(c^2*x^2-1)^(1/2))-3/
40/c^5*b/((c^2*x^2-1)/c^2/x^2)^(1/2)*e+3/40/c^6*b*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x*e*ln(c*x+(c^
2*x^2-1)^(1/2))

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Maxima [A]  time = 0.986601, size = 313, normalized size = 1.94 \begin{align*} \frac{1}{5} \, a e x^{5} + \frac{1}{3} \, a d x^{3} + \frac{1}{12} \,{\left (4 \, x^{3} \operatorname{arccsc}\left (c x\right ) + \frac{\frac{2 \, \sqrt{-\frac{1}{c^{2} x^{2}} + 1}}{c^{2}{\left (\frac{1}{c^{2} x^{2}} - 1\right )} + c^{2}} + \frac{\log \left (\sqrt{-\frac{1}{c^{2} x^{2}} + 1} + 1\right )}{c^{2}} - \frac{\log \left (\sqrt{-\frac{1}{c^{2} x^{2}} + 1} - 1\right )}{c^{2}}}{c}\right )} b d + \frac{1}{80} \,{\left (16 \, x^{5} \operatorname{arccsc}\left (c x\right ) - \frac{\frac{2 \,{\left (3 \,{\left (-\frac{1}{c^{2} x^{2}} + 1\right )}^{\frac{3}{2}} - 5 \, \sqrt{-\frac{1}{c^{2} x^{2}} + 1}\right )}}{c^{4}{\left (\frac{1}{c^{2} x^{2}} - 1\right )}^{2} + 2 \, c^{4}{\left (\frac{1}{c^{2} x^{2}} - 1\right )} + c^{4}} - \frac{3 \, \log \left (\sqrt{-\frac{1}{c^{2} x^{2}} + 1} + 1\right )}{c^{4}} + \frac{3 \, \log \left (\sqrt{-\frac{1}{c^{2} x^{2}} + 1} - 1\right )}{c^{4}}}{c}\right )} b e \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*e*x^5 + 1/3*a*d*x^3 + 1/12*(4*x^3*arccsc(c*x) + (2*sqrt(-1/(c^2*x^2) + 1)/(c^2*(1/(c^2*x^2) - 1) + c^2)
+ log(sqrt(-1/(c^2*x^2) + 1) + 1)/c^2 - log(sqrt(-1/(c^2*x^2) + 1) - 1)/c^2)/c)*b*d + 1/80*(16*x^5*arccsc(c*x)
- (2*(3*(-1/(c^2*x^2) + 1)^(3/2) - 5*sqrt(-1/(c^2*x^2) + 1))/(c^4*(1/(c^2*x^2) - 1)^2 + 2*c^4*(1/(c^2*x^2) -
1) + c^4) - 3*log(sqrt(-1/(c^2*x^2) + 1) + 1)/c^4 + 3*log(sqrt(-1/(c^2*x^2) + 1) - 1)/c^4)/c)*b*e

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Fricas [A]  time = 3.56358, size = 398, normalized size = 2.47 \begin{align*} \frac{24 \, a c^{5} e x^{5} + 40 \, a c^{5} d x^{3} + 8 \,{\left (3 \, b c^{5} e x^{5} + 5 \, b c^{5} d x^{3} - 5 \, b c^{5} d - 3 \, b c^{5} e\right )} \operatorname{arccsc}\left (c x\right ) - 16 \,{\left (5 \, b c^{5} d + 3 \, b c^{5} e\right )} \arctan \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) -{\left (20 \, b c^{2} d + 9 \, b e\right )} \log \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) +{\left (6 \, b c^{3} e x^{3} +{\left (20 \, b c^{3} d + 9 \, b c e\right )} x\right )} \sqrt{c^{2} x^{2} - 1}}{120 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(24*a*c^5*e*x^5 + 40*a*c^5*d*x^3 + 8*(3*b*c^5*e*x^5 + 5*b*c^5*d*x^3 - 5*b*c^5*d - 3*b*c^5*e)*arccsc(c*x)
- 16*(5*b*c^5*d + 3*b*c^5*e)*arctan(-c*x + sqrt(c^2*x^2 - 1)) - (20*b*c^2*d + 9*b*e)*log(-c*x + sqrt(c^2*x^2
- 1)) + (6*b*c^3*e*x^3 + (20*b*c^3*d + 9*b*c*e)*x)*sqrt(c^2*x^2 - 1))/c^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(e*x**2+d)*(a+b*acsc(c*x)),x)

[Out]

Integral(x**2*(a + b*acsc(c*x))*(d + e*x**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{2} + d\right )}{\left (b \operatorname{arccsc}\left (c x\right ) + a\right )} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x^2 + d)*(b*arccsc(c*x) + a)*x^2, x)