3.163 \(\int \frac {x^2 (a+b \csc ^{-1}(c x))}{(d+e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=276 \[ \frac {x^3 \left (a+b \csc ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {b c x^2 \sqrt {c^2 x^2-1}}{3 d \sqrt {c^2 x^2} \left (c^2 d+e\right ) \sqrt {d+e x^2}}+\frac {b x \sqrt {1-c^2 x^2} \sqrt {\frac {e x^2}{d}+1} F\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{3 d e \sqrt {c^2 x^2} \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}-\frac {b c^2 x \sqrt {1-c^2 x^2} \sqrt {d+e x^2} E\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{3 d e \sqrt {c^2 x^2} \sqrt {c^2 x^2-1} \left (c^2 d+e\right ) \sqrt {\frac {e x^2}{d}+1}} \]

[Out]

1/3*x^3*(a+b*arccsc(c*x))/d/(e*x^2+d)^(3/2)+1/3*b*c*x^2*(c^2*x^2-1)^(1/2)/d/(c^2*d+e)/(c^2*x^2)^(1/2)/(e*x^2+d
)^(1/2)-1/3*b*c^2*x*EllipticE(c*x,(-e/c^2/d)^(1/2))*(-c^2*x^2+1)^(1/2)*(e*x^2+d)^(1/2)/d/e/(c^2*d+e)/(c^2*x^2)
^(1/2)/(c^2*x^2-1)^(1/2)/(1+e*x^2/d)^(1/2)+1/3*b*x*EllipticF(c*x,(-e/c^2/d)^(1/2))*(-c^2*x^2+1)^(1/2)*(1+e*x^2
/d)^(1/2)/d/e/(c^2*x^2)^(1/2)/(c^2*x^2-1)^(1/2)/(e*x^2+d)^(1/2)

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Rubi [A]  time = 0.28, antiderivative size = 276, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {264, 5239, 12, 471, 423, 427, 426, 424, 421, 419} \[ \frac {x^3 \left (a+b \csc ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {b c x^2 \sqrt {c^2 x^2-1}}{3 d \sqrt {c^2 x^2} \left (c^2 d+e\right ) \sqrt {d+e x^2}}+\frac {b x \sqrt {1-c^2 x^2} \sqrt {\frac {e x^2}{d}+1} F\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{3 d e \sqrt {c^2 x^2} \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}-\frac {b c^2 x \sqrt {1-c^2 x^2} \sqrt {d+e x^2} E\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{3 d e \sqrt {c^2 x^2} \sqrt {c^2 x^2-1} \left (c^2 d+e\right ) \sqrt {\frac {e x^2}{d}+1}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*ArcCsc[c*x]))/(d + e*x^2)^(5/2),x]

[Out]

(b*c*x^2*Sqrt[-1 + c^2*x^2])/(3*d*(c^2*d + e)*Sqrt[c^2*x^2]*Sqrt[d + e*x^2]) + (x^3*(a + b*ArcCsc[c*x]))/(3*d*
(d + e*x^2)^(3/2)) - (b*c^2*x*Sqrt[1 - c^2*x^2]*Sqrt[d + e*x^2]*EllipticE[ArcSin[c*x], -(e/(c^2*d))])/(3*d*e*(
c^2*d + e)*Sqrt[c^2*x^2]*Sqrt[-1 + c^2*x^2]*Sqrt[1 + (e*x^2)/d]) + (b*x*Sqrt[1 - c^2*x^2]*Sqrt[1 + (e*x^2)/d]*
EllipticF[ArcSin[c*x], -(e/(c^2*d))])/(3*d*e*Sqrt[c^2*x^2]*Sqrt[-1 + c^2*x^2]*Sqrt[d + e*x^2])

Rule 12

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

Rule 264

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

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 421

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

Rule 423

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

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 426

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[a + b*x^2]/Sqrt[1 + (b*x^2)/a]
, Int[Sqrt[1 + (b*x^2)/a]/Sqrt[c + d*x^2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ
[a, 0]

Rule 427

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 + (d*x^2)/c]/Sqrt[c + d*x^2]
, Int[Sqrt[a + b*x^2]/Sqrt[1 + (d*x^2)/c], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] &&  !GtQ[c, 0]

Rule 471

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

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]))

Rubi steps

\begin {align*} \int \frac {x^2 \left (a+b \csc ^{-1}(c x)\right )}{\left (d+e x^2\right )^{5/2}} \, dx &=\frac {x^3 \left (a+b \csc ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {(b c x) \int \frac {x^2}{3 d \sqrt {-1+c^2 x^2} \left (d+e x^2\right )^{3/2}} \, dx}{\sqrt {c^2 x^2}}\\ &=\frac {x^3 \left (a+b \csc ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {(b c x) \int \frac {x^2}{\sqrt {-1+c^2 x^2} \left (d+e x^2\right )^{3/2}} \, dx}{3 d \sqrt {c^2 x^2}}\\ &=\frac {b c x^2 \sqrt {-1+c^2 x^2}}{3 d \left (c^2 d+e\right ) \sqrt {c^2 x^2} \sqrt {d+e x^2}}+\frac {x^3 \left (a+b \csc ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}-\frac {(b c x) \int \frac {\sqrt {-1+c^2 x^2}}{\sqrt {d+e x^2}} \, dx}{3 d \left (c^2 d+e\right ) \sqrt {c^2 x^2}}\\ &=\frac {b c x^2 \sqrt {-1+c^2 x^2}}{3 d \left (c^2 d+e\right ) \sqrt {c^2 x^2} \sqrt {d+e x^2}}+\frac {x^3 \left (a+b \csc ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {(b c x) \int \frac {1}{\sqrt {-1+c^2 x^2} \sqrt {d+e x^2}} \, dx}{3 d e \sqrt {c^2 x^2}}-\frac {\left (b c^3 x\right ) \int \frac {\sqrt {d+e x^2}}{\sqrt {-1+c^2 x^2}} \, dx}{3 d e \left (c^2 d+e\right ) \sqrt {c^2 x^2}}\\ &=\frac {b c x^2 \sqrt {-1+c^2 x^2}}{3 d \left (c^2 d+e\right ) \sqrt {c^2 x^2} \sqrt {d+e x^2}}+\frac {x^3 \left (a+b \csc ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}-\frac {\left (b c^3 x \sqrt {1-c^2 x^2}\right ) \int \frac {\sqrt {d+e x^2}}{\sqrt {1-c^2 x^2}} \, dx}{3 d e \left (c^2 d+e\right ) \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2}}+\frac {\left (b c x \sqrt {1+\frac {e x^2}{d}}\right ) \int \frac {1}{\sqrt {-1+c^2 x^2} \sqrt {1+\frac {e x^2}{d}}} \, dx}{3 d e \sqrt {c^2 x^2} \sqrt {d+e x^2}}\\ &=\frac {b c x^2 \sqrt {-1+c^2 x^2}}{3 d \left (c^2 d+e\right ) \sqrt {c^2 x^2} \sqrt {d+e x^2}}+\frac {x^3 \left (a+b \csc ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}-\frac {\left (b c^3 x \sqrt {1-c^2 x^2} \sqrt {d+e x^2}\right ) \int \frac {\sqrt {1+\frac {e x^2}{d}}}{\sqrt {1-c^2 x^2}} \, dx}{3 d e \left (c^2 d+e\right ) \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2} \sqrt {1+\frac {e x^2}{d}}}+\frac {\left (b c x \sqrt {1-c^2 x^2} \sqrt {1+\frac {e x^2}{d}}\right ) \int \frac {1}{\sqrt {1-c^2 x^2} \sqrt {1+\frac {e x^2}{d}}} \, dx}{3 d e \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}\\ &=\frac {b c x^2 \sqrt {-1+c^2 x^2}}{3 d \left (c^2 d+e\right ) \sqrt {c^2 x^2} \sqrt {d+e x^2}}+\frac {x^3 \left (a+b \csc ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}-\frac {b c^2 x \sqrt {1-c^2 x^2} \sqrt {d+e x^2} E\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{3 d e \left (c^2 d+e\right ) \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2} \sqrt {1+\frac {e x^2}{d}}}+\frac {b x \sqrt {1-c^2 x^2} \sqrt {1+\frac {e x^2}{d}} F\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{3 d e \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 185, normalized size = 0.67 \[ \frac {x^2 \left (a x \left (c^2 d+e\right )+b c \sqrt {1-\frac {1}{c^2 x^2}} \left (d+e x^2\right )+b x \left (c^2 d+e\right ) \csc ^{-1}(c x)\right )}{3 d \left (c^2 d+e\right ) \left (d+e x^2\right )^{3/2}}-\frac {b c x \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {\frac {e x^2}{d}+1} E\left (\sin ^{-1}\left (\sqrt {-\frac {e}{d}} x\right )|-\frac {c^2 d}{e}\right )}{3 d \sqrt {1-c^2 x^2} \sqrt {-\frac {e}{d}} \left (c^2 d+e\right ) \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*ArcCsc[c*x]))/(d + e*x^2)^(5/2),x]

[Out]

(x^2*(a*(c^2*d + e)*x + b*c*Sqrt[1 - 1/(c^2*x^2)]*(d + e*x^2) + b*(c^2*d + e)*x*ArcCsc[c*x]))/(3*d*(c^2*d + e)
*(d + e*x^2)^(3/2)) - (b*c*Sqrt[1 - 1/(c^2*x^2)]*x*Sqrt[1 + (e*x^2)/d]*EllipticE[ArcSin[Sqrt[-(e/d)]*x], -((c^
2*d)/e)])/(3*d*Sqrt[-(e/d)]*(c^2*d + e)*Sqrt[1 - c^2*x^2]*Sqrt[d + e*x^2])

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} \operatorname {arccsc}\left (c x\right ) + a x^{2}\right )} \sqrt {e x^{2} + d}}{e^{3} x^{6} + 3 \, d e^{2} x^{4} + 3 \, d^{2} e x^{2} + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arccsc(c*x))/(e*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral((b*x^2*arccsc(c*x) + a*x^2)*sqrt(e*x^2 + d)/(e^3*x^6 + 3*d*e^2*x^4 + 3*d^2*e*x^2 + d^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arccsc}\left (c x\right ) + a\right )} x^{2}}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arccsc(c*x))/(e*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arccsc(c*x) + a)*x^2/(e*x^2 + d)^(5/2), x)

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maple [F]  time = 6.91, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (a +b \,\mathrm {arccsc}\left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arccsc(c*x))/(e*x^2+d)^(5/2),x)

[Out]

int(x^2*(a+b*arccsc(c*x))/(e*x^2+d)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{3} \, a {\left (\frac {x}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} e} - \frac {x}{\sqrt {e x^{2} + d} d e}\right )} + b \int \frac {x^{2} \arctan \left (1, \sqrt {c x + 1} \sqrt {c x - 1}\right )}{{\left (e^{2} x^{4} + 2 \, d e x^{2} + d^{2}\right )} \sqrt {e x^{2} + d}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arccsc(c*x))/(e*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

-1/3*a*(x/((e*x^2 + d)^(3/2)*e) - x/(sqrt(e*x^2 + d)*d*e)) + b*integrate(x^2*arctan2(1, sqrt(c*x + 1)*sqrt(c*x
- 1))/((e^2*x^4 + 2*d*e*x^2 + d^2)*sqrt(e*x^2 + d)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2\,\left (a+b\,\mathrm {asin}\left (\frac {1}{c\,x}\right )\right )}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(a + b*asin(1/(c*x))))/(d + e*x^2)^(5/2),x)

[Out]

int((x^2*(a + b*asin(1/(c*x))))/(d + e*x^2)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*acsc(c*x))/(e*x**2+d)**(5/2),x)

[Out]

Timed out

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