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

Optimal. Leaf size=359 \[ \frac {x^3 \left (a+b \text {csch}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {b x \sqrt {d+e x^2} F\left (\tan ^{-1}(c x)|1-\frac {e}{c^2 d}\right )}{3 d^2 \sqrt {-c^2 x^2} \sqrt {-c^2 x^2-1} \left (c^2 d-e\right ) \sqrt {\frac {d+e x^2}{d \left (c^2 x^2+1\right )}}}+\frac {b c x^2 \sqrt {-c^2 x^2-1}}{3 d \sqrt {-c^2 x^2} \left (c^2 d-e\right ) \sqrt {d+e x^2}}-\frac {b c^2 x \sqrt {d+e x^2} E\left (\tan ^{-1}(c x)|1-\frac {e}{c^2 d}\right )}{3 d e \sqrt {-c^2 x^2} \sqrt {-c^2 x^2-1} \left (c^2 d-e\right ) \sqrt {\frac {d+e x^2}{d \left (c^2 x^2+1\right )}}}+\frac {b c^3 x^2 \sqrt {d+e x^2}}{3 d e \sqrt {-c^2 x^2} \sqrt {-c^2 x^2-1} \left (c^2 d-e\right )} \]

[Out]

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

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Rubi [A]  time = 0.31, antiderivative size = 359, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {264, 6302, 12, 471, 422, 418, 492, 411} \[ \frac {x^3 \left (a+b \text {csch}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {b x \sqrt {d+e x^2} F\left (\tan ^{-1}(c x)|1-\frac {e}{c^2 d}\right )}{3 d^2 \sqrt {-c^2 x^2} \sqrt {-c^2 x^2-1} \left (c^2 d-e\right ) \sqrt {\frac {d+e x^2}{d \left (c^2 x^2+1\right )}}}+\frac {b c^3 x^2 \sqrt {d+e x^2}}{3 d e \sqrt {-c^2 x^2} \sqrt {-c^2 x^2-1} \left (c^2 d-e\right )}+\frac {b c x^2 \sqrt {-c^2 x^2-1}}{3 d \sqrt {-c^2 x^2} \left (c^2 d-e\right ) \sqrt {d+e x^2}}-\frac {b c^2 x \sqrt {d+e x^2} E\left (\tan ^{-1}(c x)|1-\frac {e}{c^2 d}\right )}{3 d e \sqrt {-c^2 x^2} \sqrt {-c^2 x^2-1} \left (c^2 d-e\right ) \sqrt {\frac {d+e x^2}{d \left (c^2 x^2+1\right )}}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*ArcCsch[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]) + (b*c^3*x^2*Sqrt[d + e*x^2])/
(3*d*(c^2*d - e)*e*Sqrt[-(c^2*x^2)]*Sqrt[-1 - c^2*x^2]) + (x^3*(a + b*ArcCsch[c*x]))/(3*d*(d + e*x^2)^(3/2)) -
(b*c^2*x*Sqrt[d + e*x^2]*EllipticE[ArcTan[c*x], 1 - e/(c^2*d)])/(3*d*(c^2*d - e)*e*Sqrt[-(c^2*x^2)]*Sqrt[-1 -
c^2*x^2]*Sqrt[(d + e*x^2)/(d*(1 + c^2*x^2))]) + (b*x*Sqrt[d + e*x^2]*EllipticF[ArcTan[c*x], 1 - e/(c^2*d)])/(
3*d^2*(c^2*d - e)*Sqrt[-(c^2*x^2)]*Sqrt[-1 - c^2*x^2]*Sqrt[(d + e*x^2)/(d*(1 + c^2*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 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rule 418

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

Rule 422

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

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 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rule 6302

Int[((a_.) + ArcCsch[(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*ArcCsch[c*x], u, x] - Dist[(b*c*x)/Sqrt[-(c^2*x^2)], Int[Simp
lifyIntegrand[u/(x*Sqrt[-1 - c^2*x^2]), 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]))
|| (ILtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))

Rubi steps

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

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Mathematica [A]  time = 0.30, size = 189, normalized size = 0.53 \[ \frac {x^2 \left (a x \left (c^2 d-e\right )+b c \sqrt {\frac {1}{c^2 x^2}+1} \left (d+e x^2\right )+b x \left (c^2 d-e\right ) \text {csch}^{-1}(c x)\right )}{3 d \left (c^2 d-e\right ) \left (d+e x^2\right )^{3/2}}-\frac {b c x \sqrt {\frac {1}{c^2 x^2}+1} \sqrt {\frac {e x^2}{d}+1} E\left (\sin ^{-1}\left (\sqrt {-\frac {e}{d}} x\right )|\frac {c^2 d}{e}\right )}{3 d \sqrt {c^2 x^2+1} \sqrt {-\frac {e}{d}} \left (c^2 d-e\right ) \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*ArcCsch[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*ArcCsch[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*(c^2*d - e)*Sqrt[-(e/d)]*Sqrt[1 + c^2*x^2]*Sqrt[d + e*x^2])

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} \operatorname {arcsch}\left (c x\right ) + a x^{2}\right )} \sqrt {e x^{2} + d}}{e^{3} x^{6} + 3 \, d e^{2} x^{4} + 3 \, d^{2} e x^{2} + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2*arccsch(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 \[ \int \frac {{\left (b \operatorname {arcsch}\left (c x\right ) + a\right )} x^{2}}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.43, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (a +b \,\mathrm {arccsch}\left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{3} \, a {\left (\frac {x}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} e} - \frac {x}{\sqrt {e x^{2} + d} d e}\right )} + b \int \frac {x^{2} \log \left (\sqrt {\frac {1}{c^{2} x^{2}} + 1} + \frac {1}{c x}\right )}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arccsch(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*log(sqrt(1/(c^2*x^2) + 1) + 1/(c*
x))/(e*x^2 + d)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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