3.157 \(\int \frac {x^3 (a+b \text {csch}^{-1}(c x))}{(d+e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=169 \[ -\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {2 b c x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-c^2 x^2-1}}\right )}{3 \sqrt {d} e^2 \sqrt {-c^2 x^2}}-\frac {b c x \sqrt {-c^2 x^2-1}}{3 e \sqrt {-c^2 x^2} \left (c^2 d-e\right ) \sqrt {d+e x^2}} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 169, 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 = {266, 43, 6302, 12, 573, 152, 93, 204} \[ -\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {2 b c x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-c^2 x^2-1}}\right )}{3 \sqrt {d} e^2 \sqrt {-c^2 x^2}}-\frac {b c x \sqrt {-c^2 x^2-1}}{3 e \sqrt {-c^2 x^2} \left (c^2 d-e\right ) \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcCsch[c*x]))/(d + e*x^2)^(5/2),x]

[Out]

-(b*c*x*Sqrt[-1 - c^2*x^2])/(3*(c^2*d - e)*e*Sqrt[-(c^2*x^2)]*Sqrt[d + e*x^2]) + (d*(a + b*ArcCsch[c*x]))/(3*e
^2*(d + e*x^2)^(3/2)) - (a + b*ArcCsch[c*x])/(e^2*Sqrt[d + e*x^2]) - (2*b*c*x*ArcTan[Sqrt[d + e*x^2]/(Sqrt[d]*
Sqrt[-1 - c^2*x^2])])/(3*Sqrt[d]*e^2*Sqrt[-(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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 573

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x
_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q*(e + f*x)^r, x], x, x^n],
x] /; FreeQ[{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[Simplify[(m + 1)/n]]

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^3 \left (a+b \text {csch}^{-1}(c x)\right )}{\left (d+e x^2\right )^{5/2}} \, dx &=\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}-\frac {(b c x) \int \frac {-2 d-3 e x^2}{3 e^2 x \sqrt {-1-c^2 x^2} \left (d+e x^2\right )^{3/2}} \, dx}{\sqrt {-c^2 x^2}}\\ &=\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}-\frac {(b c x) \int \frac {-2 d-3 e x^2}{x \sqrt {-1-c^2 x^2} \left (d+e x^2\right )^{3/2}} \, dx}{3 e^2 \sqrt {-c^2 x^2}}\\ &=\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}-\frac {(b c x) \operatorname {Subst}\left (\int \frac {-2 d-3 e x}{x \sqrt {-1-c^2 x} (d+e x)^{3/2}} \, dx,x,x^2\right )}{6 e^2 \sqrt {-c^2 x^2}}\\ &=-\frac {b c x \sqrt {-1-c^2 x^2}}{3 \left (c^2 d-e\right ) e \sqrt {-c^2 x^2} \sqrt {d+e x^2}}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {(b c x) \operatorname {Subst}\left (\int \frac {d \left (c^2 d-e\right )}{x \sqrt {-1-c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{3 d \left (c^2 d-e\right ) e^2 \sqrt {-c^2 x^2}}\\ &=-\frac {b c x \sqrt {-1-c^2 x^2}}{3 \left (c^2 d-e\right ) e \sqrt {-c^2 x^2} \sqrt {d+e x^2}}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {(b c x) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {-1-c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{3 e^2 \sqrt {-c^2 x^2}}\\ &=-\frac {b c x \sqrt {-1-c^2 x^2}}{3 \left (c^2 d-e\right ) e \sqrt {-c^2 x^2} \sqrt {d+e x^2}}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {(2 b c x) \operatorname {Subst}\left (\int \frac {1}{-d-x^2} \, dx,x,\frac {\sqrt {d+e x^2}}{\sqrt {-1-c^2 x^2}}\right )}{3 e^2 \sqrt {-c^2 x^2}}\\ &=-\frac {b c x \sqrt {-1-c^2 x^2}}{3 \left (c^2 d-e\right ) e \sqrt {-c^2 x^2} \sqrt {d+e x^2}}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}-\frac {2 b c x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-1-c^2 x^2}}\right )}{3 \sqrt {d} e^2 \sqrt {-c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 201, normalized size = 1.19 \[ \frac {a \left (c^2 d-e\right ) \left (2 d+3 e x^2\right )+b c e x \sqrt {\frac {1}{c^2 x^2}+1} \left (d+e x^2\right )+b \left (c^2 d-e\right ) \text {csch}^{-1}(c x) \left (2 d+3 e x^2\right )}{3 e^2 \left (e-c^2 d\right ) \left (d+e x^2\right )^{3/2}}+\frac {2 b c x \sqrt {\frac {1}{c^2 x^2}+1} \sqrt {-d-e x^2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {c^2 x^2+1}}{\sqrt {-d-e x^2}}\right )}{3 \sqrt {d} e^2 \sqrt {c^2 x^2+1} \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*ArcCsch[c*x]))/(d + e*x^2)^(5/2),x]

[Out]

(b*c*e*Sqrt[1 + 1/(c^2*x^2)]*x*(d + e*x^2) + a*(c^2*d - e)*(2*d + 3*e*x^2) + b*(c^2*d - e)*(2*d + 3*e*x^2)*Arc
Csch[c*x])/(3*e^2*(-(c^2*d) + e)*(d + e*x^2)^(3/2)) + (2*b*c*Sqrt[1 + 1/(c^2*x^2)]*x*Sqrt[-d - e*x^2]*ArcTan[(
Sqrt[d]*Sqrt[1 + c^2*x^2])/Sqrt[-d - e*x^2]])/(3*Sqrt[d]*e^2*Sqrt[1 + c^2*x^2]*Sqrt[d + e*x^2])

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fricas [B]  time = 0.94, size = 786, normalized size = 4.65 \[ \left [-\frac {2 \, {\left (2 \, b c^{2} d^{3} - 2 \, b d^{2} e + 3 \, {\left (b c^{2} d^{2} e - b d e^{2}\right )} x^{2}\right )} \sqrt {e x^{2} + d} \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) - {\left (b c^{2} d^{3} + {\left (b c^{2} d e^{2} - b e^{3}\right )} x^{4} - b d^{2} e + 2 \, {\left (b c^{2} d^{2} e - b d e^{2}\right )} x^{2}\right )} \sqrt {d} \log \left (\frac {{\left (c^{4} d^{2} + 6 \, c^{2} d e + e^{2}\right )} x^{4} + 8 \, {\left (c^{2} d^{2} + d e\right )} x^{2} + 4 \, {\left ({\left (c^{3} d + c e\right )} x^{3} + 2 \, c d x\right )} \sqrt {e x^{2} + d} \sqrt {d} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 8 \, d^{2}}{x^{4}}\right ) + 2 \, {\left (2 \, a c^{2} d^{3} - 2 \, a d^{2} e + 3 \, {\left (a c^{2} d^{2} e - a d e^{2}\right )} x^{2} + {\left (b c d e^{2} x^{3} + b c d^{2} e x\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}}\right )} \sqrt {e x^{2} + d}}{6 \, {\left (c^{2} d^{4} e^{2} - d^{3} e^{3} + {\left (c^{2} d^{2} e^{4} - d e^{5}\right )} x^{4} + 2 \, {\left (c^{2} d^{3} e^{3} - d^{2} e^{4}\right )} x^{2}\right )}}, -\frac {{\left (b c^{2} d^{3} + {\left (b c^{2} d e^{2} - b e^{3}\right )} x^{4} - b d^{2} e + 2 \, {\left (b c^{2} d^{2} e - b d e^{2}\right )} x^{2}\right )} \sqrt {-d} \arctan \left (\frac {{\left ({\left (c^{3} d + c e\right )} x^{3} + 2 \, c d x\right )} \sqrt {e x^{2} + d} \sqrt {-d} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}}}{2 \, {\left (c^{2} d e x^{4} + {\left (c^{2} d^{2} + d e\right )} x^{2} + d^{2}\right )}}\right ) + {\left (2 \, b c^{2} d^{3} - 2 \, b d^{2} e + 3 \, {\left (b c^{2} d^{2} e - b d e^{2}\right )} x^{2}\right )} \sqrt {e x^{2} + d} \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) + {\left (2 \, a c^{2} d^{3} - 2 \, a d^{2} e + 3 \, {\left (a c^{2} d^{2} e - a d e^{2}\right )} x^{2} + {\left (b c d e^{2} x^{3} + b c d^{2} e x\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}}\right )} \sqrt {e x^{2} + d}}{3 \, {\left (c^{2} d^{4} e^{2} - d^{3} e^{3} + {\left (c^{2} d^{2} e^{4} - d e^{5}\right )} x^{4} + 2 \, {\left (c^{2} d^{3} e^{3} - d^{2} e^{4}\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccsch(c*x))/(e*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

[-1/6*(2*(2*b*c^2*d^3 - 2*b*d^2*e + 3*(b*c^2*d^2*e - b*d*e^2)*x^2)*sqrt(e*x^2 + d)*log((c*x*sqrt((c^2*x^2 + 1)
/(c^2*x^2)) + 1)/(c*x)) - (b*c^2*d^3 + (b*c^2*d*e^2 - b*e^3)*x^4 - b*d^2*e + 2*(b*c^2*d^2*e - b*d*e^2)*x^2)*sq
rt(d)*log(((c^4*d^2 + 6*c^2*d*e + e^2)*x^4 + 8*(c^2*d^2 + d*e)*x^2 + 4*((c^3*d + c*e)*x^3 + 2*c*d*x)*sqrt(e*x^
2 + d)*sqrt(d)*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 8*d^2)/x^4) + 2*(2*a*c^2*d^3 - 2*a*d^2*e + 3*(a*c^2*d^2*e - a*d
*e^2)*x^2 + (b*c*d*e^2*x^3 + b*c*d^2*e*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2)))*sqrt(e*x^2 + d))/(c^2*d^4*e^2 - d^3*e
^3 + (c^2*d^2*e^4 - d*e^5)*x^4 + 2*(c^2*d^3*e^3 - d^2*e^4)*x^2), -1/3*((b*c^2*d^3 + (b*c^2*d*e^2 - b*e^3)*x^4
- b*d^2*e + 2*(b*c^2*d^2*e - b*d*e^2)*x^2)*sqrt(-d)*arctan(1/2*((c^3*d + c*e)*x^3 + 2*c*d*x)*sqrt(e*x^2 + d)*s
qrt(-d)*sqrt((c^2*x^2 + 1)/(c^2*x^2))/(c^2*d*e*x^4 + (c^2*d^2 + d*e)*x^2 + d^2)) + (2*b*c^2*d^3 - 2*b*d^2*e +
3*(b*c^2*d^2*e - b*d*e^2)*x^2)*sqrt(e*x^2 + d)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) + (2*a*c^2*d
^3 - 2*a*d^2*e + 3*(a*c^2*d^2*e - a*d*e^2)*x^2 + (b*c*d*e^2*x^3 + b*c*d^2*e*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2)))*
sqrt(e*x^2 + d))/(c^2*d^4*e^2 - d^3*e^3 + (c^2*d^2*e^4 - d*e^5)*x^4 + 2*(c^2*d^3*e^3 - d^2*e^4)*x^2)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arcsch}\left (c x\right ) + a\right )} x^{3}}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccsch(c*x))/(e*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arccsch(c*x) + a)*x^3/(e*x^2 + d)^(5/2), x)

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maple [F]  time = 0.45, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \left (a +b \,\mathrm {arccsch}\left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arccsch(c*x))/(e*x^2+d)^(5/2),x)

[Out]

int(x^3*(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 \[ -\frac {1}{3} \, a {\left (\frac {3 \, x^{2}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} e} + \frac {2 \, d}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} e^{2}}\right )} + b \int \frac {x^{3} \log \left (\sqrt {\frac {1}{c^{2} x^{2}} + 1} + \frac {1}{c x}\right )}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccsch(c*x))/(e*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

-1/3*a*(3*x^2/((e*x^2 + d)^(3/2)*e) + 2*d/((e*x^2 + d)^(3/2)*e^2)) + b*integrate(x^3*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 = -0.01 \[ \int \frac {x^3\,\left (a+b\,\mathrm {asinh}\left (\frac {1}{c\,x}\right )\right )}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*asinh(1/(c*x))))/(d + e*x^2)^(5/2),x)

[Out]

int((x^3*(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**3*(a+b*acsch(c*x))/(e*x**2+d)**(5/2),x)

[Out]

Timed out

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