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

Optimal. Leaf size=159 \[ \frac {1}{4} d x^4 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{6} e x^6 \left (a+b \text {csch}^{-1}(c x)\right )-\frac {b x \left (-c^2 x^2-1\right )^{3/2} \left (3 c^2 d-4 e\right )}{36 c^5 \sqrt {-c^2 x^2}}-\frac {b x \sqrt {-c^2 x^2-1} \left (3 c^2 d-2 e\right )}{12 c^5 \sqrt {-c^2 x^2}}+\frac {b e x \left (-c^2 x^2-1\right )^{5/2}}{30 c^5 \sqrt {-c^2 x^2}} \]

[Out]

1/4*d*x^4*(a+b*arccsch(c*x))+1/6*e*x^6*(a+b*arccsch(c*x))-1/36*b*(3*c^2*d-4*e)*x*(-c^2*x^2-1)^(3/2)/c^5/(-c^2*
x^2)^(1/2)+1/30*b*e*x*(-c^2*x^2-1)^(5/2)/c^5/(-c^2*x^2)^(1/2)-1/12*b*(3*c^2*d-2*e)*x*(-c^2*x^2-1)^(1/2)/c^5/(-
c^2*x^2)^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {14, 6302, 12, 446, 77} \[ \frac {1}{4} d x^4 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{6} e x^6 \left (a+b \text {csch}^{-1}(c x)\right )-\frac {b x \left (-c^2 x^2-1\right )^{3/2} \left (3 c^2 d-4 e\right )}{36 c^5 \sqrt {-c^2 x^2}}-\frac {b x \sqrt {-c^2 x^2-1} \left (3 c^2 d-2 e\right )}{12 c^5 \sqrt {-c^2 x^2}}+\frac {b e x \left (-c^2 x^2-1\right )^{5/2}}{30 c^5 \sqrt {-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(3*c^2*d - 2*e)*x*Sqrt[-1 - c^2*x^2])/(12*c^5*Sqrt[-(c^2*x^2)]) - (b*(3*c^2*d - 4*e)*x*(-1 - c^2*x^2)^(3/2
))/(36*c^5*Sqrt[-(c^2*x^2)]) + (b*e*x*(-1 - c^2*x^2)^(5/2))/(30*c^5*Sqrt[-(c^2*x^2)]) + (d*x^4*(a + b*ArcCsch[
c*x]))/4 + (e*x^6*(a + b*ArcCsch[c*x]))/6

Rule 12

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

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && 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 x^3 \left (d+e x^2\right ) \left (a+b \text {csch}^{-1}(c x)\right ) \, dx &=\frac {1}{4} d x^4 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{6} e x^6 \left (a+b \text {csch}^{-1}(c x)\right )-\frac {(b c x) \int \frac {x^3 \left (3 d+2 e x^2\right )}{12 \sqrt {-1-c^2 x^2}} \, dx}{\sqrt {-c^2 x^2}}\\ &=\frac {1}{4} d x^4 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{6} e x^6 \left (a+b \text {csch}^{-1}(c x)\right )-\frac {(b c x) \int \frac {x^3 \left (3 d+2 e x^2\right )}{\sqrt {-1-c^2 x^2}} \, dx}{12 \sqrt {-c^2 x^2}}\\ &=\frac {1}{4} d x^4 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{6} e x^6 \left (a+b \text {csch}^{-1}(c x)\right )-\frac {(b c x) \operatorname {Subst}\left (\int \frac {x (3 d+2 e x)}{\sqrt {-1-c^2 x}} \, dx,x,x^2\right )}{24 \sqrt {-c^2 x^2}}\\ &=\frac {1}{4} d x^4 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{6} e x^6 \left (a+b \text {csch}^{-1}(c x)\right )-\frac {(b c x) \operatorname {Subst}\left (\int \left (\frac {-3 c^2 d+2 e}{c^4 \sqrt {-1-c^2 x}}+\frac {\left (-3 c^2 d+4 e\right ) \sqrt {-1-c^2 x}}{c^4}+\frac {2 e \left (-1-c^2 x\right )^{3/2}}{c^4}\right ) \, dx,x,x^2\right )}{24 \sqrt {-c^2 x^2}}\\ &=-\frac {b \left (3 c^2 d-2 e\right ) x \sqrt {-1-c^2 x^2}}{12 c^5 \sqrt {-c^2 x^2}}-\frac {b \left (3 c^2 d-4 e\right ) x \left (-1-c^2 x^2\right )^{3/2}}{36 c^5 \sqrt {-c^2 x^2}}+\frac {b e x \left (-1-c^2 x^2\right )^{5/2}}{30 c^5 \sqrt {-c^2 x^2}}+\frac {1}{4} d x^4 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{6} e x^6 \left (a+b \text {csch}^{-1}(c x)\right )\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 97, normalized size = 0.61 \[ \frac {1}{180} x \left (15 a x^3 \left (3 d+2 e x^2\right )+\frac {b \sqrt {\frac {1}{c^2 x^2}+1} \left (3 c^4 \left (5 d x^2+2 e x^4\right )-2 c^2 \left (15 d+4 e x^2\right )+16 e\right )}{c^5}+15 b x^3 \text {csch}^{-1}(c x) \left (3 d+2 e x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(15*a*x^3*(3*d + 2*e*x^2) + (b*Sqrt[1 + 1/(c^2*x^2)]*(16*e - 2*c^2*(15*d + 4*e*x^2) + 3*c^4*(5*d*x^2 + 2*e*
x^4)))/c^5 + 15*b*x^3*(3*d + 2*e*x^2)*ArcCsch[c*x]))/180

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fricas [A]  time = 0.89, size = 144, normalized size = 0.91 \[ \frac {30 \, a c^{5} e x^{6} + 45 \, a c^{5} d x^{4} + 15 \, {\left (2 \, b c^{5} e x^{6} + 3 \, b c^{5} d x^{4}\right )} \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) + {\left (6 \, b c^{4} e x^{5} + {\left (15 \, b c^{4} d - 8 \, b c^{2} e\right )} x^{3} - 2 \, {\left (15 \, b c^{2} d - 8 \, b e\right )} x\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}}}{180 \, c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/180*(30*a*c^5*e*x^6 + 45*a*c^5*d*x^4 + 15*(2*b*c^5*e*x^6 + 3*b*c^5*d*x^4)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x
^2)) + 1)/(c*x)) + (6*b*c^4*e*x^5 + (15*b*c^4*d - 8*b*c^2*e)*x^3 - 2*(15*b*c^2*d - 8*b*e)*x)*sqrt((c^2*x^2 + 1
)/(c^2*x^2)))/c^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 134, normalized size = 0.84 \[ \frac {\frac {a \left (\frac {1}{6} c^{6} e \,x^{6}+\frac {1}{4} c^{6} d \,x^{4}\right )}{c^{2}}+\frac {b \left (\frac {\mathrm {arccsch}\left (c x \right ) c^{6} x^{6} e}{6}+\frac {\mathrm {arccsch}\left (c x \right ) c^{6} x^{4} d}{4}+\frac {\left (c^{2} x^{2}+1\right ) \left (6 c^{4} e \,x^{4}+15 c^{4} d \,x^{2}-8 c^{2} x^{2} e -30 c^{2} d +16 e \right )}{180 \sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, c x}\right )}{c^{2}}}{c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^4*(a/c^2*(1/6*c^6*e*x^6+1/4*c^6*d*x^4)+b/c^2*(1/6*arccsch(c*x)*c^6*x^6*e+1/4*arccsch(c*x)*c^6*x^4*d+1/180*
(c^2*x^2+1)*(6*c^4*e*x^4+15*c^4*d*x^2-8*c^2*e*x^2-30*c^2*d+16*e)/((c^2*x^2+1)/c^2/x^2)^(1/2)/c/x))

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maxima [A]  time = 0.31, size = 137, normalized size = 0.86 \[ \frac {1}{6} \, a e x^{6} + \frac {1}{4} \, a d x^{4} + \frac {1}{12} \, {\left (3 \, x^{4} \operatorname {arcsch}\left (c x\right ) + \frac {c^{2} x^{3} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} - 3 \, x \sqrt {\frac {1}{c^{2} x^{2}} + 1}}{c^{3}}\right )} b d + \frac {1}{90} \, {\left (15 \, x^{6} \operatorname {arcsch}\left (c x\right ) + \frac {3 \, c^{4} x^{5} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {5}{2}} - 10 \, c^{2} x^{3} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} + 15 \, x \sqrt {\frac {1}{c^{2} x^{2}} + 1}}{c^{5}}\right )} b e \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a*e*x^6 + 1/4*a*d*x^4 + 1/12*(3*x^4*arccsch(c*x) + (c^2*x^3*(1/(c^2*x^2) + 1)^(3/2) - 3*x*sqrt(1/(c^2*x^2)
+ 1))/c^3)*b*d + 1/90*(15*x^6*arccsch(c*x) + (3*c^4*x^5*(1/(c^2*x^2) + 1)^(5/2) - 10*c^2*x^3*(1/(c^2*x^2) + 1
)^(3/2) + 15*x*sqrt(1/(c^2*x^2) + 1))/c^5)*b*e

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\left (e\,x^2+d\right )\,\left (a+b\,\mathrm {asinh}\left (\frac {1}{c\,x}\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(e*x**2+d)*(a+b*acsch(c*x)),x)

[Out]

Integral(x**3*(a + b*acsch(c*x))*(d + e*x**2), x)

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