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

Optimal. Leaf size=146 \[ \frac {\left (d+e x^2\right )^2 \left (a+b \text {csch}^{-1}(c x)\right )}{4 e}-\frac {b c d^2 x \tan ^{-1}\left (\sqrt {-c^2 x^2-1}\right )}{4 e \sqrt {-c^2 x^2}}+\frac {b x \sqrt {-c^2 x^2-1} \left (2 c^2 d-e\right )}{4 c^3 \sqrt {-c^2 x^2}}-\frac {b e x \left (-c^2 x^2-1\right )^{3/2}}{12 c^3 \sqrt {-c^2 x^2}} \]

[Out]

1/4*(e*x^2+d)^2*(a+b*arccsch(c*x))/e-1/12*b*e*x*(-c^2*x^2-1)^(3/2)/c^3/(-c^2*x^2)^(1/2)-1/4*b*c*d^2*x*arctan((
-c^2*x^2-1)^(1/2))/e/(-c^2*x^2)^(1/2)+1/4*b*(2*c^2*d-e)*x*(-c^2*x^2-1)^(1/2)/c^3/(-c^2*x^2)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {6300, 446, 88, 63, 205} \[ \frac {\left (d+e x^2\right )^2 \left (a+b \text {csch}^{-1}(c x)\right )}{4 e}-\frac {b c d^2 x \tan ^{-1}\left (\sqrt {-c^2 x^2-1}\right )}{4 e \sqrt {-c^2 x^2}}+\frac {b x \sqrt {-c^2 x^2-1} \left (2 c^2 d-e\right )}{4 c^3 \sqrt {-c^2 x^2}}-\frac {b e x \left (-c^2 x^2-1\right )^{3/2}}{12 c^3 \sqrt {-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 205

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

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 6300

Int[((a_.) + ArcCsch[(c_.)*(x_)]*(b_.))*(x_)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^(p +
1)*(a + b*ArcCsch[c*x]))/(2*e*(p + 1)), x] - Dist[(b*c*x)/(2*e*(p + 1)*Sqrt[-(c^2*x^2)]), Int[(d + e*x^2)^(p
+ 1)/(x*Sqrt[-1 - c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]

Rubi steps

\begin {align*} \int x \left (d+e x^2\right ) \left (a+b \text {csch}^{-1}(c x)\right ) \, dx &=\frac {\left (d+e x^2\right )^2 \left (a+b \text {csch}^{-1}(c x)\right )}{4 e}-\frac {(b c x) \int \frac {\left (d+e x^2\right )^2}{x \sqrt {-1-c^2 x^2}} \, dx}{4 e \sqrt {-c^2 x^2}}\\ &=\frac {\left (d+e x^2\right )^2 \left (a+b \text {csch}^{-1}(c x)\right )}{4 e}-\frac {(b c x) \operatorname {Subst}\left (\int \frac {(d+e x)^2}{x \sqrt {-1-c^2 x}} \, dx,x,x^2\right )}{8 e \sqrt {-c^2 x^2}}\\ &=\frac {\left (d+e x^2\right )^2 \left (a+b \text {csch}^{-1}(c x)\right )}{4 e}-\frac {(b c x) \operatorname {Subst}\left (\int \left (-\frac {e \left (-2 c^2 d+e\right )}{c^2 \sqrt {-1-c^2 x}}+\frac {d^2}{x \sqrt {-1-c^2 x}}-\frac {e^2 \sqrt {-1-c^2 x}}{c^2}\right ) \, dx,x,x^2\right )}{8 e \sqrt {-c^2 x^2}}\\ &=\frac {b \left (2 c^2 d-e\right ) x \sqrt {-1-c^2 x^2}}{4 c^3 \sqrt {-c^2 x^2}}-\frac {b e x \left (-1-c^2 x^2\right )^{3/2}}{12 c^3 \sqrt {-c^2 x^2}}+\frac {\left (d+e x^2\right )^2 \left (a+b \text {csch}^{-1}(c x)\right )}{4 e}-\frac {\left (b c d^2 x\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {-1-c^2 x}} \, dx,x,x^2\right )}{8 e \sqrt {-c^2 x^2}}\\ &=\frac {b \left (2 c^2 d-e\right ) x \sqrt {-1-c^2 x^2}}{4 c^3 \sqrt {-c^2 x^2}}-\frac {b e x \left (-1-c^2 x^2\right )^{3/2}}{12 c^3 \sqrt {-c^2 x^2}}+\frac {\left (d+e x^2\right )^2 \left (a+b \text {csch}^{-1}(c x)\right )}{4 e}+\frac {\left (b d^2 x\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {1}{c^2}-\frac {x^2}{c^2}} \, dx,x,\sqrt {-1-c^2 x^2}\right )}{4 c e \sqrt {-c^2 x^2}}\\ &=\frac {b \left (2 c^2 d-e\right ) x \sqrt {-1-c^2 x^2}}{4 c^3 \sqrt {-c^2 x^2}}-\frac {b e x \left (-1-c^2 x^2\right )^{3/2}}{12 c^3 \sqrt {-c^2 x^2}}+\frac {\left (d+e x^2\right )^2 \left (a+b \text {csch}^{-1}(c x)\right )}{4 e}-\frac {b c d^2 x \tan ^{-1}\left (\sqrt {-1-c^2 x^2}\right )}{4 e \sqrt {-c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 77, normalized size = 0.53 \[ \frac {x \left (3 a c^3 x \left (2 d+e x^2\right )+3 b c^3 x \text {csch}^{-1}(c x) \left (2 d+e x^2\right )+b \sqrt {\frac {1}{c^2 x^2}+1} \left (c^2 \left (6 d+e x^2\right )-2 e\right )\right )}{12 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(3*a*c^3*x*(2*d + e*x^2) + b*Sqrt[1 + 1/(c^2*x^2)]*(-2*e + c^2*(6*d + e*x^2)) + 3*b*c^3*x*(2*d + e*x^2)*Arc
Csch[c*x]))/(12*c^3)

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fricas [A]  time = 1.13, size = 123, normalized size = 0.84 \[ \frac {3 \, a c^{3} e x^{4} + 6 \, a c^{3} d x^{2} + 3 \, {\left (b c^{3} e x^{4} + 2 \, b c^{3} d x^{2}\right )} \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) + {\left (b c^{2} e x^{3} + 2 \, {\left (3 \, b c^{2} d - b e\right )} x\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}}}{12 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a*c^3*e*x^4 + 6*a*c^3*d*x^2 + 3*(b*c^3*e*x^4 + 2*b*c^3*d*x^2)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) +
1)/(c*x)) + (b*c^2*e*x^3 + 2*(3*b*c^2*d - b*e)*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2)))/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 115, normalized size = 0.79 \[ \frac {\frac {a \left (\frac {1}{4} c^{4} e \,x^{4}+\frac {1}{2} c^{4} d \,x^{2}\right )}{c^{2}}+\frac {b \left (\frac {\mathrm {arccsch}\left (c x \right ) c^{4} x^{4} e}{4}+\frac {\mathrm {arccsch}\left (c x \right ) c^{4} x^{2} d}{2}+\frac {\left (c^{2} x^{2}+1\right ) \left (c^{2} x^{2} e +6 c^{2} d -2 e \right )}{12 \sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, c x}\right )}{c^{2}}}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 95, normalized size = 0.65 \[ \frac {1}{4} \, a e x^{4} + \frac {1}{2} \, a d x^{2} + \frac {1}{2} \, {\left (x^{2} \operatorname {arcsch}\left (c x\right ) + \frac {x \sqrt {\frac {1}{c^{2} x^{2}} + 1}}{c}\right )} b d + \frac {1}{12} \, {\left (3 \, x^{4} \operatorname {arcsch}\left (c x\right ) + \frac {c^{2} x^{3} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} - 3 \, x \sqrt {\frac {1}{c^{2} x^{2}} + 1}}{c^{3}}\right )} b e \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*e*x^4 + 1/2*a*d*x^2 + 1/2*(x^2*arccsch(c*x) + x*sqrt(1/(c^2*x^2) + 1)/c)*b*d + 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*e

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x*(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 \[ \int x \left (a + b \operatorname {acsch}{\left (c x \right )}\right ) \left (d + e x^{2}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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