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

Optimal. Leaf size=203 \[ \frac {\left (d+e x^2\right )^3 \left (a+b \text {csch}^{-1}(c x)\right )}{6 e}-\frac {b c d^3 x \tan ^{-1}\left (\sqrt {-c^2 x^2-1}\right )}{6 e \sqrt {-c^2 x^2}}-\frac {b e x \left (-c^2 x^2-1\right )^{3/2} \left (3 c^2 d-2 e\right )}{18 c^5 \sqrt {-c^2 x^2}}+\frac {b e^2 x \left (-c^2 x^2-1\right )^{5/2}}{30 c^5 \sqrt {-c^2 x^2}}+\frac {b x \sqrt {-c^2 x^2-1} \left (3 c^4 d^2-3 c^2 d e+e^2\right )}{6 c^5 \sqrt {-c^2 x^2}} \]

[Out]

1/6*(e*x^2+d)^3*(a+b*arccsch(c*x))/e-1/18*b*(3*c^2*d-2*e)*e*x*(-c^2*x^2-1)^(3/2)/c^5/(-c^2*x^2)^(1/2)+1/30*b*e
^2*x*(-c^2*x^2-1)^(5/2)/c^5/(-c^2*x^2)^(1/2)-1/6*b*c*d^3*x*arctan((-c^2*x^2-1)^(1/2))/e/(-c^2*x^2)^(1/2)+1/6*b
*(3*c^4*d^2-3*c^2*d*e+e^2)*x*(-c^2*x^2-1)^(1/2)/c^5/(-c^2*x^2)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {6300, 446, 88, 63, 205} \[ \frac {\left (d+e x^2\right )^3 \left (a+b \text {csch}^{-1}(c x)\right )}{6 e}+\frac {b x \sqrt {-c^2 x^2-1} \left (3 c^4 d^2-3 c^2 d e+e^2\right )}{6 c^5 \sqrt {-c^2 x^2}}-\frac {b c d^3 x \tan ^{-1}\left (\sqrt {-c^2 x^2-1}\right )}{6 e \sqrt {-c^2 x^2}}-\frac {b e x \left (-c^2 x^2-1\right )^{3/2} \left (3 c^2 d-2 e\right )}{18 c^5 \sqrt {-c^2 x^2}}+\frac {b e^2 x \left (-c^2 x^2-1\right )^{5/2}}{30 c^5 \sqrt {-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(3*c^4*d^2 - 3*c^2*d*e + e^2)*x*Sqrt[-1 - c^2*x^2])/(6*c^5*Sqrt[-(c^2*x^2)]) - (b*(3*c^2*d - 2*e)*e*x*(-1 -
c^2*x^2)^(3/2))/(18*c^5*Sqrt[-(c^2*x^2)]) + (b*e^2*x*(-1 - c^2*x^2)^(5/2))/(30*c^5*Sqrt[-(c^2*x^2)]) + ((d +
e*x^2)^3*(a + b*ArcCsch[c*x]))/(6*e) - (b*c*d^3*x*ArcTan[Sqrt[-1 - c^2*x^2]])/(6*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 )^2 \left (a+b \text {csch}^{-1}(c x)\right ) \, dx &=\frac {\left (d+e x^2\right )^3 \left (a+b \text {csch}^{-1}(c x)\right )}{6 e}-\frac {(b c x) \int \frac {\left (d+e x^2\right )^3}{x \sqrt {-1-c^2 x^2}} \, dx}{6 e \sqrt {-c^2 x^2}}\\ &=\frac {\left (d+e x^2\right )^3 \left (a+b \text {csch}^{-1}(c x)\right )}{6 e}-\frac {(b c x) \operatorname {Subst}\left (\int \frac {(d+e x)^3}{x \sqrt {-1-c^2 x}} \, dx,x,x^2\right )}{12 e \sqrt {-c^2 x^2}}\\ &=\frac {\left (d+e x^2\right )^3 \left (a+b \text {csch}^{-1}(c x)\right )}{6 e}-\frac {(b c x) \operatorname {Subst}\left (\int \left (\frac {e \left (3 c^4 d^2-3 c^2 d e+e^2\right )}{c^4 \sqrt {-1-c^2 x}}+\frac {d^3}{x \sqrt {-1-c^2 x}}-\frac {\left (3 c^2 d-2 e\right ) e^2 \sqrt {-1-c^2 x}}{c^4}+\frac {e^3 \left (-1-c^2 x\right )^{3/2}}{c^4}\right ) \, dx,x,x^2\right )}{12 e \sqrt {-c^2 x^2}}\\ &=\frac {b \left (3 c^4 d^2-3 c^2 d e+e^2\right ) x \sqrt {-1-c^2 x^2}}{6 c^5 \sqrt {-c^2 x^2}}-\frac {b \left (3 c^2 d-2 e\right ) e x \left (-1-c^2 x^2\right )^{3/2}}{18 c^5 \sqrt {-c^2 x^2}}+\frac {b e^2 x \left (-1-c^2 x^2\right )^{5/2}}{30 c^5 \sqrt {-c^2 x^2}}+\frac {\left (d+e x^2\right )^3 \left (a+b \text {csch}^{-1}(c x)\right )}{6 e}-\frac {\left (b c d^3 x\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {-1-c^2 x}} \, dx,x,x^2\right )}{12 e \sqrt {-c^2 x^2}}\\ &=\frac {b \left (3 c^4 d^2-3 c^2 d e+e^2\right ) x \sqrt {-1-c^2 x^2}}{6 c^5 \sqrt {-c^2 x^2}}-\frac {b \left (3 c^2 d-2 e\right ) e x \left (-1-c^2 x^2\right )^{3/2}}{18 c^5 \sqrt {-c^2 x^2}}+\frac {b e^2 x \left (-1-c^2 x^2\right )^{5/2}}{30 c^5 \sqrt {-c^2 x^2}}+\frac {\left (d+e x^2\right )^3 \left (a+b \text {csch}^{-1}(c x)\right )}{6 e}+\frac {\left (b d^3 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 )}{6 c e \sqrt {-c^2 x^2}}\\ &=\frac {b \left (3 c^4 d^2-3 c^2 d e+e^2\right ) x \sqrt {-1-c^2 x^2}}{6 c^5 \sqrt {-c^2 x^2}}-\frac {b \left (3 c^2 d-2 e\right ) e x \left (-1-c^2 x^2\right )^{3/2}}{18 c^5 \sqrt {-c^2 x^2}}+\frac {b e^2 x \left (-1-c^2 x^2\right )^{5/2}}{30 c^5 \sqrt {-c^2 x^2}}+\frac {\left (d+e x^2\right )^3 \left (a+b \text {csch}^{-1}(c x)\right )}{6 e}-\frac {b c d^3 x \tan ^{-1}\left (\sqrt {-1-c^2 x^2}\right )}{6 e \sqrt {-c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 123, normalized size = 0.61 \[ \frac {1}{90} x \left (15 a x \left (3 d^2+3 d e x^2+e^2 x^4\right )+\frac {b \sqrt {\frac {1}{c^2 x^2}+1} \left (3 c^4 \left (15 d^2+5 d e x^2+e^2 x^4\right )-2 c^2 e \left (15 d+2 e x^2\right )+8 e^2\right )}{c^5}+15 b x \text {csch}^{-1}(c x) \left (3 d^2+3 d e x^2+e^2 x^4\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(15*a*x*(3*d^2 + 3*d*e*x^2 + e^2*x^4) + (b*Sqrt[1 + 1/(c^2*x^2)]*(8*e^2 - 2*c^2*e*(15*d + 2*e*x^2) + 3*c^4*
(15*d^2 + 5*d*e*x^2 + e^2*x^4)))/c^5 + 15*b*x*(3*d^2 + 3*d*e*x^2 + e^2*x^4)*ArcCsch[c*x]))/90

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fricas [A]  time = 1.16, size = 189, normalized size = 0.93 \[ \frac {15 \, a c^{5} e^{2} x^{6} + 45 \, a c^{5} d e x^{4} + 45 \, a c^{5} d^{2} x^{2} + 15 \, {\left (b c^{5} e^{2} x^{6} + 3 \, b c^{5} d e x^{4} + 3 \, b c^{5} d^{2} x^{2}\right )} \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) + {\left (3 \, b c^{4} e^{2} x^{5} + {\left (15 \, b c^{4} d e - 4 \, b c^{2} e^{2}\right )} x^{3} + {\left (45 \, b c^{4} d^{2} - 30 \, b c^{2} d e + 8 \, b e^{2}\right )} x\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}}}{90 \, c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/90*(15*a*c^5*e^2*x^6 + 45*a*c^5*d*e*x^4 + 45*a*c^5*d^2*x^2 + 15*(b*c^5*e^2*x^6 + 3*b*c^5*d*e*x^4 + 3*b*c^5*d
^2*x^2)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) + (3*b*c^4*e^2*x^5 + (15*b*c^4*d*e - 4*b*c^2*e^2)*x
^3 + (45*b*c^4*d^2 - 30*b*c^2*d*e + 8*b*e^2)*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 \[ \int {\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arcsch}\left (c x\right ) + a\right )} x\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 182, normalized size = 0.90 \[ \frac {\frac {a \left (\frac {1}{6} c^{6} e^{2} x^{6}+\frac {1}{2} c^{6} d e \,x^{4}+\frac {1}{2} c^{6} d^{2} x^{2}\right )}{c^{4}}+\frac {b \left (\frac {\mathrm {arccsch}\left (c x \right ) e^{2} c^{6} x^{6}}{6}+\frac {\mathrm {arccsch}\left (c x \right ) c^{6} d e \,x^{4}}{2}+\frac {\mathrm {arccsch}\left (c x \right ) c^{6} x^{2} d^{2}}{2}+\frac {\left (c^{2} x^{2}+1\right ) \left (3 c^{4} e^{2} x^{4}+15 c^{4} d e \,x^{2}+45 d^{2} c^{4}-4 c^{2} e^{2} x^{2}-30 c^{2} d e +8 e^{2}\right )}{90 \sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, c x}\right )}{c^{4}}}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*(a/c^4*(1/6*c^6*e^2*x^6+1/2*c^6*d*e*x^4+1/2*c^6*d^2*x^2)+b/c^4*(1/6*arccsch(c*x)*e^2*c^6*x^6+1/2*arccsch
(c*x)*c^6*d*e*x^4+1/2*arccsch(c*x)*c^6*x^2*d^2+1/90*(c^2*x^2+1)*(3*c^4*e^2*x^4+15*c^4*d*e*x^2+45*c^4*d^2-4*c^2
*e^2*x^2-30*c^2*d*e+8*e^2)/((c^2*x^2+1)/c^2/x^2)^(1/2)/c/x))

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maxima [A]  time = 0.37, size = 183, normalized size = 0.90 \[ \frac {1}{6} \, a e^{2} x^{6} + \frac {1}{2} \, a d e x^{4} + \frac {1}{2} \, a d^{2} x^{2} + \frac {1}{2} \, {\left (x^{2} \operatorname {arcsch}\left (c x\right ) + \frac {x \sqrt {\frac {1}{c^{2} x^{2}} + 1}}{c}\right )} b d^{2} + \frac {1}{6} \, {\left (3 \, x^{4} \operatorname {arcsch}\left (c x\right ) + \frac {c^{2} x^{3} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} - 3 \, x \sqrt {\frac {1}{c^{2} x^{2}} + 1}}{c^{3}}\right )} b d e + \frac {1}{90} \, {\left (15 \, x^{6} \operatorname {arcsch}\left (c x\right ) + \frac {3 \, c^{4} x^{5} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {5}{2}} - 10 \, c^{2} x^{3} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} + 15 \, x \sqrt {\frac {1}{c^{2} x^{2}} + 1}}{c^{5}}\right )} b e^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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