3.1.97 \(\int x (d+e x^2) (a+b \text {sech}^{-1}(c x)) \, dx\) [97]

Optimal. Leaf size=164 \[ -\frac {b \left (2 c^2 d+e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{4 c^4}+\frac {b e \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \left (1-c^2 x^2\right )^{3/2}}{12 c^4}+\frac {\left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right )}{4 e}-\frac {b d^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{4 e} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {6434, 531, 457, 90, 65, 214} \begin {gather*} \frac {\left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right )}{4 e}-\frac {b d^2 \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{4 e}-\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2} \left (2 c^2 d+e\right )}{4 c^4}+\frac {b e \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \left (1-c^2 x^2\right )^{3/2}}{12 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(b*(2*c^2*d + e)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/c^4 + (b*e*Sqrt[(1 + c*x)^(-1)]*Sq
rt[1 + c*x]*(1 - c^2*x^2)^(3/2))/(12*c^4) + ((d + e*x^2)^2*(a + b*ArcSech[c*x]))/(4*e) - (b*d^2*Sqrt[(1 + c*x)
^(-1)]*Sqrt[1 + c*x]*ArcTanh[Sqrt[1 - c^2*x^2]])/(4*e)

Rule 65

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 90

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 214

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

Rule 457

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 531

Int[(u_.)*((c_) + (d_.)*(x_)^(n_.))^(q_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.)*(x_)^(non2_.))^
(p_.), x_Symbol] :> Int[u*(a1*a2 + b1*b2*x^n)^p*(c + d*x^n)^q, x] /; FreeQ[{a1, b1, a2, b2, c, d, n, p, q}, x]
&& EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] && (IntegerQ[p] || (GtQ[a1, 0] && GtQ[a2, 0]))

Rule 6434

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))*(x_)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x^2)^(p +
1)*((a + b*ArcSech[c*x])/(2*e*(p + 1))), x] + Dist[b*(Sqrt[1 + c*x]/(2*e*(p + 1)))*Sqrt[1/(1 + c*x)], Int[(d +
e*x^2)^(p + 1)/(x*Sqrt[1 - c*x]*Sqrt[1 + c*x]), 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 {sech}^{-1}(c x)\right ) \, dx &=\frac {\left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right )}{4 e}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {\left (d+e x^2\right )^2}{x \sqrt {1-c x} \sqrt {1+c x}} \, dx}{4 e}\\ &=\frac {\left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right )}{4 e}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {\left (d+e x^2\right )^2}{x \sqrt {1-c^2 x^2}} \, dx}{4 e}\\ &=\frac {\left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right )}{4 e}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \text {Subst}\left (\int \frac {(d+e x)^2}{x \sqrt {1-c^2 x}} \, dx,x,x^2\right )}{8 e}\\ &=\frac {\left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right )}{4 e}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \text {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}\\ &=-\frac {b \left (2 c^2 d+e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{4 c^4}+\frac {b e \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \left (1-c^2 x^2\right )^{3/2}}{12 c^4}+\frac {\left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right )}{4 e}+\frac {\left (b d^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1-c^2 x}} \, dx,x,x^2\right )}{8 e}\\ &=-\frac {b \left (2 c^2 d+e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{4 c^4}+\frac {b e \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \left (1-c^2 x^2\right )^{3/2}}{12 c^4}+\frac {\left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right )}{4 e}-\frac {\left (b d^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \text {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^2 e}\\ &=-\frac {b \left (2 c^2 d+e\right ) \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{4 c^4}+\frac {b e \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \left (1-c^2 x^2\right )^{3/2}}{12 c^4}+\frac {\left (d+e x^2\right )^2 \left (a+b \text {sech}^{-1}(c x)\right )}{4 e}-\frac {b d^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{4 e}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 85, normalized size = 0.52 \begin {gather*} \frac {1}{12} \left (3 a x^2 \left (2 d+e x^2\right )-\frac {b \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (2 e+c^2 \left (6 d+e x^2\right )\right )}{c^4}+3 b x^2 \left (2 d+e x^2\right ) \text {sech}^{-1}(c x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.33, size = 206, normalized size = 1.26

method	result	size
derivativedivides	\(\frac {\frac {\left (e \,c^{2} x^{2}+c^{2} d \right )^{2} a}{4 c^{2} e}+\frac {b \left (\frac {\mathrm {arcsech}\left (c x \right ) c^{4} d^{2}}{4 e}+\frac {\mathrm {arcsech}\left (c x \right ) c^{4} d \,x^{2}}{2}+\frac {e \,\mathrm {arcsech}\left (c x \right ) c^{4} x^{4}}{4}-\frac {\sqrt {-\frac {c x -1}{c x}}\, c x \sqrt {\frac {c x +1}{c x}}\, \left (3 c^{4} d^{2} \arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )+6 c^{2} d e \sqrt {-c^{2} x^{2}+1}+e^{2} \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}+2 e^{2} \sqrt {-c^{2} x^{2}+1}\right )}{12 e \sqrt {-c^{2} x^{2}+1}}\right )}{c^{2}}}{c^{2}}\)	\(206\)
default	\(\frac {\frac {\left (e \,c^{2} x^{2}+c^{2} d \right )^{2} a}{4 c^{2} e}+\frac {b \left (\frac {\mathrm {arcsech}\left (c x \right ) c^{4} d^{2}}{4 e}+\frac {\mathrm {arcsech}\left (c x \right ) c^{4} d \,x^{2}}{2}+\frac {e \,\mathrm {arcsech}\left (c x \right ) c^{4} x^{4}}{4}-\frac {\sqrt {-\frac {c x -1}{c x}}\, c x \sqrt {\frac {c x +1}{c x}}\, \left (3 c^{4} d^{2} \arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )+6 c^{2} d e \sqrt {-c^{2} x^{2}+1}+e^{2} \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}+2 e^{2} \sqrt {-c^{2} x^{2}+1}\right )}{12 e \sqrt {-c^{2} x^{2}+1}}\right )}{c^{2}}}{c^{2}}\)	\(206\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(e*x^2+d)*(a+b*arcsech(c*x)),x,method=_RETURNVERBOSE)

[Out]

1/c^2*(1/4*(c^2*e*x^2+c^2*d)^2*a/c^2/e+b/c^2*(1/4/e*arcsech(c*x)*c^4*d^2+1/2*arcsech(c*x)*c^4*d*x^2+1/4*e*arcs
ech(c*x)*c^4*x^4-1/12/e*(-(c*x-1)/c/x)^(1/2)*c*x*((c*x+1)/c/x)^(1/2)*(3*c^4*d^2*arctanh(1/(-c^2*x^2+1)^(1/2))+
6*c^2*d*e*(-c^2*x^2+1)^(1/2)+e^2*(-c^2*x^2+1)^(1/2)*c^2*x^2+2*e^2*(-c^2*x^2+1)^(1/2))/(-c^2*x^2+1)^(1/2)))

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Maxima [A]
time = 0.27, size = 98, normalized size = 0.60 \begin {gather*} \frac {1}{4} \, a x^{4} e + \frac {1}{2} \, a d x^{2} + \frac {1}{2} \, {\left (x^{2} \operatorname {arsech}\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 {arsech}\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*x^4*e + 1/2*a*d*x^2 + 1/2*(x^2*arcsech(c*x) - x*sqrt(1/(c^2*x^2) - 1)/c)*b*d + 1/12*(3*x^4*arcsech(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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Fricas [A]
time = 0.38, size = 165, normalized size = 1.01 \begin {gather*} \frac {3 \, a c^{3} x^{4} \cosh \left (1\right ) + 3 \, a c^{3} x^{4} \sinh \left (1\right ) + 6 \, a c^{3} d x^{2} + 3 \, {\left (b c^{3} x^{4} \cosh \left (1\right ) + b c^{3} x^{4} \sinh \left (1\right ) + 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 (6 \, b c^{2} d x + {\left (b c^{2} x^{3} + 2 \, b x\right )} \cosh \left (1\right ) + {\left (b c^{2} x^{3} + 2 \, b x\right )} \sinh \left (1\right )\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}}{12 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.44, size = 126, normalized size = 0.77 \begin {gather*} \begin {cases} \frac {a d x^{2}}{2} + \frac {a e x^{4}}{4} + \frac {b d x^{2} \operatorname {asech}{\left (c x \right )}}{2} + \frac {b e x^{4} \operatorname {asech}{\left (c x \right )}}{4} - \frac {b d \sqrt {- c^{2} x^{2} + 1}}{2 c^{2}} - \frac {b e x^{2} \sqrt {- c^{2} x^{2} + 1}}{12 c^{2}} - \frac {b e \sqrt {- c^{2} x^{2} + 1}}{6 c^{4}} & \text {for}\: c \neq 0 \\\left (a + \infty b\right ) \left (\frac {d x^{2}}{2} + \frac {e x^{4}}{4}\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*d*x**2/2 + a*e*x**4/4 + b*d*x**2*asech(c*x)/2 + b*e*x**4*asech(c*x)/4 - b*d*sqrt(-c**2*x**2 + 1)/
(2*c**2) - b*e*x**2*sqrt(-c**2*x**2 + 1)/(12*c**2) - b*e*sqrt(-c**2*x**2 + 1)/(6*c**4), Ne(c, 0)), ((a + oo*b)
*(d*x**2/2 + e*x**4/4), True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,\left (e\,x^2+d\right )\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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