3.76 \(\int (d+e x) (a+b \text{sech}^{-1}(c x)) \, dx\)

Optimal. Leaf size=142 \[ \frac{(d+e x)^2 \left (a+b \text{sech}^{-1}(c x)\right )}{2 e}-\frac{b d^2 \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{2 e}-\frac{b e \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2}}{2 c^2}+\frac{b d \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sin ^{-1}(c x)}{c} \]

[Out]

-(b*e*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(2*c^2) + ((d + e*x)^2*(a + b*ArcSech[c*x]))/(2*e)
+ (b*d*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcSin[c*x])/c - (b*d^2*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTanh[
Sqrt[1 - c^2*x^2]])/(2*e)

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Rubi [A]  time = 0.117831, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6288, 1809, 844, 216, 266, 63, 208} \[ \frac{(d+e x)^2 \left (a+b \text{sech}^{-1}(c x)\right )}{2 e}-\frac{b d^2 \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{2 e}-\frac{b e \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2}}{2 c^2}+\frac{b d \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sin ^{-1}(c x)}{c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*(a + b*ArcSech[c*x]),x]

[Out]

-(b*e*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(2*c^2) + ((d + e*x)^2*(a + b*ArcSech[c*x]))/(2*e)
+ (b*d*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcSin[c*x])/c - (b*d^2*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTanh[
Sqrt[1 - c^2*x^2]])/(2*e)

Rule 6288

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

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

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 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 208

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

Rubi steps

\begin{align*} \int (d+e x) \left (a+b \text{sech}^{-1}(c x)\right ) \, dx &=\frac{(d+e x)^2 \left (a+b \text{sech}^{-1}(c x)\right )}{2 e}+\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{(d+e x)^2}{x \sqrt{1-c^2 x^2}} \, dx}{2 e}\\ &=-\frac{b e \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{2 c^2}+\frac{(d+e x)^2 \left (a+b \text{sech}^{-1}(c x)\right )}{2 e}-\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{-c^2 d^2-2 c^2 d e x}{x \sqrt{1-c^2 x^2}} \, dx}{2 c^2 e}\\ &=-\frac{b e \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{2 c^2}+\frac{(d+e x)^2 \left (a+b \text{sech}^{-1}(c x)\right )}{2 e}+\left (b d \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{1}{\sqrt{1-c^2 x^2}} \, dx+\frac{\left (b d^2 \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{1}{x \sqrt{1-c^2 x^2}} \, dx}{2 e}\\ &=-\frac{b e \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{2 c^2}+\frac{(d+e x)^2 \left (a+b \text{sech}^{-1}(c x)\right )}{2 e}+\frac{b d \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sin ^{-1}(c x)}{c}+\frac{\left (b d^2 \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-c^2 x}} \, dx,x,x^2\right )}{4 e}\\ &=-\frac{b e \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{2 c^2}+\frac{(d+e x)^2 \left (a+b \text{sech}^{-1}(c x)\right )}{2 e}+\frac{b d \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sin ^{-1}(c x)}{c}-\frac{\left (b d^2 \sqrt{\frac{1}{1+c x}} \sqrt{1+c 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 )}{2 c^2 e}\\ &=-\frac{b e \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{2 c^2}+\frac{(d+e x)^2 \left (a+b \text{sech}^{-1}(c x)\right )}{2 e}+\frac{b d \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sin ^{-1}(c x)}{c}-\frac{b d^2 \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{2 e}\\ \end{align*}

Mathematica [A]  time = 0.34379, size = 122, normalized size = 0.86 \[ a d x+\frac{1}{2} a e x^2-\frac{b d \sqrt{\frac{1-c x}{c x+1}} \sqrt{1-c^2 x^2} \sin ^{-1}(c x)}{c (c x-1)}+b e \left (-\frac{1}{2 c^2}-\frac{x}{2 c}\right ) \sqrt{\frac{1-c x}{c x+1}}+b d x \text{sech}^{-1}(c x)+\frac{1}{2} b e x^2 \text{sech}^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*d*x + (a*e*x^2)/2 + b*e*(-1/(2*c^2) - x/(2*c))*Sqrt[(1 - c*x)/(1 + c*x)] + b*d*x*ArcSech[c*x] + (b*e*x^2*Arc
Sech[c*x])/2 - (b*d*Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[1 - c^2*x^2]*ArcSin[c*x])/(c*(-1 + c*x))

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Maple [A]  time = 0.193, size = 125, normalized size = 0.9 \begin{align*}{\frac{1}{c} \left ({\frac{a}{c} \left ({\frac{{c}^{2}{x}^{2}e}{2}}+{c}^{2}dx \right ) }+{\frac{b}{c} \left ({\frac{{\rm arcsech} \left (cx\right ){c}^{2}{x}^{2}e}{2}}+{\rm arcsech} \left (cx\right ){c}^{2}xd+{\frac{cx}{2}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}} \left ( 2\,cd\arcsin \left ( cx \right ) -e\sqrt{-{c}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(a+b*arcsech(c*x)),x)

[Out]

1/c*(a/c*(1/2*c^2*x^2*e+c^2*d*x)+b/c*(1/2*arcsech(c*x)*c^2*x^2*e+arcsech(c*x)*c^2*x*d+1/2*(-(c*x-1)/c/x)^(1/2)
*c*x*((c*x+1)/c/x)^(1/2)*(2*c*d*arcsin(c*x)-e*(-c^2*x^2+1)^(1/2))/(-c^2*x^2+1)^(1/2)))

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Maxima [A]  time = 0.989862, size = 95, normalized size = 0.67 \begin{align*} \frac{1}{2} \, a e 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 e + a d x + \frac{{\left (c x \operatorname{arsech}\left (c x\right ) - \arctan \left (\sqrt{\frac{1}{c^{2} x^{2}} - 1}\right )\right )} b d}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.85227, size = 401, normalized size = 2.82 \begin{align*} \frac{a c e x^{2} + 2 \, a c d x - b e x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - 4 \, b d \arctan \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{c x}\right ) -{\left (2 \, b c d + b c e\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) +{\left (b c e x^{2} + 2 \, b c d x - 2 \, b c d - b c e\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \operatorname{asech}{\left (c x \right )}\right ) \left (d + e x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}{\left (b \operatorname{arsech}\left (c x\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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