3.1.76 \(\int (d+e x) (a+b \text {sech}^{-1}(c x)) \, dx\) [76]

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

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6423, 1823, 858, 222, 272, 65, 214} \begin {gather*} \frac {(d+e x)^2 \left (a+b \text {sech}^{-1}(c x)\right )}{2 e}+\frac {b d \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \text {ArcSin}(c x)}{c}-\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*(a + b*ArcSech[c*x]),x]

[Out]

-1/2*(b*e*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^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 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 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 222

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 272

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 858

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 1823

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 6423

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]/(e*(m + 1)))*Sqrt[1/(1 + c*x)], 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]

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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.24, size = 139, normalized size = 0.98 \begin {gather*} a d x+\frac {1}{2} a e x^2+b e \left (-\frac {1}{2 c^2}-\frac {x}{2 c}\right ) \sqrt {\frac {1-c x}{1+c x}}+b d x \text {sech}^{-1}(c x)+\frac {1}{2} b e x^2 \text {sech}^{-1}(c x)-\frac {2 b d \sqrt {\frac {1-c x}{1+c x}} \sqrt {1-c^2 x^2} \text {ArcTan}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c-c^2 x} \end {gather*}

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*1/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 - (2*b*d*Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[1 - c^2*x^2]*ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]])/(c - c^
2*x)

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Maple [A]
time = 0.18, size = 125, normalized size = 0.88

method	result	size
derivativedivides	\(\frac {\frac {a \left (d \,c^{2} x +\frac {1}{2} e \,c^{2} x^{2}\right )}{c}+\frac {b \left (\mathrm {arcsech}\left (c x \right ) d \,c^{2} x +\frac {\mathrm {arcsech}\left (c x \right ) e \,c^{2} x^{2}}{2}+\frac {\sqrt {-\frac {c x -1}{c x}}\, c x \sqrt {\frac {c x +1}{c x}}\, \left (2 d c \arcsin \left (c x \right )-e \sqrt {-c^{2} x^{2}+1}\right )}{2 \sqrt {-c^{2} x^{2}+1}}\right )}{c}}{c}\)	\(125\)
default	\(\frac {\frac {a \left (d \,c^{2} x +\frac {1}{2} e \,c^{2} x^{2}\right )}{c}+\frac {b \left (\mathrm {arcsech}\left (c x \right ) d \,c^{2} x +\frac {\mathrm {arcsech}\left (c x \right ) e \,c^{2} x^{2}}{2}+\frac {\sqrt {-\frac {c x -1}{c x}}\, c x \sqrt {\frac {c x +1}{c x}}\, \left (2 d c \arcsin \left (c x \right )-e \sqrt {-c^{2} x^{2}+1}\right )}{2 \sqrt {-c^{2} x^{2}+1}}\right )}{c}}{c}\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(a/c*(d*c^2*x+1/2*e*c^2*x^2)+b/c*(arcsech(c*x)*d*c^2*x+1/2*arcsech(c*x)*e*c^2*x^2+1/2*(-(c*x-1)/c/x)^(1/2)
*c*x*((c*x+1)/c/x)^(1/2)*(2*d*c*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 = 0.27, size = 72, normalized size = 0.51 \begin {gather*} \frac {1}{2} \, a x^{2} e + a d x + \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 + \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 {gather*}

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (72) = 144\).
time = 0.38, size = 216, normalized size = 1.52 \begin {gather*} \frac {a c x^{2} \cosh \left (1\right ) + a c x^{2} \sinh \left (1\right ) + 2 \, a c d x - 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 \cosh \left (1\right ) + b c \sinh \left (1\right )\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) + {\left (2 \, b c d x - 2 \, b c d + {\left (b c x^{2} - b c\right )} \cosh \left (1\right ) + {\left (b c x^{2} - b c\right )} \sinh \left (1\right )\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) - {\left (b x \cosh \left (1\right ) + b x \sinh \left (1\right )\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}}{2 \, c} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \operatorname {asech}{\left (c x \right )}\right ) \left (d + e x\right )\, dx \end {gather*}

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 = 0.00, 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((e*x+d)*(a+b*arcsech(c*x)),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.53, size = 99, normalized size = 0.70 \begin {gather*} \frac {a\,x\,\left (2\,d+e\,x\right )}{2}+\frac {b\,d\,\mathrm {atan}\left (\frac {1}{\sqrt {\frac {1}{c\,x}-1}\,\sqrt {\frac {1}{c\,x}+1}}\right )}{c}+\frac {b\,e\,x^2\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )}{2}+b\,d\,x\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )-\frac {b\,e\,x\,\sqrt {\frac {1}{c\,x}-1}\,\sqrt {\frac {1}{c\,x}+1}}{2\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*x*(2*d + e*x))/2 + (b*d*atan(1/((1/(c*x) - 1)^(1/2)*(1/(c*x) + 1)^(1/2))))/c + (b*e*x^2*acosh(1/(c*x)))/2 +
b*d*x*acosh(1/(c*x)) - (b*e*x*(1/(c*x) - 1)^(1/2)*(1/(c*x) + 1)^(1/2))/(2*c)

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