3.152 \(\int \frac{x (a+b \text{sech}^{-1}(c x))}{\sqrt{d+e x^2}} \, dx\)

Optimal. Leaf size=153 \[ \frac{\sqrt{d+e x^2} \left (a+b \text{sech}^{-1}(c x)\right )}{e}-\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{1-c^2 x^2}}{c \sqrt{d+e x^2}}\right )}{c \sqrt{e}}-\frac{b \sqrt{d} \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{e} \]

[Out]

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

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Rubi [A]  time = 0.295115, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {6299, 517, 446, 105, 63, 217, 203, 93, 207} \[ \frac{\sqrt{d+e x^2} \left (a+b \text{sech}^{-1}(c x)\right )}{e}-\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{1-c^2 x^2}}{c \sqrt{d+e x^2}}\right )}{c \sqrt{e}}-\frac{b \sqrt{d} \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6299

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]*Sqrt[1/(1 + c*x)])/(2*e*(p + 1)), 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]

Rule 517

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 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 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -1])))

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 217

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

Rule 203

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 207

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

Rubi steps

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

Mathematica [A]  time = 0.536362, size = 239, normalized size = 1.56 \[ \frac{\sqrt{d+e x^2} \left (a+b \text{sech}^{-1}(c x)\right )}{e}+\frac{b \sqrt{\frac{1-c x}{c x+1}} \sqrt{1-c^2 x^2} \left (\sqrt{-c^2} \sqrt{e} \sqrt{c^2 (-d)-e} \sqrt{\frac{c^2 \left (d+e x^2\right )}{c^2 d+e}} \sin ^{-1}\left (\frac{c \sqrt{e} \sqrt{1-c^2 x^2}}{\sqrt{-c^2} \sqrt{c^2 (-d)-e}}\right )+c^3 \sqrt{d} \sqrt{-d-e x^2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{1-c^2 x^2}}{\sqrt{-d-e x^2}}\right )\right )}{c^3 e (c x-1) \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.265, size = 0, normalized size = 0. \begin{align*} \int{x \left ( a+b{\rm arcsech} \left (cx\right ) \right ){\frac{1}{\sqrt{e{x}^{2}+d}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b{\left (\frac{\sqrt{e x^{2} + d} \log \left (\sqrt{c x + 1} \sqrt{-c x + 1} + 1\right )}{e} - \int \frac{2 \,{\left (c^{2} e x^{2} - e\right )} x \log \left (\sqrt{x}\right ) +{\left (c^{2} e x^{2} \log \left (c\right ) - e \log \left (c\right )\right )} x +{\left (2 \,{\left (c^{2} e x^{2} - e\right )} x \log \left (\sqrt{x}\right ) +{\left ({\left (e \log \left (c\right ) + e\right )} c^{2} x^{2} + c^{2} d - e \log \left (c\right )\right )} x\right )} e^{\left (\frac{1}{2} \, \log \left (c x + 1\right ) + \frac{1}{2} \, \log \left (-c x + 1\right )\right )}}{{\left (c^{2} e x^{2} +{\left (c^{2} e x^{2} - e\right )} e^{\left (\frac{1}{2} \, \log \left (c x + 1\right ) + \frac{1}{2} \, \log \left (-c x + 1\right )\right )} - e\right )} \sqrt{e x^{2} + d}}\,{d x}\right )} + \frac{\sqrt{e x^{2} + d} a}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsech(c*x))/(e*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 3.58199, size = 2419, normalized size = 15.81 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsech(c*x))/(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

[1/4*(4*sqrt(e*x^2 + d)*b*c*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) + b*c*sqrt(d)*log(((c^4*d^2 -
6*c^2*d*e + e^2)*x^4 - 8*(c^2*d^2 - d*e)*x^2 + 4*((c^3*d - c*e)*x^3 - 2*c*d*x)*sqrt(e*x^2 + d)*sqrt(d)*sqrt(-(
c^2*x^2 - 1)/(c^2*x^2)) + 8*d^2)/x^4) + 4*sqrt(e*x^2 + d)*a*c - b*sqrt(-e)*log(8*c^4*e^2*x^4 + c^4*d^2 - 6*c^2
*d*e + 8*(c^4*d*e - c^2*e^2)*x^2 - 4*(2*c^4*e*x^3 + (c^4*d - c^2*e)*x)*sqrt(e*x^2 + d)*sqrt(-e)*sqrt(-(c^2*x^2
- 1)/(c^2*x^2)) + e^2))/(c*e), 1/4*(4*sqrt(e*x^2 + d)*b*c*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x))
+ b*c*sqrt(d)*log(((c^4*d^2 - 6*c^2*d*e + e^2)*x^4 - 8*(c^2*d^2 - d*e)*x^2 + 4*((c^3*d - c*e)*x^3 - 2*c*d*x)*
sqrt(e*x^2 + d)*sqrt(d)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 8*d^2)/x^4) + 4*sqrt(e*x^2 + d)*a*c - 2*b*sqrt(e)*arc
tan(1/2*(2*c^2*e*x^3 + (c^2*d - e)*x)*sqrt(e*x^2 + d)*sqrt(e)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2))/(c^2*e^2*x^4 + (c
^2*d*e - e^2)*x^2 - d*e)))/(c*e), -1/4*(2*b*c*sqrt(-d)*arctan(-1/2*((c^3*d - c*e)*x^3 - 2*c*d*x)*sqrt(e*x^2 +
d)*sqrt(-d)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2))/(c^2*d*e*x^4 + (c^2*d^2 - d*e)*x^2 - d^2)) - 4*sqrt(e*x^2 + d)*b*c*
log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - 4*sqrt(e*x^2 + d)*a*c + b*sqrt(-e)*log(8*c^4*e^2*x^4 + c
^4*d^2 - 6*c^2*d*e + 8*(c^4*d*e - c^2*e^2)*x^2 - 4*(2*c^4*e*x^3 + (c^4*d - c^2*e)*x)*sqrt(e*x^2 + d)*sqrt(-e)*
sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + e^2))/(c*e), -1/2*(b*c*sqrt(-d)*arctan(-1/2*((c^3*d - c*e)*x^3 - 2*c*d*x)*sqr
t(e*x^2 + d)*sqrt(-d)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2))/(c^2*d*e*x^4 + (c^2*d^2 - d*e)*x^2 - d^2)) - 2*sqrt(e*x^2
+ d)*b*c*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - 2*sqrt(e*x^2 + d)*a*c + b*sqrt(e)*arctan(1/2*(
2*c^2*e*x^3 + (c^2*d - e)*x)*sqrt(e*x^2 + d)*sqrt(e)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2))/(c^2*e^2*x^4 + (c^2*d*e -
e^2)*x^2 - d*e)))/(c*e)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \left (a + b \operatorname{asech}{\left (c x \right )}\right )}{\sqrt{d + e x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*asech(c*x))/(e*x**2+d)**(1/2),x)

[Out]

Integral(x*(a + b*asech(c*x))/sqrt(d + e*x**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsech}\left (c x\right ) + a\right )} x}{\sqrt{e x^{2} + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsech(c*x))/(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

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