3.7 \(\int \frac {(d-c^2 d x^2) (a+b \cosh ^{-1}(c x))}{x^2} \, dx\)

Optimal. Leaf size=76 \[ c^2 (-d) x \left (a+b \cosh ^{-1}(c x)\right )-\frac {d \left (a+b \cosh ^{-1}(c x)\right )}{x}+b c d \sqrt {c x-1} \sqrt {c x+1}+b c d \tan ^{-1}\left (\sqrt {c x-1} \sqrt {c x+1}\right ) \]

[Out]

-d*(a+b*arccosh(c*x))/x-c^2*d*x*(a+b*arccosh(c*x))+b*c*d*arctan((c*x-1)^(1/2)*(c*x+1)^(1/2))+b*c*d*(c*x-1)^(1/
2)*(c*x+1)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {14, 5731, 12, 460, 92, 205} \[ c^2 (-d) x \left (a+b \cosh ^{-1}(c x)\right )-\frac {d \left (a+b \cosh ^{-1}(c x)\right )}{x}+b c d \sqrt {c x-1} \sqrt {c x+1}+b c d \tan ^{-1}\left (\sqrt {c x-1} \sqrt {c x+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[((d - c^2*d*x^2)*(a + b*ArcCosh[c*x]))/x^2,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 92

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

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 460

Int[((e_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.)*(x_)^(non2_.))^(p_.)*((c_) + (d_.)
*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m + 1)*(a1 + b1*x^(n/2))^(p + 1)*(a2 + b2*x^(n/2))^(p + 1))/(b1*b2*e*
(m + n*(p + 1) + 1)), x] - Dist[(a1*a2*d*(m + 1) - b1*b2*c*(m + n*(p + 1) + 1))/(b1*b2*(m + n*(p + 1) + 1)), I
nt[(e*x)^m*(a1 + b1*x^(n/2))^p*(a2 + b2*x^(n/2))^p, x], x] /; FreeQ[{a1, b1, a2, b2, c, d, e, m, n, p}, x] &&
EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] && NeQ[m + n*(p + 1) + 1, 0]

Rule 5731

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcCosh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(Sqrt[1
+ c*x]*Sqrt[-1 + c*x]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 110, normalized size = 1.45 \[ -a c^2 d x-\frac {a d}{x}+\frac {b c d \sqrt {c^2 x^2-1} \tan ^{-1}\left (\sqrt {c^2 x^2-1}\right )}{\sqrt {c x-1} \sqrt {c x+1}}-b c^2 d x \cosh ^{-1}(c x)+b c d \sqrt {c x-1} \sqrt {c x+1}-\frac {b d \cosh ^{-1}(c x)}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[((d - c^2*d*x^2)*(a + b*ArcCosh[c*x]))/x^2,x]

[Out]

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

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fricas [A]  time = 0.51, size = 127, normalized size = 1.67 \[ -\frac {a c^{2} d x^{2} - 2 \, b c d x \arctan \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) - \sqrt {c^{2} x^{2} - 1} b c d x - {\left (b c^{2} + b\right )} d x \log \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) + a d + {\left (b c^{2} d x^{2} - {\left (b c^{2} + b\right )} d x + b d\right )} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)*(a+b*arccosh(c*x))/x^2,x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)*(a+b*arccosh(c*x))/x^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.02, size = 100, normalized size = 1.32 \[ -d a \,c^{2} x -\frac {d a}{x}-d b \,\mathrm {arccosh}\left (c x \right ) c^{2} x -\frac {d b \,\mathrm {arccosh}\left (c x \right )}{x}+b c d \sqrt {c x -1}\, \sqrt {c x +1}-\frac {c d b \sqrt {c x -1}\, \sqrt {c x +1}\, \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right )}{\sqrt {c^{2} x^{2}-1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-c^2*d*x^2+d)*(a+b*arccosh(c*x))/x^2,x)

[Out]

-d*a*c^2*x-d*a/x-d*b*arccosh(c*x)*c^2*x-d*b*arccosh(c*x)/x+b*c*d*(c*x-1)^(1/2)*(c*x+1)^(1/2)-c*d*b*(c*x-1)^(1/
2)*(c*x+1)^(1/2)/(c^2*x^2-1)^(1/2)*arctan(1/(c^2*x^2-1)^(1/2))

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maxima [A]  time = 0.45, size = 66, normalized size = 0.87 \[ -a c^{2} d x - {\left (c x \operatorname {arcosh}\left (c x\right ) - \sqrt {c^{2} x^{2} - 1}\right )} b c d - {\left (c \arcsin \left (\frac {1}{c {\left | x \right |}}\right ) + \frac {\operatorname {arcosh}\left (c x\right )}{x}\right )} b d - \frac {a d}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)*(a+b*arccosh(c*x))/x^2,x, algorithm="maxima")

[Out]

-a*c^2*d*x - (c*x*arccosh(c*x) - sqrt(c^2*x^2 - 1))*b*c*d - (c*arcsin(1/(c*abs(x))) + arccosh(c*x)/x)*b*d - a*
d/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c**2*d*x**2+d)*(a+b*acosh(c*x))/x**2,x)

[Out]

-d*(Integral(a*c**2, x) + Integral(-a/x**2, x) + Integral(b*c**2*acosh(c*x), x) + Integral(-b*acosh(c*x)/x**2,
x))

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