3.1.9 \(\int \frac {(d-c^2 d x^2) (a+b \cosh ^{-1}(c x))}{x^4} \, dx\) [9]

Optimal. Leaf size=90 \[ \frac {b c d \sqrt {-1+c x} \sqrt {1+c x}}{6 x^2}-\frac {d \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}+\frac {c^2 d \left (a+b \cosh ^{-1}(c x)\right )}{x}-\frac {5}{6} b c^3 d \text {ArcTan}\left (\sqrt {-1+c x} \sqrt {1+c x}\right ) \]

[Out]

-1/3*d*(a+b*arccosh(c*x))/x^3+c^2*d*(a+b*arccosh(c*x))/x-5/6*b*c^3*d*arctan((c*x-1)^(1/2)*(c*x+1)^(1/2))+1/6*b
*c*d*(c*x-1)^(1/2)*(c*x+1)^(1/2)/x^2

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Rubi [A]
time = 0.09, antiderivative size = 90, 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, 5921, 12, 465, 94, 211} \begin {gather*} \frac {c^2 d \left (a+b \cosh ^{-1}(c x)\right )}{x}-\frac {d \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac {5}{6} b c^3 d \text {ArcTan}\left (\sqrt {c x-1} \sqrt {c x+1}\right )+\frac {b c d \sqrt {c x-1} \sqrt {c x+1}}{6 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d - c^2*d*x^2)*(a + b*ArcCosh[c*x]))/x^4,x]

[Out]

(b*c*d*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(6*x^2) - (d*(a + b*ArcCosh[c*x]))/(3*x^3) + (c^2*d*(a + b*ArcCosh[c*x]))
/x - (5*b*c^3*d*ArcTan[Sqrt[-1 + c*x]*Sqrt[1 + c*x]])/6

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 94

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 211

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

Rule 465

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

Rule 5921

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^4} \, dx &=-\frac {d \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}+\frac {c^2 d \left (a+b \cosh ^{-1}(c x)\right )}{x}-(b c) \int \frac {d \left (-1+3 c^2 x^2\right )}{3 x^3 \sqrt {-1+c x} \sqrt {1+c x}} \, dx\\ &=-\frac {d \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}+\frac {c^2 d \left (a+b \cosh ^{-1}(c x)\right )}{x}-\frac {1}{3} (b c d) \int \frac {-1+3 c^2 x^2}{x^3 \sqrt {-1+c x} \sqrt {1+c x}} \, dx\\ &=\frac {b c d \sqrt {-1+c x} \sqrt {1+c x}}{6 x^2}-\frac {d \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}+\frac {c^2 d \left (a+b \cosh ^{-1}(c x)\right )}{x}-\frac {1}{6} \left (5 b c^3 d\right ) \int \frac {1}{x \sqrt {-1+c x} \sqrt {1+c x}} \, dx\\ &=\frac {b c d \sqrt {-1+c x} \sqrt {1+c x}}{6 x^2}-\frac {d \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}+\frac {c^2 d \left (a+b \cosh ^{-1}(c x)\right )}{x}-\frac {1}{6} \left (5 b c^4 d\right ) \text {Subst}\left (\int \frac {1}{c+c x^2} \, dx,x,\sqrt {-1+c x} \sqrt {1+c x}\right )\\ &=\frac {b c d \sqrt {-1+c x} \sqrt {1+c x}}{6 x^2}-\frac {d \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}+\frac {c^2 d \left (a+b \cosh ^{-1}(c x)\right )}{x}-\frac {5}{6} b c^3 d \tan ^{-1}\left (\sqrt {-1+c x} \sqrt {1+c x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 127, normalized size = 1.41 \begin {gather*} -\frac {a d}{3 x^3}+\frac {a c^2 d}{x}+\frac {b c d \sqrt {-1+c x} \sqrt {1+c x}}{6 x^2}-\frac {b d \cosh ^{-1}(c x)}{3 x^3}+\frac {b c^2 d \cosh ^{-1}(c x)}{x}-\frac {5 b c^3 d \sqrt {-1+c^2 x^2} \text {ArcTan}\left (\sqrt {-1+c^2 x^2}\right )}{6 \sqrt {-1+c x} \sqrt {1+c x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d - c^2*d*x^2)*(a + b*ArcCosh[c*x]))/x^4,x]

[Out]

-1/3*(a*d)/x^3 + (a*c^2*d)/x + (b*c*d*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(6*x^2) - (b*d*ArcCosh[c*x])/(3*x^3) + (b*
c^2*d*ArcCosh[c*x])/x - (5*b*c^3*d*Sqrt[-1 + c^2*x^2]*ArcTan[Sqrt[-1 + c^2*x^2]])/(6*Sqrt[-1 + c*x]*Sqrt[1 + c
*x])

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Maple [A]
time = 1.99, size = 119, normalized size = 1.32

method	result	size
derivativedivides	\(c^{3} \left (-a d \left (-\frac {1}{c x}+\frac {1}{3 c^{3} x^{3}}\right )+\frac {b d \,\mathrm {arccosh}\left (c x \right )}{c x}-\frac {b d \,\mathrm {arccosh}\left (c x \right )}{3 c^{3} x^{3}}+\frac {5 b d \sqrt {c x -1}\, \sqrt {c x +1}\, \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right )}{6 \sqrt {c^{2} x^{2}-1}}+\frac {b d \sqrt {c x -1}\, \sqrt {c x +1}}{6 c^{2} x^{2}}\right )\)	\(119\)
default	\(c^{3} \left (-a d \left (-\frac {1}{c x}+\frac {1}{3 c^{3} x^{3}}\right )+\frac {b d \,\mathrm {arccosh}\left (c x \right )}{c x}-\frac {b d \,\mathrm {arccosh}\left (c x \right )}{3 c^{3} x^{3}}+\frac {5 b d \sqrt {c x -1}\, \sqrt {c x +1}\, \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right )}{6 \sqrt {c^{2} x^{2}-1}}+\frac {b d \sqrt {c x -1}\, \sqrt {c x +1}}{6 c^{2} x^{2}}\right )\)	\(119\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-c^2*d*x^2+d)*(a+b*arccosh(c*x))/x^4,x,method=_RETURNVERBOSE)

[Out]

c^3*(-a*d*(-1/c/x+1/3/c^3/x^3)+b*d*arccosh(c*x)/c/x-1/3*b*d*arccosh(c*x)/c^3/x^3+5/6*b*d*(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))+1/6*b*d*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c^2/x^2)

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Maxima [A]
time = 0.47, size = 89, normalized size = 0.99 \begin {gather*} {\left (c \arcsin \left (\frac {1}{c {\left | x \right |}}\right ) + \frac {\operatorname {arcosh}\left (c x\right )}{x}\right )} b c^{2} d - \frac {1}{6} \, {\left ({\left (c^{2} \arcsin \left (\frac {1}{c {\left | x \right |}}\right ) - \frac {\sqrt {c^{2} x^{2} - 1}}{x^{2}}\right )} c + \frac {2 \, \operatorname {arcosh}\left (c x\right )}{x^{3}}\right )} b d + \frac {a c^{2} d}{x} - \frac {a d}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)*(a+b*arccosh(c*x))/x^4,x, algorithm="maxima")

[Out]

(c*arcsin(1/(c*abs(x))) + arccosh(c*x)/x)*b*c^2*d - 1/6*((c^2*arcsin(1/(c*abs(x))) - sqrt(c^2*x^2 - 1)/x^2)*c
+ 2*arccosh(c*x)/x^3)*b*d + a*c^2*d/x - 1/3*a*d/x^3

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Fricas [A]
time = 0.39, size = 146, normalized size = 1.62 \begin {gather*} -\frac {10 \, b c^{3} d x^{3} \arctan \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) - 6 \, a c^{2} d x^{2} + 2 \, {\left (3 \, b c^{2} - b\right )} d x^{3} \log \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) - \sqrt {c^{2} x^{2} - 1} b c d x + 2 \, a d - 2 \, {\left (3 \, b c^{2} d x^{2} - {\left (3 \, b c^{2} - b\right )} d x^{3} - b d\right )} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right )}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)*(a+b*arccosh(c*x))/x^4,x, algorithm="fricas")

[Out]

-1/6*(10*b*c^3*d*x^3*arctan(-c*x + sqrt(c^2*x^2 - 1)) - 6*a*c^2*d*x^2 + 2*(3*b*c^2 - b)*d*x^3*log(-c*x + sqrt(
c^2*x^2 - 1)) - sqrt(c^2*x^2 - 1)*b*c*d*x + 2*a*d - 2*(3*b*c^2*d*x^2 - (3*b*c^2 - b)*d*x^3 - b*d)*log(c*x + sq
rt(c^2*x^2 - 1)))/x^3

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - d \left (\int \left (- \frac {a}{x^{4}}\right )\, dx + \int \frac {a c^{2}}{x^{2}}\, dx + \int \left (- \frac {b \operatorname {acosh}{\left (c x \right )}}{x^{4}}\right )\, dx + \int \frac {b c^{2} \operatorname {acosh}{\left (c x \right )}}{x^{2}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c**2*d*x**2+d)*(a+b*acosh(c*x))/x**4,x)

[Out]

-d*(Integral(-a/x**4, x) + Integral(a*c**2/x**2, x) + Integral(-b*acosh(c*x)/x**4, x) + Integral(b*c**2*acosh(
c*x)/x**2, x))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)*(a+b*arccosh(c*x))/x^4,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,\left (d-c^2\,d\,x^2\right )}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*acosh(c*x))*(d - c^2*d*x^2))/x^4,x)

[Out]

int(((a + b*acosh(c*x))*(d - c^2*d*x^2))/x^4, x)

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