3.1.29 \(\int \frac {x^3 (a+b \cosh ^{-1}(c x))}{d-c^2 d x^2} \, dx\) [29]

Optimal. Leaf size=140 \[ \frac {b x \sqrt {-1+c x} \sqrt {1+c x}}{4 c^3 d}+\frac {b \cosh ^{-1}(c x)}{4 c^4 d}-\frac {x^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 c^2 d}+\frac {\left (a+b \cosh ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac {\left (a+b \cosh ^{-1}(c x)\right ) \log \left (1-e^{2 \cosh ^{-1}(c x)}\right )}{c^4 d}-\frac {b \text {PolyLog}\left (2,e^{2 \cosh ^{-1}(c x)}\right )}{2 c^4 d} \]

[Out]

1/4*b*arccosh(c*x)/c^4/d-1/2*x^2*(a+b*arccosh(c*x))/c^2/d+1/2*(a+b*arccosh(c*x))^2/b/c^4/d-(a+b*arccosh(c*x))*
ln(1-(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2)/c^4/d-1/2*b*polylog(2,(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2)/c^4/d+1/4
*b*x*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c^3/d

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Rubi [A]
time = 0.14, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {5938, 5913, 3797, 2221, 2317, 2438, 92, 54} \begin {gather*} \frac {\left (a+b \cosh ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac {\log \left (1-e^{2 \cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{c^4 d}-\frac {x^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 c^2 d}-\frac {b \text {Li}_2\left (e^{2 \cosh ^{-1}(c x)}\right )}{2 c^4 d}+\frac {b \cosh ^{-1}(c x)}{4 c^4 d}+\frac {b x \sqrt {c x-1} \sqrt {c x+1}}{4 c^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcCosh[c*x]))/(d - c^2*d*x^2),x]

[Out]

(b*x*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(4*c^3*d) + (b*ArcCosh[c*x])/(4*c^4*d) - (x^2*(a + b*ArcCosh[c*x]))/(2*c^2*
d) + (a + b*ArcCosh[c*x])^2/(2*b*c^4*d) - ((a + b*ArcCosh[c*x])*Log[1 - E^(2*ArcCosh[c*x])])/(c^4*d) - (b*Poly
Log[2, E^(2*ArcCosh[c*x])])/(2*c^4*d)

Rule 54

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ArcCosh[b*(x/a)]/b, x] /; FreeQ[{a,
b, c, d}, x] && EqQ[a + c, 0] && EqQ[b - d, 0] && GtQ[a, 0]

Rule 92

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a + b*x
)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3797

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> Simp[(-I)*((
c + d*x)^(m + 1)/(d*(m + 1))), x] + Dist[2*I, Int[((c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*
fz*x))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 5913

Int[(((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/e, Subst[Int[(
a + b*x)^n*Coth[x], x], x, ArcCosh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0]

Rule 5938

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcCosh[c*x])^n/(e*(m + 2*p + 1))), x] + (Dist[f^2*((m - 1)/(c^2*(
m + 2*p + 1))), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcCosh[c*x])^n, x], x] - Dist[b*f*(n/(c*(m + 2*p + 1))
)*Simp[(d + e*x^2)^p/((1 + c*x)^p*(-1 + c*x)^p)], Int[(f*x)^(m - 1)*(1 + c*x)^(p + 1/2)*(-1 + c*x)^(p + 1/2)*(
a + b*ArcCosh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && I
GtQ[m, 1] && NeQ[m + 2*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{d-c^2 d x^2} \, dx &=-\frac {x^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 c^2 d}+\frac {\int \frac {x \left (a+b \cosh ^{-1}(c x)\right )}{d-c^2 d x^2} \, dx}{c^2}+\frac {b \int \frac {x^2}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{2 c d}\\ &=\frac {b x \sqrt {-1+c x} \sqrt {1+c x}}{4 c^3 d}-\frac {x^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 c^2 d}-\frac {\text {Subst}\left (\int (a+b x) \coth (x) \, dx,x,\cosh ^{-1}(c x)\right )}{c^4 d}+\frac {b \int \frac {1}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{4 c^3 d}\\ &=\frac {b x \sqrt {-1+c x} \sqrt {1+c x}}{4 c^3 d}+\frac {b \cosh ^{-1}(c x)}{4 c^4 d}-\frac {x^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 c^2 d}+\frac {\left (a+b \cosh ^{-1}(c x)\right )^2}{2 b c^4 d}+\frac {2 \text {Subst}\left (\int \frac {e^{2 x} (a+b x)}{1-e^{2 x}} \, dx,x,\cosh ^{-1}(c x)\right )}{c^4 d}\\ &=\frac {b x \sqrt {-1+c x} \sqrt {1+c x}}{4 c^3 d}+\frac {b \cosh ^{-1}(c x)}{4 c^4 d}-\frac {x^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 c^2 d}+\frac {\left (a+b \cosh ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac {\left (a+b \cosh ^{-1}(c x)\right ) \log \left (1-e^{2 \cosh ^{-1}(c x)}\right )}{c^4 d}+\frac {b \text {Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\cosh ^{-1}(c x)\right )}{c^4 d}\\ &=\frac {b x \sqrt {-1+c x} \sqrt {1+c x}}{4 c^3 d}+\frac {b \cosh ^{-1}(c x)}{4 c^4 d}-\frac {x^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 c^2 d}+\frac {\left (a+b \cosh ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac {\left (a+b \cosh ^{-1}(c x)\right ) \log \left (1-e^{2 \cosh ^{-1}(c x)}\right )}{c^4 d}+\frac {b \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 \cosh ^{-1}(c x)}\right )}{2 c^4 d}\\ &=\frac {b x \sqrt {-1+c x} \sqrt {1+c x}}{4 c^3 d}+\frac {b \cosh ^{-1}(c x)}{4 c^4 d}-\frac {x^2 \left (a+b \cosh ^{-1}(c x)\right )}{2 c^2 d}+\frac {\left (a+b \cosh ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac {\left (a+b \cosh ^{-1}(c x)\right ) \log \left (1-e^{2 \cosh ^{-1}(c x)}\right )}{c^4 d}-\frac {b \text {Li}_2\left (e^{2 \cosh ^{-1}(c x)}\right )}{2 c^4 d}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 151, normalized size = 1.08 \begin {gather*} -\frac {2 c^2 x^2 \left (a+b \cosh ^{-1}(c x)\right )-\frac {2 \left (a+b \cosh ^{-1}(c x)\right )^2}{b}-b \left (c x \sqrt {-1+c x} \sqrt {1+c x}+2 \tanh ^{-1}\left (\sqrt {\frac {-1+c x}{1+c x}}\right )\right )+4 \left (a+b \cosh ^{-1}(c x)\right ) \log \left (1-e^{\cosh ^{-1}(c x)}\right )+4 \left (a+b \cosh ^{-1}(c x)\right ) \log \left (1+e^{\cosh ^{-1}(c x)}\right )+4 b \text {PolyLog}\left (2,-e^{\cosh ^{-1}(c x)}\right )+4 b \text {PolyLog}\left (2,e^{\cosh ^{-1}(c x)}\right )}{4 c^4 d} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^3*(a + b*ArcCosh[c*x]))/(d - c^2*d*x^2),x]

[Out]

-1/4*(2*c^2*x^2*(a + b*ArcCosh[c*x]) - (2*(a + b*ArcCosh[c*x])^2)/b - b*(c*x*Sqrt[-1 + c*x]*Sqrt[1 + c*x] + 2*
ArcTanh[Sqrt[(-1 + c*x)/(1 + c*x)]]) + 4*(a + b*ArcCosh[c*x])*Log[1 - E^ArcCosh[c*x]] + 4*(a + b*ArcCosh[c*x])
*Log[1 + E^ArcCosh[c*x]] + 4*b*PolyLog[2, -E^ArcCosh[c*x]] + 4*b*PolyLog[2, E^ArcCosh[c*x]])/(c^4*d)

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Maple [A]
time = 4.57, size = 222, normalized size = 1.59

method	result	size
derivativedivides	\(\frac {-\frac {a \,c^{2} x^{2}}{2 d}-\frac {a \ln \left (c x -1\right )}{2 d}-\frac {a \ln \left (c x +1\right )}{2 d}+\frac {b \mathrm {arccosh}\left (c x \right )^{2}}{2 d}-\frac {b \,\mathrm {arccosh}\left (c x \right ) c^{2} x^{2}}{2 d}+\frac {b \sqrt {c x +1}\, \sqrt {c x -1}\, c x}{4 d}+\frac {b \,\mathrm {arccosh}\left (c x \right )}{4 d}-\frac {b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1+c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d}-\frac {b \polylog \left (2, -c x -\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d}-\frac {b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1-c x -\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d}-\frac {b \polylog \left (2, c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d}}{c^{4}}\)	\(222\)
default	\(\frac {-\frac {a \,c^{2} x^{2}}{2 d}-\frac {a \ln \left (c x -1\right )}{2 d}-\frac {a \ln \left (c x +1\right )}{2 d}+\frac {b \mathrm {arccosh}\left (c x \right )^{2}}{2 d}-\frac {b \,\mathrm {arccosh}\left (c x \right ) c^{2} x^{2}}{2 d}+\frac {b \sqrt {c x +1}\, \sqrt {c x -1}\, c x}{4 d}+\frac {b \,\mathrm {arccosh}\left (c x \right )}{4 d}-\frac {b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1+c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d}-\frac {b \polylog \left (2, -c x -\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d}-\frac {b \,\mathrm {arccosh}\left (c x \right ) \ln \left (1-c x -\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d}-\frac {b \polylog \left (2, c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{d}}{c^{4}}\)	\(222\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arccosh(c*x))/(-c^2*d*x^2+d),x,method=_RETURNVERBOSE)

[Out]

1/c^4*(-1/2*a/d*c^2*x^2-1/2*a/d*ln(c*x-1)-1/2*a/d*ln(c*x+1)+1/2*b/d*arccosh(c*x)^2-1/2*b/d*arccosh(c*x)*c^2*x^
2+1/4*b/d*(c*x+1)^(1/2)*(c*x-1)^(1/2)*c*x+1/4*b/d*arccosh(c*x)-b/d*arccosh(c*x)*ln(1+c*x+(c*x-1)^(1/2)*(c*x+1)
^(1/2))-b/d*polylog(2,-c*x-(c*x-1)^(1/2)*(c*x+1)^(1/2))-b/d*arccosh(c*x)*ln(1-c*x-(c*x-1)^(1/2)*(c*x+1)^(1/2))
-b/d*polylog(2,c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccosh(c*x))/(-c^2*d*x^2+d),x, algorithm="maxima")

[Out]

-1/2*a*(x^2/(c^2*d) + log(c^2*x^2 - 1)/(c^4*d)) + 1/8*b*((2*c^2*x^2 - 4*(c^2*x^2 + log(c*x + 1) + log(c*x - 1)
)*log(c*x + sqrt(c*x + 1)*sqrt(c*x - 1)) + 2*(log(c*x - 1) + 1)*log(c*x + 1) + log(c*x + 1)^2 + log(c*x - 1)^2
+ 2*log(c*x - 1))/(c^4*d) - 8*integrate(1/2*(c^2*x^2 + log(c*x + 1) + log(c*x - 1))/(c^6*d*x^3 - c^4*d*x + (c
^5*d*x^2 - c^3*d)*e^(1/2*log(c*x + 1) + 1/2*log(c*x - 1))), x))

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Fricas [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(x^3*(a+b*arccosh(c*x))/(-c^2*d*x^2+d),x, algorithm="fricas")

[Out]

integral(-(b*x^3*arccosh(c*x) + a*x^3)/(c^2*d*x^2 - d), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {a x^{3}}{c^{2} x^{2} - 1}\, dx + \int \frac {b x^{3} \operatorname {acosh}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*acosh(c*x))/(-c**2*d*x**2+d),x)

[Out]

-(Integral(a*x**3/(c**2*x**2 - 1), x) + Integral(b*x**3*acosh(c*x)/(c**2*x**2 - 1), x))/d

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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(x^3*(a+b*arccosh(c*x))/(-c^2*d*x^2+d),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 {x^3\,\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )}{d-c^2\,d\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*acosh(c*x)))/(d - c^2*d*x^2),x)

[Out]

int((x^3*(a + b*acosh(c*x)))/(d - c^2*d*x^2), x)

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