3.2.17 \(\int \frac {x^2 (a+b \cosh ^{-1}(c x))}{(d-c^2 d x^2)^{3/2}} \, dx\) [117]

Optimal. Leaf size=143 \[ \frac {x \left (a+b \cosh ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {\sqrt {-1+c x} \sqrt {1+c x} \left (a+b \cosh ^{-1}(c x)\right )^2}{2 b c^3 d \sqrt {d-c^2 d x^2}}-\frac {b \sqrt {-1+c x} \sqrt {1+c x} \log \left (1-c^2 x^2\right )}{2 c^3 d \sqrt {d-c^2 d x^2}} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {5934, 5892, 74, 266} \begin {gather*} \frac {x \left (a+b \cosh ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {\sqrt {c x-1} \sqrt {c x+1} \left (a+b \cosh ^{-1}(c x)\right )^2}{2 b c^3 d \sqrt {d-c^2 d x^2}}-\frac {b \sqrt {c x-1} \sqrt {c x+1} \log \left (1-c^2 x^2\right )}{2 c^3 d \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 74

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 5892

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*S
imp[Sqrt[1 + c*x]*(Sqrt[-1 + c*x]/Sqrt[d + e*x^2])]*(a + b*ArcCosh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e,
n}, x] && EqQ[c^2*d + e, 0] && NeQ[n, -1]

Rule 5934

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/(2*e*(p + 1))), x] + (-Dist[f^2*((m - 1)/(2*e*(p +
1))), Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcCosh[c*x])^n, x], x] - Dist[b*f*(n/(2*c*(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*A
rcCosh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1]
&& IGtQ[m, 1]

Rubi steps

\begin {align*} \int \frac {x^2 \left (a+b \cosh ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{3/2}} \, dx &=-\frac {\left (\sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {x^2 \left (a+b \cosh ^{-1}(c x)\right )}{(-1+c x)^{3/2} (1+c x)^{3/2}} \, dx}{d \sqrt {d-c^2 d x^2}}\\ &=\frac {x \left (a+b \cosh ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {\left (\sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {a+b \cosh ^{-1}(c x)}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {\left (b \sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {x}{-1+c^2 x^2} \, dx}{c d \sqrt {d-c^2 d x^2}}\\ &=\frac {x \left (a+b \cosh ^{-1}(c x)\right )}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {\sqrt {-1+c x} \sqrt {1+c x} \left (a+b \cosh ^{-1}(c x)\right )^2}{2 b c^3 d \sqrt {d-c^2 d x^2}}-\frac {b \sqrt {-1+c x} \sqrt {1+c x} \log \left (1-c^2 x^2\right )}{2 c^3 d \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.42, size = 159, normalized size = 1.11 \begin {gather*} \frac {2 a c d x+2 a \sqrt {d} \sqrt {d-c^2 d x^2} \text {ArcTan}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (-1+c^2 x^2\right )}\right )+b d \left (2 c x \cosh ^{-1}(c x)-\sqrt {\frac {-1+c x}{1+c x}} (1+c x) \left (\cosh ^{-1}(c x)^2+2 \log \left (\sqrt {\frac {-1+c x}{1+c x}} (1+c x)\right )\right )\right )}{2 c^3 d^2 \sqrt {d-c^2 d x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(278\) vs. \(2(125)=250\).
time = 5.05, size = 279, normalized size = 1.95

method	result	size
default	\(\frac {a x}{c^{2} d \sqrt {-c^{2} d \,x^{2}+d}}-\frac {a \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{c^{2} d \sqrt {c^{2} d}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x -1}\, \sqrt {c x +1}\, \mathrm {arccosh}\left (c x \right )^{2}}{2 d^{2} c^{3} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x -1}\, \sqrt {c x +1}\, \mathrm {arccosh}\left (c x \right )}{d^{2} c^{3} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \mathrm {arccosh}\left (c x \right ) x}{d^{2} c^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x -1}\, \sqrt {c x +1}\, \ln \left (\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}-1\right )}{d^{2} c^{3} \left (c^{2} x^{2}-1\right )}\)	\(279\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

a*(x/(sqrt(-c^2*d*x^2 + d)*c^2*d) - arcsin(c*x)/(c^3*d^(3/2))) + b*integrate(x^2*log(c*x + sqrt(c*x + 1)*sqrt(
c*x - 1))/(-c^2*d*x^2 + d)^(3/2), 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^2*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(-c^2*d*x^2 + d)*(b*x^2*arccosh(c*x) + a*x^2)/(c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*(a + b*acosh(c*x))/(-d*(c*x - 1)*(c*x + 1))**(3/2), 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(x^2*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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