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

Optimal. Leaf size=124 \[ \frac {1}{2} x \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )-\frac {\sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )^2}{4 b c \sqrt {c x-1} \sqrt {c x+1}}-\frac {b c x^2 \sqrt {d-c^2 d x^2}}{4 \sqrt {c x-1} \sqrt {c x+1}} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5713, 5683, 5676, 30} \[ \frac {1}{2} x \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )-\frac {\sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )^2}{4 b c \sqrt {c x-1} \sqrt {c x+1}}-\frac {b c x^2 \sqrt {d-c^2 d x^2}}{4 \sqrt {c x-1} \sqrt {c x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*x^2*Sqrt[d - c^2*d*x^2])/(4*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) + (x*Sqrt[d - c^2*d*x^2]*(a + b*ArcCosh[c*x]))
/2 - (Sqrt[d - c^2*d*x^2]*(a + b*ArcCosh[c*x])^2)/(4*b*c*Sqrt[-1 + c*x]*Sqrt[1 + c*x])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 5676

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_.)*(x_)]), x_Symbol]
:> Simp[(a + b*ArcCosh[c*x])^(n + 1)/(b*c*Sqrt[-(d1*d2)]*(n + 1)), x] /; FreeQ[{a, b, c, d1, e1, d2, e2, n},
x] && EqQ[e1, c*d1] && EqQ[e2, -(c*d2)] && GtQ[d1, 0] && LtQ[d2, 0] && NeQ[n, -1]

Rule 5683

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_.)*(x_)], x_Symbol] :
> Simp[(x*Sqrt[d1 + e1*x]*Sqrt[d2 + e2*x]*(a + b*ArcCosh[c*x])^n)/2, x] + (-Dist[(Sqrt[d1 + e1*x]*Sqrt[d2 + e2
*x])/(2*Sqrt[1 + c*x]*Sqrt[-1 + c*x]), Int[(a + b*ArcCosh[c*x])^n/(Sqrt[1 + c*x]*Sqrt[-1 + c*x]), x], x] - Dis
t[(b*c*n*Sqrt[d1 + e1*x]*Sqrt[d2 + e2*x])/(2*Sqrt[1 + c*x]*Sqrt[-1 + c*x]), Int[x*(a + b*ArcCosh[c*x])^(n - 1)
, x], x]) /; FreeQ[{a, b, c, d1, e1, d2, e2}, x] && EqQ[e1, c*d1] && EqQ[e2, -(c*d2)] && GtQ[n, 0]

Rule 5713

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Dist[((-d)^IntPart[p]*(
d + e*x^2)^FracPart[p])/((1 + c*x)^FracPart[p]*(-1 + c*x)^FracPart[p]), Int[(1 + c*x)^p*(-1 + c*x)^p*(a + b*Ar
cCosh[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[c^2*d + e, 0] &&  !IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.61, size = 144, normalized size = 1.16 \[ \frac {1}{8} \left (4 a x \sqrt {d-c^2 d x^2}-\frac {4 a \sqrt {d} \tan ^{-1}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (c^2 x^2-1\right )}\right )}{c}-\frac {b \sqrt {d-c^2 d x^2} \left (2 \cosh ^{-1}(c x)^2+\cosh \left (2 \cosh ^{-1}(c x)\right )-2 \cosh ^{-1}(c x) \sinh \left (2 \cosh ^{-1}(c x)\right )\right )}{c \sqrt {\frac {c x-1}{c x+1}} (c x+1)}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*a*x*Sqrt[d - c^2*d*x^2] - (4*a*Sqrt[d]*ArcTan[(c*x*Sqrt[d - c^2*d*x^2])/(Sqrt[d]*(-1 + c^2*x^2))])/c - (b*S
qrt[d - c^2*d*x^2]*(2*ArcCosh[c*x]^2 + Cosh[2*ArcCosh[c*x]] - 2*ArcCosh[c*x]*Sinh[2*ArcCosh[c*x]]))/(c*Sqrt[(-
1 + c*x)/(1 + c*x)]*(1 + c*x)))/8

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {-c^{2} d x^{2} + d} {\left (b \operatorname {arcosh}\left (c x\right ) + a\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-c^2*d*x^2 + d)*(b*arccosh(c*x) + a), 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((a+b*arccosh(c*x))*(-c^2*d*x^2+d)^(1/2),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 [B]  time = 0.24, size = 239, normalized size = 1.93 \[ \frac {a x \sqrt {-c^{2} d \,x^{2}+d}}{2}+\frac {a d \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{2 \sqrt {c^{2} d}}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \mathrm {arccosh}\left (c x \right )^{2}}{4 \sqrt {c x -1}\, \sqrt {c x +1}\, c}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c^{2} \mathrm {arccosh}\left (c x \right ) x^{3}}{2 \left (c x +1\right ) \left (c x -1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c \,x^{2}}{4 \sqrt {c x +1}\, \sqrt {c x -1}}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \mathrm {arccosh}\left (c x \right ) x}{2 \left (c x +1\right ) \left (c x -1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}}{8 \sqrt {c x +1}\, \sqrt {c x -1}\, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x*(-c^2*d*x^2+d)^(1/2)+1/2*a*d/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))-1/4*b*(-d*(c^2
*x^2-1))^(1/2)/(c*x-1)^(1/2)/(c*x+1)^(1/2)/c*arccosh(c*x)^2+1/2*b*(-d*(c^2*x^2-1))^(1/2)/(c*x+1)/(c*x-1)*c^2*a
rccosh(c*x)*x^3-1/4*b*(-d*(c^2*x^2-1))^(1/2)/(c*x+1)^(1/2)/(c*x-1)^(1/2)*c*x^2-1/2*b*(-d*(c^2*x^2-1))^(1/2)/(c
*x+1)/(c*x-1)*arccosh(c*x)*x+1/8*b*(-d*(c^2*x^2-1))^(1/2)/(c*x+1)^(1/2)/(c*x-1)^(1/2)/c

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, {\left (\sqrt {-c^{2} d x^{2} + d} x + \frac {\sqrt {d} \arcsin \left (c x\right )}{c}\right )} a + b \int \sqrt {-c^{2} d x^{2} + d} \log \left (c x + \sqrt {c x + 1} \sqrt {c x - 1}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(c*x))*(d - c^2*d*x^2)^(1/2),x)

[Out]

int((a + b*acosh(c*x))*(d - c^2*d*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )} \left (a + b \operatorname {acosh}{\left (c x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-d*(c*x - 1)*(c*x + 1))*(a + b*acosh(c*x)), x)

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