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

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

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {5798, 5718, 207} \[ \frac {a+b \cosh ^{-1}(c x)}{c^2 d \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {c x-1} \sqrt {c x+1} \tanh ^{-1}(c x)}{c^2 d \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 5718

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

Rule 5798

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_.)*((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[(f*x)^m*(1 + c*
x)^p*(-1 + c*x)^p*(a + b*ArcCosh[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[c^2*d + e, 0]
&&  !IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.24, size = 90, normalized size = 1.18 \[ -\frac {\sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{c^2 d^2 \left (c^2 x^2-1\right )}-\frac {b \sqrt {-d \left (c^2 x^2-1\right )} \tanh ^{-1}(c x)}{c^2 d^2 \sqrt {c x-1} \sqrt {c x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.78, size = 327, normalized size = 4.30 \[ \left [-\frac {4 \, \sqrt {-c^{2} d x^{2} + d} b \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) + {\left (b c^{2} x^{2} - b\right )} \sqrt {-d} \log \left (-\frac {c^{6} d x^{6} + 5 \, c^{4} d x^{4} - 5 \, c^{2} d x^{2} - 4 \, {\left (c^{3} x^{3} + c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} \sqrt {-d} - d}{c^{6} x^{6} - 3 \, c^{4} x^{4} + 3 \, c^{2} x^{2} - 1}\right ) + 4 \, \sqrt {-c^{2} d x^{2} + d} a}{4 \, {\left (c^{4} d^{2} x^{2} - c^{2} d^{2}\right )}}, -\frac {{\left (b c^{2} x^{2} - b\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} c \sqrt {d} x}{c^{4} d x^{4} - d}\right ) + 2 \, \sqrt {-c^{2} d x^{2} + d} b \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) + 2 \, \sqrt {-c^{2} d x^{2} + d} a}{2 \, {\left (c^{4} d^{2} x^{2} - c^{2} d^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(4*sqrt(-c^2*d*x^2 + d)*b*log(c*x + sqrt(c^2*x^2 - 1)) + (b*c^2*x^2 - b)*sqrt(-d)*log(-(c^6*d*x^6 + 5*c^
4*d*x^4 - 5*c^2*d*x^2 - 4*(c^3*x^3 + c*x)*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*sqrt(-d) - d)/(c^6*x^6 - 3*c^
4*x^4 + 3*c^2*x^2 - 1)) + 4*sqrt(-c^2*d*x^2 + d)*a)/(c^4*d^2*x^2 - c^2*d^2), -1/2*((b*c^2*x^2 - b)*sqrt(d)*arc
tan(2*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*c*sqrt(d)*x/(c^4*d*x^4 - d)) + 2*sqrt(-c^2*d*x^2 + d)*b*log(c*x +
sqrt(c^2*x^2 - 1)) + 2*sqrt(-c^2*d*x^2 + d)*a)/(c^4*d^2*x^2 - c^2*d^2)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.24, size = 198, normalized size = 2.61 \[ \frac {a}{c^{2} d \sqrt {-c^{2} d \,x^{2}+d}}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \mathrm {arccosh}\left (c x \right )}{c^{2} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x -1}\, \sqrt {c x +1}\, \ln \left (c x +\sqrt {c x -1}\, \sqrt {c x +1}-1\right )}{c^{2} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x -1}\, \sqrt {c x +1}\, \ln \left (1+c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )}{c^{2} d^{2} \left (c^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a/c^2/d/(-c^2*d*x^2+d)^(1/2)-b*(-d*(c^2*x^2-1))^(1/2)/c^2/d^2/(c^2*x^2-1)*arccosh(c*x)+b*(-d*(c^2*x^2-1))^(1/2
)*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c^2/d^2/(c^2*x^2-1)*ln(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2)-1)-b*(-d*(c^2*x^2-1))^(1/
2)*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c^2/d^2/(c^2*x^2-1)*ln(1+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 \[ b {\left (\frac {\frac {{\left (c \sqrt {d} x + \sqrt {c x + 1} \sqrt {c x - 1} \sqrt {d}\right )} \log \left (c x + \sqrt {c x + 1} \sqrt {c x - 1}\right )}{\sqrt {-c x + 1}} + \frac {\sqrt {c x + 1} \sqrt {c x - 1} \sqrt {d}}{\sqrt {-c x + 1}}}{\sqrt {c x + 1} c^{3} d^{2} x + {\left (c x + 1\right )} \sqrt {c x - 1} c^{2} d^{2}} - \int \frac {c^{2} x^{3} + c x^{2} e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (c x - 1\right )\right )} - x}{\sqrt {-c x + 1} {\left ({\left (c^{2} d^{\frac {3}{2}} x^{2} - d^{\frac {3}{2}}\right )} e^{\left (\frac {3}{2} \, \log \left (c x + 1\right ) + \log \left (c x - 1\right )\right )} + 2 \, {\left (c^{3} d^{\frac {3}{2}} x^{3} - c d^{\frac {3}{2}} x\right )} e^{\left (\log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (c x - 1\right )\right )} + {\left (c^{4} d^{\frac {3}{2}} x^{4} - c^{2} d^{\frac {3}{2}} x^{2}\right )} \sqrt {c x + 1}\right )}}\,{d x}\right )} + \frac {a}{\sqrt {-c^{2} d x^{2} + d} c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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