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

Optimal. Leaf size=150 \[ \frac{\sqrt{d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{c^4 d^2}+\frac{a+b \cosh ^{-1}(c x)}{c^4 d \sqrt{d-c^2 d x^2}}-\frac{b x \sqrt{d-c^2 d x^2}}{c^3 d^2 \sqrt{c x-1} \sqrt{c x+1}}-\frac{b \sqrt{d-c^2 d x^2} \tanh ^{-1}(c x)}{c^4 d^2 \sqrt{c x-1} \sqrt{c x+1}} \]

[Out]

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

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Rubi [A]  time = 0.385017, antiderivative size = 163, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {5798, 98, 21, 74, 5733, 388, 208} \[ \frac{x^2 \left (a+b \cosh ^{-1}(c x)\right )}{c^2 d \sqrt{d-c^2 d x^2}}+\frac{2 (1-c x) (c x+1) \left (a+b \cosh ^{-1}(c x)\right )}{c^4 d \sqrt{d-c^2 d x^2}}+\frac{b x \sqrt{c x-1} \sqrt{c x+1}}{c^3 d \sqrt{d-c^2 d x^2}}+\frac{b \sqrt{c x-1} \sqrt{c x+1} \tanh ^{-1}(c x)}{c^4 d \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 98

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 74

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

Rule 5733

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d1_) + (e1_.)*(x_))^(p_)*((d2_) + (e2_.)*(x_))^(p_), x_Sym
bol] :> With[{u = IntHide[x^m*(1 + c*x)^p*(-1 + c*x)^p, x]}, Dist[(-(d1*d2))^p*(a + b*ArcCosh[c*x]), u, x] - D
ist[b*c*(-(d1*d2))^p, Int[SimplifyIntegrand[u/(Sqrt[1 + c*x]*Sqrt[-1 + c*x]), x], x], x]] /; FreeQ[{a, b, c, d
1, e1, d2, e2}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2, 0] && IntegerQ[p - 1/2] && (IGtQ[(m + 1)/2, 0] || IL
tQ[(m + 2*p + 3)/2, 0]) && NeQ[p, -2^(-1)] && GtQ[d1, 0] && LtQ[d2, 0]

Rule 388

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

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.0690195, size = 97, normalized size = 0.65 \[ \frac{-a c^2 x^2+2 a+b \left (2-c^2 x^2\right ) \cosh ^{-1}(c x)+b c x \sqrt{c x-1} \sqrt{c x+1}+b \sqrt{c x-1} \sqrt{c x+1} \tanh ^{-1}(c x)}{c^4 d \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.218, size = 313, normalized size = 2.1 \begin{align*} -{\frac{a{x}^{2}}{{c}^{2}d}{\frac{1}{\sqrt{-{c}^{2}d{x}^{2}+d}}}}+2\,{\frac{a}{d{c}^{4}\sqrt{-{c}^{2}d{x}^{2}+d}}}+{\frac{b{x}^{2}{\rm arccosh} \left (cx\right )}{{c}^{2}{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{bx}{{c}^{3}{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{cx+1}\sqrt{cx-1}}-2\,{\frac{b\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }{\rm arccosh} \left (cx\right )}{{d}^{2}{c}^{4} \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{b}{{d}^{2}{c}^{4} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{cx+1}\sqrt{cx-1}\ln \left ( cx+\sqrt{cx-1}\sqrt{cx+1}-1 \right ) }-{\frac{b}{{d}^{2}{c}^{4} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{cx+1}\sqrt{cx-1}\ln \left ( 1+cx+\sqrt{cx-1}\sqrt{cx+1} \right ) } \end{align*}

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)^(3/2),x)

[Out]

-a*x^2/c^2/d/(-c^2*d*x^2+d)^(1/2)+2*a/d/c^4/(-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)*x^2-b*(-d*(c^2*x^2-1))^(1/2)/c^3/d^2/(c^2*x^2-1)*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x-2*b*(-d*(c^2*x^2-1
))^(1/2)/c^4/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^4/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^4/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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.54527, size = 915, normalized size = 6.1 \begin{align*} \left [-\frac{4 \, \sqrt{-c^{2} d x^{2} + d} \sqrt{c^{2} x^{2} - 1} b c x - 4 \,{\left (b c^{2} x^{2} - 2 \, b\right )} \sqrt{-c^{2} d x^{2} + d} \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 \,{\left (a c^{2} x^{2} - 2 \, a\right )} \sqrt{-c^{2} d x^{2} + d}}{4 \,{\left (c^{6} d^{2} x^{2} - c^{4} d^{2}\right )}}, -\frac{2 \, \sqrt{-c^{2} d x^{2} + d} \sqrt{c^{2} x^{2} - 1} b c x +{\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 \,{\left (b c^{2} x^{2} - 2 \, b\right )} \sqrt{-c^{2} d x^{2} + d} \log \left (c x + \sqrt{c^{2} x^{2} - 1}\right ) - 2 \,{\left (a c^{2} x^{2} - 2 \, a\right )} \sqrt{-c^{2} d x^{2} + d}}{2 \,{\left (c^{6} d^{2} x^{2} - c^{4} d^{2}\right )}}\right ] \end{align*}

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)^(3/2),x, algorithm="fricas")

[Out]

[-1/4*(4*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*b*c*x - 4*(b*c^2*x^2 - 2*b)*sqrt(-c^2*d*x^2 + d)*log(c*x + sqr
t(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)*sqr
t(-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*(a*c^2*x^2 - 2*a)
*sqrt(-c^2*d*x^2 + d))/(c^6*d^2*x^2 - c^4*d^2), -1/2*(2*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*b*c*x + (b*c^2*
x^2 - b)*sqrt(d)*arctan(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*(b*c^2*x^2 -
2*b)*sqrt(-c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 - 1)) - 2*(a*c^2*x^2 - 2*a)*sqrt(-c^2*d*x^2 + d))/(c^6*d^2*x
^2 - c^4*d^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \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{align*}

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)**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arcosh}\left (c x\right ) + a\right )} x^{3}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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)^(3/2),x, algorithm="giac")

[Out]

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