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

Optimal. Leaf size=143 \[ \frac{1}{5} c^4 d^2 x^5 \left (a+b \cosh ^{-1}(c x)\right )-\frac{2}{3} c^2 d^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+d^2 x \left (a+b \cosh ^{-1}(c x)\right )-\frac{b d^2 (c x-1)^{5/2} (c x+1)^{5/2}}{25 c}+\frac{4 b d^2 (c x-1)^{3/2} (c x+1)^{3/2}}{45 c}-\frac{8 b d^2 \sqrt{c x-1} \sqrt{c x+1}}{15 c} \]

[Out]

(-8*b*d^2*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(15*c) + (4*b*d^2*(-1 + c*x)^(3/2)*(1 + c*x)^(3/2))/(45*c) - (b*d^2*(-
1 + c*x)^(5/2)*(1 + c*x)^(5/2))/(25*c) + d^2*x*(a + b*ArcCosh[c*x]) - (2*c^2*d^2*x^3*(a + b*ArcCosh[c*x]))/3 +
(c^4*d^2*x^5*(a + b*ArcCosh[c*x]))/5

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Rubi [A]  time = 0.151904, antiderivative size = 177, normalized size of antiderivative = 1.24, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {194, 5680, 12, 520, 1247, 698} \[ \frac{1}{5} c^4 d^2 x^5 \left (a+b \cosh ^{-1}(c x)\right )-\frac{2}{3} c^2 d^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+d^2 x \left (a+b \cosh ^{-1}(c x)\right )+\frac{b d^2 \left (1-c^2 x^2\right )^3}{25 c \sqrt{c x-1} \sqrt{c x+1}}+\frac{4 b d^2 \left (1-c^2 x^2\right )^2}{45 c \sqrt{c x-1} \sqrt{c x+1}}+\frac{8 b d^2 \left (1-c^2 x^2\right )}{15 c \sqrt{c x-1} \sqrt{c x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b*d^2*(1 - c^2*x^2))/(15*c*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) + (4*b*d^2*(1 - c^2*x^2)^2)/(45*c*Sqrt[-1 + c*x]*S
qrt[1 + c*x]) + (b*d^2*(1 - c^2*x^2)^3)/(25*c*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) + d^2*x*(a + b*ArcCosh[c*x]) - (2*
c^2*d^2*x^3*(a + b*ArcCosh[c*x]))/3 + (c^4*d^2*x^5*(a + b*ArcCosh[c*x]))/5

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 5680

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2
)^p, x]}, Dist[a + b*ArcCosh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(Sqrt[1 + c*x]*Sqrt[-1 + c*x]), x
], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 520

Int[(u_.)*((c_) + (d_.)*(x_)^(n_.) + (e_.)*(x_)^(n2_.))^(q_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b
2_.)*(x_)^(non2_.))^(p_.), x_Symbol] :> Dist[((a1 + b1*x^(n/2))^FracPart[p]*(a2 + b2*x^(n/2))^FracPart[p])/(a1
*a2 + b1*b2*x^n)^FracPart[p], Int[u*(a1*a2 + b1*b2*x^n)^p*(c + d*x^n + e*x^(2*n))^q, x], x] /; FreeQ[{a1, b1,
a2, b2, c, d, e, n, p, q}, x] && EqQ[non2, n/2] && EqQ[n2, 2*n] && EqQ[a2*b1 + a1*b2, 0]

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin{align*} \int \left (d-c^2 d x^2\right )^2 \left (a+b \cosh ^{-1}(c x)\right ) \, dx &=d^2 x \left (a+b \cosh ^{-1}(c x)\right )-\frac{2}{3} c^2 d^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{5} c^4 d^2 x^5 \left (a+b \cosh ^{-1}(c x)\right )-(b c) \int \frac{d^2 x \left (15-10 c^2 x^2+3 c^4 x^4\right )}{15 \sqrt{-1+c x} \sqrt{1+c x}} \, dx\\ &=d^2 x \left (a+b \cosh ^{-1}(c x)\right )-\frac{2}{3} c^2 d^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{5} c^4 d^2 x^5 \left (a+b \cosh ^{-1}(c x)\right )-\frac{1}{15} \left (b c d^2\right ) \int \frac{x \left (15-10 c^2 x^2+3 c^4 x^4\right )}{\sqrt{-1+c x} \sqrt{1+c x}} \, dx\\ &=d^2 x \left (a+b \cosh ^{-1}(c x)\right )-\frac{2}{3} c^2 d^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{5} c^4 d^2 x^5 \left (a+b \cosh ^{-1}(c x)\right )-\frac{\left (b c d^2 \sqrt{-1+c^2 x^2}\right ) \int \frac{x \left (15-10 c^2 x^2+3 c^4 x^4\right )}{\sqrt{-1+c^2 x^2}} \, dx}{15 \sqrt{-1+c x} \sqrt{1+c x}}\\ &=d^2 x \left (a+b \cosh ^{-1}(c x)\right )-\frac{2}{3} c^2 d^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{5} c^4 d^2 x^5 \left (a+b \cosh ^{-1}(c x)\right )-\frac{\left (b c d^2 \sqrt{-1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{15-10 c^2 x+3 c^4 x^2}{\sqrt{-1+c^2 x}} \, dx,x,x^2\right )}{30 \sqrt{-1+c x} \sqrt{1+c x}}\\ &=d^2 x \left (a+b \cosh ^{-1}(c x)\right )-\frac{2}{3} c^2 d^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{5} c^4 d^2 x^5 \left (a+b \cosh ^{-1}(c x)\right )-\frac{\left (b c d^2 \sqrt{-1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \left (\frac{8}{\sqrt{-1+c^2 x}}-4 \sqrt{-1+c^2 x}+3 \left (-1+c^2 x\right )^{3/2}\right ) \, dx,x,x^2\right )}{30 \sqrt{-1+c x} \sqrt{1+c x}}\\ &=\frac{8 b d^2 \left (1-c^2 x^2\right )}{15 c \sqrt{-1+c x} \sqrt{1+c x}}+\frac{4 b d^2 \left (1-c^2 x^2\right )^2}{45 c \sqrt{-1+c x} \sqrt{1+c x}}+\frac{b d^2 \left (1-c^2 x^2\right )^3}{25 c \sqrt{-1+c x} \sqrt{1+c x}}+d^2 x \left (a+b \cosh ^{-1}(c x)\right )-\frac{2}{3} c^2 d^2 x^3 \left (a+b \cosh ^{-1}(c x)\right )+\frac{1}{5} c^4 d^2 x^5 \left (a+b \cosh ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.15438, size = 99, normalized size = 0.69 \[ \frac{d^2 \left (15 a c x \left (3 c^4 x^4-10 c^2 x^2+15\right )+b \sqrt{c x-1} \sqrt{c x+1} \left (-9 c^4 x^4+38 c^2 x^2-149\right )+15 b c x \left (3 c^4 x^4-10 c^2 x^2+15\right ) \cosh ^{-1}(c x)\right )}{225 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(b*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(-149 + 38*c^2*x^2 - 9*c^4*x^4) + 15*a*c*x*(15 - 10*c^2*x^2 + 3*c^4*x^4)
+ 15*b*c*x*(15 - 10*c^2*x^2 + 3*c^4*x^4)*ArcCosh[c*x]))/(225*c)

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Maple [A]  time = 0.013, size = 102, normalized size = 0.7 \begin{align*}{\frac{1}{c} \left ({d}^{2}a \left ({\frac{{c}^{5}{x}^{5}}{5}}-{\frac{2\,{c}^{3}{x}^{3}}{3}}+cx \right ) +{d}^{2}b \left ({\frac{{\rm arccosh} \left (cx\right ){c}^{5}{x}^{5}}{5}}-{\frac{2\,{c}^{3}{x}^{3}{\rm arccosh} \left (cx\right )}{3}}+cx{\rm arccosh} \left (cx\right )-{\frac{9\,{c}^{4}{x}^{4}-38\,{c}^{2}{x}^{2}+149}{225}\sqrt{cx-1}\sqrt{cx+1}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(d^2*a*(1/5*c^5*x^5-2/3*c^3*x^3+c*x)+d^2*b*(1/5*arccosh(c*x)*c^5*x^5-2/3*c^3*x^3*arccosh(c*x)+c*x*arccosh(
c*x)-1/225*(c*x-1)^(1/2)*(c*x+1)^(1/2)*(9*c^4*x^4-38*c^2*x^2+149)))

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Maxima [A]  time = 1.17515, size = 262, normalized size = 1.83 \begin{align*} \frac{1}{5} \, a c^{4} d^{2} x^{5} + \frac{1}{75} \,{\left (15 \, x^{5} \operatorname{arcosh}\left (c x\right ) -{\left (\frac{3 \, \sqrt{c^{2} x^{2} - 1} x^{4}}{c^{2}} + \frac{4 \, \sqrt{c^{2} x^{2} - 1} x^{2}}{c^{4}} + \frac{8 \, \sqrt{c^{2} x^{2} - 1}}{c^{6}}\right )} c\right )} b c^{4} d^{2} - \frac{2}{3} \, a c^{2} d^{2} x^{3} - \frac{2}{9} \,{\left (3 \, x^{3} \operatorname{arcosh}\left (c x\right ) - c{\left (\frac{\sqrt{c^{2} x^{2} - 1} x^{2}}{c^{2}} + \frac{2 \, \sqrt{c^{2} x^{2} - 1}}{c^{4}}\right )}\right )} b c^{2} d^{2} + a d^{2} x + \frac{{\left (c x \operatorname{arcosh}\left (c x\right ) - \sqrt{c^{2} x^{2} - 1}\right )} b d^{2}}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*c^4*d^2*x^5 + 1/75*(15*x^5*arccosh(c*x) - (3*sqrt(c^2*x^2 - 1)*x^4/c^2 + 4*sqrt(c^2*x^2 - 1)*x^2/c^4 + 8
*sqrt(c^2*x^2 - 1)/c^6)*c)*b*c^4*d^2 - 2/3*a*c^2*d^2*x^3 - 2/9*(3*x^3*arccosh(c*x) - c*(sqrt(c^2*x^2 - 1)*x^2/
c^2 + 2*sqrt(c^2*x^2 - 1)/c^4))*b*c^2*d^2 + a*d^2*x + (c*x*arccosh(c*x) - sqrt(c^2*x^2 - 1))*b*d^2/c

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Fricas [A]  time = 1.86214, size = 296, normalized size = 2.07 \begin{align*} \frac{45 \, a c^{5} d^{2} x^{5} - 150 \, a c^{3} d^{2} x^{3} + 225 \, a c d^{2} x + 15 \,{\left (3 \, b c^{5} d^{2} x^{5} - 10 \, b c^{3} d^{2} x^{3} + 15 \, b c d^{2} x\right )} \log \left (c x + \sqrt{c^{2} x^{2} - 1}\right ) -{\left (9 \, b c^{4} d^{2} x^{4} - 38 \, b c^{2} d^{2} x^{2} + 149 \, b d^{2}\right )} \sqrt{c^{2} x^{2} - 1}}{225 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/225*(45*a*c^5*d^2*x^5 - 150*a*c^3*d^2*x^3 + 225*a*c*d^2*x + 15*(3*b*c^5*d^2*x^5 - 10*b*c^3*d^2*x^3 + 15*b*c*
d^2*x)*log(c*x + sqrt(c^2*x^2 - 1)) - (9*b*c^4*d^2*x^4 - 38*b*c^2*d^2*x^2 + 149*b*d^2)*sqrt(c^2*x^2 - 1))/c

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Sympy [A]  time = 3.44075, size = 172, normalized size = 1.2 \begin{align*} \begin{cases} \frac{a c^{4} d^{2} x^{5}}{5} - \frac{2 a c^{2} d^{2} x^{3}}{3} + a d^{2} x + \frac{b c^{4} d^{2} x^{5} \operatorname{acosh}{\left (c x \right )}}{5} - \frac{b c^{3} d^{2} x^{4} \sqrt{c^{2} x^{2} - 1}}{25} - \frac{2 b c^{2} d^{2} x^{3} \operatorname{acosh}{\left (c x \right )}}{3} + \frac{38 b c d^{2} x^{2} \sqrt{c^{2} x^{2} - 1}}{225} + b d^{2} x \operatorname{acosh}{\left (c x \right )} - \frac{149 b d^{2} \sqrt{c^{2} x^{2} - 1}}{225 c} & \text{for}\: c \neq 0 \\d^{2} x \left (a + \frac{i \pi b}{2}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**4*d**2*x**5/5 - 2*a*c**2*d**2*x**3/3 + a*d**2*x + b*c**4*d**2*x**5*acosh(c*x)/5 - b*c**3*d**2*
x**4*sqrt(c**2*x**2 - 1)/25 - 2*b*c**2*d**2*x**3*acosh(c*x)/3 + 38*b*c*d**2*x**2*sqrt(c**2*x**2 - 1)/225 + b*d
**2*x*acosh(c*x) - 149*b*d**2*sqrt(c**2*x**2 - 1)/(225*c), Ne(c, 0)), (d**2*x*(a + I*pi*b/2), True))

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Giac [A]  time = 1.46325, size = 281, normalized size = 1.97 \begin{align*} \frac{1}{5} \, a c^{4} d^{2} x^{5} + \frac{1}{75} \,{\left (15 \, x^{5} \log \left (c x + \sqrt{c^{2} x^{2} - 1}\right ) - \frac{3 \,{\left (c^{2} x^{2} - 1\right )}^{\frac{5}{2}} + 10 \,{\left (c^{2} x^{2} - 1\right )}^{\frac{3}{2}} + 15 \, \sqrt{c^{2} x^{2} - 1}}{c^{5}}\right )} b c^{4} d^{2} - \frac{2}{3} \, a c^{2} d^{2} x^{3} - \frac{2}{9} \,{\left (3 \, x^{3} \log \left (c x + \sqrt{c^{2} x^{2} - 1}\right ) - \frac{{\left (c^{2} x^{2} - 1\right )}^{\frac{3}{2}} + 3 \, \sqrt{c^{2} x^{2} - 1}}{c^{3}}\right )} b c^{2} d^{2} +{\left (x \log \left (c x + \sqrt{c^{2} x^{2} - 1}\right ) - \frac{\sqrt{c^{2} x^{2} - 1}}{c}\right )} b d^{2} + a d^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*c^4*d^2*x^5 + 1/75*(15*x^5*log(c*x + sqrt(c^2*x^2 - 1)) - (3*(c^2*x^2 - 1)^(5/2) + 10*(c^2*x^2 - 1)^(3/2
) + 15*sqrt(c^2*x^2 - 1))/c^5)*b*c^4*d^2 - 2/3*a*c^2*d^2*x^3 - 2/9*(3*x^3*log(c*x + sqrt(c^2*x^2 - 1)) - ((c^2
*x^2 - 1)^(3/2) + 3*sqrt(c^2*x^2 - 1))/c^3)*b*c^2*d^2 + (x*log(c*x + sqrt(c^2*x^2 - 1)) - sqrt(c^2*x^2 - 1)/c)
*b*d^2 + a*d^2*x