3.478 \(\int \frac{(d+e x^2)^2 (a+b \cosh ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=184 \[ -\frac{d^2 \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac{2 d e \left (a+b \cosh ^{-1}(c x)\right )}{x}+e^2 x \left (a+b \cosh ^{-1}(c x)\right )-\frac{b c d^2 \left (1-c^2 x^2\right )}{6 x^2 \sqrt{c x-1} \sqrt{c x+1}}+\frac{b c d \sqrt{c^2 x^2-1} \left (c^2 d+12 e\right ) \tan ^{-1}\left (\sqrt{c^2 x^2-1}\right )}{6 \sqrt{c x-1} \sqrt{c x+1}}+\frac{b e^2 \left (1-c^2 x^2\right )}{c \sqrt{c x-1} \sqrt{c x+1}} \]

[Out]

(b*e^2*(1 - c^2*x^2))/(c*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) - (b*c*d^2*(1 - c^2*x^2))/(6*x^2*Sqrt[-1 + c*x]*Sqrt[1
+ c*x]) - (d^2*(a + b*ArcCosh[c*x]))/(3*x^3) - (2*d*e*(a + b*ArcCosh[c*x]))/x + e^2*x*(a + b*ArcCosh[c*x]) + (
b*c*d*(c^2*d + 12*e)*Sqrt[-1 + c^2*x^2]*ArcTan[Sqrt[-1 + c^2*x^2]])/(6*Sqrt[-1 + c*x]*Sqrt[1 + c*x])

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Rubi [A]  time = 0.278789, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {270, 5790, 520, 1251, 897, 1157, 388, 205} \[ -\frac{d^2 \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac{2 d e \left (a+b \cosh ^{-1}(c x)\right )}{x}+e^2 x \left (a+b \cosh ^{-1}(c x)\right )-\frac{b c d^2 \left (1-c^2 x^2\right )}{6 x^2 \sqrt{c x-1} \sqrt{c x+1}}+\frac{b c d \sqrt{c^2 x^2-1} \left (c^2 d+12 e\right ) \tan ^{-1}\left (\sqrt{c^2 x^2-1}\right )}{6 \sqrt{c x-1} \sqrt{c x+1}}+\frac{b e^2 \left (1-c^2 x^2\right )}{c \sqrt{c x-1} \sqrt{c x+1}} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^2*(a + b*ArcCosh[c*x]))/x^4,x]

[Out]

(b*e^2*(1 - c^2*x^2))/(c*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) - (b*c*d^2*(1 - c^2*x^2))/(6*x^2*Sqrt[-1 + c*x]*Sqrt[1
+ c*x]) - (d^2*(a + b*ArcCosh[c*x]))/(3*x^3) - (2*d*e*(a + b*ArcCosh[c*x]))/x + e^2*x*(a + b*ArcCosh[c*x]) + (
b*c*d*(c^2*d + 12*e)*Sqrt[-1 + c^2*x^2]*ArcTan[Sqrt[-1 + c^2*x^2]])/(6*Sqrt[-1 + c*x]*Sqrt[1 + c*x])

Rule 270

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

Rule 5790

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(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, f, m}, x] && NeQ[c^2*d + e, 0] && IntegerQ[p] &
& (GtQ[p, 0] || (IGtQ[(m - 1)/2, 0] && LeQ[m + p, 0]))

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 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(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] &&
IntegerQ[(m - 1)/2]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

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 205

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

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\right )^2 \left (a+b \cosh ^{-1}(c x)\right )}{x^4} \, dx &=-\frac{d^2 \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac{2 d e \left (a+b \cosh ^{-1}(c x)\right )}{x}+e^2 x \left (a+b \cosh ^{-1}(c x)\right )-(b c) \int \frac{-\frac{d^2}{3}-2 d e x^2+e^2 x^4}{x^3 \sqrt{-1+c x} \sqrt{1+c x}} \, dx\\ &=-\frac{d^2 \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac{2 d e \left (a+b \cosh ^{-1}(c x)\right )}{x}+e^2 x \left (a+b \cosh ^{-1}(c x)\right )-\frac{\left (b c \sqrt{-1+c^2 x^2}\right ) \int \frac{-\frac{d^2}{3}-2 d e x^2+e^2 x^4}{x^3 \sqrt{-1+c^2 x^2}} \, dx}{\sqrt{-1+c x} \sqrt{1+c x}}\\ &=-\frac{d^2 \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac{2 d e \left (a+b \cosh ^{-1}(c x)\right )}{x}+e^2 x \left (a+b \cosh ^{-1}(c x)\right )-\frac{\left (b c \sqrt{-1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{-\frac{d^2}{3}-2 d e x+e^2 x^2}{x^2 \sqrt{-1+c^2 x}} \, dx,x,x^2\right )}{2 \sqrt{-1+c x} \sqrt{1+c x}}\\ &=-\frac{d^2 \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac{2 d e \left (a+b \cosh ^{-1}(c x)\right )}{x}+e^2 x \left (a+b \cosh ^{-1}(c x)\right )-\frac{\left (b \sqrt{-1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\frac{-\frac{1}{3} c^4 d^2-2 c^2 d e+e^2}{c^4}-\frac{\left (2 c^2 d e-2 e^2\right ) x^2}{c^4}+\frac{e^2 x^4}{c^4}}{\left (\frac{1}{c^2}+\frac{x^2}{c^2}\right )^2} \, dx,x,\sqrt{-1+c^2 x^2}\right )}{c \sqrt{-1+c x} \sqrt{1+c x}}\\ &=-\frac{b c d^2 \left (1-c^2 x^2\right )}{6 x^2 \sqrt{-1+c x} \sqrt{1+c x}}-\frac{d^2 \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac{2 d e \left (a+b \cosh ^{-1}(c x)\right )}{x}+e^2 x \left (a+b \cosh ^{-1}(c x)\right )+\frac{\left (b c \sqrt{-1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\frac{1}{3} \left (d^2+\frac{12 d e}{c^2}-\frac{6 e^2}{c^4}\right )-\frac{2 e^2 x^2}{c^4}}{\frac{1}{c^2}+\frac{x^2}{c^2}} \, dx,x,\sqrt{-1+c^2 x^2}\right )}{2 \sqrt{-1+c x} \sqrt{1+c x}}\\ &=\frac{b e^2 \left (1-c^2 x^2\right )}{c \sqrt{-1+c x} \sqrt{1+c x}}-\frac{b c d^2 \left (1-c^2 x^2\right )}{6 x^2 \sqrt{-1+c x} \sqrt{1+c x}}-\frac{d^2 \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac{2 d e \left (a+b \cosh ^{-1}(c x)\right )}{x}+e^2 x \left (a+b \cosh ^{-1}(c x)\right )+\frac{\left (b c d \left (d+\frac{12 e}{c^2}\right ) \sqrt{-1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{c^2}+\frac{x^2}{c^2}} \, dx,x,\sqrt{-1+c^2 x^2}\right )}{6 \sqrt{-1+c x} \sqrt{1+c x}}\\ &=\frac{b e^2 \left (1-c^2 x^2\right )}{c \sqrt{-1+c x} \sqrt{1+c x}}-\frac{b c d^2 \left (1-c^2 x^2\right )}{6 x^2 \sqrt{-1+c x} \sqrt{1+c x}}-\frac{d^2 \left (a+b \cosh ^{-1}(c x)\right )}{3 x^3}-\frac{2 d e \left (a+b \cosh ^{-1}(c x)\right )}{x}+e^2 x \left (a+b \cosh ^{-1}(c x)\right )+\frac{b c d \left (c^2 d+12 e\right ) \sqrt{-1+c^2 x^2} \tan ^{-1}\left (\sqrt{-1+c^2 x^2}\right )}{6 \sqrt{-1+c x} \sqrt{1+c x}}\\ \end{align*}

Mathematica [A]  time = 0.237185, size = 133, normalized size = 0.72 \[ -\frac{a d^2}{3 x^3}-\frac{2 a d e}{x}+a e^2 x-\frac{1}{6} b c d \left (c^2 d+12 e\right ) \tan ^{-1}\left (\frac{1}{\sqrt{c x-1} \sqrt{c x+1}}\right )+b \sqrt{c x-1} \sqrt{c x+1} \left (\frac{c d^2}{6 x^2}-\frac{e^2}{c}\right )-\frac{b \cosh ^{-1}(c x) \left (d^2+6 d e x^2-3 e^2 x^4\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^2*(a + b*ArcCosh[c*x]))/x^4,x]

[Out]

-(a*d^2)/(3*x^3) - (2*a*d*e)/x + a*e^2*x + b*(-(e^2/c) + (c*d^2)/(6*x^2))*Sqrt[-1 + c*x]*Sqrt[1 + c*x] - (b*(d
^2 + 6*d*e*x^2 - 3*e^2*x^4)*ArcCosh[c*x])/(3*x^3) - (b*c*d*(c^2*d + 12*e)*ArcTan[1/(Sqrt[-1 + c*x]*Sqrt[1 + c*
x])])/6

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Maple [A]  time = 0.02, size = 196, normalized size = 1.1 \begin{align*} ax{e}^{2}-2\,{\frac{ade}{x}}-{\frac{a{d}^{2}}{3\,{x}^{3}}}+b{\rm arccosh} \left (cx\right )x{e}^{2}-2\,{\frac{bd{\rm arccosh} \left (cx\right )e}{x}}-{\frac{b{d}^{2}{\rm arccosh} \left (cx\right )}{3\,{x}^{3}}}-{\frac{{c}^{3}b{d}^{2}}{6}\sqrt{cx-1}\sqrt{cx+1}\arctan \left ({\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}}}-2\,{\frac{cb\sqrt{cx-1}\sqrt{cx+1}\arctan \left ({\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}} \right ) de}{\sqrt{{c}^{2}{x}^{2}-1}}}+{\frac{cb{d}^{2}}{6\,{x}^{2}}\sqrt{cx-1}\sqrt{cx+1}}-{\frac{b{e}^{2}}{c}\sqrt{cx-1}\sqrt{cx+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^2*(a+b*arccosh(c*x))/x^4,x)

[Out]

a*x*e^2-2*a*d*e/x-1/3*a*d^2/x^3+b*arccosh(c*x)*x*e^2-2*b*arccosh(c*x)*d*e/x-1/3*b*arccosh(c*x)*d^2/x^3-1/6*c^3
*b*(c*x-1)^(1/2)*(c*x+1)^(1/2)/(c^2*x^2-1)^(1/2)*d^2*arctan(1/(c^2*x^2-1)^(1/2))-2*c*b*(c*x-1)^(1/2)*(c*x+1)^(
1/2)/(c^2*x^2-1)^(1/2)*arctan(1/(c^2*x^2-1)^(1/2))*d*e+1/6*b*c*d^2*(c*x-1)^(1/2)*(c*x+1)^(1/2)/x^2-1/c*b*(c*x-
1)^(1/2)*(c*x+1)^(1/2)*e^2

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Maxima [A]  time = 1.68929, size = 176, normalized size = 0.96 \begin{align*} -\frac{1}{6} \,{\left ({\left (c^{2} \arcsin \left (\frac{1}{\sqrt{c^{2}}{\left | x \right |}}\right ) - \frac{\sqrt{c^{2} x^{2} - 1}}{x^{2}}\right )} c + \frac{2 \, \operatorname{arcosh}\left (c x\right )}{x^{3}}\right )} b d^{2} - 2 \,{\left (c \arcsin \left (\frac{1}{\sqrt{c^{2}}{\left | x \right |}}\right ) + \frac{\operatorname{arcosh}\left (c x\right )}{x}\right )} b d e + a e^{2} x + \frac{{\left (c x \operatorname{arcosh}\left (c x\right ) - \sqrt{c^{2} x^{2} - 1}\right )} b e^{2}}{c} - \frac{2 \, a d e}{x} - \frac{a d^{2}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*arccosh(c*x))/x^4,x, algorithm="maxima")

[Out]

-1/6*((c^2*arcsin(1/(sqrt(c^2)*abs(x))) - sqrt(c^2*x^2 - 1)/x^2)*c + 2*arccosh(c*x)/x^3)*b*d^2 - 2*(c*arcsin(1
/(sqrt(c^2)*abs(x))) + arccosh(c*x)/x)*b*d*e + a*e^2*x + (c*x*arccosh(c*x) - sqrt(c^2*x^2 - 1))*b*e^2/c - 2*a*
d*e/x - 1/3*a*d^2/x^3

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Fricas [A]  time = 3.34796, size = 487, normalized size = 2.65 \begin{align*} \frac{6 \, a c e^{2} x^{4} - 12 \, a c d e x^{2} + 2 \,{\left (b c^{4} d^{2} + 12 \, b c^{2} d e\right )} x^{3} \arctan \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) + 2 \,{\left (b c d^{2} + 6 \, b c d e - 3 \, b c e^{2}\right )} x^{3} \log \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) - 2 \, a c d^{2} + 2 \,{\left (3 \, b c e^{2} x^{4} - 6 \, b c d e x^{2} - b c d^{2} +{\left (b c d^{2} + 6 \, b c d e - 3 \, b c e^{2}\right )} x^{3}\right )} \log \left (c x + \sqrt{c^{2} x^{2} - 1}\right ) +{\left (b c^{2} d^{2} x - 6 \, b e^{2} x^{3}\right )} \sqrt{c^{2} x^{2} - 1}}{6 \, c x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*arccosh(c*x))/x^4,x, algorithm="fricas")

[Out]

1/6*(6*a*c*e^2*x^4 - 12*a*c*d*e*x^2 + 2*(b*c^4*d^2 + 12*b*c^2*d*e)*x^3*arctan(-c*x + sqrt(c^2*x^2 - 1)) + 2*(b
*c*d^2 + 6*b*c*d*e - 3*b*c*e^2)*x^3*log(-c*x + sqrt(c^2*x^2 - 1)) - 2*a*c*d^2 + 2*(3*b*c*e^2*x^4 - 6*b*c*d*e*x
^2 - b*c*d^2 + (b*c*d^2 + 6*b*c*d*e - 3*b*c*e^2)*x^3)*log(c*x + sqrt(c^2*x^2 - 1)) + (b*c^2*d^2*x - 6*b*e^2*x^
3)*sqrt(c^2*x^2 - 1))/(c*x^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{acosh}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**2*(a+b*acosh(c*x))/x**4,x)

[Out]

Integral((a + b*acosh(c*x))*(d + e*x**2)**2/x**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )}^{2}{\left (b \operatorname{arcosh}\left (c x\right ) + a\right )}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*arccosh(c*x))/x^4,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^2*(b*arccosh(c*x) + a)/x^4, x)