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]

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

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Rubi [A]  time = 0.28, antiderivative size = 184, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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 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 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]))

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*}

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Mathematica [A]  time = 0.27, 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]

-1/3*(a*d^2)/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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fricas [A]  time = 0.77, size = 216, normalized size = 1.17 \[ \frac {6 \, a c e^{2} x^{4} - 12 \, a c d e x^{2} + 2 \, {\left (b c^{4} d^{2} + 12 \, b c^{2} d e\right )} x^{3} \arctan \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) + 2 \, {\left (b c d^{2} + 6 \, b c d e - 3 \, b c e^{2}\right )} x^{3} \log \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) - 2 \, a c d^{2} + 2 \, {\left (3 \, b c e^{2} x^{4} - 6 \, b c d e x^{2} - b c d^{2} + {\left (b c d^{2} + 6 \, b c d e - 3 \, b c e^{2}\right )} x^{3}\right )} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) + {\left (b c^{2} d^{2} x - 6 \, b e^{2} x^{3}\right )} \sqrt {c^{2} x^{2} - 1}}{6 \, c x^{3}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \operatorname {arcosh}\left (c x\right ) + a\right )}}{x^{4}}\,{d x} \]

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)

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maple [A]  time = 0.02, size = 196, normalized size = 1.07 \[ a x \,e^{2}-\frac {2 a d e}{x}-\frac {a \,d^{2}}{3 x^{3}}+b \,\mathrm {arccosh}\left (c x \right ) x \,e^{2}-\frac {2 b \,\mathrm {arccosh}\left (c x \right ) d e}{x}-\frac {b \,\mathrm {arccosh}\left (c x \right ) d^{2}}{3 x^{3}}-\frac {c^{3} b \sqrt {c x -1}\, \sqrt {c x +1}\, d^{2} \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right )}{6 \sqrt {c^{2} x^{2}-1}}+\frac {b c \,d^{2} \sqrt {c x -1}\, \sqrt {c x +1}}{6 x^{2}}-\frac {2 c b \sqrt {c x -1}\, \sqrt {c x +1}\, \arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right ) d e}{\sqrt {c^{2} x^{2}-1}}-\frac {b \sqrt {c x -1}\, \sqrt {c x +1}\, e^{2}}{c} \]

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))+1/6*b*c*d^2*(c*x-1)^(1/2)*(c*
x+1)^(1/2)/x^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/c*b*(c*x-
1)^(1/2)*(c*x+1)^(1/2)*e^2

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maxima [A]  time = 0.55, size = 126, normalized size = 0.68 \[ -\frac {1}{6} \, {\left ({\left (c^{2} \arcsin \left (\frac {1}{c {\left | x \right |}}\right ) - \frac {\sqrt {c^{2} x^{2} - 1}}{x^{2}}\right )} c + \frac {2 \, \operatorname {arcosh}\left (c x\right )}{x^{3}}\right )} b d^{2} - 2 \, {\left (c \arcsin \left (\frac {1}{c {\left | x \right |}}\right ) + \frac {\operatorname {arcosh}\left (c x\right )}{x}\right )} b d e + a e^{2} x + \frac {{\left (c x \operatorname {arcosh}\left (c x\right ) - \sqrt {c^{2} x^{2} - 1}\right )} b e^{2}}{c} - \frac {2 \, a d e}{x} - \frac {a d^{2}}{3 \, x^{3}} \]

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/(c*abs(x))) - sqrt(c^2*x^2 - 1)/x^2)*c + 2*arccosh(c*x)/x^3)*b*d^2 - 2*(c*arcsin(1/(c*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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^2}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {acosh}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}}{x^{4}}\, dx \]

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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