3.127 \(\int \frac {x^2 (a+b \cosh ^{-1}(c x))}{(d-c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=133 \[ \frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {b \sqrt {c x-1} \sqrt {c x+1} \log \left (1-c^2 x^2\right )}{6 c^3 d^2 \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {c x-1} \sqrt {c x+1}}{6 c^3 d \left (d-c^2 d x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.40, antiderivative size = 160, normalized size of antiderivative = 1.20, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {5798, 5724, 266, 43} \[ \frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 d^2 (1-c x) (c x+1) \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {c x-1} \sqrt {c x+1}}{6 c^3 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {c x-1} \sqrt {c x+1} \log \left (1-c^2 x^2\right )}{6 c^3 d^2 \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*ArcCosh[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

(b*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(6*c^3*d^2*(1 - c^2*x^2)*Sqrt[d - c^2*d*x^2]) + (x^3*(a + b*ArcCosh[c*x]))/(3
*d^2*(1 - c*x)*(1 + c*x)*Sqrt[d - c^2*d*x^2]) + (b*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*Log[1 - c^2*x^2])/(6*c^3*d^2*S
qrt[d - c^2*d*x^2])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5724

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d1_) + (e1_.)*(x_))^(p_.)*((d2_) + (e2_.)*(x
_))^(p_.), x_Symbol] :> Simp[((f*x)^(m + 1)*(d1 + e1*x)^(p + 1)*(d2 + e2*x)^(p + 1)*(a + b*ArcCosh[c*x])^n)/(d
1*d2*f*(m + 1)), x] + Dist[(b*c*n*(-(d1*d2))^IntPart[p]*(d1 + e1*x)^FracPart[p]*(d2 + e2*x)^FracPart[p])/(f*(m
+ 1)*(1 + c*x)^FracPart[p]*(-1 + c*x)^FracPart[p]), Int[(f*x)^(m + 1)*(-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, f, m, p}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2,
0] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -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^2 \left (a+b \cosh ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^{5/2}} \, dx &=\frac {\left (\sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {x^2 \left (a+b \cosh ^{-1}(c x)\right )}{(-1+c x)^{5/2} (1+c x)^{5/2}} \, dx}{d^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 d^2 (1-c x) (1+c x) \sqrt {d-c^2 d x^2}}+\frac {\left (b c \sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {x^3}{\left (-1+c^2 x^2\right )^2} \, dx}{3 d^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 d^2 (1-c x) (1+c x) \sqrt {d-c^2 d x^2}}+\frac {\left (b c \sqrt {-1+c x} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \frac {x}{\left (-1+c^2 x\right )^2} \, dx,x,x^2\right )}{6 d^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 d^2 (1-c x) (1+c x) \sqrt {d-c^2 d x^2}}+\frac {\left (b c \sqrt {-1+c x} \sqrt {1+c x}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{c^2 \left (-1+c^2 x\right )^2}+\frac {1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{6 d^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {b \sqrt {-1+c x} \sqrt {1+c x}}{6 c^3 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}+\frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 d^2 (1-c x) (1+c x) \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {-1+c x} \sqrt {1+c x} \log \left (1-c^2 x^2\right )}{6 c^3 d^2 \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 101, normalized size = 0.76 \[ \frac {\sqrt {c x-1} \sqrt {c x+1} \left (\frac {b \left (\frac {1}{1-c^2 x^2}+\log \left (1-c^2 x^2\right )\right )}{c^3}-\frac {2 x^3 \left (a+b \cosh ^{-1}(c x)\right )}{(c x-1)^{3/2} (c x+1)^{3/2}}\right )}{6 d^2 \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*ArcCosh[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

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

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b x^{2} \operatorname {arcosh}\left (c x\right ) + a x^{2}\right )}}{c^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-c^2*d*x^2 + d)*(b*x^2*arccosh(c*x) + a*x^2)/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3
), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} x^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arccosh(c*x) + a)*x^2/(-c^2*d*x^2 + d)^(5/2), x)

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maple [B]  time = 0.54, size = 1228, normalized size = 9.23 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(5/2),x)

[Out]

1/3*a/c^2/d*x/(-c^2*d*x^2+d)^(3/2)-1/3*a/c^2/d^2*x/(-c^2*d*x^2+d)^(1/2)+2/3*b*(-d*(c^2*x^2-1))^(1/2)*(c*x-1)^(
1/2)*(c*x+1)^(1/2)/d^3/c^3/(c^2*x^2-1)*arccosh(c*x)-b*(-d*(c^2*x^2-1))^(1/2)/(3*c^8*x^8-9*c^6*x^6+10*c^4*x^4-5
*c^2*x^2+1)*c^3/d^3*arccosh(c*x)*(c*x-1)^(1/2)*(c*x+1)^(1/2)*x^6+b*(-d*(c^2*x^2-1))^(1/2)/(3*c^8*x^8-9*c^6*x^6
+10*c^4*x^4-5*c^2*x^2+1)*c^4/d^3*arccosh(c*x)*x^7-1/6*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^8*x^8-9*c^6*x^6+10*c^4*x^4
-5*c^2*x^2+1)*c^2/d^3*(c*x-1)*(c*x+1)*x^5+1/6*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^8*x^8-9*c^6*x^6+10*c^4*x^4-5*c^2*x
^2+1)*c^4/d^3*x^7+2*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^8*x^8-9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*c/d^3*arccosh(c*x)*(
c*x-1)^(1/2)*(c*x+1)^(1/2)*x^4-b*(-d*(c^2*x^2-1))^(1/2)/(3*c^8*x^8-9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*c^2/d^3*a
rccosh(c*x)*x^5+1/6*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^8*x^8-9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)/d^3*(c*x-1)*(c*x+1)*
x^3+1/2*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^8*x^8-9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*c/d^3*(c*x-1)^(1/2)*(c*x+1)^(1/2
)*x^4-1/3*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^8*x^8-9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)*c^2/d^3*x^5-4/3*b*(-d*(c^2*x^2
-1))^(1/2)/(3*c^8*x^8-9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)/c/d^3*arccosh(c*x)*(c*x-1)^(1/2)*(c*x+1)^(1/2)*x^2+1/3
*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^8*x^8-9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)/d^3*arccosh(c*x)*x^3-1/2*b*(-d*(c^2*x^2
-1))^(1/2)/(3*c^8*x^8-9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)/c/d^3*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^2+1/6*b*(-d*(c^2*x
^2-1))^(1/2)/(3*c^8*x^8-9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)/d^3*x^3+1/3*b*(-d*(c^2*x^2-1))^(1/2)/(3*c^8*x^8-9*c^
6*x^6+10*c^4*x^4-5*c^2*x^2+1)/c^3/d^3*arccosh(c*x)*(c*x-1)^(1/2)*(c*x+1)^(1/2)+1/6*b*(-d*(c^2*x^2-1))^(1/2)/(3
*c^8*x^8-9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)/c^3/d^3*(c*x-1)^(1/2)*(c*x+1)^(1/2)-1/3*b*(-d*(c^2*x^2-1))^(1/2)*(c
*x-1)^(1/2)*(c*x+1)^(1/2)/d^3/c^3/(c^2*x^2-1)*ln((c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2-1)

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maxima [A]  time = 0.68, size = 169, normalized size = 1.27 \[ \frac {1}{6} \, b c {\left (\frac {\sqrt {-d}}{c^{6} d^{3} x^{2} - c^{4} d^{3}} - \frac {\sqrt {-d} \log \left (c x + 1\right )}{c^{4} d^{3}} - \frac {\sqrt {-d} \log \left (c x - 1\right )}{c^{4} d^{3}}\right )} - \frac {1}{3} \, b {\left (\frac {x}{\sqrt {-c^{2} d x^{2} + d} c^{2} d^{2}} - \frac {x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d}\right )} \operatorname {arcosh}\left (c x\right ) - \frac {1}{3} \, a {\left (\frac {x}{\sqrt {-c^{2} d x^{2} + d} c^{2} d^{2}} - \frac {x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(a + b*acosh(c*x)))/(d - c^2*d*x^2)^(5/2),x)

[Out]

int((x^2*(a + b*acosh(c*x)))/(d - c^2*d*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (a + b \operatorname {acosh}{\left (c x \right )}\right )}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*acosh(c*x))/(-c**2*d*x**2+d)**(5/2),x)

[Out]

Integral(x**2*(a + b*acosh(c*x))/(-d*(c*x - 1)*(c*x + 1))**(5/2), x)

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