3.125 \(\int \frac {x^4 (a+b \cosh ^{-1}(c x))}{(d-c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=224 \[ \frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {\sqrt {c x-1} \sqrt {c x+1} \left (a+b \cosh ^{-1}(c x)\right )^2}{2 b c^5 d^2 \sqrt {d-c^2 d x^2}}-\frac {x \left (a+b \cosh ^{-1}(c x)\right )}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {2 b \sqrt {c x-1} \sqrt {c x+1} \log \left (1-c^2 x^2\right )}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {c x-1} \sqrt {c x+1}}{6 c^5 d \left (d-c^2 d x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.76, antiderivative size = 251, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5798, 5752, 5676, 260, 266, 43} \[ \frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 c^2 d^2 (1-c x) (c x+1) \sqrt {d-c^2 d x^2}}-\frac {x \left (a+b \cosh ^{-1}(c x)\right )}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\sqrt {c x-1} \sqrt {c x+1} \left (a+b \cosh ^{-1}(c x)\right )^2}{2 b c^5 d^2 \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {c x-1} \sqrt {c x+1}}{6 c^5 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}+\frac {2 b \sqrt {c x-1} \sqrt {c x+1} \log \left (1-c^2 x^2\right )}{3 c^5 d^2 \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(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^5*d^2*(1 - c^2*x^2)*Sqrt[d - c^2*d*x^2]) - (x*(a + b*ArcCosh[c*x]))/(c^4
*d^2*Sqrt[d - c^2*d*x^2]) + (x^3*(a + b*ArcCosh[c*x]))/(3*c^2*d^2*(1 - c*x)*(1 + c*x)*Sqrt[d - c^2*d*x^2]) + (
Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(a + b*ArcCosh[c*x])^2)/(2*b*c^5*d^2*Sqrt[d - c^2*d*x^2]) + (2*b*Sqrt[-1 + c*x]*S
qrt[1 + c*x]*Log[1 - c^2*x^2])/(3*c^5*d^2*Sqrt[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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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 5676

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_.)*(x_)]), x_Symbol]
:> Simp[(a + b*ArcCosh[c*x])^(n + 1)/(b*c*Sqrt[-(d1*d2)]*(n + 1)), x] /; FreeQ[{a, b, c, d1, e1, d2, e2, n},
x] && EqQ[e1, c*d1] && EqQ[e2, -(c*d2)] && GtQ[d1, 0] && LtQ[d2, 0] && NeQ[n, -1]

Rule 5752

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d1_) + (e1_.)*(x_))^(p_)*((d2_) + (e2_.)*(x_
))^(p_), x_Symbol] :> Simp[(f*(f*x)^(m - 1)*(d1 + e1*x)^(p + 1)*(d2 + e2*x)^(p + 1)*(a + b*ArcCosh[c*x])^n)/(2
*e1*e2*(p + 1)), x] + (-Dist[(f^2*(m - 1))/(2*e1*e2*(p + 1)), Int[(f*x)^(m - 2)*(d1 + e1*x)^(p + 1)*(d2 + e2*x
)^(p + 1)*(a + b*ArcCosh[c*x])^n, x], x] - Dist[(b*f*n*(-(d1*d2))^IntPart[p]*(d1 + e1*x)^FracPart[p]*(d2 + e2*
x)^FracPart[p])/(2*c*(p + 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}, x] && EqQ[e1 - c*d1, 0]
&& EqQ[e2 + c*d2, 0] && GtQ[n, 0] && LtQ[p, -1] && GtQ[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^4 \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^4 \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 c^2 d^2 (1-c x) (1+c x) \sqrt {d-c^2 d x^2}}+\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)^{3/2} (1+c x)^{3/2}} \, dx}{c^2 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (b \sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {x^3}{\left (-1+c^2 x^2\right )^2} \, dx}{3 c d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {x \left (a+b \cosh ^{-1}(c x)\right )}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 c^2 d^2 (1-c x) (1+c x) \sqrt {d-c^2 d x^2}}+\frac {\left (\sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {a+b \cosh ^{-1}(c x)}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (b \sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {x}{-1+c^2 x^2} \, dx}{c^3 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (b \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 c d^2 \sqrt {d-c^2 d x^2}}\\ &=-\frac {x \left (a+b \cosh ^{-1}(c x)\right )}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 c^2 d^2 (1-c x) (1+c x) \sqrt {d-c^2 d x^2}}+\frac {\sqrt {-1+c x} \sqrt {1+c x} \left (a+b \cosh ^{-1}(c x)\right )^2}{2 b c^5 d^2 \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 )}{2 c^5 d^2 \sqrt {d-c^2 d x^2}}+\frac {\left (b \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 c d^2 \sqrt {d-c^2 d x^2}}\\ &=\frac {b \sqrt {-1+c x} \sqrt {1+c x}}{6 c^5 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}-\frac {x \left (a+b \cosh ^{-1}(c x)\right )}{c^4 d^2 \sqrt {d-c^2 d x^2}}+\frac {x^3 \left (a+b \cosh ^{-1}(c x)\right )}{3 c^2 d^2 (1-c x) (1+c x) \sqrt {d-c^2 d x^2}}+\frac {\sqrt {-1+c x} \sqrt {1+c x} \left (a+b \cosh ^{-1}(c x)\right )^2}{2 b c^5 d^2 \sqrt {d-c^2 d x^2}}+\frac {2 b \sqrt {-1+c x} \sqrt {1+c x} \log \left (1-c^2 x^2\right )}{3 c^5 d^2 \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.76, size = 225, normalized size = 1.00 \[ \frac {\frac {2 a c x \left (4 c^2 x^2-3\right ) \sqrt {d-c^2 d x^2}}{\left (c^2 x^2-1\right )^2}-6 a \sqrt {d} \tan ^{-1}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (c^2 x^2-1\right )}\right )+\frac {b d \left (-\frac {\sqrt {\frac {c x-1}{c x+1}} (c x+1)+2 c x \cosh ^{-1}(c x)}{c^2 x^2-1}-8 c x \cosh ^{-1}(c x)+\sqrt {\frac {c x-1}{c x+1}} (c x+1) \left (8 \log \left (\sqrt {\frac {c x-1}{c x+1}} (c x+1)\right )+3 \cosh ^{-1}(c x)^2\right )\right )}{\sqrt {d-c^2 d x^2}}}{6 c^5 d^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^4*(a + b*ArcCosh[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

((2*a*c*x*(-3 + 4*c^2*x^2)*Sqrt[d - c^2*d*x^2])/(-1 + c^2*x^2)^2 - 6*a*Sqrt[d]*ArcTan[(c*x*Sqrt[d - c^2*d*x^2]
)/(Sqrt[d]*(-1 + c^2*x^2))] + (b*d*(-8*c*x*ArcCosh[c*x] - (Sqrt[(-1 + c*x)/(1 + c*x)]*(1 + c*x) + 2*c*x*ArcCos
h[c*x])/(-1 + c^2*x^2) + Sqrt[(-1 + c*x)/(1 + c*x)]*(1 + c*x)*(3*ArcCosh[c*x]^2 + 8*Log[Sqrt[(-1 + c*x)/(1 + c
*x)]*(1 + c*x)])))/Sqrt[d - c^2*d*x^2])/(6*c^5*d^3)

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (b x^{4} \operatorname {arcosh}\left (c x\right ) + a x^{4}\right )} \sqrt {-c^{2} d x^{2} + d}}{c^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral(-(b*x^4*arccosh(c*x) + a*x^4)*sqrt(-c^2*d*x^2 + d)/(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 \[ \int \frac {{\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} x^{4}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arccosh(c*x) + a)*x^4/(-c^2*d*x^2 + d)^(5/2), x)

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maple [B]  time = 0.84, size = 1519, normalized size = 6.78 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(5/2),x)

[Out]

1/3*a*x^3/c^2/d/(-c^2*d*x^2+d)^(3/2)-a/c^4/d^2*x/(-c^2*d*x^2+d)^(1/2)+a/c^4/d^2/(c^2*d)^(1/2)*arctan((c^2*d)^(
1/2)*x/(-c^2*d*x^2+d)^(1/2))-1/2*b*(-d*(c^2*x^2-1))^(1/2)*(c*x-1)^(1/2)*(c*x+1)^(1/2)/d^3/c^5/(c^2*x^2-1)*arcc
osh(c*x)^2+8/3*b*(-d*(c^2*x^2-1))^(1/2)*(c*x-1)^(1/2)*(c*x+1)^(1/2)/d^3/c^5/(c^2*x^2-1)*arccosh(c*x)-32*b*(-d*
(c^2*x^2-1))^(1/2)/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)*c/d^3*arccosh(c*x)*(c*x+1)^(1/2)*(c*x-1)^
(1/2)*x^6+32*b*(-d*(c^2*x^2-1))^(1/2)/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)*c^2/d^3*arccosh(c*x)*x
^7-8/3*b*(-d*(c^2*x^2-1))^(1/2)/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/d^3*(c*x+1)*(c*x-1)*x^5+8/3*
b*(-d*(c^2*x^2-1))^(1/2)/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)*c^2/d^3*x^7+84*b*(-d*(c^2*x^2-1))^(
1/2)/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c/d^3*arccosh(c*x)*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^4-76*b
*(-d*(c^2*x^2-1))^(1/2)/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/d^3*arccosh(c*x)*x^5+14/3*b*(-d*(c^2
*x^2-1))^(1/2)/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c^2/d^3*(c*x+1)*(c*x-1)*x^3+4*b*(-d*(c^2*x^2-
1))^(1/2)/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c/d^3*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^4-22/3*b*(-d*(
c^2*x^2-1))^(1/2)/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/d^3*x^5-220/3*b*(-d*(c^2*x^2-1))^(1/2)/(24
*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c^3/d^3*arccosh(c*x)*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^2+181/3*b*(-
d*(c^2*x^2-1))^(1/2)/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c^2/d^3*arccosh(c*x)*x^3-2*b*(-d*(c^2*x
^2-1))^(1/2)/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c^4/d^3*(c*x+1)*(c*x-1)*x-13/2*b*(-d*(c^2*x^2-1
))^(1/2)/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c^3/d^3*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^2+20/3*b*(-d*
(c^2*x^2-1))^(1/2)/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c^2/d^3*x^3+64/3*b*(-d*(c^2*x^2-1))^(1/2)
/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c^5/d^3*arccosh(c*x)*(c*x+1)^(1/2)*(c*x-1)^(1/2)-16*b*(-d*(
c^2*x^2-1))^(1/2)/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c^4/d^3*arccosh(c*x)*x+8/3*b*(-d*(c^2*x^2-
1))^(1/2)/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c^5/d^3*(c*x-1)^(1/2)*(c*x+1)^(1/2)-2*b*(-d*(c^2*x
^2-1))^(1/2)/(24*c^8*x^8-87*c^6*x^6+118*c^4*x^4-71*c^2*x^2+16)/c^4/d^3*x-4/3*b*(-d*(c^2*x^2-1))^(1/2)*(c*x-1)^
(1/2)*(c*x+1)^(1/2)/d^3/c^5/(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 [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, {\left (x {\left (\frac {3 \, x^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d} - \frac {2}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4} d}\right )} - \frac {x}{\sqrt {-c^{2} d x^{2} + d} c^{4} d^{2}} + \frac {3 \, \arcsin \left (c x\right )}{c^{5} d^{\frac {5}{2}}}\right )} a + b \int \frac {x^{4} \log \left (c x + \sqrt {c x + 1} \sqrt {c x - 1}\right )}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

1/3*(x*(3*x^2/((-c^2*d*x^2 + d)^(3/2)*c^2*d) - 2/((-c^2*d*x^2 + d)^(3/2)*c^4*d)) - x/(sqrt(-c^2*d*x^2 + d)*c^4
*d^2) + 3*arcsin(c*x)/(c^5*d^(5/2)))*a + b*integrate(x^4*log(c*x + sqrt(c*x + 1)*sqrt(c*x - 1))/(-c^2*d*x^2 +
d)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + b*acosh(c*x)))/(d - c^2*d*x^2)^(5/2),x)

[Out]

int((x^4*(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 \[ \int \frac {x^{4} \left (a + b \operatorname {acosh}{\left (c x \right )}\right )}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*acosh(c*x))/(-c**2*d*x**2+d)**(5/2),x)

[Out]

Integral(x**4*(a + b*acosh(c*x))/(-d*(c*x - 1)*(c*x + 1))**(5/2), x)

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