3.2.5 \(\int \frac {x^4 (a+b \cosh ^{-1}(c x))}{\sqrt {d-c^2 d x^2}} \, dx\) [105]

Optimal. Leaf size=212 \[ -\frac {3 b x^2 \sqrt {-1+c x} \sqrt {1+c x}}{16 c^3 \sqrt {d-c^2 d x^2}}-\frac {b x^4 \sqrt {-1+c x} \sqrt {1+c x}}{16 c \sqrt {d-c^2 d x^2}}-\frac {3 x \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{8 c^4 d}-\frac {x^3 \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{4 c^2 d}+\frac {3 \sqrt {-1+c x} \sqrt {1+c x} \left (a+b \cosh ^{-1}(c x)\right )^2}{16 b c^5 \sqrt {d-c^2 d x^2}} \]

[Out]

-3/16*b*x^2*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c^3/(-c^2*d*x^2+d)^(1/2)-1/16*b*x^4*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c/(-c^
2*d*x^2+d)^(1/2)+3/16*(a+b*arccosh(c*x))^2*(c*x-1)^(1/2)*(c*x+1)^(1/2)/b/c^5/(-c^2*d*x^2+d)^(1/2)-3/8*x*(a+b*a
rccosh(c*x))*(-c^2*d*x^2+d)^(1/2)/c^4/d-1/4*x^3*(a+b*arccosh(c*x))*(-c^2*d*x^2+d)^(1/2)/c^2/d

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Rubi [A]
time = 0.17, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {5938, 5892, 30} \begin {gather*} -\frac {x^3 \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{4 c^2 d}+\frac {3 \sqrt {c x-1} \sqrt {c x+1} \left (a+b \cosh ^{-1}(c x)\right )^2}{16 b c^5 \sqrt {d-c^2 d x^2}}-\frac {3 x \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{8 c^4 d}-\frac {b x^4 \sqrt {c x-1} \sqrt {c x+1}}{16 c \sqrt {d-c^2 d x^2}}-\frac {3 b x^2 \sqrt {c x-1} \sqrt {c x+1}}{16 c^3 \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*x^2*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(16*c^3*Sqrt[d - c^2*d*x^2]) - (b*x^4*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(1
6*c*Sqrt[d - c^2*d*x^2]) - (3*x*Sqrt[d - c^2*d*x^2]*(a + b*ArcCosh[c*x]))/(8*c^4*d) - (x^3*Sqrt[d - c^2*d*x^2]
*(a + b*ArcCosh[c*x]))/(4*c^2*d) + (3*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(a + b*ArcCosh[c*x])^2)/(16*b*c^5*Sqrt[d -
c^2*d*x^2])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 5892

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*S
imp[Sqrt[1 + c*x]*(Sqrt[-1 + c*x]/Sqrt[d + e*x^2])]*(a + b*ArcCosh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e,
n}, x] && EqQ[c^2*d + e, 0] && NeQ[n, -1]

Rule 5938

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcCosh[c*x])^n/(e*(m + 2*p + 1))), x] + (Dist[f^2*((m - 1)/(c^2*(
m + 2*p + 1))), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcCosh[c*x])^n, x], x] - Dist[b*f*(n/(c*(m + 2*p + 1))
)*Simp[(d + e*x^2)^p/((1 + c*x)^p*(-1 + c*x)^p)], Int[(f*x)^(m - 1)*(1 + c*x)^(p + 1/2)*(-1 + c*x)^(p + 1/2)*(
a + b*ArcCosh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && I
GtQ[m, 1] && NeQ[m + 2*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {x^4 \left (a+b \cosh ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^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 )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{\sqrt {d-c^2 d x^2}}\\ &=-\frac {x^3 (1-c x) (1+c x) \left (a+b \cosh ^{-1}(c x)\right )}{4 c^2 \sqrt {d-c^2 d x^2}}+\frac {\left (3 \sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {x^2 \left (a+b \cosh ^{-1}(c x)\right )}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{4 c^2 \sqrt {d-c^2 d x^2}}-\frac {\left (b \sqrt {-1+c x} \sqrt {1+c x}\right ) \int x^3 \, dx}{4 c \sqrt {d-c^2 d x^2}}\\ &=-\frac {b x^4 \sqrt {-1+c x} \sqrt {1+c x}}{16 c \sqrt {d-c^2 d x^2}}-\frac {3 x (1-c x) (1+c x) \left (a+b \cosh ^{-1}(c x)\right )}{8 c^4 \sqrt {d-c^2 d x^2}}-\frac {x^3 (1-c x) (1+c x) \left (a+b \cosh ^{-1}(c x)\right )}{4 c^2 \sqrt {d-c^2 d x^2}}+\frac {\left (3 \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}{8 c^4 \sqrt {d-c^2 d x^2}}-\frac {\left (3 b \sqrt {-1+c x} \sqrt {1+c x}\right ) \int x \, dx}{8 c^3 \sqrt {d-c^2 d x^2}}\\ &=-\frac {3 b x^2 \sqrt {-1+c x} \sqrt {1+c x}}{16 c^3 \sqrt {d-c^2 d x^2}}-\frac {b x^4 \sqrt {-1+c x} \sqrt {1+c x}}{16 c \sqrt {d-c^2 d x^2}}-\frac {3 x (1-c x) (1+c x) \left (a+b \cosh ^{-1}(c x)\right )}{8 c^4 \sqrt {d-c^2 d x^2}}-\frac {x^3 (1-c x) (1+c x) \left (a+b \cosh ^{-1}(c x)\right )}{4 c^2 \sqrt {d-c^2 d x^2}}+\frac {3 \sqrt {-1+c x} \sqrt {1+c x} \left (a+b \cosh ^{-1}(c x)\right )^2}{16 b c^5 \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.59, size = 171, normalized size = 0.81 \begin {gather*} \frac {-\frac {16 a c x \left (3+2 c^2 x^2\right ) \sqrt {d-c^2 d x^2}}{d}-\frac {48 a \text {ArcTan}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (-1+c^2 x^2\right )}\right )}{\sqrt {d}}+\frac {b \sqrt {\frac {-1+c x}{1+c x}} (1+c x) \left (-16 \cosh \left (2 \cosh ^{-1}(c x)\right )-\cosh \left (4 \cosh ^{-1}(c x)\right )+4 \cosh ^{-1}(c x) \left (6 \cosh ^{-1}(c x)+8 \sinh \left (2 \cosh ^{-1}(c x)\right )+\sinh \left (4 \cosh ^{-1}(c x)\right )\right )\right )}{\sqrt {d-c^2 d x^2}}}{128 c^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-16*a*c*x*(3 + 2*c^2*x^2)*Sqrt[d - c^2*d*x^2])/d - (48*a*ArcTan[(c*x*Sqrt[d - c^2*d*x^2])/(Sqrt[d]*(-1 + c^2
*x^2))])/Sqrt[d] + (b*Sqrt[(-1 + c*x)/(1 + c*x)]*(1 + c*x)*(-16*Cosh[2*ArcCosh[c*x]] - Cosh[4*ArcCosh[c*x]] +
4*ArcCosh[c*x]*(6*ArcCosh[c*x] + 8*Sinh[2*ArcCosh[c*x]] + Sinh[4*ArcCosh[c*x]])))/Sqrt[d - c^2*d*x^2])/(128*c^
5)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(567\) vs. \(2(180)=360\).
time = 6.20, size = 568, normalized size = 2.68

method	result	size
default	\(-\frac {a \,x^{3} \sqrt {-c^{2} d \,x^{2}+d}}{4 c^{2} d}-\frac {3 a x \sqrt {-c^{2} d \,x^{2}+d}}{8 c^{4} d}+\frac {3 a \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{8 c^{4} \sqrt {c^{2} d}}+b \left (-\frac {3 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {c x -1}\, \sqrt {c x +1}\, \mathrm {arccosh}\left (c x \right )^{2}}{16 d \,c^{5} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (8 c^{5} x^{5}-12 c^{3} x^{3}+8 \sqrt {c x +1}\, \sqrt {c x -1}\, x^{4} c^{4}+4 c x -8 \sqrt {c x +1}\, \sqrt {c x -1}\, x^{2} c^{2}+\sqrt {c x -1}\, \sqrt {c x +1}\right ) \left (-1+4 \,\mathrm {arccosh}\left (c x \right )\right )}{256 d \,c^{5} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (2 c^{3} x^{3}-2 c x +2 \sqrt {c x +1}\, \sqrt {c x -1}\, x^{2} c^{2}-\sqrt {c x -1}\, \sqrt {c x +1}\right ) \left (-1+2 \,\mathrm {arccosh}\left (c x \right )\right )}{16 d \,c^{5} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-2 \sqrt {c x +1}\, \sqrt {c x -1}\, x^{2} c^{2}+2 c^{3} x^{3}+\sqrt {c x -1}\, \sqrt {c x +1}-2 c x \right ) \left (1+2 \,\mathrm {arccosh}\left (c x \right )\right )}{16 d \,c^{5} \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-8 \sqrt {c x +1}\, \sqrt {c x -1}\, x^{4} c^{4}+8 c^{5} x^{5}+8 \sqrt {c x +1}\, \sqrt {c x -1}\, x^{2} c^{2}-12 c^{3} x^{3}-\sqrt {c x -1}\, \sqrt {c x +1}+4 c x \right ) \left (1+4 \,\mathrm {arccosh}\left (c x \right )\right )}{256 d \,c^{5} \left (c^{2} x^{2}-1\right )}\right )\)	\(568\)

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)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/4*a*x^3/c^2/d*(-c^2*d*x^2+d)^(1/2)-3/8*a/c^4*x/d*(-c^2*d*x^2+d)^(1/2)+3/8*a/c^4/(c^2*d)^(1/2)*arctan((c^2*d
)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))+b*(-3/16*(-d*(c^2*x^2-1))^(1/2)*(c*x-1)^(1/2)*(c*x+1)^(1/2)/d/c^5/(c^2*x^2-1)*
arccosh(c*x)^2-1/256*(-d*(c^2*x^2-1))^(1/2)*(8*c^5*x^5-12*c^3*x^3+8*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^4*c^4+4*c*x-
8*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^2*c^2+(c*x-1)^(1/2)*(c*x+1)^(1/2))*(-1+4*arccosh(c*x))/d/c^5/(c^2*x^2-1)-1/16*
(-d*(c^2*x^2-1))^(1/2)*(2*c^3*x^3-2*c*x+2*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^2*c^2-(c*x-1)^(1/2)*(c*x+1)^(1/2))*(-1
+2*arccosh(c*x))/d/c^5/(c^2*x^2-1)-1/16*(-d*(c^2*x^2-1))^(1/2)*(-2*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^2*c^2+2*c^3*x
^3+(c*x-1)^(1/2)*(c*x+1)^(1/2)-2*c*x)*(1+2*arccosh(c*x))/d/c^5/(c^2*x^2-1)-1/256*(-d*(c^2*x^2-1))^(1/2)*(-8*(c
*x+1)^(1/2)*(c*x-1)^(1/2)*x^4*c^4+8*c^5*x^5+8*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^2*c^2-12*c^3*x^3-(c*x-1)^(1/2)*(c*
x+1)^(1/2)+4*c*x)*(1+4*arccosh(c*x))/d/c^5/(c^2*x^2-1))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)^(1/2),x, algorithm="maxima")

[Out]

-1/8*a*(2*sqrt(-c^2*d*x^2 + d)*x^3/(c^2*d) + 3*sqrt(-c^2*d*x^2 + d)*x/(c^4*d) - 3*arcsin(c*x)/(c^5*sqrt(d))) +
b*integrate(x^4*log(c*x + sqrt(c*x + 1)*sqrt(c*x - 1))/sqrt(-c^2*d*x^2 + d), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} \left (a + b \operatorname {acosh}{\left (c x \right )}\right )}{\sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )}}\, dx \end {gather*}

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)**(1/2),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^4\,\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )}{\sqrt {d-c^2\,d\,x^2}} \,d x \end {gather*}

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)^(1/2),x)

[Out]

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

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