3.83 \(\int x (d-c^2 d x^2)^{3/2} (a+b \cosh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=165 \[ -\frac {\left (d-c^2 d x^2\right )^{5/2} \left (a+b \cosh ^{-1}(c x)\right )}{5 c^2 d}+\frac {b d x \sqrt {d-c^2 d x^2}}{5 c \sqrt {c x-1} \sqrt {c x+1}}-\frac {2 b c d x^3 \sqrt {d-c^2 d x^2}}{15 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b c^3 d x^5 \sqrt {d-c^2 d x^2}}{25 \sqrt {c x-1} \sqrt {c x+1}} \]

[Out]

-1/5*(-c^2*d*x^2+d)^(5/2)*(a+b*arccosh(c*x))/c^2/d+1/5*b*d*x*(-c^2*d*x^2+d)^(1/2)/c/(c*x-1)^(1/2)/(c*x+1)^(1/2
)-2/15*b*c*d*x^3*(-c^2*d*x^2+d)^(1/2)/(c*x-1)^(1/2)/(c*x+1)^(1/2)+1/25*b*c^3*d*x^5*(-c^2*d*x^2+d)^(1/2)/(c*x-1
)^(1/2)/(c*x+1)^(1/2)

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Rubi [A]  time = 0.27, antiderivative size = 178, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {5798, 5718, 194} \[ -\frac {d (1-c x)^2 (c x+1)^2 \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{5 c^2}+\frac {b c^3 d x^5 \sqrt {d-c^2 d x^2}}{25 \sqrt {c x-1} \sqrt {c x+1}}-\frac {2 b c d x^3 \sqrt {d-c^2 d x^2}}{15 \sqrt {c x-1} \sqrt {c x+1}}+\frac {b d x \sqrt {d-c^2 d x^2}}{5 c \sqrt {c x-1} \sqrt {c x+1}} \]

Antiderivative was successfully verified.

[In]

Int[x*(d - c^2*d*x^2)^(3/2)*(a + b*ArcCosh[c*x]),x]

[Out]

(b*d*x*Sqrt[d - c^2*d*x^2])/(5*c*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) - (2*b*c*d*x^3*Sqrt[d - c^2*d*x^2])/(15*Sqrt[-1
+ c*x]*Sqrt[1 + c*x]) + (b*c^3*d*x^5*Sqrt[d - c^2*d*x^2])/(25*Sqrt[-1 + c*x]*Sqrt[1 + c*x]) - (d*(1 - c*x)^2*
(1 + c*x)^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcCosh[c*x]))/(5*c^2)

Rule 194

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

Rule 5718

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d1_) + (e1_.)*(x_))^(p_.)*((d2_) + (e2_.)*(x_))^(p_.), x_
Symbol] :> Simp[((d1 + e1*x)^(p + 1)*(d2 + e2*x)^(p + 1)*(a + b*ArcCosh[c*x])^n)/(2*e1*e2*(p + 1)), x] - Dist[
(b*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[(-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, p}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2, 0] && GtQ[n, 0] && NeQ[p, -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 x \left (d-c^2 d x^2\right )^{3/2} \left (a+b \cosh ^{-1}(c x)\right ) \, dx &=-\frac {\left (d \sqrt {d-c^2 d x^2}\right ) \int x (-1+c x)^{3/2} (1+c x)^{3/2} \left (a+b \cosh ^{-1}(c x)\right ) \, dx}{\sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {d (1-c x)^2 (1+c x)^2 \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{5 c^2}+\frac {\left (b d \sqrt {d-c^2 d x^2}\right ) \int \left (-1+c^2 x^2\right )^2 \, dx}{5 c \sqrt {-1+c x} \sqrt {1+c x}}\\ &=-\frac {d (1-c x)^2 (1+c x)^2 \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{5 c^2}+\frac {\left (b d \sqrt {d-c^2 d x^2}\right ) \int \left (1-2 c^2 x^2+c^4 x^4\right ) \, dx}{5 c \sqrt {-1+c x} \sqrt {1+c x}}\\ &=\frac {b d x \sqrt {d-c^2 d x^2}}{5 c \sqrt {-1+c x} \sqrt {1+c x}}-\frac {2 b c d x^3 \sqrt {d-c^2 d x^2}}{15 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b c^3 d x^5 \sqrt {d-c^2 d x^2}}{25 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {d (1-c x)^2 (1+c x)^2 \sqrt {d-c^2 d x^2} \left (a+b \cosh ^{-1}(c x)\right )}{5 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 107, normalized size = 0.65 \[ -\frac {d \sqrt {d-c^2 d x^2} \left (15 a \left (c^2 x^2-1\right )^3+15 b \left (c^2 x^2-1\right )^3 \cosh ^{-1}(c x)+b c x \sqrt {c x-1} \sqrt {c x+1} \left (-3 c^4 x^4+10 c^2 x^2-15\right )\right )}{75 c^2 \left (c^2 x^2-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(d - c^2*d*x^2)^(3/2)*(a + b*ArcCosh[c*x]),x]

[Out]

-1/75*(d*Sqrt[d - c^2*d*x^2]*(15*a*(-1 + c^2*x^2)^3 + b*c*x*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(-15 + 10*c^2*x^2 - 3
*c^4*x^4) + 15*b*(-1 + c^2*x^2)^3*ArcCosh[c*x]))/(c^2*(-1 + c^2*x^2))

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fricas [A]  time = 0.57, size = 185, normalized size = 1.12 \[ -\frac {15 \, {\left (b c^{6} d x^{6} - 3 \, b c^{4} d x^{4} + 3 \, b c^{2} d x^{2} - b d\right )} \sqrt {-c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) - {\left (3 \, b c^{5} d x^{5} - 10 \, b c^{3} d x^{3} + 15 \, b c d x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} + 15 \, {\left (a c^{6} d x^{6} - 3 \, a c^{4} d x^{4} + 3 \, a c^{2} d x^{2} - a d\right )} \sqrt {-c^{2} d x^{2} + d}}{75 \, {\left (c^{4} x^{2} - c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-c^2*d*x^2+d)^(3/2)*(a+b*arccosh(c*x)),x, algorithm="fricas")

[Out]

-1/75*(15*(b*c^6*d*x^6 - 3*b*c^4*d*x^4 + 3*b*c^2*d*x^2 - b*d)*sqrt(-c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 - 1)
) - (3*b*c^5*d*x^5 - 10*b*c^3*d*x^3 + 15*b*c*d*x)*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1) + 15*(a*c^6*d*x^6 - 3
*a*c^4*d*x^4 + 3*a*c^2*d*x^2 - a*d)*sqrt(-c^2*d*x^2 + d))/(c^4*x^2 - c^2)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-c^2*d*x^2+d)^(3/2)*(a+b*arccosh(c*x)),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.31, size = 620, normalized size = 3.76 \[ -\frac {a \left (-c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}{5 c^{2} d}+b \left (-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (16 c^{6} x^{6}-28 c^{4} x^{4}+16 \sqrt {c x +1}\, \sqrt {c x -1}\, x^{5} c^{5}+13 c^{2} x^{2}-20 \sqrt {c x +1}\, \sqrt {c x -1}\, x^{3} c^{3}+5 \sqrt {c x +1}\, \sqrt {c x -1}\, x c -1\right ) \left (-1+5 \,\mathrm {arccosh}\left (c x \right )\right ) d}{800 \left (c x +1\right ) c^{2} \left (c x -1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (4 c^{4} x^{4}-5 c^{2} x^{2}+4 \sqrt {c x +1}\, \sqrt {c x -1}\, x^{3} c^{3}-3 \sqrt {c x +1}\, \sqrt {c x -1}\, x c +1\right ) \left (-1+3 \,\mathrm {arccosh}\left (c x \right )\right ) d}{96 \left (c x +1\right ) c^{2} \left (c x -1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (\sqrt {c x +1}\, \sqrt {c x -1}\, x c +c^{2} x^{2}-1\right ) \left (-1+\mathrm {arccosh}\left (c x \right )\right ) d}{16 \left (c x +1\right ) c^{2} \left (c x -1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-\sqrt {c x +1}\, \sqrt {c x -1}\, x c +c^{2} x^{2}-1\right ) \left (1+\mathrm {arccosh}\left (c x \right )\right ) d}{16 \left (c x +1\right ) c^{2} \left (c x -1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-4 \sqrt {c x +1}\, \sqrt {c x -1}\, x^{3} c^{3}+4 c^{4} x^{4}+3 \sqrt {c x +1}\, \sqrt {c x -1}\, x c -5 c^{2} x^{2}+1\right ) \left (1+3 \,\mathrm {arccosh}\left (c x \right )\right ) d}{96 \left (c x +1\right ) c^{2} \left (c x -1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (-16 \sqrt {c x +1}\, \sqrt {c x -1}\, x^{5} c^{5}+16 c^{6} x^{6}+20 \sqrt {c x +1}\, \sqrt {c x -1}\, x^{3} c^{3}-28 c^{4} x^{4}-5 \sqrt {c x +1}\, \sqrt {c x -1}\, x c +13 c^{2} x^{2}-1\right ) \left (1+5 \,\mathrm {arccosh}\left (c x \right )\right ) d}{800 \left (c x +1\right ) c^{2} \left (c x -1\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(-c^2*d*x^2+d)^(3/2)*(a+b*arccosh(c*x)),x)

[Out]

-1/5*a/c^2/d*(-c^2*d*x^2+d)^(5/2)+b*(-1/800*(-d*(c^2*x^2-1))^(1/2)*(16*c^6*x^6-28*c^4*x^4+16*(c*x+1)^(1/2)*(c*
x-1)^(1/2)*x^5*c^5+13*c^2*x^2-20*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^3*c^3+5*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x*c-1)*(-1+
5*arccosh(c*x))*d/(c*x+1)/c^2/(c*x-1)+1/96*(-d*(c^2*x^2-1))^(1/2)*(4*c^4*x^4-5*c^2*x^2+4*(c*x+1)^(1/2)*(c*x-1)
^(1/2)*x^3*c^3-3*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x*c+1)*(-1+3*arccosh(c*x))*d/(c*x+1)/c^2/(c*x-1)-1/16*(-d*(c^2*x^
2-1))^(1/2)*((c*x+1)^(1/2)*(c*x-1)^(1/2)*x*c+c^2*x^2-1)*(-1+arccosh(c*x))*d/(c*x+1)/c^2/(c*x-1)-1/16*(-d*(c^2*
x^2-1))^(1/2)*(-(c*x+1)^(1/2)*(c*x-1)^(1/2)*x*c+c^2*x^2-1)*(1+arccosh(c*x))*d/(c*x+1)/c^2/(c*x-1)+1/96*(-d*(c^
2*x^2-1))^(1/2)*(-4*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^3*c^3+4*c^4*x^4+3*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x*c-5*c^2*x^2+
1)*(1+3*arccosh(c*x))*d/(c*x+1)/c^2/(c*x-1)-1/800*(-d*(c^2*x^2-1))^(1/2)*(-16*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^5*
c^5+16*c^6*x^6+20*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x^3*c^3-28*c^4*x^4-5*(c*x+1)^(1/2)*(c*x-1)^(1/2)*x*c+13*c^2*x^2-
1)*(1+5*arccosh(c*x))*d/(c*x+1)/c^2/(c*x-1))

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maxima [A]  time = 0.77, size = 102, normalized size = 0.62 \[ -\frac {{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} b \operatorname {arcosh}\left (c x\right )}{5 \, c^{2} d} - \frac {{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} a}{5 \, c^{2} d} + \frac {{\left (3 \, c^{4} \sqrt {-d} d^{2} x^{5} - 10 \, c^{2} \sqrt {-d} d^{2} x^{3} + 15 \, \sqrt {-d} d^{2} x\right )} b}{75 \, c d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-c^2*d*x^2+d)^(3/2)*(a+b*arccosh(c*x)),x, algorithm="maxima")

[Out]

-1/5*(-c^2*d*x^2 + d)^(5/2)*b*arccosh(c*x)/(c^2*d) - 1/5*(-c^2*d*x^2 + d)^(5/2)*a/(c^2*d) + 1/75*(3*c^4*sqrt(-
d)*d^2*x^5 - 10*c^2*sqrt(-d)*d^2*x^3 + 15*sqrt(-d)*d^2*x)*b/(c*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*acosh(c*x))*(d - c^2*d*x^2)^(3/2),x)

[Out]

int(x*(a + b*acosh(c*x))*(d - c^2*d*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {acosh}{\left (c x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-c**2*d*x**2+d)**(3/2)*(a+b*acosh(c*x)),x)

[Out]

Integral(x*(-d*(c*x - 1)*(c*x + 1))**(3/2)*(a + b*acosh(c*x)), x)

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