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

Optimal. Leaf size=127 \[ \frac{a+b \cosh ^{-1}(c x)}{3 c^2 d \left (d-c^2 d x^2\right )^{3/2}}+\frac{b \sqrt{c x-1} \sqrt{c x+1} \tanh ^{-1}(c x)}{6 c^2 d^2 \sqrt{d-c^2 d x^2}}+\frac{b x \sqrt{c x-1} \sqrt{c x+1}}{6 c d \left (d-c^2 d x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.275824, antiderivative size = 154, normalized size of antiderivative = 1.21, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {5798, 5718, 199, 207} \[ \frac{a+b \cosh ^{-1}(c x)}{3 c^2 d^2 (1-c x) (c x+1) \sqrt{d-c^2 d x^2}}+\frac{b x \sqrt{c x-1} \sqrt{c x+1}}{6 c d^2 \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}}+\frac{b \sqrt{c x-1} \sqrt{c x+1} \tanh ^{-1}(c x)}{6 c^2 d^2 \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x \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 \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{a+b \cosh ^{-1}(c x)}{3 c^2 d^2 (1-c x) (1+c x) \sqrt{d-c^2 d x^2}}+\frac{\left (b \sqrt{-1+c x} \sqrt{1+c x}\right ) \int \frac{1}{\left (-1+c^2 x^2\right )^2} \, dx}{3 c d^2 \sqrt{d-c^2 d x^2}}\\ &=\frac{b x \sqrt{-1+c x} \sqrt{1+c x}}{6 c d^2 \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}}+\frac{a+b \cosh ^{-1}(c x)}{3 c^2 d^2 (1-c x) (1+c x) \sqrt{d-c^2 d x^2}}-\frac{\left (b \sqrt{-1+c x} \sqrt{1+c x}\right ) \int \frac{1}{-1+c^2 x^2} \, dx}{6 c d^2 \sqrt{d-c^2 d x^2}}\\ &=\frac{b x \sqrt{-1+c x} \sqrt{1+c x}}{6 c d^2 \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2}}+\frac{a+b \cosh ^{-1}(c x)}{3 c^2 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} \tanh ^{-1}(c x)}{6 c^2 d^2 \sqrt{d-c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.255342, size = 119, normalized size = 0.94 \[ \frac{\sqrt{d-c^2 d x^2} \left (2 a+b c x \sqrt{c x-1} \sqrt{c x+1}+2 b \cosh ^{-1}(c x)\right )}{6 c^2 d^3 \left (c^2 x^2-1\right )^2}-\frac{b \sqrt{-d \left (c^2 x^2-1\right )} \tanh ^{-1}(c x)}{6 c^2 d^3 \sqrt{c x-1} \sqrt{c x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.164, size = 249, normalized size = 2. \begin{align*}{\frac{a}{3\,{c}^{2}d} \left ( -{c}^{2}d{x}^{2}+d \right ) ^{-{\frac{3}{2}}}}+{\frac{bx}{6\,{d}^{3} \left ({c}^{2}{x}^{2}-1 \right ) ^{2}c}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{cx+1}\sqrt{cx-1}}+{\frac{b{\rm arccosh} \left (cx\right )}{3\,{d}^{3} \left ({c}^{2}{x}^{2}-1 \right ) ^{2}{c}^{2}}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{b}{6\,{c}^{2}{d}^{3} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{cx+1}\sqrt{cx-1}\ln \left ( cx+\sqrt{cx-1}\sqrt{cx+1}-1 \right ) }-{\frac{b}{6\,{c}^{2}{d}^{3} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{cx+1}\sqrt{cx-1}\ln \left ( 1+cx+\sqrt{cx-1}\sqrt{cx+1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b \int \frac{x \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} + \frac{a}{3 \,{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}} c^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.59786, size = 910, normalized size = 7.17 \begin{align*} \left [\frac{4 \, \sqrt{-c^{2} d x^{2} + d} \sqrt{c^{2} x^{2} - 1} b c x + 8 \, \sqrt{-c^{2} d x^{2} + d} b \log \left (c x + \sqrt{c^{2} x^{2} - 1}\right ) -{\left (b c^{4} x^{4} - 2 \, b c^{2} x^{2} + b\right )} \sqrt{-d} \log \left (-\frac{c^{6} d x^{6} + 5 \, c^{4} d x^{4} - 5 \, c^{2} d x^{2} - 4 \,{\left (c^{3} x^{3} + c x\right )} \sqrt{-c^{2} d x^{2} + d} \sqrt{c^{2} x^{2} - 1} \sqrt{-d} - d}{c^{6} x^{6} - 3 \, c^{4} x^{4} + 3 \, c^{2} x^{2} - 1}\right ) + 8 \, \sqrt{-c^{2} d x^{2} + d} a}{24 \,{\left (c^{6} d^{3} x^{4} - 2 \, c^{4} d^{3} x^{2} + c^{2} d^{3}\right )}}, \frac{2 \, \sqrt{-c^{2} d x^{2} + d} \sqrt{c^{2} x^{2} - 1} b c x -{\left (b c^{4} x^{4} - 2 \, b c^{2} x^{2} + b\right )} \sqrt{d} \arctan \left (\frac{2 \, \sqrt{-c^{2} d x^{2} + d} \sqrt{c^{2} x^{2} - 1} c \sqrt{d} x}{c^{4} d x^{4} - d}\right ) + 4 \, \sqrt{-c^{2} d x^{2} + d} b \log \left (c x + \sqrt{c^{2} x^{2} - 1}\right ) + 4 \, \sqrt{-c^{2} d x^{2} + d} a}{12 \,{\left (c^{6} d^{3} x^{4} - 2 \, c^{4} d^{3} x^{2} + c^{2} d^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(4*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*b*c*x + 8*sqrt(-c^2*d*x^2 + d)*b*log(c*x + sqrt(c^2*x^2 - 1))
- (b*c^4*x^4 - 2*b*c^2*x^2 + b)*sqrt(-d)*log(-(c^6*d*x^6 + 5*c^4*d*x^4 - 5*c^2*d*x^2 - 4*(c^3*x^3 + c*x)*sqrt(
-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*sqrt(-d) - d)/(c^6*x^6 - 3*c^4*x^4 + 3*c^2*x^2 - 1)) + 8*sqrt(-c^2*d*x^2 + d
)*a)/(c^6*d^3*x^4 - 2*c^4*d^3*x^2 + c^2*d^3), 1/12*(2*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*b*c*x - (b*c^4*x^
4 - 2*b*c^2*x^2 + b)*sqrt(d)*arctan(2*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*c*sqrt(d)*x/(c^4*d*x^4 - d)) + 4*
sqrt(-c^2*d*x^2 + d)*b*log(c*x + sqrt(c^2*x^2 - 1)) + 4*sqrt(-c^2*d*x^2 + d)*a)/(c^6*d^3*x^4 - 2*c^4*d^3*x^2 +
c^2*d^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arcosh}\left (c x\right ) + a\right )} x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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