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

Optimal. Leaf size=92 \[ \frac {b \sqrt {c+d x-1} \sqrt {c+d x+1} \left (a+b \cosh ^{-1}(c+d x)\right )}{d e^3 (c+d x)}-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}-\frac {b^2 \log (c+d x)}{d e^3} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {5866, 12, 5662, 5724, 29} \[ \frac {b \sqrt {c+d x-1} \sqrt {c+d x+1} \left (a+b \cosh ^{-1}(c+d x)\right )}{d e^3 (c+d x)}-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}-\frac {b^2 \log (c+d x)}{d e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 5662

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcC
osh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCosh[c*x])^(n - 1))/(Sqr
t[-1 + c*x]*Sqrt[1 + c*x]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

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 5866

Int[((a_.) + ArcCosh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcCosh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rubi steps

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

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Mathematica [A]  time = 0.21, size = 81, normalized size = 0.88 \[ \frac {b \left (\frac {\sqrt {c+d x-1} \sqrt {c+d x+1} \left (a+b \cosh ^{-1}(c+d x)\right )}{c+d x}-b \log (c+d x)\right )-\frac {\left (a+b \cosh ^{-1}(c+d x)\right )^2}{2 (c+d x)^2}}{d e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.68, size = 320, normalized size = 3.48 \[ \frac {2 \, a b c^{2} d^{2} x^{2} + 4 \, a b c^{3} d x + 2 \, a b c^{4} - b^{2} c^{2} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\right )^{2} - a^{2} c^{2} + 2 \, {\left (a b d^{2} x^{2} + 2 \, a b c d x + {\left (b^{2} c^{2} d x + b^{2} c^{3}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\right ) - 2 \, {\left (b^{2} c^{2} d^{2} x^{2} + 2 \, b^{2} c^{3} d x + b^{2} c^{4}\right )} \log \left (d x + c\right ) + 2 \, {\left (a b d^{2} x^{2} + 2 \, a b c d x + a b c^{2}\right )} \log \left (-d x - c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\right ) + 2 \, {\left (a b c^{2} d x + a b c^{3}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} - 1}}{2 \, {\left (c^{2} d^{3} e^{3} x^{2} + 2 \, c^{3} d^{2} e^{3} x + c^{4} d e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arcosh}\left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arccosh(d*x + c) + a)^2/(d*e*x + c*e)^3, x)

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maple [B]  time = 0.27, size = 194, normalized size = 2.11 \[ -\frac {a^{2}}{2 d \,e^{3} \left (d x +c \right )^{2}}+\frac {b^{2} \mathrm {arccosh}\left (d x +c \right )}{d \,e^{3}}+\frac {b^{2} \mathrm {arccosh}\left (d x +c \right ) \sqrt {d x +c +1}\, \sqrt {d x +c -1}}{d \,e^{3} \left (d x +c \right )}-\frac {b^{2} \mathrm {arccosh}\left (d x +c \right )^{2}}{2 d \,e^{3} \left (d x +c \right )^{2}}-\frac {b^{2} \ln \left (1+\left (d x +c +\sqrt {d x +c -1}\, \sqrt {d x +c +1}\right )^{2}\right )}{d \,e^{3}}-\frac {a b \,\mathrm {arccosh}\left (d x +c \right )}{d \,e^{3} \left (d x +c \right )^{2}}+\frac {a b \sqrt {d x +c -1}\, \sqrt {d x +c +1}}{d \,e^{3} \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.69, size = 229, normalized size = 2.49 \[ {\left (\frac {\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} - 1} d \operatorname {arcosh}\left (d x + c\right )}{d^{3} e^{3} x + c d^{2} e^{3}} - \frac {\log \left (d x + c\right )}{d e^{3}}\right )} b^{2} + a b {\left (\frac {\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} - 1} d}{d^{3} e^{3} x + c d^{2} e^{3}} - \frac {\operatorname {arcosh}\left (d x + c\right )}{d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}}\right )} - \frac {b^{2} \operatorname {arcosh}\left (d x + c\right )^{2}}{2 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} - \frac {a^{2}}{2 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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