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

Optimal. Leaf size=138 \[ -\frac {a+b \cosh ^{-1}(c x)}{2 e (d+e x)^2}+\frac {b c^3 d \tanh ^{-1}\left (\frac {\sqrt {c x+1} \sqrt {c d+e}}{\sqrt {c x-1} \sqrt {c d-e}}\right )}{e (c d-e)^{3/2} (c d+e)^{3/2}}-\frac {b c \sqrt {c x-1} \sqrt {c x+1}}{2 \left (c^2 d^2-e^2\right ) (d+e x)} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5802, 96, 93, 208} \[ -\frac {a+b \cosh ^{-1}(c x)}{2 e (d+e x)^2}-\frac {b c \sqrt {c x-1} \sqrt {c x+1}}{2 \left (c^2 d^2-e^2\right ) (d+e x)}+\frac {b c^3 d \tanh ^{-1}\left (\frac {\sqrt {c x+1} \sqrt {c d+e}}{\sqrt {c x-1} \sqrt {c d-e}}\right )}{e (c d-e)^{3/2} (c d+e)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 208

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

Rule 5802

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

Rubi steps

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

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Mathematica [A]  time = 0.39, size = 184, normalized size = 1.33 \[ \frac {1}{2} \left (-\frac {a}{e (d+e x)^2}-\frac {b c \sqrt {c x-1} \sqrt {c x+1}}{\left (c^2 d^2-e^2\right ) (d+e x)}+\frac {b c^3 d \log (d+e x)}{e \left (c^2 d^2-e^2\right )^{3/2}}-\frac {b c^3 d \log \left (-\sqrt {c x-1} \sqrt {c x+1} \sqrt {c^2 d^2-e^2}+c^2 d x+e\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}-\frac {b \cosh ^{-1}(c x)}{e (d+e x)^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.91, size = 1132, normalized size = 8.20 \[ \left [-\frac {{\left (a + b\right )} c^{4} d^{6} - {\left (2 \, a + b\right )} c^{2} d^{4} e^{2} + a d^{2} e^{4} + {\left (b c^{4} d^{4} e^{2} - b c^{2} d^{2} e^{4}\right )} x^{2} + {\left (b c^{3} d^{3} e^{2} x^{2} + 2 \, b c^{3} d^{4} e x + b c^{3} d^{5}\right )} \sqrt {c^{2} d^{2} - e^{2}} \log \left (\frac {c^{3} d^{2} x + c d e - \sqrt {c^{2} d^{2} - e^{2}} {\left (c^{2} d x + e\right )} + {\left (c^{2} d^{2} - \sqrt {c^{2} d^{2} - e^{2}} c d - e^{2}\right )} \sqrt {c^{2} x^{2} - 1}}{e x + d}\right ) + 2 \, {\left (b c^{4} d^{5} e - b c^{2} d^{3} e^{3}\right )} x - {\left ({\left (b c^{4} d^{4} e^{2} - 2 \, b c^{2} d^{2} e^{4} + b e^{6}\right )} x^{2} + 2 \, {\left (b c^{4} d^{5} e - 2 \, b c^{2} d^{3} e^{3} + b d e^{5}\right )} x\right )} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) - {\left (b c^{4} d^{6} - 2 \, b c^{2} d^{4} e^{2} + b d^{2} e^{4} + {\left (b c^{4} d^{4} e^{2} - 2 \, b c^{2} d^{2} e^{4} + b e^{6}\right )} x^{2} + 2 \, {\left (b c^{4} d^{5} e - 2 \, b c^{2} d^{3} e^{3} + b d e^{5}\right )} x\right )} \log \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) + {\left (b c^{3} d^{5} e - b c d^{3} e^{3} + {\left (b c^{3} d^{4} e^{2} - b c d^{2} e^{4}\right )} x\right )} \sqrt {c^{2} x^{2} - 1}}{2 \, {\left (c^{4} d^{8} e - 2 \, c^{2} d^{6} e^{3} + d^{4} e^{5} + {\left (c^{4} d^{6} e^{3} - 2 \, c^{2} d^{4} e^{5} + d^{2} e^{7}\right )} x^{2} + 2 \, {\left (c^{4} d^{7} e^{2} - 2 \, c^{2} d^{5} e^{4} + d^{3} e^{6}\right )} x\right )}}, -\frac {{\left (a + b\right )} c^{4} d^{6} - {\left (2 \, a + b\right )} c^{2} d^{4} e^{2} + a d^{2} e^{4} + {\left (b c^{4} d^{4} e^{2} - b c^{2} d^{2} e^{4}\right )} x^{2} + 2 \, {\left (b c^{3} d^{3} e^{2} x^{2} + 2 \, b c^{3} d^{4} e x + b c^{3} d^{5}\right )} \sqrt {-c^{2} d^{2} + e^{2}} \arctan \left (-\frac {\sqrt {-c^{2} d^{2} + e^{2}} \sqrt {c^{2} x^{2} - 1} e - \sqrt {-c^{2} d^{2} + e^{2}} {\left (c e x + c d\right )}}{c^{2} d^{2} - e^{2}}\right ) + 2 \, {\left (b c^{4} d^{5} e - b c^{2} d^{3} e^{3}\right )} x - {\left ({\left (b c^{4} d^{4} e^{2} - 2 \, b c^{2} d^{2} e^{4} + b e^{6}\right )} x^{2} + 2 \, {\left (b c^{4} d^{5} e - 2 \, b c^{2} d^{3} e^{3} + b d e^{5}\right )} x\right )} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) - {\left (b c^{4} d^{6} - 2 \, b c^{2} d^{4} e^{2} + b d^{2} e^{4} + {\left (b c^{4} d^{4} e^{2} - 2 \, b c^{2} d^{2} e^{4} + b e^{6}\right )} x^{2} + 2 \, {\left (b c^{4} d^{5} e - 2 \, b c^{2} d^{3} e^{3} + b d e^{5}\right )} x\right )} \log \left (-c x + \sqrt {c^{2} x^{2} - 1}\right ) + {\left (b c^{3} d^{5} e - b c d^{3} e^{3} + {\left (b c^{3} d^{4} e^{2} - b c d^{2} e^{4}\right )} x\right )} \sqrt {c^{2} x^{2} - 1}}{2 \, {\left (c^{4} d^{8} e - 2 \, c^{2} d^{6} e^{3} + d^{4} e^{5} + {\left (c^{4} d^{6} e^{3} - 2 \, c^{2} d^{4} e^{5} + d^{2} e^{7}\right )} x^{2} + 2 \, {\left (c^{4} d^{7} e^{2} - 2 \, c^{2} d^{5} e^{4} + d^{3} e^{6}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((a + b)*c^4*d^6 - (2*a + b)*c^2*d^4*e^2 + a*d^2*e^4 + (b*c^4*d^4*e^2 - b*c^2*d^2*e^4)*x^2 + (b*c^3*d^3*
e^2*x^2 + 2*b*c^3*d^4*e*x + b*c^3*d^5)*sqrt(c^2*d^2 - e^2)*log((c^3*d^2*x + c*d*e - sqrt(c^2*d^2 - e^2)*(c^2*d
*x + e) + (c^2*d^2 - sqrt(c^2*d^2 - e^2)*c*d - e^2)*sqrt(c^2*x^2 - 1))/(e*x + d)) + 2*(b*c^4*d^5*e - b*c^2*d^3
*e^3)*x - ((b*c^4*d^4*e^2 - 2*b*c^2*d^2*e^4 + b*e^6)*x^2 + 2*(b*c^4*d^5*e - 2*b*c^2*d^3*e^3 + b*d*e^5)*x)*log(
c*x + sqrt(c^2*x^2 - 1)) - (b*c^4*d^6 - 2*b*c^2*d^4*e^2 + b*d^2*e^4 + (b*c^4*d^4*e^2 - 2*b*c^2*d^2*e^4 + b*e^6
)*x^2 + 2*(b*c^4*d^5*e - 2*b*c^2*d^3*e^3 + b*d*e^5)*x)*log(-c*x + sqrt(c^2*x^2 - 1)) + (b*c^3*d^5*e - b*c*d^3*
e^3 + (b*c^3*d^4*e^2 - b*c*d^2*e^4)*x)*sqrt(c^2*x^2 - 1))/(c^4*d^8*e - 2*c^2*d^6*e^3 + d^4*e^5 + (c^4*d^6*e^3
- 2*c^2*d^4*e^5 + d^2*e^7)*x^2 + 2*(c^4*d^7*e^2 - 2*c^2*d^5*e^4 + d^3*e^6)*x), -1/2*((a + b)*c^4*d^6 - (2*a +
b)*c^2*d^4*e^2 + a*d^2*e^4 + (b*c^4*d^4*e^2 - b*c^2*d^2*e^4)*x^2 + 2*(b*c^3*d^3*e^2*x^2 + 2*b*c^3*d^4*e*x + b*
c^3*d^5)*sqrt(-c^2*d^2 + e^2)*arctan(-(sqrt(-c^2*d^2 + e^2)*sqrt(c^2*x^2 - 1)*e - sqrt(-c^2*d^2 + e^2)*(c*e*x
+ c*d))/(c^2*d^2 - e^2)) + 2*(b*c^4*d^5*e - b*c^2*d^3*e^3)*x - ((b*c^4*d^4*e^2 - 2*b*c^2*d^2*e^4 + b*e^6)*x^2
+ 2*(b*c^4*d^5*e - 2*b*c^2*d^3*e^3 + b*d*e^5)*x)*log(c*x + sqrt(c^2*x^2 - 1)) - (b*c^4*d^6 - 2*b*c^2*d^4*e^2 +
b*d^2*e^4 + (b*c^4*d^4*e^2 - 2*b*c^2*d^2*e^4 + b*e^6)*x^2 + 2*(b*c^4*d^5*e - 2*b*c^2*d^3*e^3 + b*d*e^5)*x)*lo
g(-c*x + sqrt(c^2*x^2 - 1)) + (b*c^3*d^5*e - b*c*d^3*e^3 + (b*c^3*d^4*e^2 - b*c*d^2*e^4)*x)*sqrt(c^2*x^2 - 1))
/(c^4*d^8*e - 2*c^2*d^6*e^3 + d^4*e^5 + (c^4*d^6*e^3 - 2*c^2*d^4*e^5 + d^2*e^7)*x^2 + 2*(c^4*d^7*e^2 - 2*c^2*d
^5*e^4 + d^3*e^6)*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 361, normalized size = 2.62 \[ -\frac {c^{2} a}{2 \left (c x e +c d \right )^{2} e}-\frac {c^{2} b \,\mathrm {arccosh}\left (c x \right )}{2 \left (c x e +c d \right )^{2} e}-\frac {c^{4} b \sqrt {c x +1}\, \sqrt {c x -1}\, \ln \left (-\frac {2 \left (c^{2} d x -\sqrt {c^{2} x^{2}-1}\, \sqrt {\frac {c^{2} d^{2}-e^{2}}{e^{2}}}\, e +e \right )}{c x e +c d}\right ) x d}{2 e \sqrt {c^{2} x^{2}-1}\, \left (c d +e \right ) \left (c d -e \right ) \left (c x e +c d \right ) \sqrt {\frac {c^{2} d^{2}-e^{2}}{e^{2}}}}-\frac {c^{4} b \sqrt {c x +1}\, \sqrt {c x -1}\, \ln \left (-\frac {2 \left (c^{2} d x -\sqrt {c^{2} x^{2}-1}\, \sqrt {\frac {c^{2} d^{2}-e^{2}}{e^{2}}}\, e +e \right )}{c x e +c d}\right ) d^{2}}{2 e^{2} \sqrt {c^{2} x^{2}-1}\, \left (c d +e \right ) \left (c d -e \right ) \left (c x e +c d \right ) \sqrt {\frac {c^{2} d^{2}-e^{2}}{e^{2}}}}-\frac {c^{2} b \sqrt {c x +1}\, \sqrt {c x -1}}{2 \left (c d +e \right ) \left (c d -e \right ) \left (c x e +c d \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e-c*d>0)', see `assume?` for m
ore details)Is e-c*d positive, negative or zero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(c*x))/(d + e*x)^3,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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