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

Optimal. Leaf size=104 \[ -\frac {a+b \cosh ^{-1}(c+d x)}{4 d e^5 (c+d x)^4}+\frac {b \sqrt {c+d x-1} \sqrt {c+d x+1}}{6 d e^5 (c+d x)}+\frac {b \sqrt {c+d x-1} \sqrt {c+d x+1}}{12 d e^5 (c+d x)^3} \]

[Out]

1/4*(-a-b*arccosh(d*x+c))/d/e^5/(d*x+c)^4+1/12*b*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)/d/e^5/(d*x+c)^3+1/6*b*(d*x+c-
1)^(1/2)*(d*x+c+1)^(1/2)/d/e^5/(d*x+c)

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Rubi [A]  time = 0.07, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5866, 12, 5662, 103, 95} \[ -\frac {a+b \cosh ^{-1}(c+d x)}{4 d e^5 (c+d x)^4}+\frac {b \sqrt {c+d x-1} \sqrt {c+d x+1}}{6 d e^5 (c+d x)}+\frac {b \sqrt {c+d x-1} \sqrt {c+d x+1}}{12 d e^5 (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x])/(12*d*e^5*(c + d*x)^3) + (b*Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x])/(6*
d*e^5*(c + d*x)) - (a + b*ArcCosh[c + d*x])/(4*d*e^5*(c + d*x)^4)

Rule 12

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

Rule 95

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] /; FreeQ[{a, b, c, d,
e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && EqQ[a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1), 0
] && NeQ[m, -1]

Rule 103

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[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*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

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 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 {a+b \cosh ^{-1}(c+d x)}{(c e+d e x)^5} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a+b \cosh ^{-1}(x)}{e^5 x^5} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {a+b \cosh ^{-1}(x)}{x^5} \, dx,x,c+d x\right )}{d e^5}\\ &=-\frac {a+b \cosh ^{-1}(c+d x)}{4 d e^5 (c+d x)^4}+\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x} x^4 \sqrt {1+x}} \, dx,x,c+d x\right )}{4 d e^5}\\ &=\frac {b \sqrt {-1+c+d x} \sqrt {1+c+d x}}{12 d e^5 (c+d x)^3}-\frac {a+b \cosh ^{-1}(c+d x)}{4 d e^5 (c+d x)^4}+\frac {b \operatorname {Subst}\left (\int \frac {2}{\sqrt {-1+x} x^2 \sqrt {1+x}} \, dx,x,c+d x\right )}{12 d e^5}\\ &=\frac {b \sqrt {-1+c+d x} \sqrt {1+c+d x}}{12 d e^5 (c+d x)^3}-\frac {a+b \cosh ^{-1}(c+d x)}{4 d e^5 (c+d x)^4}+\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x} x^2 \sqrt {1+x}} \, dx,x,c+d x\right )}{6 d e^5}\\ &=\frac {b \sqrt {-1+c+d x} \sqrt {1+c+d x}}{12 d e^5 (c+d x)^3}+\frac {b \sqrt {-1+c+d x} \sqrt {1+c+d x}}{6 d e^5 (c+d x)}-\frac {a+b \cosh ^{-1}(c+d x)}{4 d e^5 (c+d x)^4}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 86, normalized size = 0.83 \[ \frac {-3 a+b \sqrt {c+d x-1} \sqrt {c+d x+1} \left (2 c^3+6 c^2 d x+6 c d^2 x^2+c+2 d^3 x^3+d x\right )-3 b \cosh ^{-1}(c+d x)}{12 d e^5 (c+d x)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a + b*Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x]*(c + 2*c^3 + d*x + 6*c^2*d*x + 6*c*d^2*x^2 + 2*d^3*x^3) - 3*b*A
rcCosh[c + d*x])/(12*d*e^5*(c + d*x)^4)

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fricas [B]  time = 0.51, size = 208, normalized size = 2.00 \[ \frac {3 \, a d^{4} x^{4} + 12 \, a c d^{3} x^{3} + 18 \, a c^{2} d^{2} x^{2} + 12 \, a c^{3} d x - 3 \, b c^{4} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\right ) + {\left (2 \, b c^{4} d^{3} x^{3} + 6 \, b c^{5} d^{2} x^{2} + 2 \, b c^{7} + b c^{5} + {\left (6 \, b c^{6} + b c^{4}\right )} d x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} - 1}}{12 \, {\left (c^{4} d^{5} e^{5} x^{4} + 4 \, c^{5} d^{4} e^{5} x^{3} + 6 \, c^{6} d^{3} e^{5} x^{2} + 4 \, c^{7} d^{2} e^{5} x + c^{8} d e^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a*d^4*x^4 + 12*a*c*d^3*x^3 + 18*a*c^2*d^2*x^2 + 12*a*c^3*d*x - 3*b*c^4*log(d*x + c + sqrt(d^2*x^2 + 2*
c*d*x + c^2 - 1)) + (2*b*c^4*d^3*x^3 + 6*b*c^5*d^2*x^2 + 2*b*c^7 + b*c^5 + (6*b*c^6 + b*c^4)*d*x)*sqrt(d^2*x^2
+ 2*c*d*x + c^2 - 1))/(c^4*d^5*e^5*x^4 + 4*c^5*d^4*e^5*x^3 + 6*c^6*d^3*e^5*x^2 + 4*c^7*d^2*e^5*x + c^8*d*e^5)

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.01, size = 76, normalized size = 0.73 \[ \frac {-\frac {a}{4 e^{5} \left (d x +c \right )^{4}}+\frac {b \left (-\frac {\mathrm {arccosh}\left (d x +c \right )}{4 \left (d x +c \right )^{4}}+\frac {\sqrt {d x +c -1}\, \sqrt {d x +c +1}\, \left (2 \left (d x +c \right )^{2}+1\right )}{12 \left (d x +c \right )^{3}}\right )}{e^{5}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-1/4*a/e^5/(d*x+c)^4+b/e^5*(-1/4/(d*x+c)^4*arccosh(d*x+c)+1/12*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)*(2*(d*x+c)
^2+1)/(d*x+c)^3))

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maxima [B]  time = 0.56, size = 260, normalized size = 2.50 \[ \frac {1}{12} \, b {\left (\frac {{\left (2 \, d^{4} x^{4} + 8 \, c d^{3} x^{3} + 2 \, c^{4} + {\left (12 \, c^{2} d^{2} - d^{2}\right )} x^{2} - c^{2} + 2 \, {\left (4 \, c^{3} d - c d\right )} x - 1\right )} d}{{\left (d^{5} e^{5} x^{3} + 3 \, c d^{4} e^{5} x^{2} + 3 \, c^{2} d^{3} e^{5} x + c^{3} d^{2} e^{5}\right )} \sqrt {d x + c + 1} \sqrt {d x + c - 1}} - \frac {3 \, \operatorname {arcosh}\left (d x + c\right )}{d^{5} e^{5} x^{4} + 4 \, c d^{4} e^{5} x^{3} + 6 \, c^{2} d^{3} e^{5} x^{2} + 4 \, c^{3} d^{2} e^{5} x + c^{4} d e^{5}}\right )} - \frac {a}{4 \, {\left (d^{5} e^{5} x^{4} + 4 \, c d^{4} e^{5} x^{3} + 6 \, c^{2} d^{3} e^{5} x^{2} + 4 \, c^{3} d^{2} e^{5} x + c^{4} d e^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*b*((2*d^4*x^4 + 8*c*d^3*x^3 + 2*c^4 + (12*c^2*d^2 - d^2)*x^2 - c^2 + 2*(4*c^3*d - c*d)*x - 1)*d/((d^5*e^5
*x^3 + 3*c*d^4*e^5*x^2 + 3*c^2*d^3*e^5*x + c^3*d^2*e^5)*sqrt(d*x + c + 1)*sqrt(d*x + c - 1)) - 3*arccosh(d*x +
c)/(d^5*e^5*x^4 + 4*c*d^4*e^5*x^3 + 6*c^2*d^3*e^5*x^2 + 4*c^3*d^2*e^5*x + c^4*d*e^5)) - 1/4*a/(d^5*e^5*x^4 +
4*c*d^4*e^5*x^3 + 6*c^2*d^3*e^5*x^2 + 4*c^3*d^2*e^5*x + c^4*d*e^5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {acosh}\left (c+d\,x\right )}{{\left (c\,e+d\,e\,x\right )}^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(c + d*x))/(c*e + d*e*x)^5,x)

[Out]

int((a + b*acosh(c + d*x))/(c*e + d*e*x)^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{c^{5} + 5 c^{4} d x + 10 c^{3} d^{2} x^{2} + 10 c^{2} d^{3} x^{3} + 5 c d^{4} x^{4} + d^{5} x^{5}}\, dx + \int \frac {b \operatorname {acosh}{\left (c + d x \right )}}{c^{5} + 5 c^{4} d x + 10 c^{3} d^{2} x^{2} + 10 c^{2} d^{3} x^{3} + 5 c d^{4} x^{4} + d^{5} x^{5}}\, dx}{e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c**5 + 5*c**4*d*x + 10*c**3*d**2*x**2 + 10*c**2*d**3*x**3 + 5*c*d**4*x**4 + d**5*x**5), x) + Inte
gral(b*acosh(c + d*x)/(c**5 + 5*c**4*d*x + 10*c**3*d**2*x**2 + 10*c**2*d**3*x**3 + 5*c*d**4*x**4 + d**5*x**5),
x))/e**5

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