3.101 \(\int \frac{a+b \cosh ^{-1}(c+d x)}{(c e+d e x)^4} \, dx\)

Optimal. Leaf size=99 \[ -\frac{a+b \cosh ^{-1}(c+d x)}{3 d e^4 (c+d x)^3}+\frac{b \sqrt{c+d x-1} \sqrt{c+d x+1}}{6 d e^4 (c+d x)^2}+\frac{b \tan ^{-1}\left (\sqrt{c+d x-1} \sqrt{c+d x+1}\right )}{6 d e^4} \]

[Out]

(b*Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x])/(6*d*e^4*(c + d*x)^2) - (a + b*ArcCosh[c + d*x])/(3*d*e^4*(c + d*x)^3
) + (b*ArcTan[Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x]])/(6*d*e^4)

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Rubi [A]  time = 0.0650914, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5866, 12, 5662, 103, 92, 203} \[ -\frac{a+b \cosh ^{-1}(c+d x)}{3 d e^4 (c+d x)^3}+\frac{b \sqrt{c+d x-1} \sqrt{c+d x+1}}{6 d e^4 (c+d x)^2}+\frac{b \tan ^{-1}\left (\sqrt{c+d x-1} \sqrt{c+d x+1}\right )}{6 d e^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCosh[c + d*x])/(c*e + d*e*x)^4,x]

[Out]

(b*Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x])/(6*d*e^4*(c + d*x)^2) - (a + b*ArcCosh[c + d*x])/(3*d*e^4*(c + d*x)^3
) + (b*ArcTan[Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x]])/(6*d*e^4)

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
]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 92

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 203

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

Rubi steps

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

Mathematica [A]  time = 0.206815, size = 101, normalized size = 1.02 \[ \frac{\frac{b \left (\frac{(c+d x-1) (c+d x+1)}{(c+d x)^2}+\sqrt{(c+d x)^2-1} \tan ^{-1}\left (\sqrt{(c+d x)^2-1}\right )\right )}{\sqrt{c+d x-1} \sqrt{c+d x+1}}-\frac{2 \left (a+b \cosh ^{-1}(c+d x)\right )}{(c+d x)^3}}{6 d e^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCosh[c + d*x])/(c*e + d*e*x)^4,x]

[Out]

((-2*(a + b*ArcCosh[c + d*x]))/(c + d*x)^3 + (b*(((-1 + c + d*x)*(1 + c + d*x))/(c + d*x)^2 + Sqrt[-1 + (c + d
*x)^2]*ArcTan[Sqrt[-1 + (c + d*x)^2]]))/(Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x]))/(6*d*e^4)

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Maple [A]  time = 0.004, size = 120, normalized size = 1.2 \begin{align*} -{\frac{a}{3\,d{e}^{4} \left ( dx+c \right ) ^{3}}}-{\frac{b{\rm arccosh} \left (dx+c\right )}{3\,d{e}^{4} \left ( dx+c \right ) ^{3}}}-{\frac{b}{6\,d{e}^{4}}\sqrt{dx+c-1}\sqrt{dx+c+1}\arctan \left ({\frac{1}{\sqrt{ \left ( dx+c \right ) ^{2}-1}}} \right ){\frac{1}{\sqrt{ \left ( dx+c \right ) ^{2}-1}}}}+{\frac{b}{6\,d{e}^{4} \left ( dx+c \right ) ^{2}}\sqrt{dx+c-1}\sqrt{dx+c+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccosh(d*x+c))/(d*e*x+c*e)^4,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{6} \, b{\left (\frac{2 \, d^{2} x^{2} + 4 \, c d x + 2 \, c^{2} -{\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )} \log \left (d x + c + 1\right ) +{\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )} \log \left (d x + c - 1\right ) - 2 \, \log \left (d x + \sqrt{d x + c + 1} \sqrt{d x + c - 1} + c\right )}{d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}} - 6 \, \int \frac{1}{3 \,{\left (d^{6} e^{4} x^{6} + 6 \, c d^{5} e^{4} x^{5} + c^{6} e^{4} - c^{4} e^{4} +{\left (15 \, c^{2} d^{4} e^{4} - d^{4} e^{4}\right )} x^{4} + 4 \,{\left (5 \, c^{3} d^{3} e^{4} - c d^{3} e^{4}\right )} x^{3} + 3 \,{\left (5 \, c^{4} d^{2} e^{4} - 2 \, c^{2} d^{2} e^{4}\right )} x^{2} + 2 \,{\left (3 \, c^{5} d e^{4} - 2 \, c^{3} d e^{4}\right )} x +{\left (d^{5} e^{4} x^{5} + 5 \, c d^{4} e^{4} x^{4} + c^{5} e^{4} - c^{3} e^{4} +{\left (10 \, c^{2} d^{3} e^{4} - d^{3} e^{4}\right )} x^{3} +{\left (10 \, c^{3} d^{2} e^{4} - 3 \, c d^{2} e^{4}\right )} x^{2} +{\left (5 \, c^{4} d e^{4} - 3 \, c^{2} d e^{4}\right )} x\right )} e^{\left (\frac{1}{2} \, \log \left (d x + c + 1\right ) + \frac{1}{2} \, \log \left (d x + c - 1\right )\right )}\right )}}\,{d x}\right )} - \frac{a}{3 \,{\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x+c))/(d*e*x+c*e)^4,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.78078, size = 610, normalized size = 6.16 \begin{align*} -\frac{2 \, a c^{3} - 2 \,{\left (b c^{3} d^{3} x^{3} + 3 \, b c^{4} d^{2} x^{2} + 3 \, b c^{5} d x + b c^{6}\right )} \arctan \left (-d x - c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\right ) - 2 \,{\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\right ) - 2 \,{\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (-d x - c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\right ) -{\left (b c^{3} d x + b c^{4}\right )} \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}}{6 \,{\left (c^{3} d^{4} e^{4} x^{3} + 3 \, c^{4} d^{3} e^{4} x^{2} + 3 \, c^{5} d^{2} e^{4} x + c^{6} d e^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x+c))/(d*e*x+c*e)^4,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac{b \operatorname{acosh}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acosh(d*x+c))/(d*e*x+c*e)**4,x)

[Out]

(Integral(a/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(b*acosh(c + d*x)
/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x))/e**4

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{arcosh}\left (d x + c\right ) + a}{{\left (d e x + c e\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(d*x+c))/(d*e*x+c*e)^4,x, algorithm="giac")

[Out]

integrate((b*arccosh(d*x + c) + a)/(d*e*x + c*e)^4, x)