3.1.99 \(\int \frac {a+b \cosh ^{-1}(c+d x)}{(c e+d e x)^2} \, dx\) [99]

Optimal. Leaf size=56 \[ -\frac {a+b \cosh ^{-1}(c+d x)}{d e^2 (c+d x)}+\frac {b \text {ArcTan}\left (\sqrt {-1+c+d x} \sqrt {1+c+d x}\right )}{d e^2} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5996, 12, 5883, 94, 209} \begin {gather*} \frac {b \text {ArcTan}\left (\sqrt {c+d x-1} \sqrt {c+d x+1}\right )}{d e^2}-\frac {a+b \cosh ^{-1}(c+d x)}{d e^2 (c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 94

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 209

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

Rule 5883

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)/(Sqrt
[1 + c*x]*Sqrt[-1 + c*x])), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5996

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

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Mathematica [A]
time = 0.09, size = 78, normalized size = 1.39 \begin {gather*} \frac {\frac {-a-b \cosh ^{-1}(c+d x)}{c+d x}+\frac {b \sqrt {-1+(c+d x)^2} \text {ArcTan}\left (\sqrt {-1+(c+d x)^2}\right )}{\sqrt {-1+c+d x} \sqrt {1+c+d x}}}{d e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 4.03, size = 83, normalized size = 1.48

method	result	size
derivativedivides	\(\frac {-\frac {a}{e^{2} \left (d x +c \right )}-\frac {b \,\mathrm {arccosh}\left (d x +c \right )}{e^{2} \left (d x +c \right )}-\frac {b \sqrt {d x +c -1}\, \sqrt {d x +c +1}\, \arctan \left (\frac {1}{\sqrt {\left (d x +c \right )^{2}-1}}\right )}{e^{2} \sqrt {\left (d x +c \right )^{2}-1}}}{d}\)	\(83\)
default	\(\frac {-\frac {a}{e^{2} \left (d x +c \right )}-\frac {b \,\mathrm {arccosh}\left (d x +c \right )}{e^{2} \left (d x +c \right )}-\frac {b \sqrt {d x +c -1}\, \sqrt {d x +c +1}\, \arctan \left (\frac {1}{\sqrt {\left (d x +c \right )^{2}-1}}\right )}{e^{2} \sqrt {\left (d x +c \right )^{2}-1}}}{d}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 72, normalized size = 1.29 \begin {gather*} -b {\left (\frac {\arcsin \left (\frac {d e^{2}}{{\left | d^{2} x e^{2} + c d e^{2} \right |}}\right ) e^{\left (-2\right )}}{d} + \frac {\operatorname {arcosh}\left (d x + c\right )}{d^{2} x e^{2} + c d e^{2}}\right )} - \frac {a}{d^{2} x e^{2} + c d e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*(arcsin(d*e^2/abs(d^2*x*e^2 + c*d*e^2))*e^(-2)/d + arccosh(d*x + c)/(d^2*x*e^2 + c*d*e^2)) - a/(d^2*x*e^2 +
c*d*e^2)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (50) = 100\).
time = 0.39, size = 168, normalized size = 3.00 \begin {gather*} \frac {b d x \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\right ) - a c + 2 \, {\left (b c d x + b c^{2}\right )} \arctan \left (-d x - c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\right ) + {\left (b d x + b c\right )} \log \left (-d x - c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\right )}{{\left (c d^{2} x + c^{2} d\right )} \cosh \left (1\right )^{2} + 2 \, {\left (c d^{2} x + c^{2} d\right )} \cosh \left (1\right ) \sinh \left (1\right ) + {\left (c d^{2} x + c^{2} d\right )} \sinh \left (1\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*d*x*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 - 1)) - a*c + 2*(b*c*d*x + b*c^2)*arctan(-d*x - c + sqrt(d^2
*x^2 + 2*c*d*x + c^2 - 1)) + (b*d*x + b*c)*log(-d*x - c + sqrt(d^2*x^2 + 2*c*d*x + c^2 - 1)))/((c*d^2*x + c^2*
d)*cosh(1)^2 + 2*(c*d^2*x + c^2*d)*cosh(1)*sinh(1) + (c*d^2*x + c^2*d)*sinh(1)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(t_

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {a+b\,\mathrm {acosh}\left (c+d\,x\right )}{{\left (c\,e+d\,e\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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