3.3.94 \(\int \cosh ^{-1}(\frac {c}{a+b x}) \, dx\) [294]

Optimal. Leaf size=58 \[ \frac {(a+b x) \text {sech}^{-1}\left (\frac {a}{c}+\frac {b x}{c}\right )}{b}-\frac {2 c \text {ArcTan}\left (\sqrt {\frac {\left (1-\frac {a}{c}\right ) c-b x}{a+c+b x}}\right )}{b} \]

[Out]

(b*x+a)*arcsech(a/c+b*x/c)/b-2*c*arctan((((1-a/c)*c-b*x)/(b*x+a+c))^(1/2))/b

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Rubi [A]
time = 0.07, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6012, 6448, 1983, 12, 209} \begin {gather*} \frac {(a+b x) \text {sech}^{-1}\left (\frac {a}{c}+\frac {b x}{c}\right )}{b}-\frac {2 c \text {ArcTan}\left (\sqrt {\frac {c \left (1-\frac {a}{c}\right )-b x}{a+b x+c}}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcCosh[c/(a + b*x)],x]

[Out]

((a + b*x)*ArcSech[a/c + (b*x)/c])/b - (2*c*ArcTan[Sqrt[((1 - a/c)*c - b*x)/(a + c + b*x)]])/b

Rule 12

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

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 1983

Int[(u_)^(r_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> With[{q = Den
ominator[p]}, Dist[q*e*((b*c - a*d)/n), Subst[Int[SimplifyIntegrand[x^(q*(p + 1) - 1)*(((-a)*e + c*x^q)^(1/n -
1)/(b*e - d*x^q)^(1/n + 1))*(u /. x -> ((-a)*e + c*x^q)^(1/n)/(b*e - d*x^q)^(1/n))^r, x], x], x, (e*((a + b*x
^n)/(c + d*x^n)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e}, x] && PolynomialQ[u, x] && FractionQ[p] && IntegerQ[1/
n] && IntegerQ[r]

Rule 6012

Int[ArcCosh[(c_.)/((a_.) + (b_.)*(x_)^(n_.))]^(m_.)*(u_.), x_Symbol] :> Int[u*ArcSech[a/c + b*(x^n/c)]^m, x] /
; FreeQ[{a, b, c, n, m}, x]

Rule 6448

Int[ArcSech[(c_) + (d_.)*(x_)], x_Symbol] :> Simp[(c + d*x)*(ArcSech[c + d*x]/d), x] + Int[Sqrt[(1 - c - d*x)/
(1 + c + d*x)]/(1 - c - d*x), x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [A]
time = 0.24, size = 116, normalized size = 2.00 \begin {gather*} x \cosh ^{-1}\left (\frac {c}{a+b x}\right )+\frac {2 \sqrt {-\frac {a-c+b x}{a+c+b x}} \sqrt {a+c+b x} \left (a \text {ArcTan}\left (\frac {\sqrt {a-c+b x}}{\sqrt {a+c+b x}}\right )-c \tanh ^{-1}\left (\frac {\sqrt {a-c+b x}}{\sqrt {a+c+b x}}\right )\right )}{b \sqrt {a-c+b x}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCosh[c/(a + b*x)],x]

[Out]

x*ArcCosh[c/(a + b*x)] + (2*Sqrt[-((a - c + b*x)/(a + c + b*x))]*Sqrt[a + c + b*x]*(a*ArcTan[Sqrt[a - c + b*x]
/Sqrt[a + c + b*x]] - c*ArcTanh[Sqrt[a - c + b*x]/Sqrt[a + c + b*x]]))/(b*Sqrt[a - c + b*x])

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Maple [A]
time = 0.16, size = 87, normalized size = 1.50

method	result	size
derivativedivides	\(-\frac {c \left (-\frac {\left (b x +a \right ) \mathrm {arccosh}\left (\frac {c}{b x +a}\right )}{c}-\frac {\sqrt {\frac {c}{b x +a}-1}\, \sqrt {\frac {c}{b x +a}+1}\, \arctan \left (\frac {1}{\sqrt {\frac {c^{2}}{\left (b x +a \right )^{2}}-1}}\right )}{\sqrt {\frac {c^{2}}{\left (b x +a \right )^{2}}-1}}\right )}{b}\)	\(87\)
default	\(-\frac {c \left (-\frac {\left (b x +a \right ) \mathrm {arccosh}\left (\frac {c}{b x +a}\right )}{c}-\frac {\sqrt {\frac {c}{b x +a}-1}\, \sqrt {\frac {c}{b x +a}+1}\, \arctan \left (\frac {1}{\sqrt {\frac {c^{2}}{\left (b x +a \right )^{2}}-1}}\right )}{\sqrt {\frac {c^{2}}{\left (b x +a \right )^{2}}-1}}\right )}{b}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccosh(c/(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 276 vs. \(2 (55) = 110\).
time = 0.43, size = 276, normalized size = 4.76 \begin {gather*} \frac {2 \, b x \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - c^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + c}{b x + a}\right ) - 2 \, c \arctan \left (\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - c^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{b^{2} x^{2} + 2 \, a b x + a^{2} - c^{2}}\right ) + a \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - c^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + c}{x}\right ) - a \log \left (\frac {{\left (b x + a\right )} \sqrt {-\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - c^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - c}{x}\right )}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acosh(c/(a + b*x)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (55) = 110\).
time = 2.29, size = 119, normalized size = 2.05 \begin {gather*} \frac {c \arcsin \left (-\frac {b x + a}{c}\right ) \mathrm {sgn}\left (b\right ) \mathrm {sgn}\left (c\right )}{{\left | b \right |}} + x \log \left (\sqrt {\frac {c}{b x + a} + 1} \sqrt {\frac {c}{b x + a} - 1} + \frac {c}{b x + a}\right ) - \frac {a \log \left (\frac {{\left | -2 \, b c - 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + c^{2}} {\left | b \right |} \right |}}{{\left | -2 \, b^{2} x - 2 \, a b \right |}}\right )}{{\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*arcsin(-(b*x + a)/c)*sgn(b)*sgn(c)/abs(b) + x*log(sqrt(c/(b*x + a) + 1)*sqrt(c/(b*x + a) - 1) + c/(b*x + a))
- a*log(abs(-2*b*c - 2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + c^2)*abs(b))/abs(-2*b^2*x - 2*a*b))/abs(b)

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Mupad [B]
time = 1.51, size = 53, normalized size = 0.91 \begin {gather*} \frac {\mathrm {acosh}\left (\frac {c}{a+b\,x}\right )\,\left (a+b\,x\right )}{b}+\frac {c\,\mathrm {atan}\left (\frac {1}{\sqrt {\frac {c}{a+b\,x}-1}\,\sqrt {\frac {c}{a+b\,x}+1}}\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acosh(c/(a + b*x)),x)

[Out]

(acosh(c/(a + b*x))*(a + b*x))/b + (c*atan(1/((c/(a + b*x) - 1)^(1/2)*(c/(a + b*x) + 1)^(1/2))))/b

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