3.114 \(\int \cos ^{-1}(\frac {c}{a+b x}) \, dx\)

Optimal. Leaf size=48 \[ \frac {(a+b x) \sec ^{-1}\left (\frac {a}{c}+\frac {b x}{c}\right )}{b}-\frac {c \tanh ^{-1}\left (\sqrt {1-\frac {c^2}{(a+b x)^2}}\right )}{b} \]

[Out]

(b*x+a)*arcsec(a/c+b*x/c)/b-c*arctanh((1-c^2/(b*x+a)^2)^(1/2))/b

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Rubi [A]  time = 0.03, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4833, 5250, 372, 266, 63, 206} \[ \frac {(a+b x) \sec ^{-1}\left (\frac {a}{c}+\frac {b x}{c}\right )}{b}-\frac {c \tanh ^{-1}\left (\sqrt {1-\frac {c^2}{(a+b x)^2}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[c/(a + b*x)],x]

[Out]

((a + b*x)*ArcSec[a/c + (b*x)/c])/b - (c*ArcTanh[Sqrt[1 - c^2/(a + b*x)^2]])/b

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 372

Int[(u_)^(m_.)*((a_) + (b_.)*(v_)^(n_))^(p_.), x_Symbol] :> Dist[u^m/(Coefficient[v, x, 1]*v^m), Subst[Int[x^m
*(a + b*x^n)^p, x], x, v], x] /; FreeQ[{a, b, m, n, p}, x] && LinearPairQ[u, v, x]

Rule 4833

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

Rule 5250

Int[ArcSec[(c_) + (d_.)*(x_)], x_Symbol] :> Simp[((c + d*x)*ArcSec[c + d*x])/d, x] - Int[1/((c + d*x)*Sqrt[1 -
1/(c + d*x)^2]), x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [B]  time = 0.17, size = 141, normalized size = 2.94 \[ x \cos ^{-1}\left (\frac {c}{a+b x}\right )-\frac {(a+b x) \sqrt {\frac {a^2+2 a b x+b^2 x^2-c^2}{(a+b x)^2}} \left (c \tanh ^{-1}\left (\frac {a+b x}{\sqrt {a^2+2 a b x+b^2 x^2-c^2}}\right )-a \tan ^{-1}\left (\frac {\sqrt {(a+b x)^2-c^2}}{c}\right )\right )}{b \sqrt {a^2+2 a b x+b^2 x^2-c^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[c/(a + b*x)],x]

[Out]

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

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fricas [B]  time = 0.48, size = 140, normalized size = 2.92 \[ \frac {b x \arccos \left (\frac {c}{b x + a}\right ) + 2 \, a \arctan \left (-\frac {b x - {\left (b x + a\right )} \sqrt {\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - c^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + a}{c}\right ) + c \log \left (-b x + {\left (b x + a\right )} \sqrt {\frac {b^{2} x^{2} + 2 \, a b x + a^{2} - c^{2}}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - a\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x*arccos(c/(b*x + a)) + 2*a*arctan(-(b*x - (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)) + a)/c) + c*log(-b*x + (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)) - a
))/b

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giac [B]  time = 0.22, size = 95, normalized size = 1.98 \[ -\frac {b {\left (\frac {c^{2} {\left (\log \left (\sqrt {-\frac {c^{2}}{{\left (b x + a\right )}^{2}} + 1} + 1\right ) - \log \left (-\sqrt {-\frac {c^{2}}{{\left (b x + a\right )}^{2}} + 1} + 1\right )\right )}}{b^{2}} - \frac {2 \, {\left (b x + a\right )} c \arccos \left (-\frac {c}{{\left (b x + a\right )} {\left (\frac {a}{b x + a} - 1\right )} - a}\right )}{b^{2}}\right )}}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 45, normalized size = 0.94 \[ -\frac {c \left (-\frac {\left (b x +a \right ) \arccos \left (\frac {c}{b x +a}\right )}{c}+\arctanh \left (\frac {1}{\sqrt {1-\frac {c^{2}}{\left (b x +a \right )^{2}}}}\right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(c/(b*x+a)),x)

[Out]

-1/b*c*(-1/c*(b*x+a)*arccos(c/(b*x+a))+arctanh(1/(1-c^2/(b*x+a)^2)^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ x \arctan \left (\frac {\sqrt {b x + a + c} \sqrt {b x + a - c}}{c}\right ) - \int \frac {{\left (b^{2} c x^{2} + a b c x\right )} e^{\left (\frac {1}{2} \, \log \left (b x + a + c\right ) + \frac {1}{2} \, \log \left (b x + a - c\right )\right )}}{b^{2} c^{2} x^{2} + 2 \, a b c^{2} x + a^{2} c^{2} - c^{4} + {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - c^{2}\right )} e^{\left (\log \left (b x + a + c\right ) + \log \left (b x + a - c\right )\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arctan(sqrt(b*x + a + c)*sqrt(b*x + a - c)/c) - integrate((b^2*c*x^2 + a*b*c*x)*e^(1/2*log(b*x + a + c) + 1/
2*log(b*x + a - c))/(b^2*c^2*x^2 + 2*a*b*c^2*x + a^2*c^2 - c^4 + (b^2*x^2 + 2*a*b*x + a^2 - c^2)*e^(log(b*x +
a + c) + log(b*x + a - c))), x)

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mupad [B]  time = 0.63, size = 43, normalized size = 0.90 \[ \frac {\mathrm {acos}\left (\frac {c}{a+b\,x}\right )\,\left (a+b\,x\right )}{b}-\frac {c\,\mathrm {atanh}\left (\frac {1}{\sqrt {1-\frac {c^2}{{\left (a+b\,x\right )}^2}}}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(c/(a + b*x)),x)

[Out]

(acos(c/(a + b*x))*(a + b*x))/b - (c*atanh(1/(1 - c^2/(a + b*x)^2)^(1/2)))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acos(c/(a + b*x)), x)

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