3.147 \(\int \frac{a+b \cos ^{-1}(c x)}{x^4} \, dx\)
Optimal. Leaf size=62 \[ -\frac{a+b \cos ^{-1}(c x)}{3 x^3}+\frac{b c \sqrt{1-c^2 x^2}}{6 x^2}+\frac{1}{6} b c^3 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right ) \]
[Out]
(b*c*Sqrt[1 - c^2*x^2])/(6*x^2) - (a + b*ArcCos[c*x])/(3*x^3) + (b*c^3*ArcTanh[Sqrt[1 - c^2*x^2]])/6
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Rubi [A] time = 0.0367287, antiderivative size = 62, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used =
{4628, 266, 51, 63, 208} \[ -\frac{a+b \cos ^{-1}(c x)}{3 x^3}+\frac{b c \sqrt{1-c^2 x^2}}{6 x^2}+\frac{1}{6} b c^3 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right ) \]
Antiderivative was successfully verified.
[In]
Int[(a + b*ArcCos[c*x])/x^4,x]
[Out]
(b*c*Sqrt[1 - c^2*x^2])/(6*x^2) - (a + b*ArcCos[c*x])/(3*x^3) + (b*c^3*ArcTanh[Sqrt[1 - c^2*x^2]])/6
Rule 4628
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcCo
s[c*x])^n)/(d*(m + 1)), x] + Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCos[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]
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 51
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]
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 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \frac{a+b \cos ^{-1}(c x)}{x^4} \, dx &=-\frac{a+b \cos ^{-1}(c x)}{3 x^3}-\frac{1}{3} (b c) \int \frac{1}{x^3 \sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{a+b \cos ^{-1}(c x)}{3 x^3}-\frac{1}{6} (b c) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1-c^2 x}} \, dx,x,x^2\right )\\ &=\frac{b c \sqrt{1-c^2 x^2}}{6 x^2}-\frac{a+b \cos ^{-1}(c x)}{3 x^3}-\frac{1}{12} \left (b c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-c^2 x}} \, dx,x,x^2\right )\\ &=\frac{b c \sqrt{1-c^2 x^2}}{6 x^2}-\frac{a+b \cos ^{-1}(c x)}{3 x^3}+\frac{1}{6} (b c) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{c^2}-\frac{x^2}{c^2}} \, dx,x,\sqrt{1-c^2 x^2}\right )\\ &=\frac{b c \sqrt{1-c^2 x^2}}{6 x^2}-\frac{a+b \cos ^{-1}(c x)}{3 x^3}+\frac{1}{6} b c^3 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )\\ \end{align*}
Mathematica [A] time = 0.0247396, size = 79, normalized size = 1.27 \[ -\frac{a}{3 x^3}+\frac{b c \sqrt{1-c^2 x^2}}{6 x^2}+\frac{1}{6} b c^3 \log \left (\sqrt{1-c^2 x^2}+1\right )-\frac{1}{6} b c^3 \log (x)-\frac{b \cos ^{-1}(c x)}{3 x^3} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*ArcCos[c*x])/x^4,x]
[Out]
-a/(3*x^3) + (b*c*Sqrt[1 - c^2*x^2])/(6*x^2) - (b*ArcCos[c*x])/(3*x^3) - (b*c^3*Log[x])/6 + (b*c^3*Log[1 + Sqr
t[1 - c^2*x^2]])/6
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Maple [A] time = 0.004, size = 65, normalized size = 1.1 \begin{align*}{c}^{3} \left ( -{\frac{a}{3\,{c}^{3}{x}^{3}}}+b \left ( -{\frac{\arccos \left ( cx \right ) }{3\,{c}^{3}{x}^{3}}}+{\frac{1}{6\,{c}^{2}{x}^{2}}\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{1}{6}{\it Artanh} \left ({\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}} \right ) } \right ) \right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*arccos(c*x))/x^4,x)
[Out]
c^3*(-1/3*a/c^3/x^3+b*(-1/3/c^3/x^3*arccos(c*x)+1/6/c^2/x^2*(-c^2*x^2+1)^(1/2)+1/6*arctanh(1/(-c^2*x^2+1)^(1/2
))))
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Maxima [A] time = 1.43471, size = 93, normalized size = 1.5 \begin{align*} \frac{1}{6} \,{\left ({\left (c^{2} \log \left (\frac{2 \, \sqrt{-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) + \frac{\sqrt{-c^{2} x^{2} + 1}}{x^{2}}\right )} c - \frac{2 \, \arccos \left (c x\right )}{x^{3}}\right )} b - \frac{a}{3 \, x^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arccos(c*x))/x^4,x, algorithm="maxima")
[Out]
1/6*((c^2*log(2*sqrt(-c^2*x^2 + 1)/abs(x) + 2/abs(x)) + sqrt(-c^2*x^2 + 1)/x^2)*c - 2*arccos(c*x)/x^3)*b - 1/3
*a/x^3
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Fricas [B] time = 3.18666, size = 281, normalized size = 4.53 \begin{align*} \frac{b c^{3} x^{3} \log \left (\sqrt{-c^{2} x^{2} + 1} + 1\right ) - b c^{3} x^{3} \log \left (\sqrt{-c^{2} x^{2} + 1} - 1\right ) - 4 \, b x^{3} \arctan \left (\frac{\sqrt{-c^{2} x^{2} + 1} c x}{c^{2} x^{2} - 1}\right ) + 2 \, \sqrt{-c^{2} x^{2} + 1} b c x + 4 \,{\left (b x^{3} - b\right )} \arccos \left (c x\right ) - 4 \, a}{12 \, x^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arccos(c*x))/x^4,x, algorithm="fricas")
[Out]
1/12*(b*c^3*x^3*log(sqrt(-c^2*x^2 + 1) + 1) - b*c^3*x^3*log(sqrt(-c^2*x^2 + 1) - 1) - 4*b*x^3*arctan(sqrt(-c^2
*x^2 + 1)*c*x/(c^2*x^2 - 1)) + 2*sqrt(-c^2*x^2 + 1)*b*c*x + 4*(b*x^3 - b)*arccos(c*x) - 4*a)/x^3
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Sympy [A] time = 3.2627, size = 121, normalized size = 1.95 \begin{align*} - \frac{a}{3 x^{3}} - \frac{b c \left (\begin{cases} - \frac{c^{2} \operatorname{acosh}{\left (\frac{1}{c x} \right )}}{2} - \frac{c \sqrt{-1 + \frac{1}{c^{2} x^{2}}}}{2 x} & \text{for}\: \frac{1}{\left |{c^{2} x^{2}}\right |} > 1 \\\frac{i c^{2} \operatorname{asin}{\left (\frac{1}{c x} \right )}}{2} - \frac{i c}{2 x \sqrt{1 - \frac{1}{c^{2} x^{2}}}} + \frac{i}{2 c x^{3} \sqrt{1 - \frac{1}{c^{2} x^{2}}}} & \text{otherwise} \end{cases}\right )}{3} - \frac{b \operatorname{acos}{\left (c x \right )}}{3 x^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*acos(c*x))/x**4,x)
[Out]
-a/(3*x**3) - b*c*Piecewise((-c**2*acosh(1/(c*x))/2 - c*sqrt(-1 + 1/(c**2*x**2))/(2*x), 1/Abs(c**2*x**2) > 1),
(I*c**2*asin(1/(c*x))/2 - I*c/(2*x*sqrt(1 - 1/(c**2*x**2))) + I/(2*c*x**3*sqrt(1 - 1/(c**2*x**2))), True))/3
- b*acos(c*x)/(3*x**3)
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Giac [B] time = 2.42751, size = 2206, normalized size = 35.58 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arccos(c*x))/x^4,x, algorithm="giac")
[Out]
-1/3*b*c^3*arccos(c*x)/(3*(c^2*x^2 - 1)/(c*x + 1)^2 + 3*(c^2*x^2 - 1)^2/(c*x + 1)^4 + (c^2*x^2 - 1)^3/(c*x + 1
)^6 + 1) + 1/6*b*c^3*log(abs(c*x + sqrt(-c^2*x^2 + 1) + 1))/(3*(c^2*x^2 - 1)/(c*x + 1)^2 + 3*(c^2*x^2 - 1)^2/(
c*x + 1)^4 + (c^2*x^2 - 1)^3/(c*x + 1)^6 + 1) - 1/6*b*c^3*log(abs(-c*x + sqrt(-c^2*x^2 + 1) - 1))/(3*(c^2*x^2
- 1)/(c*x + 1)^2 + 3*(c^2*x^2 - 1)^2/(c*x + 1)^4 + (c^2*x^2 - 1)^3/(c*x + 1)^6 + 1) - 1/3*a*c^3/(3*(c^2*x^2 -
1)/(c*x + 1)^2 + 3*(c^2*x^2 - 1)^2/(c*x + 1)^4 + (c^2*x^2 - 1)^3/(c*x + 1)^6 + 1) + (c^2*x^2 - 1)*b*c^3*arccos
(c*x)/((c*x + 1)^2*(3*(c^2*x^2 - 1)/(c*x + 1)^2 + 3*(c^2*x^2 - 1)^2/(c*x + 1)^4 + (c^2*x^2 - 1)^3/(c*x + 1)^6
+ 1)) + 1/2*(c^2*x^2 - 1)*b*c^3*log(abs(c*x + sqrt(-c^2*x^2 + 1) + 1))/((c*x + 1)^2*(3*(c^2*x^2 - 1)/(c*x + 1)
^2 + 3*(c^2*x^2 - 1)^2/(c*x + 1)^4 + (c^2*x^2 - 1)^3/(c*x + 1)^6 + 1)) - 1/2*(c^2*x^2 - 1)*b*c^3*log(abs(-c*x
+ sqrt(-c^2*x^2 + 1) - 1))/((c*x + 1)^2*(3*(c^2*x^2 - 1)/(c*x + 1)^2 + 3*(c^2*x^2 - 1)^2/(c*x + 1)^4 + (c^2*x^
2 - 1)^3/(c*x + 1)^6 + 1)) + 1/3*sqrt(-c^2*x^2 + 1)*b*c^3/((c*x + 1)*(3*(c^2*x^2 - 1)/(c*x + 1)^2 + 3*(c^2*x^2
- 1)^2/(c*x + 1)^4 + (c^2*x^2 - 1)^3/(c*x + 1)^6 + 1)) + (c^2*x^2 - 1)*a*c^3/((c*x + 1)^2*(3*(c^2*x^2 - 1)/(c
*x + 1)^2 + 3*(c^2*x^2 - 1)^2/(c*x + 1)^4 + (c^2*x^2 - 1)^3/(c*x + 1)^6 + 1)) - (c^2*x^2 - 1)^2*b*c^3*arccos(c
*x)/((c*x + 1)^4*(3*(c^2*x^2 - 1)/(c*x + 1)^2 + 3*(c^2*x^2 - 1)^2/(c*x + 1)^4 + (c^2*x^2 - 1)^3/(c*x + 1)^6 +
1)) + 1/2*(c^2*x^2 - 1)^2*b*c^3*log(abs(c*x + sqrt(-c^2*x^2 + 1) + 1))/((c*x + 1)^4*(3*(c^2*x^2 - 1)/(c*x + 1)
^2 + 3*(c^2*x^2 - 1)^2/(c*x + 1)^4 + (c^2*x^2 - 1)^3/(c*x + 1)^6 + 1)) - 1/2*(c^2*x^2 - 1)^2*b*c^3*log(abs(-c*
x + sqrt(-c^2*x^2 + 1) - 1))/((c*x + 1)^4*(3*(c^2*x^2 - 1)/(c*x + 1)^2 + 3*(c^2*x^2 - 1)^2/(c*x + 1)^4 + (c^2*
x^2 - 1)^3/(c*x + 1)^6 + 1)) - (c^2*x^2 - 1)^2*a*c^3/((c*x + 1)^4*(3*(c^2*x^2 - 1)/(c*x + 1)^2 + 3*(c^2*x^2 -
1)^2/(c*x + 1)^4 + (c^2*x^2 - 1)^3/(c*x + 1)^6 + 1)) + 1/3*(c^2*x^2 - 1)^3*b*c^3*arccos(c*x)/((c*x + 1)^6*(3*(
c^2*x^2 - 1)/(c*x + 1)^2 + 3*(c^2*x^2 - 1)^2/(c*x + 1)^4 + (c^2*x^2 - 1)^3/(c*x + 1)^6 + 1)) + 1/6*(c^2*x^2 -
1)^3*b*c^3*log(abs(c*x + sqrt(-c^2*x^2 + 1) + 1))/((c*x + 1)^6*(3*(c^2*x^2 - 1)/(c*x + 1)^2 + 3*(c^2*x^2 - 1)^
2/(c*x + 1)^4 + (c^2*x^2 - 1)^3/(c*x + 1)^6 + 1)) - 1/6*(c^2*x^2 - 1)^3*b*c^3*log(abs(-c*x + sqrt(-c^2*x^2 + 1
) - 1))/((c*x + 1)^6*(3*(c^2*x^2 - 1)/(c*x + 1)^2 + 3*(c^2*x^2 - 1)^2/(c*x + 1)^4 + (c^2*x^2 - 1)^3/(c*x + 1)^
6 + 1)) - 1/3*(c^2*x^2 - 1)^2*sqrt(-c^2*x^2 + 1)*b*c^3/((c*x + 1)^5*(3*(c^2*x^2 - 1)/(c*x + 1)^2 + 3*(c^2*x^2
- 1)^2/(c*x + 1)^4 + (c^2*x^2 - 1)^3/(c*x + 1)^6 + 1)) + 1/3*(c^2*x^2 - 1)^3*a*c^3/((c*x + 1)^6*(3*(c^2*x^2 -
1)/(c*x + 1)^2 + 3*(c^2*x^2 - 1)^2/(c*x + 1)^4 + (c^2*x^2 - 1)^3/(c*x + 1)^6 + 1))