Integrand size = 12, antiderivative size = 62 \[ \int \frac {a+b \arcsin (c x)}{x^4} \, dx=-\frac {b c \sqrt {1-c^2 x^2}}{6 x^2}-\frac {a+b \arcsin (c x)}{3 x^3}-\frac {1}{6} b c^3 \text {arctanh}\left (\sqrt {1-c^2 x^2}\right ) \] Output:
-1/6*b*c*(-c^2*x^2+1)^(1/2)/x^2-1/3*(a+b*arcsin(c*x))/x^3-1/6*b*c^3*arctan h((-c^2*x^2+1)^(1/2))
Time = 0.01 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.08 \[ \int \frac {a+b \arcsin (c x)}{x^4} \, dx=-\frac {a}{3 x^3}-\frac {b c \sqrt {1-c^2 x^2}}{6 x^2}-\frac {b \arcsin (c x)}{3 x^3}-\frac {1}{6} b c^3 \text {arctanh}\left (\sqrt {1-c^2 x^2}\right ) \] Input:
Integrate[(a + b*ArcSin[c*x])/x^4,x]
Output:
-1/3*a/x^3 - (b*c*Sqrt[1 - c^2*x^2])/(6*x^2) - (b*ArcSin[c*x])/(3*x^3) - ( b*c^3*ArcTanh[Sqrt[1 - c^2*x^2]])/6
Time = 0.21 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5138, 243, 52, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {a+b \arcsin (c x)}{x^4} \, dx\) |
\(\Big \downarrow \) 5138 |
\(\displaystyle \frac {1}{3} b c \int \frac {1}{x^3 \sqrt {1-c^2 x^2}}dx-\frac {a+b \arcsin (c x)}{3 x^3}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{6} b c \int \frac {1}{x^4 \sqrt {1-c^2 x^2}}dx^2-\frac {a+b \arcsin (c x)}{3 x^3}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{6} b c \left (\frac {1}{2} c^2 \int \frac {1}{x^2 \sqrt {1-c^2 x^2}}dx^2-\frac {\sqrt {1-c^2 x^2}}{x^2}\right )-\frac {a+b \arcsin (c x)}{3 x^3}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{6} b c \left (-\int \frac {1}{\frac {1}{c^2}-\frac {x^4}{c^2}}d\sqrt {1-c^2 x^2}-\frac {\sqrt {1-c^2 x^2}}{x^2}\right )-\frac {a+b \arcsin (c x)}{3 x^3}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{6} b c \left (c^2 \left (-\text {arctanh}\left (\sqrt {1-c^2 x^2}\right )\right )-\frac {\sqrt {1-c^2 x^2}}{x^2}\right )-\frac {a+b \arcsin (c x)}{3 x^3}\) |
Input:
Int[(a + b*ArcSin[c*x])/x^4,x]
Output:
-1/3*(a + b*ArcSin[c*x])/x^3 + (b*c*(-(Sqrt[1 - c^2*x^2]/x^2) - c^2*ArcTan h[Sqrt[1 - c^2*x^2]]))/6
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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^n/(d*(m + 1))), x] - Simp[b*c*(n /(d*(m + 1))) Int[(d*x)^(m + 1)*((a + b*ArcSin[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]
Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.98
method | result | size |
parts | \(-\frac {a}{3 x^{3}}+b \,c^{3} \left (-\frac {\arcsin \left (c x \right )}{3 c^{3} x^{3}}-\frac {\sqrt {-c^{2} x^{2}+1}}{6 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{6}\right )\) | \(61\) |
derivativedivides | \(c^{3} \left (-\frac {a}{3 c^{3} x^{3}}+b \left (-\frac {\arcsin \left (c x \right )}{3 c^{3} x^{3}}-\frac {\sqrt {-c^{2} x^{2}+1}}{6 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{6}\right )\right )\) | \(65\) |
default | \(c^{3} \left (-\frac {a}{3 c^{3} x^{3}}+b \left (-\frac {\arcsin \left (c x \right )}{3 c^{3} x^{3}}-\frac {\sqrt {-c^{2} x^{2}+1}}{6 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{6}\right )\right )\) | \(65\) |
Input:
int((a+b*arcsin(c*x))/x^4,x,method=_RETURNVERBOSE)
Output:
-1/3*a/x^3+b*c^3*(-1/3/c^3/x^3*arcsin(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)))
Time = 0.11 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.29 \[ \int \frac {a+b \arcsin (c x)}{x^4} \, dx=-\frac {b c^{3} x^{3} \log \left (\sqrt {-c^{2} x^{2} + 1} + 1\right ) - b c^{3} x^{3} \log \left (\sqrt {-c^{2} x^{2} + 1} - 1\right ) + 2 \, \sqrt {-c^{2} x^{2} + 1} b c x + 4 \, b \arcsin \left (c x\right ) + 4 \, a}{12 \, x^{3}} \] Input:
integrate((a+b*arcsin(c*x))/x^4,x, algorithm="fricas")
Output:
-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) + 2*sqrt(-c^2*x^2 + 1)*b*c*x + 4*b*arcsin(c*x) + 4*a)/x^3
Time = 1.61 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.89 \[ \int \frac {a+b \arcsin (c x)}{x^4} \, dx=- \frac {a}{3 x^{3}} + \frac {b c \left (\begin {cases} - \frac {c^{2} \operatorname {acosh}{\left (\frac {1}{c x} \right )}}{2} + \frac {c}{2 x \sqrt {-1 + \frac {1}{c^{2} x^{2}}}} - \frac {1}{2 c x^{3} \sqrt {-1 + \frac {1}{c^{2} x^{2}}}} & \text {for}\: \frac {1}{\left |{c^{2} x^{2}}\right |} > 1 \\\frac {i c^{2} \operatorname {asin}{\left (\frac {1}{c x} \right )}}{2} - \frac {i c \sqrt {1 - \frac {1}{c^{2} x^{2}}}}{2 x} & \text {otherwise} \end {cases}\right )}{3} - \frac {b \operatorname {asin}{\left (c x \right )}}{3 x^{3}} \] Input:
integrate((a+b*asin(c*x))/x**4,x)
Output:
-a/(3*x**3) + b*c*Piecewise((-c**2*acosh(1/(c*x))/2 + c/(2*x*sqrt(-1 + 1/( c**2*x**2))) - 1/(2*c*x**3*sqrt(-1 + 1/(c**2*x**2))), 1/Abs(c**2*x**2) > 1 ), (I*c**2*asin(1/(c*x))/2 - I*c*sqrt(1 - 1/(c**2*x**2))/(2*x), True))/3 - b*asin(c*x)/(3*x**3)
Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.11 \[ \int \frac {a+b \arcsin (c x)}{x^4} \, dx=-\frac {1}{6} \, {\left ({\left (c^{2} \log \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\sqrt {-c^{2} x^{2} + 1}}{x^{2}}\right )} c + \frac {2 \, \arcsin \left (c x\right )}{x^{3}}\right )} b - \frac {a}{3 \, x^{3}} \] Input:
integrate((a+b*arcsin(c*x))/x^4,x, algorithm="maxima")
Output:
-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*arcsin(c*x)/x^3)*b - 1/3*a/x^3
Leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (52) = 104\).
Time = 0.30 (sec) , antiderivative size = 284, normalized size of antiderivative = 4.58 \[ \int \frac {a+b \arcsin (c x)}{x^4} \, dx=-\frac {b c^{6} x^{3} \arcsin \left (c x\right )}{24 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{3}} - \frac {a c^{6} x^{3}}{24 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{3}} + \frac {b c^{5} x^{2}}{24 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{2}} - \frac {b c^{4} x \arcsin \left (c x\right )}{8 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}} - \frac {a c^{4} x}{8 \, {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}} + \frac {1}{6} \, b c^{3} \log \left ({\left | c \right |} {\left | x \right |}\right ) - \frac {1}{6} \, b c^{3} \log \left (\sqrt {-c^{2} x^{2} + 1} + 1\right ) - \frac {b c^{2} {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )} \arcsin \left (c x\right )}{8 \, x} - \frac {a c^{2} {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}}{8 \, x} - \frac {b c {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{2}}{24 \, x^{2}} - \frac {b {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{3} \arcsin \left (c x\right )}{24 \, x^{3}} - \frac {a {\left (\sqrt {-c^{2} x^{2} + 1} + 1\right )}^{3}}{24 \, x^{3}} \] Input:
integrate((a+b*arcsin(c*x))/x^4,x, algorithm="giac")
Output:
-1/24*b*c^6*x^3*arcsin(c*x)/(sqrt(-c^2*x^2 + 1) + 1)^3 - 1/24*a*c^6*x^3/(s qrt(-c^2*x^2 + 1) + 1)^3 + 1/24*b*c^5*x^2/(sqrt(-c^2*x^2 + 1) + 1)^2 - 1/8 *b*c^4*x*arcsin(c*x)/(sqrt(-c^2*x^2 + 1) + 1) - 1/8*a*c^4*x/(sqrt(-c^2*x^2 + 1) + 1) + 1/6*b*c^3*log(abs(c)*abs(x)) - 1/6*b*c^3*log(sqrt(-c^2*x^2 + 1) + 1) - 1/8*b*c^2*(sqrt(-c^2*x^2 + 1) + 1)*arcsin(c*x)/x - 1/8*a*c^2*(sq rt(-c^2*x^2 + 1) + 1)/x - 1/24*b*c*(sqrt(-c^2*x^2 + 1) + 1)^2/x^2 - 1/24*b *(sqrt(-c^2*x^2 + 1) + 1)^3*arcsin(c*x)/x^3 - 1/24*a*(sqrt(-c^2*x^2 + 1) + 1)^3/x^3
Timed out. \[ \int \frac {a+b \arcsin (c x)}{x^4} \, dx=\int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{x^4} \,d x \] Input:
int((a + b*asin(c*x))/x^4,x)
Output:
int((a + b*asin(c*x))/x^4, x)
Time = 0.19 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.77 \[ \int \frac {a+b \arcsin (c x)}{x^4} \, dx=\frac {-2 \mathit {asin} \left (c x \right ) b -\sqrt {-c^{2} x^{2}+1}\, b c x +\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (c x \right )}{2}\right )\right ) b \,c^{3} x^{3}-2 a}{6 x^{3}} \] Input:
int((a+b*asin(c*x))/x^4,x)
Output:
( - 2*asin(c*x)*b - sqrt( - c**2*x**2 + 1)*b*c*x + log(tan(asin(c*x)/2))*b *c**3*x**3 - 2*a)/(6*x**3)