Integrand size = 19, antiderivative size = 85 \[ \int \frac {\left (d+e x^2\right ) (a+b \arccos (c x))}{x^4} \, dx=-\frac {b c d \sqrt {1-c^2 x^2}}{6 x^2}-\frac {d (a+b \arccos (c x))}{3 x^3}-\frac {e (a+b \arccos (c x))}{x}-\frac {1}{6} b c \left (c^2 d+6 e\right ) \text {arctanh}\left (\sqrt {1-c^2 x^2}\right ) \] Output:
-1/6*b*c*d*(-c^2*x^2+1)^(1/2)/x^2-1/3*d*(a+b*arccos(c*x))/x^3-e*(a+b*arcco s(c*x))/x-1/6*b*c*(c^2*d+6*e)*arctanh((-c^2*x^2+1)^(1/2))
Time = 0.04 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.53 \[ \int \frac {\left (d+e x^2\right ) (a+b \arccos (c x))}{x^4} \, dx=-\frac {a d}{3 x^3}-\frac {a e}{x}+\frac {b c d \sqrt {1-c^2 x^2}}{6 x^2}-\frac {b d \arccos (c x)}{3 x^3}-\frac {b e \arccos (c x)}{x}-\frac {1}{6} b c^3 d \log (x)-b c e \log (x)+\frac {1}{6} b c^3 d \log \left (1+\sqrt {1-c^2 x^2}\right )+b c e \log \left (1+\sqrt {1-c^2 x^2}\right ) \] Input:
Integrate[((d + e*x^2)*(a + b*ArcCos[c*x]))/x^4,x]
Output:
-1/3*(a*d)/x^3 - (a*e)/x + (b*c*d*Sqrt[1 - c^2*x^2])/(6*x^2) - (b*d*ArcCos [c*x])/(3*x^3) - (b*e*ArcCos[c*x])/x - (b*c^3*d*Log[x])/6 - b*c*e*Log[x] + (b*c^3*d*Log[1 + Sqrt[1 - c^2*x^2]])/6 + b*c*e*Log[1 + Sqrt[1 - c^2*x^2]]
Time = 0.30 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {5231, 27, 354, 87, 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.
\(\displaystyle \int \frac {\left (d+e x^2\right ) (a+b \arccos (c x))}{x^4} \, dx\) |
\(\Big \downarrow \) 5231 |
\(\displaystyle b c \int -\frac {3 e x^2+d}{3 x^3 \sqrt {1-c^2 x^2}}dx-\frac {d (a+b \arccos (c x))}{3 x^3}-\frac {e (a+b \arccos (c x))}{x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{3} b c \int \frac {3 e x^2+d}{x^3 \sqrt {1-c^2 x^2}}dx-\frac {d (a+b \arccos (c x))}{3 x^3}-\frac {e (a+b \arccos (c x))}{x}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {1}{6} b c \int \frac {3 e x^2+d}{x^4 \sqrt {1-c^2 x^2}}dx^2-\frac {d (a+b \arccos (c x))}{3 x^3}-\frac {e (a+b \arccos (c x))}{x}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle -\frac {1}{6} b c \left (\frac {1}{2} \left (c^2 d+6 e\right ) \int \frac {1}{x^2 \sqrt {1-c^2 x^2}}dx^2-\frac {d \sqrt {1-c^2 x^2}}{x^2}\right )-\frac {d (a+b \arccos (c x))}{3 x^3}-\frac {e (a+b \arccos (c x))}{x}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {1}{6} b c \left (-\frac {\left (c^2 d+6 e\right ) \int \frac {1}{\frac {1}{c^2}-\frac {x^4}{c^2}}d\sqrt {1-c^2 x^2}}{c^2}-\frac {d \sqrt {1-c^2 x^2}}{x^2}\right )-\frac {d (a+b \arccos (c x))}{3 x^3}-\frac {e (a+b \arccos (c x))}{x}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {d (a+b \arccos (c x))}{3 x^3}-\frac {e (a+b \arccos (c x))}{x}-\frac {1}{6} b c \left (-\text {arctanh}\left (\sqrt {1-c^2 x^2}\right ) \left (c^2 d+6 e\right )-\frac {d \sqrt {1-c^2 x^2}}{x^2}\right )\) |
Input:
Int[((d + e*x^2)*(a + b*ArcCos[c*x]))/x^4,x]
Output:
-1/3*(d*(a + b*ArcCos[c*x]))/x^3 - (e*(a + b*ArcCos[c*x]))/x - (b*c*(-((d* Sqrt[1 - c^2*x^2])/x^2) - (c^2*d + 6*e)*ArcTanh[Sqrt[1 - c^2*x^2]]))/6
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_ )^2)^(p_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Simp [(a + b*ArcCos[c*x]) u, x] + Simp[b*c Int[SimplifyIntegrand[u/Sqrt[1 - c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[c^2*d + e, 0] && IntegerQ[p] && (GtQ[p, 0] || (IGtQ[(m - 1)/2, 0] && LeQ[m + p, 0]))
Time = 0.18 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.33
method | result | size |
parts | \(a \left (-\frac {e}{x}-\frac {d}{3 x^{3}}\right )+b \,c^{3} \left (-\frac {\arccos \left (c x \right ) e}{c^{3} x}-\frac {\arccos \left (c x \right ) d}{3 c^{3} x^{3}}+\frac {-d \,c^{2} \left (-\frac {\sqrt {-c^{2} x^{2}+1}}{2 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{2}\right )+3 e \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{3 c^{2}}\right )\) | \(113\) |
derivativedivides | \(c^{3} \left (\frac {a \left (-\frac {e}{c x}-\frac {d}{3 c \,x^{3}}\right )}{c^{2}}+\frac {b \left (-\frac {\arccos \left (c x \right ) e}{c x}-\frac {\arccos \left (c x \right ) d}{3 c \,x^{3}}-\frac {d \,c^{2} \left (-\frac {\sqrt {-c^{2} x^{2}+1}}{2 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{2}\right )}{3}+e \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )\right )}{c^{2}}\right )\) | \(119\) |
default | \(c^{3} \left (\frac {a \left (-\frac {e}{c x}-\frac {d}{3 c \,x^{3}}\right )}{c^{2}}+\frac {b \left (-\frac {\arccos \left (c x \right ) e}{c x}-\frac {\arccos \left (c x \right ) d}{3 c \,x^{3}}-\frac {d \,c^{2} \left (-\frac {\sqrt {-c^{2} x^{2}+1}}{2 c^{2} x^{2}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{2}\right )}{3}+e \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )\right )}{c^{2}}\right )\) | \(119\) |
Input:
int((e*x^2+d)*(a+b*arccos(c*x))/x^4,x,method=_RETURNVERBOSE)
Output:
a*(-e/x-1/3*d/x^3)+b*c^3*(-1/c^3*arccos(c*x)*e/x-1/3*arccos(c*x)*d/c^3/x^3 +1/3/c^2*(-d*c^2*(-1/2/c^2/x^2*(-c^2*x^2+1)^(1/2)-1/2*arctanh(1/(-c^2*x^2+ 1)^(1/2)))+3*e*arctanh(1/(-c^2*x^2+1)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (75) = 150\).
Time = 0.15 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.98 \[ \int \frac {\left (d+e x^2\right ) (a+b \arccos (c x))}{x^4} \, dx=-\frac {4 \, {\left (b d + 3 \, b e\right )} x^{3} \arctan \left (\frac {\sqrt {-c^{2} x^{2} + 1} c x}{c^{2} x^{2} - 1}\right ) - {\left (b c^{3} d + 6 \, b c e\right )} x^{3} \log \left (\sqrt {-c^{2} x^{2} + 1} + 1\right ) + {\left (b c^{3} d + 6 \, b c e\right )} x^{3} \log \left (\sqrt {-c^{2} x^{2} + 1} - 1\right ) - 2 \, \sqrt {-c^{2} x^{2} + 1} b c d x + 12 \, a e x^{2} + 4 \, a d + 4 \, {\left (3 \, b e x^{2} - {\left (b d + 3 \, b e\right )} x^{3} + b d\right )} \arccos \left (c x\right )}{12 \, x^{3}} \] Input:
integrate((e*x^2+d)*(a+b*arccos(c*x))/x^4,x, algorithm="fricas")
Output:
-1/12*(4*(b*d + 3*b*e)*x^3*arctan(sqrt(-c^2*x^2 + 1)*c*x/(c^2*x^2 - 1)) - (b*c^3*d + 6*b*c*e)*x^3*log(sqrt(-c^2*x^2 + 1) + 1) + (b*c^3*d + 6*b*c*e)* x^3*log(sqrt(-c^2*x^2 + 1) - 1) - 2*sqrt(-c^2*x^2 + 1)*b*c*d*x + 12*a*e*x^ 2 + 4*a*d + 4*(3*b*e*x^2 - (b*d + 3*b*e)*x^3 + b*d)*arccos(c*x))/x^3
Time = 2.79 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.00 \[ \int \frac {\left (d+e x^2\right ) (a+b \arccos (c x))}{x^4} \, dx=- \frac {a d}{3 x^{3}} - \frac {a e}{x} - \frac {b c d \left (\begin {cases} - \frac {c^{2} \operatorname {acosh}{\left (\frac {1}{c x} \right )}}{2} + \frac {c}{2 x \sqrt {-1 + \frac {1}{c^{2} x^{2}}}} - \frac {1}{2 c x^{3} \sqrt {-1 + \frac {1}{c^{2} x^{2}}}} & \text {for}\: \frac {1}{\left |{c^{2} x^{2}}\right |} > 1 \\\frac {i c^{2} \operatorname {asin}{\left (\frac {1}{c x} \right )}}{2} - \frac {i c \sqrt {1 - \frac {1}{c^{2} x^{2}}}}{2 x} & \text {otherwise} \end {cases}\right )}{3} - b c e \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{c x} \right )} & \text {for}\: \frac {1}{\left |{c^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{c x} \right )} & \text {otherwise} \end {cases}\right ) - \frac {b d \operatorname {acos}{\left (c x \right )}}{3 x^{3}} - \frac {b e \operatorname {acos}{\left (c x \right )}}{x} \] Input:
integrate((e*x**2+d)*(a+b*acos(c*x))/x**4,x)
Output:
-a*d/(3*x**3) - a*e/x - b*c*d*Piecewise((-c**2*acosh(1/(c*x))/2 + c/(2*x*s qrt(-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*c*e*Piecewise((-acosh(1/(c*x)), 1/Abs(c**2*x**2) > 1), (I*a sin(1/(c*x)), True)) - b*d*acos(c*x)/(3*x**3) - b*e*acos(c*x)/x
Time = 0.12 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.40 \[ \int \frac {\left (d+e x^2\right ) (a+b \arccos (c x))}{x^4} \, dx=\frac {1}{6} \, {\left ({\left (c^{2} \log \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\sqrt {-c^{2} x^{2} + 1}}{x^{2}}\right )} c - \frac {2 \, \arccos \left (c x\right )}{x^{3}}\right )} b d + {\left (c \log \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) - \frac {\arccos \left (c x\right )}{x}\right )} b e - \frac {a e}{x} - \frac {a d}{3 \, x^{3}} \] Input:
integrate((e*x^2+d)*(a+b*arccos(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*arccos(c*x)/x^3)*b*d + (c*log(2*sqrt(-c^2*x^2 + 1)/abs(x) + 2/ abs(x)) - arccos(c*x)/x)*b*e - a*e/x - 1/3*a*d/x^3
Leaf count of result is larger than twice the leaf count of optimal. 3082 vs. \(2 (75) = 150\).
Time = 121.07 (sec) , antiderivative size = 3082, normalized size of antiderivative = 36.26 \[ \int \frac {\left (d+e x^2\right ) (a+b \arccos (c x))}{x^4} \, dx=\text {Too large to display} \] Input:
integrate((e*x^2+d)*(a+b*arccos(c*x))/x^4,x, algorithm="giac")
Output:
-1/3*b*c^3*d*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*d*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*d*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*d/(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*d*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*d*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*d*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*d/((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*d/((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*d* 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)) - b*c*e*arccos(c*x)/(3*...
Timed out. \[ \int \frac {\left (d+e x^2\right ) (a+b \arccos (c x))}{x^4} \, dx=\int \frac {\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )\,\left (e\,x^2+d\right )}{x^4} \,d x \] Input:
int(((a + b*acos(c*x))*(d + e*x^2))/x^4,x)
Output:
int(((a + b*acos(c*x))*(d + e*x^2))/x^4, x)
Time = 0.25 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.01 \[ \int \frac {\left (d+e x^2\right ) (a+b \arccos (c x))}{x^4} \, dx=\frac {-2 \mathit {acos} \left (c x \right ) b d -6 \mathit {acos} \left (c x \right ) b e \,x^{2}+\sqrt {-c^{2} x^{2}+1}\, b c d x -\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (c x \right )}{2}\right )\right ) b \,c^{3} d \,x^{3}-6 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (c x \right )}{2}\right )\right ) b c e \,x^{3}-2 a d -6 a e \,x^{2}}{6 x^{3}} \] Input:
int((e*x^2+d)*(a+b*acos(c*x))/x^4,x)
Output:
( - 2*acos(c*x)*b*d - 6*acos(c*x)*b*e*x**2 + sqrt( - c**2*x**2 + 1)*b*c*d* x - log(tan(asin(c*x)/2))*b*c**3*d*x**3 - 6*log(tan(asin(c*x)/2))*b*c*e*x* *3 - 2*a*d - 6*a*e*x**2)/(6*x**3)