3.1 \(\int \frac {x^3 (a+b \cos ^{-1}(c x))}{d-c^2 d x^2} \, dx\)

Optimal. Leaf size=144 \[ \frac {i \left (a+b \cos ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac {\log \left (1-e^{2 i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )}{c^4 d}-\frac {x^2 \left (a+b \cos ^{-1}(c x)\right )}{2 c^2 d}+\frac {i b \text {Li}_2\left (e^{2 i \cos ^{-1}(c x)}\right )}{2 c^4 d}-\frac {b \sin ^{-1}(c x)}{4 c^4 d}+\frac {b x \sqrt {1-c^2 x^2}}{4 c^3 d} \]

[Out]

-1/2*x^2*(a+b*arccos(c*x))/c^2/d+1/2*I*(a+b*arccos(c*x))^2/b/c^4/d-1/4*b*arcsin(c*x)/c^4/d-(a+b*arccos(c*x))*l
n(1-(c*x+I*(-c^2*x^2+1)^(1/2))^2)/c^4/d+1/2*I*b*polylog(2,(c*x+I*(-c^2*x^2+1)^(1/2))^2)/c^4/d+1/4*b*x*(-c^2*x^
2+1)^(1/2)/c^3/d

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Rubi [A]  time = 0.20, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {4716, 4676, 3717, 2190, 2279, 2391, 321, 216} \[ \frac {i b \text {PolyLog}\left (2,e^{2 i \cos ^{-1}(c x)}\right )}{2 c^4 d}-\frac {x^2 \left (a+b \cos ^{-1}(c x)\right )}{2 c^2 d}+\frac {i \left (a+b \cos ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac {\log \left (1-e^{2 i \cos ^{-1}(c x)}\right ) \left (a+b \cos ^{-1}(c x)\right )}{c^4 d}+\frac {b x \sqrt {1-c^2 x^2}}{4 c^3 d}-\frac {b \sin ^{-1}(c x)}{4 c^4 d} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcCos[c*x]))/(d - c^2*d*x^2),x]

[Out]

(b*x*Sqrt[1 - c^2*x^2])/(4*c^3*d) - (x^2*(a + b*ArcCos[c*x]))/(2*c^2*d) + ((I/2)*(a + b*ArcCos[c*x])^2)/(b*c^4
*d) - (b*ArcSin[c*x])/(4*c^4*d) - ((a + b*ArcCos[c*x])*Log[1 - E^((2*I)*ArcCos[c*x])])/(c^4*d) + ((I/2)*b*Poly
Log[2, E^((2*I)*ArcCos[c*x])])/(c^4*d)

Rule 216

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4676

Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/e, Subst[Int[(a
+ b*x)^n*Cot[x], x], x, ArcCos[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0]

Rule 4716

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*(a + b*ArcCos[c*x])^n)/(e*(m + 2*p + 1)), x] + (Dist[(f^2*(m - 1))/(c^2*(m
+ 2*p + 1)), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcCos[c*x])^n, x], x] - Dist[(b*f*n*d^IntPart[p]*(d + e*
x^2)^FracPart[p])/(c*(m + 2*p + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a +
b*ArcCos[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[m,
1] && NeQ[m + 2*p + 1, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {x^3 \left (a+b \cos ^{-1}(c x)\right )}{d-c^2 d x^2} \, dx &=-\frac {x^2 \left (a+b \cos ^{-1}(c x)\right )}{2 c^2 d}+\frac {\int \frac {x \left (a+b \cos ^{-1}(c x)\right )}{d-c^2 d x^2} \, dx}{c^2}-\frac {b \int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx}{2 c d}\\ &=\frac {b x \sqrt {1-c^2 x^2}}{4 c^3 d}-\frac {x^2 \left (a+b \cos ^{-1}(c x)\right )}{2 c^2 d}-\frac {\operatorname {Subst}\left (\int (a+b x) \cot (x) \, dx,x,\cos ^{-1}(c x)\right )}{c^4 d}-\frac {b \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{4 c^3 d}\\ &=\frac {b x \sqrt {1-c^2 x^2}}{4 c^3 d}-\frac {x^2 \left (a+b \cos ^{-1}(c x)\right )}{2 c^2 d}+\frac {i \left (a+b \cos ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac {b \sin ^{-1}(c x)}{4 c^4 d}+\frac {(2 i) \operatorname {Subst}\left (\int \frac {e^{2 i x} (a+b x)}{1-e^{2 i x}} \, dx,x,\cos ^{-1}(c x)\right )}{c^4 d}\\ &=\frac {b x \sqrt {1-c^2 x^2}}{4 c^3 d}-\frac {x^2 \left (a+b \cos ^{-1}(c x)\right )}{2 c^2 d}+\frac {i \left (a+b \cos ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac {b \sin ^{-1}(c x)}{4 c^4 d}-\frac {\left (a+b \cos ^{-1}(c x)\right ) \log \left (1-e^{2 i \cos ^{-1}(c x)}\right )}{c^4 d}+\frac {b \operatorname {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{c^4 d}\\ &=\frac {b x \sqrt {1-c^2 x^2}}{4 c^3 d}-\frac {x^2 \left (a+b \cos ^{-1}(c x)\right )}{2 c^2 d}+\frac {i \left (a+b \cos ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac {b \sin ^{-1}(c x)}{4 c^4 d}-\frac {\left (a+b \cos ^{-1}(c x)\right ) \log \left (1-e^{2 i \cos ^{-1}(c x)}\right )}{c^4 d}-\frac {(i b) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \cos ^{-1}(c x)}\right )}{2 c^4 d}\\ &=\frac {b x \sqrt {1-c^2 x^2}}{4 c^3 d}-\frac {x^2 \left (a+b \cos ^{-1}(c x)\right )}{2 c^2 d}+\frac {i \left (a+b \cos ^{-1}(c x)\right )^2}{2 b c^4 d}-\frac {b \sin ^{-1}(c x)}{4 c^4 d}-\frac {\left (a+b \cos ^{-1}(c x)\right ) \log \left (1-e^{2 i \cos ^{-1}(c x)}\right )}{c^4 d}+\frac {i b \text {Li}_2\left (e^{2 i \cos ^{-1}(c x)}\right )}{2 c^4 d}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 161, normalized size = 1.12 \[ -\frac {2 a c^2 x^2+2 a \log \left (1-c^2 x^2\right )-b c x \sqrt {1-c^2 x^2}+2 b c^2 x^2 \cos ^{-1}(c x)-4 i b \text {Li}_2\left (-e^{i \cos ^{-1}(c x)}\right )-4 i b \text {Li}_2\left (e^{i \cos ^{-1}(c x)}\right )+b \sin ^{-1}(c x)-2 i b \cos ^{-1}(c x)^2+4 b \cos ^{-1}(c x) \log \left (1-e^{i \cos ^{-1}(c x)}\right )+4 b \cos ^{-1}(c x) \log \left (1+e^{i \cos ^{-1}(c x)}\right )}{4 c^4 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^3*(a + b*ArcCos[c*x]))/(d - c^2*d*x^2),x]

[Out]

-1/4*(2*a*c^2*x^2 - b*c*x*Sqrt[1 - c^2*x^2] + 2*b*c^2*x^2*ArcCos[c*x] - (2*I)*b*ArcCos[c*x]^2 + b*ArcSin[c*x]
+ 4*b*ArcCos[c*x]*Log[1 - E^(I*ArcCos[c*x])] + 4*b*ArcCos[c*x]*Log[1 + E^(I*ArcCos[c*x])] + 2*a*Log[1 - c^2*x^
2] - (4*I)*b*PolyLog[2, -E^(I*ArcCos[c*x])] - (4*I)*b*PolyLog[2, E^(I*ArcCos[c*x])])/(c^4*d)

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b x^{3} \arccos \left (c x\right ) + a x^{3}}{c^{2} d x^{2} - d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccos(c*x))/(-c^2*d*x^2+d),x, algorithm="fricas")

[Out]

integral(-(b*x^3*arccos(c*x) + a*x^3)/(c^2*d*x^2 - d), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {{\left (b \arccos \left (c x\right ) + a\right )} x^{3}}{c^{2} d x^{2} - d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccos(c*x))/(-c^2*d*x^2+d),x, algorithm="giac")

[Out]

integrate(-(b*arccos(c*x) + a)*x^3/(c^2*d*x^2 - d), x)

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maple [A]  time = 0.49, size = 228, normalized size = 1.58 \[ -\frac {a \,x^{2}}{2 c^{2} d}-\frac {a \ln \left (c x -1\right )}{2 c^{4} d}-\frac {a \ln \left (c x +1\right )}{2 c^{4} d}+\frac {i b \arccos \left (c x \right )^{2}}{2 c^{4} d}-\frac {b \arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )}{c^{4} d}-\frac {b \arccos \left (c x \right ) \ln \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )}{c^{4} d}+\frac {i b \polylog \left (2, -c x -i \sqrt {-c^{2} x^{2}+1}\right )}{c^{4} d}+\frac {i b \polylog \left (2, c x +i \sqrt {-c^{2} x^{2}+1}\right )}{c^{4} d}-\frac {b \cos \left (2 \arccos \left (c x \right )\right ) \arccos \left (c x \right )}{4 c^{4} d}+\frac {b \sin \left (2 \arccos \left (c x \right )\right )}{8 c^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arccos(c*x))/(-c^2*d*x^2+d),x)

[Out]

-1/2/c^2*a/d*x^2-1/2/c^4*a/d*ln(c*x-1)-1/2/c^4*a/d*ln(c*x+1)+1/2*I/c^4*b/d*arccos(c*x)^2-1/c^4*b/d*arccos(c*x)
*ln(1+c*x+I*(-c^2*x^2+1)^(1/2))-1/c^4*b/d*arccos(c*x)*ln(1-c*x-I*(-c^2*x^2+1)^(1/2))+I/c^4*b/d*polylog(2,-c*x-
I*(-c^2*x^2+1)^(1/2))+I/c^4*b/d*polylog(2,c*x+I*(-c^2*x^2+1)^(1/2))-1/4/c^4*b/d*cos(2*arccos(c*x))*arccos(c*x)
+1/8/c^4*b/d*sin(2*arccos(c*x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, a {\left (\frac {x^{2}}{c^{2} d} + \frac {\log \left (c^{2} x^{2} - 1\right )}{c^{4} d}\right )} + \frac {{\left (c^{4} d \int \frac {c^{2} x^{2} e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (-c x + 1\right )\right )} + e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (-c x + 1\right )\right )} \log \left (c x + 1\right ) + e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (-c x + 1\right )\right )} \log \left (-c x + 1\right )}{c^{7} d x^{4} - c^{5} d x^{2} - {\left (c^{5} d x^{2} - c^{3} d\right )} {\left (c x + 1\right )} {\left (c x - 1\right )}}\,{d x} - {\left (c^{2} x^{2} + \log \left (c x + 1\right ) + \log \left (-c x + 1\right )\right )} \arctan \left (\sqrt {c x + 1} \sqrt {-c x + 1}, c x\right )\right )} b}{2 \, c^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccos(c*x))/(-c^2*d*x^2+d),x, algorithm="maxima")

[Out]

-1/2*a*(x^2/(c^2*d) + log(c^2*x^2 - 1)/(c^4*d)) + 1/2*(2*c^4*d*integrate(1/2*(c^2*x^2*e^(1/2*log(c*x + 1) + 1/
2*log(-c*x + 1)) + e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 1))*log(c*x + 1) + e^(1/2*log(c*x + 1) + 1/2*log(-c*x
+ 1))*log(-c*x + 1))/(c^7*d*x^4 - c^5*d*x^2 + (c^5*d*x^2 - c^3*d)*e^(log(c*x + 1) + log(-c*x + 1))), x) - (c^2
*x^2 + log(c*x + 1) + log(-c*x + 1))*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x))*b/(c^4*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}{d-c^2\,d\,x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*acos(c*x)))/(d - c^2*d*x^2),x)

[Out]

int((x^3*(a + b*acos(c*x)))/(d - c^2*d*x^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {a x^{3}}{c^{2} x^{2} - 1}\, dx + \int \frac {b x^{3} \operatorname {acos}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*acos(c*x))/(-c**2*d*x**2+d),x)

[Out]

-(Integral(a*x**3/(c**2*x**2 - 1), x) + Integral(b*x**3*acos(c*x)/(c**2*x**2 - 1), x))/d

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