3.1.20 \(\int \frac {(d+e x^2) (a+b \text {ArcCos}(c x))}{x} \, dx\) [20]

Optimal. Leaf size=132 \[ -\frac {b e x \sqrt {1-c^2 x^2}}{4 c}+\frac {1}{2} e x^2 (a+b \text {ArcCos}(c x))+\frac {b e \text {ArcSin}(c x)}{4 c^2}+\frac {1}{2} i b d \text {ArcSin}(c x)^2-b d \text {ArcSin}(c x) \log \left (1-e^{2 i \text {ArcSin}(c x)}\right )+d (a+b \text {ArcCos}(c x)) \log (x)+b d \text {ArcSin}(c x) \log (x)+\frac {1}{2} i b d \text {PolyLog}\left (2,e^{2 i \text {ArcSin}(c x)}\right ) \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {14, 4816, 12, 6874, 327, 222, 2363, 4721, 3798, 2221, 2317, 2438} \begin {gather*} d \log (x) (a+b \text {ArcCos}(c x))+\frac {1}{2} e x^2 (a+b \text {ArcCos}(c x))+\frac {b e \text {ArcSin}(c x)}{4 c^2}+\frac {1}{2} i b d \text {Li}_2\left (e^{2 i \text {ArcSin}(c x)}\right )+\frac {1}{2} i b d \text {ArcSin}(c x)^2-b d \text {ArcSin}(c x) \log \left (1-e^{2 i \text {ArcSin}(c x)}\right )+b d \log (x) \text {ArcSin}(c x)-\frac {b e x \sqrt {1-c^2 x^2}}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 222

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 327

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2363

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-e, 2]*(x/Sqr
t[d])]*((a + b*Log[c*x^n])/Rt[-e, 2]), x] - Dist[b*(n/Rt[-e, 2]), Int[ArcSin[Rt[-e, 2]*(x/Sqrt[d])]/x, x], x]
/; FreeQ[{a, b, c, d, e, n}, x] && GtQ[d, 0] && NegQ[e]

Rule 2438

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 3798

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 4721

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n*Cot[x], x], x, ArcSin[c*
x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 4816

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]}, Dist[a + b*ArcCos[c*x], u, x] + Dist[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]))

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]
time = 0.14, size = 124, normalized size = 0.94 \begin {gather*} \frac {1}{2} \left (a e x^2+b e x^2 \text {ArcCos}(c x)+\frac {b e \left (-\frac {1}{2} c x \sqrt {1-c^2 x^2}+\text {ArcTan}\left (\frac {c x}{-1+\sqrt {1-c^2 x^2}}\right )\right )}{c^2}+2 a d \log (x)-i b d \left (\text {ArcCos}(c x) \left (\text {ArcCos}(c x)+2 i \log \left (1+e^{2 i \text {ArcCos}(c x)}\right )\right )+\text {PolyLog}\left (2,-e^{2 i \text {ArcCos}(c x)}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.57, size = 130, normalized size = 0.98

method	result	size
derivativedivides	\(\frac {a e \,x^{2}}{2}+d a \ln \left (c x \right )-\frac {i b \arccos \left (c x \right )^{2} d}{2}-\frac {b e x \sqrt {-c^{2} x^{2}+1}}{4 c}+\frac {b \arccos \left (c x \right ) e \,x^{2}}{2}-\frac {b \arccos \left (c x \right ) e}{4 c^{2}}+b d \arccos \left (c x \right ) \ln \left (1+\left (c x +i \sqrt {-c^{2} x^{2}+1}\right )^{2}\right )-\frac {i b d \polylog \left (2, -\left (c x +i \sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{2}\)	\(130\)
default	\(\frac {a e \,x^{2}}{2}+d a \ln \left (c x \right )-\frac {i b \arccos \left (c x \right )^{2} d}{2}-\frac {b e x \sqrt {-c^{2} x^{2}+1}}{4 c}+\frac {b \arccos \left (c x \right ) e \,x^{2}}{2}-\frac {b \arccos \left (c x \right ) e}{4 c^{2}}+b d \arccos \left (c x \right ) \ln \left (1+\left (c x +i \sqrt {-c^{2} x^{2}+1}\right )^{2}\right )-\frac {i b d \polylog \left (2, -\left (c x +i \sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{2}\)	\(130\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)*(a+b*arccos(c*x))/x,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {acos}{\left (c x \right )}\right ) \left (d + e x^{2}\right )}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*acos(c*x))*(d + e*x**2)/x, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )\,\left (e\,x^2+d\right )}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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