3.2.86 \(\int \frac {x (a+b \text {ArcSin}(c x))^2}{d-c^2 d x^2} \, dx\) [186]

Optimal. Leaf size=117 \[ \frac {i (a+b \text {ArcSin}(c x))^3}{3 b c^2 d}-\frac {(a+b \text {ArcSin}(c x))^2 \log \left (1+e^{2 i \text {ArcSin}(c x)}\right )}{c^2 d}+\frac {i b (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,-e^{2 i \text {ArcSin}(c x)}\right )}{c^2 d}-\frac {b^2 \text {PolyLog}\left (3,-e^{2 i \text {ArcSin}(c x)}\right )}{2 c^2 d} \]

[Out]

1/3*I*(a+b*arcsin(c*x))^3/b/c^2/d-(a+b*arcsin(c*x))^2*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)/c^2/d+I*b*(a+b*arcsin
(c*x))*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)/c^2/d-1/2*b^2*polylog(3,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)/c^2/d

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Rubi [A]
time = 0.13, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4765, 3800, 2221, 2611, 2320, 6724} \begin {gather*} \frac {i b \text {Li}_2\left (-e^{2 i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{c^2 d}+\frac {i (a+b \text {ArcSin}(c x))^3}{3 b c^2 d}-\frac {\log \left (1+e^{2 i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))^2}{c^2 d}-\frac {b^2 \text {Li}_3\left (-e^{2 i \text {ArcSin}(c x)}\right )}{2 c^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3800

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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*(e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 4765

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

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx &=\frac {\text {Subst}\left (\int (a+b x)^2 \tan (x) \, dx,x,\sin ^{-1}(c x)\right )}{c^2 d}\\ &=\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^2 d}-\frac {(2 i) \text {Subst}\left (\int \frac {e^{2 i x} (a+b x)^2}{1+e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )}{c^2 d}\\ &=\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^2 d}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^2 d}+\frac {(2 b) \text {Subst}\left (\int (a+b x) \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^2 d}\\ &=\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^2 d}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^2 d}+\frac {i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{c^2 d}-\frac {\left (i b^2\right ) \text {Subst}\left (\int \text {Li}_2\left (-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^2 d}\\ &=\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^2 d}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^2 d}+\frac {i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{c^2 d}-\frac {b^2 \text {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 c^2 d}\\ &=\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^2 d}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^2 d}+\frac {i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{c^2 d}-\frac {b^2 \text {Li}_3\left (-e^{2 i \sin ^{-1}(c x)}\right )}{2 c^2 d}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 143, normalized size = 1.22 \begin {gather*} \frac {6 i a b \text {ArcSin}(c x)^2+2 i b^2 \text {ArcSin}(c x)^3-12 a b \text {ArcSin}(c x) \log \left (1+e^{2 i \text {ArcSin}(c x)}\right )-6 b^2 \text {ArcSin}(c x)^2 \log \left (1+e^{2 i \text {ArcSin}(c x)}\right )-3 a^2 \log \left (1-c^2 x^2\right )+6 i b (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,-e^{2 i \text {ArcSin}(c x)}\right )-3 b^2 \text {PolyLog}\left (3,-e^{2 i \text {ArcSin}(c x)}\right )}{6 c^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((6*I)*a*b*ArcSin[c*x]^2 + (2*I)*b^2*ArcSin[c*x]^3 - 12*a*b*ArcSin[c*x]*Log[1 + E^((2*I)*ArcSin[c*x])] - 6*b^2
*ArcSin[c*x]^2*Log[1 + E^((2*I)*ArcSin[c*x])] - 3*a^2*Log[1 - c^2*x^2] + (6*I)*b*(a + b*ArcSin[c*x])*PolyLog[2
, -E^((2*I)*ArcSin[c*x])] - 3*b^2*PolyLog[3, -E^((2*I)*ArcSin[c*x])])/(6*c^2*d)

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Maple [A]
time = 0.08, size = 235, normalized size = 2.01

method	result	size
derivativedivides	\(\frac {-\frac {a^{2} \ln \left (c x -1\right )}{2 d}-\frac {a^{2} \ln \left (c x +1\right )}{2 d}+\frac {i b^{2} \arcsin \left (c x \right )^{3}}{3 d}-\frac {b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}+\frac {i b^{2} \arcsin \left (c x \right ) \polylog \left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}-\frac {b^{2} \polylog \left (3, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{2 d}+\frac {i a b \arcsin \left (c x \right )^{2}}{d}-\frac {2 a b \arcsin \left (c x \right ) \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}+\frac {i a b \polylog \left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}}{c^{2}}\)	\(235\)
default	\(\frac {-\frac {a^{2} \ln \left (c x -1\right )}{2 d}-\frac {a^{2} \ln \left (c x +1\right )}{2 d}+\frac {i b^{2} \arcsin \left (c x \right )^{3}}{3 d}-\frac {b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}+\frac {i b^{2} \arcsin \left (c x \right ) \polylog \left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}-\frac {b^{2} \polylog \left (3, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{2 d}+\frac {i a b \arcsin \left (c x \right )^{2}}{d}-\frac {2 a b \arcsin \left (c x \right ) \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}+\frac {i a b \polylog \left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}}{c^{2}}\)	\(235\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*(-1/2*a^2/d*ln(c*x-1)-1/2*a^2/d*ln(c*x+1)+1/3*I*b^2/d*arcsin(c*x)^3-b^2/d*arcsin(c*x)^2*ln(1+(I*c*x+(-c^
2*x^2+1)^(1/2))^2)+I*b^2/d*arcsin(c*x)*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)-1/2*b^2*polylog(3,-(I*c*x+(-c^
2*x^2+1)^(1/2))^2)/d+I*a*b/d*arcsin(c*x)^2-2*a*b/d*arcsin(c*x)*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)+I*a*b/d*poly
log(2,-(I*c*x+(-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(x*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d),x, algorithm="maxima")

[Out]

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

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {a^{2} x}{c^{2} x^{2} - 1}\, dx + \int \frac {b^{2} x \operatorname {asin}^{2}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx + \int \frac {2 a b x \operatorname {asin}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(a**2*x/(c**2*x**2 - 1), x) + Integral(b**2*x*asin(c*x)**2/(c**2*x**2 - 1), x) + Integral(2*a*b*x*as
in(c*x)/(c**2*x**2 - 1), x))/d

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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