3.28 \(\int \frac {x^4 (a+b \sin ^{-1}(c x))}{d-c^2 d x^2} \, dx\)

Optimal. Leaf size=172 \[ -\frac {2 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^5 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d}-\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d}+\frac {i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac {i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}+\frac {b \left (1-c^2 x^2\right )^{3/2}}{9 c^5 d}-\frac {4 b \sqrt {1-c^2 x^2}}{3 c^5 d} \]

[Out]

1/9*b*(-c^2*x^2+1)^(3/2)/c^5/d-x*(a+b*arcsin(c*x))/c^4/d-1/3*x^3*(a+b*arcsin(c*x))/c^2/d-2*I*(a+b*arcsin(c*x))
*arctan(I*c*x+(-c^2*x^2+1)^(1/2))/c^5/d+I*b*polylog(2,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c^5/d-I*b*polylog(2,I*(I*
c*x+(-c^2*x^2+1)^(1/2)))/c^5/d-4/3*b*(-c^2*x^2+1)^(1/2)/c^5/d

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Rubi [A]  time = 0.24, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {4715, 4657, 4181, 2279, 2391, 261, 266, 43} \[ \frac {i b \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac {i b \text {PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d}-\frac {2 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^5 d}+\frac {b \left (1-c^2 x^2\right )^{3/2}}{9 c^5 d}-\frac {4 b \sqrt {1-c^2 x^2}}{3 c^5 d} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2),x]

[Out]

(-4*b*Sqrt[1 - c^2*x^2])/(3*c^5*d) + (b*(1 - c^2*x^2)^(3/2))/(9*c^5*d) - (x*(a + b*ArcSin[c*x]))/(c^4*d) - (x^
3*(a + b*ArcSin[c*x]))/(3*c^2*d) - ((2*I)*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/(c^5*d) + (I*b*PolyLo
g[2, (-I)*E^(I*ArcSin[c*x])])/(c^5*d) - (I*b*PolyLog[2, I*E^(I*ArcSin[c*x])])/(c^5*d)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 261

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 4181

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

Rule 4657

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

Rule 4715

Int[((a_.) + ArcSin[(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*ArcSin[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*ArcSin[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*ArcSin[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^4 \left (a+b \sin ^{-1}(c x)\right )}{d-c^2 d x^2} \, dx &=-\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d}+\frac {\int \frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{d-c^2 d x^2} \, dx}{c^2}+\frac {b \int \frac {x^3}{\sqrt {1-c^2 x^2}} \, dx}{3 c d}\\ &=-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d}-\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d}+\frac {\int \frac {a+b \sin ^{-1}(c x)}{d-c^2 d x^2} \, dx}{c^4}+\frac {b \int \frac {x}{\sqrt {1-c^2 x^2}} \, dx}{c^3 d}+\frac {b \operatorname {Subst}\left (\int \frac {x}{\sqrt {1-c^2 x}} \, dx,x,x^2\right )}{6 c d}\\ &=-\frac {b \sqrt {1-c^2 x^2}}{c^5 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d}-\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d}+\frac {\operatorname {Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{c^5 d}+\frac {b \operatorname {Subst}\left (\int \left (\frac {1}{c^2 \sqrt {1-c^2 x}}-\frac {\sqrt {1-c^2 x}}{c^2}\right ) \, dx,x,x^2\right )}{6 c d}\\ &=-\frac {4 b \sqrt {1-c^2 x^2}}{3 c^5 d}+\frac {b \left (1-c^2 x^2\right )^{3/2}}{9 c^5 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d}-\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d}-\frac {2 i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac {b \operatorname {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^5 d}+\frac {b \operatorname {Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^5 d}\\ &=-\frac {4 b \sqrt {1-c^2 x^2}}{3 c^5 d}+\frac {b \left (1-c^2 x^2\right )^{3/2}}{9 c^5 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d}-\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d}-\frac {2 i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^5 d}+\frac {(i b) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac {(i b) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c^5 d}\\ &=-\frac {4 b \sqrt {1-c^2 x^2}}{3 c^5 d}+\frac {b \left (1-c^2 x^2\right )^{3/2}}{9 c^5 d}-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{c^4 d}-\frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d}-\frac {2 i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^5 d}+\frac {i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}-\frac {i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c^5 d}\\ \end {align*}

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Mathematica [A]  time = 0.36, size = 286, normalized size = 1.66 \[ -\frac {6 a c^3 x^3+18 a c x+9 a \log (1-c x)-9 a \log (c x+1)+6 b c^3 x^3 \sin ^{-1}(c x)+2 b c^2 x^2 \sqrt {1-c^2 x^2}+22 b \sqrt {1-c^2 x^2}-18 i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )+18 i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )+18 b c x \sin ^{-1}(c x)+9 i \pi b \sin ^{-1}(c x)-18 b \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-9 \pi b \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+18 b \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-9 \pi b \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+9 \pi b \log \left (\sin \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+9 \pi b \log \left (-\cos \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{18 c^5 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^4*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2),x]

[Out]

-1/18*(18*a*c*x + 6*a*c^3*x^3 + 22*b*Sqrt[1 - c^2*x^2] + 2*b*c^2*x^2*Sqrt[1 - c^2*x^2] + (9*I)*b*Pi*ArcSin[c*x
] + 18*b*c*x*ArcSin[c*x] + 6*b*c^3*x^3*ArcSin[c*x] - 9*b*Pi*Log[1 - I*E^(I*ArcSin[c*x])] - 18*b*ArcSin[c*x]*Lo
g[1 - I*E^(I*ArcSin[c*x])] - 9*b*Pi*Log[1 + I*E^(I*ArcSin[c*x])] + 18*b*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x]
)] + 9*a*Log[1 - c*x] - 9*a*Log[1 + c*x] + 9*b*Pi*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] + 9*b*Pi*Log[Sin[(Pi + 2*A
rcSin[c*x])/4]] - (18*I)*b*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] + (18*I)*b*PolyLog[2, I*E^(I*ArcSin[c*x])])/(c^5
*d)

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b x^{4} \arcsin \left (c x\right ) + a x^{4}}{c^{2} d x^{2} - d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d),x, algorithm="fricas")

[Out]

integral(-(b*x^4*arcsin(c*x) + a*x^4)/(c^2*d*x^2 - d), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {{\left (b \arcsin \left (c x\right ) + a\right )} x^{4}}{c^{2} d x^{2} - d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d),x, algorithm="giac")

[Out]

integrate(-(b*arcsin(c*x) + a)*x^4/(c^2*d*x^2 - d), x)

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maple [A]  time = 0.27, size = 270, normalized size = 1.57 \[ -\frac {a \,x^{3}}{3 c^{2} d}-\frac {a x}{c^{4} d}-\frac {a \ln \left (c x -1\right )}{2 c^{5} d}+\frac {a \ln \left (c x +1\right )}{2 c^{5} d}-\frac {b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{5} d}+\frac {b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{5} d}+\frac {i b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{5} d}-\frac {i b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{5} d}-\frac {11 b \sqrt {-c^{2} x^{2}+1}}{9 c^{5} d}-\frac {b \arcsin \left (c x \right ) x}{c^{4} d}-\frac {b \arcsin \left (c x \right ) x^{3}}{3 c^{2} d}-\frac {b \sqrt {-c^{2} x^{2}+1}\, x^{2}}{9 c^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d),x)

[Out]

-1/3/c^2*a/d*x^3-1/c^4*a/d*x-1/2/c^5*a/d*ln(c*x-1)+1/2/c^5*a/d*ln(c*x+1)-1/c^5*b/d*arcsin(c*x)*ln(1+I*(I*c*x+(
-c^2*x^2+1)^(1/2)))+1/c^5*b/d*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))+I/c^5*b/d*dilog(1+I*(I*c*x+(-c^2*
x^2+1)^(1/2)))-I/c^5*b/d*dilog(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-11/9*b*(-c^2*x^2+1)^(1/2)/c^5/d-1/c^4*b/d*arcsi
n(c*x)*x-1/3/c^2*b/d*arcsin(c*x)*x^3-1/9/c^3*b/d*(-c^2*x^2+1)^(1/2)*x^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{6} \, a {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4} d} - \frac {3 \, \log \left (c x + 1\right )}{c^{5} d} + \frac {3 \, \log \left (c x - 1\right )}{c^{5} d}\right )} + \frac {-\frac {1}{3} \, {\left (c^{5} d {\left (\frac {2 \, {\left (c^{2} x^{2} + 2\right )} \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{5} d} + \frac {18 \, \sqrt {c x + 1} \sqrt {-c x + 1}}{c^{5} d} + 3 \, \int -\frac {3 \, \sqrt {c x + 1} \sqrt {-c x + 1} {\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )}}{c^{6} d x^{2} - c^{4} d}\,{d x}\right )} + 6 \, {\left (c^{3} x^{3} + 3 \, c x\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) - 9 \, \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (c x + 1\right ) + 9 \, \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left (-c x + 1\right )\right )} b}{6 \, c^{5} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c*x))/(-c^2*d*x^2+d),x, algorithm="maxima")

[Out]

-1/6*a*(2*(c^2*x^3 + 3*x)/(c^4*d) - 3*log(c*x + 1)/(c^5*d) + 3*log(c*x - 1)/(c^5*d)) + 1/6*(6*c^5*d*integrate(
-1/6*(2*c^3*x^3 + 6*c*x - 3*log(c*x + 1) + 3*log(-c*x + 1))*sqrt(c*x + 1)*sqrt(-c*x + 1)/(c^6*d*x^2 - c^4*d),
x) - 2*(c^3*x^3 + 3*c*x)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + 3*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x +
1))*log(c*x + 1) - 3*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(-c*x + 1))*b/(c^5*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{d-c^2\,d\,x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + b*asin(c*x)))/(d - c^2*d*x^2),x)

[Out]

int((x^4*(a + b*asin(c*x)))/(d - c^2*d*x^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {a x^{4}}{c^{2} x^{2} - 1}\, dx + \int \frac {b x^{4} \operatorname {asin}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*asin(c*x))/(-c**2*d*x**2+d),x)

[Out]

-(Integral(a*x**4/(c**2*x**2 - 1), x) + Integral(b*x**4*asin(c*x)/(c**2*x**2 - 1), x))/d

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