3.1.36 \(\int \frac {a+b \text {ArcSin}(c x)}{x^4 (d-c^2 d x^2)} \, dx\) [36]

Optimal. Leaf size=173 \[ -\frac {b c \sqrt {1-c^2 x^2}}{6 d x^2}-\frac {a+b \text {ArcSin}(c x)}{3 d x^3}-\frac {c^2 (a+b \text {ArcSin}(c x))}{d x}-\frac {2 i c^3 (a+b \text {ArcSin}(c x)) \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right )}{d}-\frac {7 b c^3 \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{6 d}+\frac {i b c^3 \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )}{d}-\frac {i b c^3 \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )}{d} \]

[Out]

1/3*(-a-b*arcsin(c*x))/d/x^3-c^2*(a+b*arcsin(c*x))/d/x-2*I*c^3*(a+b*arcsin(c*x))*arctan(I*c*x+(-c^2*x^2+1)^(1/
2))/d-7/6*b*c^3*arctanh((-c^2*x^2+1)^(1/2))/d+I*b*c^3*polylog(2,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))/d-I*b*c^3*polyl
og(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))/d-1/6*b*c*(-c^2*x^2+1)^(1/2)/d/x^2

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Rubi [A]
time = 0.17, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {4789, 4749, 4266, 2317, 2438, 272, 65, 214, 44} \begin {gather*} -\frac {2 i c^3 \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{d}-\frac {c^2 (a+b \text {ArcSin}(c x))}{d x}-\frac {a+b \text {ArcSin}(c x)}{3 d x^3}+\frac {i b c^3 \text {Li}_2\left (-i e^{i \text {ArcSin}(c x)}\right )}{d}-\frac {i b c^3 \text {Li}_2\left (i e^{i \text {ArcSin}(c x)}\right )}{d}-\frac {b c \sqrt {1-c^2 x^2}}{6 d x^2}-\frac {7 b c^3 \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{6 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(b*c*Sqrt[1 - c^2*x^2])/(d*x^2) - (a + b*ArcSin[c*x])/(3*d*x^3) - (c^2*(a + b*ArcSin[c*x]))/(d*x) - ((2*I
)*c^3*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/d - (7*b*c^3*ArcTanh[Sqrt[1 - c^2*x^2]])/(6*d) + (I*b*c^3
*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/d - (I*b*c^3*PolyLog[2, I*E^(I*ArcSin[c*x])])/d

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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]

Rule 272

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 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 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 4266

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 4749

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 4789

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
f*x)^(m + 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])^n/(d*f*(m + 1))), x] + (Dist[c^2*((m + 2*p + 3)/(f^2*(m
+ 1))), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] - Dist[b*c*(n/(f*(m + 1)))*Simp[(d + e*x
^2)^p/(1 - c^2*x^2)^p], Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; Free
Q[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && ILtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {a+b \sin ^{-1}(c x)}{x^4 \left (d-c^2 d x^2\right )} \, dx &=-\frac {a+b \sin ^{-1}(c x)}{3 d x^3}+c^2 \int \frac {a+b \sin ^{-1}(c x)}{x^2 \left (d-c^2 d x^2\right )} \, dx+\frac {(b c) \int \frac {1}{x^3 \sqrt {1-c^2 x^2}} \, dx}{3 d}\\ &=-\frac {a+b \sin ^{-1}(c x)}{3 d x^3}-\frac {c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x}+c^4 \int \frac {a+b \sin ^{-1}(c x)}{d-c^2 d x^2} \, dx+\frac {(b c) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {1-c^2 x}} \, dx,x,x^2\right )}{6 d}+\frac {\left (b c^3\right ) \int \frac {1}{x \sqrt {1-c^2 x^2}} \, dx}{d}\\ &=-\frac {b c \sqrt {1-c^2 x^2}}{6 d x^2}-\frac {a+b \sin ^{-1}(c x)}{3 d x^3}-\frac {c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x}+\frac {c^3 \text {Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{d}+\frac {\left (b c^3\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1-c^2 x}} \, dx,x,x^2\right )}{12 d}+\frac {\left (b c^3\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1-c^2 x}} \, dx,x,x^2\right )}{2 d}\\ &=-\frac {b c \sqrt {1-c^2 x^2}}{6 d x^2}-\frac {a+b \sin ^{-1}(c x)}{3 d x^3}-\frac {c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x}-\frac {2 i c^3 \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {(b c) \text {Subst}\left (\int \frac {1}{\frac {1}{c^2}-\frac {x^2}{c^2}} \, dx,x,\sqrt {1-c^2 x^2}\right )}{6 d}-\frac {(b c) \text {Subst}\left (\int \frac {1}{\frac {1}{c^2}-\frac {x^2}{c^2}} \, dx,x,\sqrt {1-c^2 x^2}\right )}{d}-\frac {\left (b c^3\right ) \text {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d}+\frac {\left (b c^3\right ) \text {Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d}\\ &=-\frac {b c \sqrt {1-c^2 x^2}}{6 d x^2}-\frac {a+b \sin ^{-1}(c x)}{3 d x^3}-\frac {c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x}-\frac {2 i c^3 \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {7 b c^3 \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{6 d}+\frac {\left (i b c^3\right ) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {\left (i b c^3\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{d}\\ &=-\frac {b c \sqrt {1-c^2 x^2}}{6 d x^2}-\frac {a+b \sin ^{-1}(c x)}{3 d x^3}-\frac {c^2 \left (a+b \sin ^{-1}(c x)\right )}{d x}-\frac {2 i c^3 \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {7 b c^3 \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{6 d}+\frac {i b c^3 \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {i b c^3 \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{d}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(350\) vs. \(2(173)=346\).
time = 0.09, size = 350, normalized size = 2.02 \begin {gather*} -\frac {2 a+6 a c^2 x^2+b c x \sqrt {1-c^2 x^2}+2 b \text {ArcSin}(c x)+6 b c^2 x^2 \text {ArcSin}(c x)+3 i b c^3 \pi x^3 \text {ArcSin}(c x)+7 b c^3 x^3 \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )-3 b c^3 \pi x^3 \log \left (1-i e^{i \text {ArcSin}(c x)}\right )-6 b c^3 x^3 \text {ArcSin}(c x) \log \left (1-i e^{i \text {ArcSin}(c x)}\right )-3 b c^3 \pi x^3 \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+6 b c^3 x^3 \text {ArcSin}(c x) \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+3 a c^3 x^3 \log (1-c x)-3 a c^3 x^3 \log (1+c x)+3 b c^3 \pi x^3 \log \left (-\cos \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )+3 b c^3 \pi x^3 \log \left (\sin \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )-6 i b c^3 x^3 \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )+6 i b c^3 x^3 \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )}{6 d x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(2*a + 6*a*c^2*x^2 + b*c*x*Sqrt[1 - c^2*x^2] + 2*b*ArcSin[c*x] + 6*b*c^2*x^2*ArcSin[c*x] + (3*I)*b*c^3*Pi
*x^3*ArcSin[c*x] + 7*b*c^3*x^3*ArcTanh[Sqrt[1 - c^2*x^2]] - 3*b*c^3*Pi*x^3*Log[1 - I*E^(I*ArcSin[c*x])] - 6*b*
c^3*x^3*ArcSin[c*x]*Log[1 - I*E^(I*ArcSin[c*x])] - 3*b*c^3*Pi*x^3*Log[1 + I*E^(I*ArcSin[c*x])] + 6*b*c^3*x^3*A
rcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])] + 3*a*c^3*x^3*Log[1 - c*x] - 3*a*c^3*x^3*Log[1 + c*x] + 3*b*c^3*Pi*x^3
*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] + 3*b*c^3*Pi*x^3*Log[Sin[(Pi + 2*ArcSin[c*x])/4]] - (6*I)*b*c^3*x^3*PolyLog
[2, (-I)*E^(I*ArcSin[c*x])] + (6*I)*b*c^3*x^3*PolyLog[2, I*E^(I*ArcSin[c*x])])/(d*x^3)

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Maple [A]
time = 0.21, size = 291, normalized size = 1.68

method	result	size
derivativedivides	\(c^{3} \left (\frac {a \ln \left (c x +1\right )}{2 d}-\frac {a \ln \left (c x -1\right )}{2 d}-\frac {a}{3 d \,c^{3} x^{3}}-\frac {a}{d c x}-\frac {b \arcsin \left (c x \right )}{d c x}-\frac {b \sqrt {-c^{2} x^{2}+1}}{6 d \,c^{2} x^{2}}-\frac {b \arcsin \left (c x \right )}{3 d \,c^{3} x^{3}}+\frac {7 b \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-1\right )}{6 d}-\frac {7 b \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{6 d}-\frac {i b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {i b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}\right )\)	\(291\)
default	\(c^{3} \left (\frac {a \ln \left (c x +1\right )}{2 d}-\frac {a \ln \left (c x -1\right )}{2 d}-\frac {a}{3 d \,c^{3} x^{3}}-\frac {a}{d c x}-\frac {b \arcsin \left (c x \right )}{d c x}-\frac {b \sqrt {-c^{2} x^{2}+1}}{6 d \,c^{2} x^{2}}-\frac {b \arcsin \left (c x \right )}{3 d \,c^{3} x^{3}}+\frac {7 b \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-1\right )}{6 d}-\frac {7 b \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{6 d}-\frac {i b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {i b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}\right )\)	\(291\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^3*(1/2*a/d*ln(c*x+1)-1/2*a/d*ln(c*x-1)-1/3*a/d/c^3/x^3-a/d/c/x-b/d*arcsin(c*x)/c/x-1/6*b/d/c^2/x^2*(-c^2*x^2
+1)^(1/2)-1/3*b/d*arcsin(c*x)/c^3/x^3+7/6*b/d*ln(I*c*x+(-c^2*x^2+1)^(1/2)-1)-7/6*b/d*ln(1+I*c*x+(-c^2*x^2+1)^(
1/2))-I*b/d*dilog(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))+b/d*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-b/d*arcsi
n(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+I*b/d*dilog(1+I*(I*c*x+(-c^2*x^2+1)^(1/2))))

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

[Out]

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

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

[Out]

integral(-(b*arcsin(c*x) + a)/(c^2*d*x^6 - d*x^4), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {a}{c^{2} x^{6} - x^{4}}\, dx + \int \frac {b \operatorname {asin}{\left (c x \right )}}{c^{2} x^{6} - x^{4}}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(a/(c**2*x**6 - x**4), x) + Integral(b*asin(c*x)/(c**2*x**6 - x**4), 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((a+b*arcsin(c*x))/x^4/(-c^2*d*x^2+d),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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