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

Optimal. Leaf size=333 \[ \frac {2 i b c^3 \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac {2 i b c^3 \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac {2 i c^3 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d}-\frac {14 b c^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 d}-\frac {b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 d x^2}-\frac {c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{d x}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{3 d x^3}+\frac {7 i b^2 c^3 \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{3 d}-\frac {7 i b^2 c^3 \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{3 d}-\frac {2 b^2 c^3 \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{d}+\frac {2 b^2 c^3 \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {b^2 c^2}{3 d x} \]

[Out]

-1/3*b^2*c^2/d/x-1/3*(a+b*arcsin(c*x))^2/d/x^3-c^2*(a+b*arcsin(c*x))^2/d/x-2*I*c^3*(a+b*arcsin(c*x))^2*arctan(
I*c*x+(-c^2*x^2+1)^(1/2))/d-14/3*b*c^3*(a+b*arcsin(c*x))*arctanh(I*c*x+(-c^2*x^2+1)^(1/2))/d+7/3*I*b^2*c^3*pol
ylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))/d+2*I*b*c^3*(a+b*arcsin(c*x))*polylog(2,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))/d-2*I
*b*c^3*(a+b*arcsin(c*x))*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))/d-7/3*I*b^2*c^3*polylog(2,I*c*x+(-c^2*x^2+1)^
(1/2))/d-2*b^2*c^3*polylog(3,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))/d+2*b^2*c^3*polylog(3,I*(I*c*x+(-c^2*x^2+1)^(1/2))
)/d-1/3*b*c*(a+b*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/d/x^2

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Rubi [A]  time = 0.65, antiderivative size = 333, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 11, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {4701, 4657, 4181, 2531, 2282, 6589, 4709, 4183, 2279, 2391, 30} \[ \frac {2 i b c^3 \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac {2 i b c^3 \text {PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}+\frac {7 i b^2 c^3 \text {PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{3 d}-\frac {7 i b^2 c^3 \text {PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{3 d}-\frac {2 b^2 c^3 \text {PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{d}+\frac {2 b^2 c^3 \text {PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 d x^2}-\frac {c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{d x}-\frac {2 i c^3 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d}-\frac {14 b c^3 \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 d}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{3 d x^3}-\frac {b^2 c^2}{3 d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^2*c^2)/(3*d*x) - (b*c*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(3*d*x^2) - (a + b*ArcSin[c*x])^2/(3*d*x^3) -
(c^2*(a + b*ArcSin[c*x])^2)/(d*x) - ((2*I)*c^3*(a + b*ArcSin[c*x])^2*ArcTan[E^(I*ArcSin[c*x])])/d - (14*b*c^3
*(a + b*ArcSin[c*x])*ArcTanh[E^(I*ArcSin[c*x])])/(3*d) + (((7*I)/3)*b^2*c^3*PolyLog[2, -E^(I*ArcSin[c*x])])/d
+ ((2*I)*b*c^3*(a + b*ArcSin[c*x])*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/d - ((2*I)*b*c^3*(a + b*ArcSin[c*x])*Po
lyLog[2, I*E^(I*ArcSin[c*x])])/d - (((7*I)/3)*b^2*c^3*PolyLog[2, E^(I*ArcSin[c*x])])/d - (2*b^2*c^3*PolyLog[3,
(-I)*E^(I*ArcSin[c*x])])/d + (2*b^2*c^3*PolyLog[3, I*E^(I*ArcSin[c*x])])/d

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2282

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 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 2531

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 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 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && 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 4701

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*d^IntPart[p]*(d + e*x^2)^F
racPart[p])/(f*(m + 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] && LtQ[m, -1] && Inte
gerQ[m]

Rule 4709

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

Rule 6589

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 {\left (a+b \sin ^{-1}(c x)\right )^2}{x^4 \left (d-c^2 d x^2\right )} \, dx &=-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{3 d x^3}+c^2 \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{x^2 \left (d-c^2 d x^2\right )} \, dx+\frac {(2 b c) \int \frac {a+b \sin ^{-1}(c x)}{x^3 \sqrt {1-c^2 x^2}} \, dx}{3 d}\\ &=-\frac {b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 d x^2}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{3 d x^3}-\frac {c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{d x}+c^4 \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx+\frac {\left (b^2 c^2\right ) \int \frac {1}{x^2} \, dx}{3 d}+\frac {\left (b c^3\right ) \int \frac {a+b \sin ^{-1}(c x)}{x \sqrt {1-c^2 x^2}} \, dx}{3 d}+\frac {\left (2 b c^3\right ) \int \frac {a+b \sin ^{-1}(c x)}{x \sqrt {1-c^2 x^2}} \, dx}{d}\\ &=-\frac {b^2 c^2}{3 d x}-\frac {b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 d x^2}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{3 d x^3}-\frac {c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{d x}+\frac {c^3 \operatorname {Subst}\left (\int (a+b x)^2 \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{d}+\frac {\left (b c^3\right ) \operatorname {Subst}\left (\int (a+b x) \csc (x) \, dx,x,\sin ^{-1}(c x)\right )}{3 d}+\frac {\left (2 b c^3\right ) \operatorname {Subst}\left (\int (a+b x) \csc (x) \, dx,x,\sin ^{-1}(c x)\right )}{d}\\ &=-\frac {b^2 c^2}{3 d x}-\frac {b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 d x^2}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{3 d x^3}-\frac {c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{d x}-\frac {2 i c^3 \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {14 b c^3 \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{3 d}-\frac {\left (2 b c^3\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d}+\frac {\left (2 b c^3\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d}-\frac {\left (b^2 c^3\right ) \operatorname {Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{3 d}+\frac {\left (b^2 c^3\right ) \operatorname {Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{3 d}-\frac {\left (2 b^2 c^3\right ) \operatorname {Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d}+\frac {\left (2 b^2 c^3\right ) \operatorname {Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d}\\ &=-\frac {b^2 c^2}{3 d x}-\frac {b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 d x^2}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{3 d x^3}-\frac {c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{d x}-\frac {2 i c^3 \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {14 b c^3 \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{3 d}+\frac {2 i b c^3 \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {2 i b c^3 \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{d}+\frac {\left (i b^2 c^3\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{3 d}-\frac {\left (i b^2 c^3\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{3 d}+\frac {\left (2 i b^2 c^3\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {\left (2 i b^2 c^3\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {\left (2 i b^2 c^3\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d}+\frac {\left (2 i b^2 c^3\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d}\\ &=-\frac {b^2 c^2}{3 d x}-\frac {b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 d x^2}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{3 d x^3}-\frac {c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{d x}-\frac {2 i c^3 \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {14 b c^3 \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{3 d}+\frac {7 i b^2 c^3 \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{3 d}+\frac {2 i b c^3 \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {2 i b c^3 \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {7 i b^2 c^3 \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{3 d}-\frac {\left (2 b^2 c^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{d}+\frac {\left (2 b^2 c^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{d}\\ &=-\frac {b^2 c^2}{3 d x}-\frac {b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 d x^2}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{3 d x^3}-\frac {c^2 \left (a+b \sin ^{-1}(c x)\right )^2}{d x}-\frac {2 i c^3 \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {14 b c^3 \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{3 d}+\frac {7 i b^2 c^3 \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{3 d}+\frac {2 i b c^3 \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {2 i b c^3 \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {7 i b^2 c^3 \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{3 d}-\frac {2 b^2 c^3 \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{d}+\frac {2 b^2 c^3 \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{d}\\ \end {align*}

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Mathematica [B]  time = 7.84, size = 849, normalized size = 2.55 \[ -\frac {a^2 \log (1-c x) c^3}{2 d}+\frac {a^2 \log (c x+1) c^3}{2 d}-\frac {b^2 \left (\frac {1}{2} c x \sin ^{-1}(c x)^2 \csc ^4\left (\frac {1}{2} \sin ^{-1}(c x)\right )+2 \sin ^{-1}(c x) \csc ^2\left (\frac {1}{2} \sin ^{-1}(c x)\right )+\frac {8 \sin ^{-1}(c x)^2 \sin ^4\left (\frac {1}{2} \sin ^{-1}(c x)\right )}{c^3 x^3}-2 \sin ^{-1}(c x) \sec ^2\left (\frac {1}{2} \sin ^{-1}(c x)\right )+14 \sin ^{-1}(c x)^2 \cot \left (\frac {1}{2} \sin ^{-1}(c x)\right )+4 \cot \left (\frac {1}{2} \sin ^{-1}(c x)\right )-56 \sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )-24 \sin ^{-1}(c x)^2 \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+24 \sin ^{-1}(c x)^2 \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+56 \sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )-56 i \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )-48 i \sin ^{-1}(c x) \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )+48 i \sin ^{-1}(c x) \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )+56 i \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )+48 \text {Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )-48 \text {Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )+14 \sin ^{-1}(c x)^2 \tan \left (\frac {1}{2} \sin ^{-1}(c x)\right )+4 \tan \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right ) c^3}{24 d}-\frac {a^2 c^2}{d x}-\frac {2 a b \left (\frac {1}{2} \left (-\frac {i \sin ^{-1}(c x)^2}{2 c}+\frac {2 \log \left (1+i e^{i \sin ^{-1}(c x)}\right ) \sin ^{-1}(c x)}{c}+\frac {3 i \pi \sin ^{-1}(c x)}{2 c}+\frac {2 \pi \log \left (1+e^{-i \sin ^{-1}(c x)}\right )}{c}-\frac {\pi \log \left (1+i e^{i \sin ^{-1}(c x)}\right )}{c}-\frac {2 \pi \log \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )}{c}+\frac {\pi \log \left (-\cos \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{c}-\frac {2 i \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c}\right ) c^4-\frac {1}{2} \left (-\frac {i \sin ^{-1}(c x)^2}{2 c}+\frac {2 \log \left (1-i e^{i \sin ^{-1}(c x)}\right ) \sin ^{-1}(c x)}{c}+\frac {i \pi \sin ^{-1}(c x)}{2 c}+\frac {2 \pi \log \left (1+e^{-i \sin ^{-1}(c x)}\right )}{c}+\frac {\pi \log \left (1-i e^{i \sin ^{-1}(c x)}\right )}{c}-\frac {2 \pi \log \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )}{c}-\frac {\pi \log \left (\sin \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{c}-\frac {2 i \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c}\right ) c^4-\left (-\frac {\sin ^{-1}(c x)}{x}-c \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )\right ) c^2+\frac {c^3 \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right ) x^3+c \sqrt {1-c^2 x^2} x+2 \sin ^{-1}(c x)}{6 x^3}\right )}{d}-\frac {a^2}{3 d x^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/3*a^2/(d*x^3) - (a^2*c^2)/(d*x) - (a^2*c^3*Log[1 - c*x])/(2*d) + (a^2*c^3*Log[1 + c*x])/(2*d) - (2*a*b*(-(c
^2*(-(ArcSin[c*x]/x) - c*ArcTanh[Sqrt[1 - c^2*x^2]])) + (c*x*Sqrt[1 - c^2*x^2] + 2*ArcSin[c*x] + c^3*x^3*ArcTa
nh[Sqrt[1 - c^2*x^2]])/(6*x^3) + (c^4*((((3*I)/2)*Pi*ArcSin[c*x])/c - ((I/2)*ArcSin[c*x]^2)/c + (2*Pi*Log[1 +
E^((-I)*ArcSin[c*x])])/c - (Pi*Log[1 + I*E^(I*ArcSin[c*x])])/c + (2*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])])/
c - (2*Pi*Log[Cos[ArcSin[c*x]/2]])/c + (Pi*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]])/c - ((2*I)*PolyLog[2, (-I)*E^(I*
ArcSin[c*x])])/c))/2 - (c^4*(((I/2)*Pi*ArcSin[c*x])/c - ((I/2)*ArcSin[c*x]^2)/c + (2*Pi*Log[1 + E^((-I)*ArcSin
[c*x])])/c + (Pi*Log[1 - I*E^(I*ArcSin[c*x])])/c + (2*ArcSin[c*x]*Log[1 - I*E^(I*ArcSin[c*x])])/c - (2*Pi*Log[
Cos[ArcSin[c*x]/2]])/c - (Pi*Log[Sin[(Pi + 2*ArcSin[c*x])/4]])/c - ((2*I)*PolyLog[2, I*E^(I*ArcSin[c*x])])/c))
/2))/d - (b^2*c^3*(4*Cot[ArcSin[c*x]/2] + 14*ArcSin[c*x]^2*Cot[ArcSin[c*x]/2] + 2*ArcSin[c*x]*Csc[ArcSin[c*x]/
2]^2 + (c*x*ArcSin[c*x]^2*Csc[ArcSin[c*x]/2]^4)/2 - 56*ArcSin[c*x]*Log[1 - E^(I*ArcSin[c*x])] - 24*ArcSin[c*x]
^2*Log[1 - I*E^(I*ArcSin[c*x])] + 24*ArcSin[c*x]^2*Log[1 + I*E^(I*ArcSin[c*x])] + 56*ArcSin[c*x]*Log[1 + E^(I*
ArcSin[c*x])] - (56*I)*PolyLog[2, -E^(I*ArcSin[c*x])] - (48*I)*ArcSin[c*x]*PolyLog[2, (-I)*E^(I*ArcSin[c*x])]
+ (48*I)*ArcSin[c*x]*PolyLog[2, I*E^(I*ArcSin[c*x])] + (56*I)*PolyLog[2, E^(I*ArcSin[c*x])] + 48*PolyLog[3, (-
I)*E^(I*ArcSin[c*x])] - 48*PolyLog[3, I*E^(I*ArcSin[c*x])] - 2*ArcSin[c*x]*Sec[ArcSin[c*x]/2]^2 + (8*ArcSin[c*
x]^2*Sin[ArcSin[c*x]/2]^4)/(c^3*x^3) + 4*Tan[ArcSin[c*x]/2] + 14*ArcSin[c*x]^2*Tan[ArcSin[c*x]/2]))/(24*d)

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fricas [F]  time = 2.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}{c^{2} d x^{6} - d x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.45, size = 725, normalized size = 2.18 \[ -\frac {c \,b^{2} \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )}{3 d \,x^{2}}-\frac {c a b \sqrt {-c^{2} x^{2}+1}}{3 d \,x^{2}}-\frac {2 c^{2} a b \arcsin \left (c x \right )}{d x}+\frac {2 c^{3} a b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 c^{3} a b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 i c^{3} b^{2} \arcsin \left (c x \right ) \polylog \left (2, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {2 i c^{3} a b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 i c^{3} a b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {2 i c^{3} b^{2} \arcsin \left (c x \right ) \polylog \left (2, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {2 a b \arcsin \left (c x \right )}{3 d \,x^{3}}-\frac {c^{2} a^{2}}{d x}-\frac {b^{2} \arcsin \left (c x \right )^{2}}{3 d \,x^{3}}-\frac {c^{3} a^{2} \ln \left (c x -1\right )}{2 d}+\frac {c^{3} a^{2} \ln \left (c x +1\right )}{2 d}-\frac {2 b^{2} c^{3} \polylog \left (3, -i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {2 b^{2} c^{3} \polylog \left (3, i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {7 c^{3} a b \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{3 d}+\frac {7 c^{3} a b \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-1\right )}{3 d}+\frac {c^{3} b^{2} \arcsin \left (c x \right )^{2} \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {c^{3} b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {7 c^{3} b^{2} \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{3 d}+\frac {7 i c^{3} b^{2} \dilog \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{3 d}+\frac {7 i c^{3} b^{2} \dilog \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )}{3 d}-\frac {c^{2} b^{2} \arcsin \left (c x \right )^{2}}{d x}-\frac {b^{2} c^{2}}{3 d x}-\frac {a^{2}}{3 d \,x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*c*b^2/d/x^2*(-c^2*x^2+1)^(1/2)*arcsin(c*x)-1/3*c*a*b/d/x^2*(-c^2*x^2+1)^(1/2)-2*c^2*a*b/d*arcsin(c*x)/x+2
*c^3*a*b/d*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-2*c^3*a*b/d*arcsin(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^(
1/2)))-2*I*c^3/d*b^2*arcsin(c*x)*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))+2*I*c^3*a*b/d*dilog(1+I*(I*c*x+(-c^2*
x^2+1)^(1/2)))-2*I*c^3*a*b/d*dilog(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))+2*I*c^3/d*b^2*arcsin(c*x)*polylog(2,-I*(I*c
*x+(-c^2*x^2+1)^(1/2)))-7/3*c^3*a*b/d*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))+7/3*c^3*a*b/d*ln(I*c*x+(-c^2*x^2+1)^(1/2)
-1)+c^3/d*b^2*arcsin(c*x)^2*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-c^3/d*b^2*arcsin(c*x)^2*ln(1+I*(I*c*x+(-c^2*x^2
+1)^(1/2)))-7/3*c^3*b^2/d*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-2/3*a*b/d*arcsin(c*x)/x^3+7/3*I*c^3*b^2/d
*dilog(1+I*c*x+(-c^2*x^2+1)^(1/2))+7/3*I*c^3*b^2/d*dilog(I*c*x+(-c^2*x^2+1)^(1/2))-c^2*a^2/d/x-1/3*b^2/d/x^3*a
rcsin(c*x)^2-1/2*c^3*a^2/d*ln(c*x-1)+1/2*c^3*a^2/d*ln(c*x+1)-c^2*b^2/d/x*arcsin(c*x)^2-2*b^2*c^3*polylog(3,-I*
(I*c*x+(-c^2*x^2+1)^(1/2)))/d+2*b^2*c^3*polylog(3,I*(I*c*x+(-c^2*x^2+1)^(1/2)))/d-1/3*b^2*c^2/d/x-1/3*a^2/d/x^
3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))^2/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^2 + 1/6*(3*b^2*c^3*x^3*arctan2
(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2*log(c*x + 1) - 3*b^2*c^3*x^3*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^
2*log(-c*x + 1) + 6*d*x^3*integrate(-1/3*(6*a*b*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) - (3*b^2*c^4*x^4*ar
ctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(c*x + 1) - 3*b^2*c^4*x^4*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1
))*log(-c*x + 1) - 2*(3*b^2*c^3*x^3 + b^2*c*x)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)))*sqrt(c*x + 1)*sqrt(
-c*x + 1))/(c^2*d*x^6 - d*x^4), x) - 2*(3*b^2*c^2*x^2 + b^2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2)/(d*
x^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(a**2/(c**2*x**6 - x**4), x) + Integral(b**2*asin(c*x)**2/(c**2*x**6 - x**4), x) + Integral(2*a*b*as
in(c*x)/(c**2*x**6 - x**4), x))/d

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