3.17 \(\int \frac {(d-c^2 d x^2)^2 (a+b \sin ^{-1}(c x))}{x^3} \, dx\)

Optimal. Leaf size=201 \[ -c^2 d^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac {d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{2 x^2}+\frac {i c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )^2}{b}-2 c^2 d^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )+i b c^2 d^2 \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )-\frac {b c d^2 \left (1-c^2 x^2\right )^{3/2}}{2 x}-\frac {1}{4} b c^2 d^2 \sin ^{-1}(c x)-\frac {1}{4} b c^3 d^2 x \sqrt {1-c^2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4685, 277, 195, 216, 4683, 4625, 3717, 2190, 2279, 2391} \[ i b c^2 d^2 \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )-c^2 d^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac {d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{2 x^2}+\frac {i c^2 d^2 \left (a+b \sin ^{-1}(c x)\right )^2}{b}-2 c^2 d^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{4} b c^3 d^2 x \sqrt {1-c^2 x^2}-\frac {b c d^2 \left (1-c^2 x^2\right )^{3/2}}{2 x}-\frac {1}{4} b c^2 d^2 \sin ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[((d - c^2*d*x^2)^2*(a + b*ArcSin[c*x]))/x^3,x]

[Out]

-(b*c^3*d^2*x*Sqrt[1 - c^2*x^2])/4 - (b*c*d^2*(1 - c^2*x^2)^(3/2))/(2*x) - (b*c^2*d^2*ArcSin[c*x])/4 - c^2*d^2
*(1 - c^2*x^2)*(a + b*ArcSin[c*x]) - (d^2*(1 - c^2*x^2)^2*(a + b*ArcSin[c*x]))/(2*x^2) + (I*c^2*d^2*(a + b*Arc
Sin[c*x])^2)/b - 2*c^2*d^2*(a + b*ArcSin[c*x])*Log[1 - E^((2*I)*ArcSin[c*x])] + I*b*c^2*d^2*PolyLog[2, E^((2*I
)*ArcSin[c*x])]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 216

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 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 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 3717

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 4625

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

Rule 4683

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

Rule 4685

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

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 162, normalized size = 0.81 \[ \frac {d^2 \left (2 a c^4 x^4-8 a c^2 x^2 \log (x)-2 a+4 i b c^2 x^2 \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )-2 b c x \sqrt {1-c^2 x^2}+4 i b c^2 x^2 \sin ^{-1}(c x)^2+b \sin ^{-1}(c x) \left (2 c^4 x^4-c^2 x^2-8 c^2 x^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-2\right )+b c^3 x^3 \sqrt {1-c^2 x^2}\right )}{4 x^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d - c^2*d*x^2)^2*(a + b*ArcSin[c*x]))/x^3,x]

[Out]

(d^2*(-2*a + 2*a*c^4*x^4 - 2*b*c*x*Sqrt[1 - c^2*x^2] + b*c^3*x^3*Sqrt[1 - c^2*x^2] + (4*I)*b*c^2*x^2*ArcSin[c*
x]^2 + b*ArcSin[c*x]*(-2 - c^2*x^2 + 2*c^4*x^4 - 8*c^2*x^2*Log[1 - E^((2*I)*ArcSin[c*x])]) - 8*a*c^2*x^2*Log[x
] + (4*I)*b*c^2*x^2*PolyLog[2, E^((2*I)*ArcSin[c*x])]))/(4*x^2)

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a c^{4} d^{2} x^{4} - 2 \, a c^{2} d^{2} x^{2} + a d^{2} + {\left (b c^{4} d^{2} x^{4} - 2 \, b c^{2} d^{2} x^{2} + b d^{2}\right )} \arcsin \left (c x\right )}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^2*(a+b*arcsin(c*x))/x^3,x, algorithm="fricas")

[Out]

integral((a*c^4*d^2*x^4 - 2*a*c^2*d^2*x^2 + a*d^2 + (b*c^4*d^2*x^4 - 2*b*c^2*d^2*x^2 + b*d^2)*arcsin(c*x))/x^3
, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c^{2} d x^{2} - d\right )}^{2} {\left (b \arcsin \left (c x\right ) + a\right )}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^2*(a+b*arcsin(c*x))/x^3,x, algorithm="giac")

[Out]

integrate((c^2*d*x^2 - d)^2*(b*arcsin(c*x) + a)/x^3, x)

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maple [A]  time = 0.57, size = 278, normalized size = 1.38 \[ \frac {c^{4} d^{2} a \,x^{2}}{2}-2 c^{2} d^{2} a \ln \left (c x \right )-\frac {d^{2} a}{2 x^{2}}+i c^{2} d^{2} b \arcsin \left (c x \right )^{2}+\frac {b \,c^{3} d^{2} x \sqrt {-c^{2} x^{2}+1}}{4}+\frac {c^{4} d^{2} b \arcsin \left (c x \right ) x^{2}}{2}-\frac {b \,c^{2} d^{2} \arcsin \left (c x \right )}{4}+\frac {i c^{2} d^{2} b}{2}-\frac {b c \,d^{2} \sqrt {-c^{2} x^{2}+1}}{2 x}-\frac {d^{2} b \arcsin \left (c x \right )}{2 x^{2}}-2 c^{2} d^{2} b \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-2 c^{2} d^{2} b \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+2 i c^{2} d^{2} b \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )+2 i c^{2} d^{2} b \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-c^2*d*x^2+d)^2*(a+b*arcsin(c*x))/x^3,x)

[Out]

1/2*c^4*d^2*a*x^2-2*c^2*d^2*a*ln(c*x)-1/2*d^2*a/x^2+I*c^2*d^2*b*arcsin(c*x)^2+1/4*b*c^3*d^2*x*(-c^2*x^2+1)^(1/
2)+1/2*c^4*d^2*b*arcsin(c*x)*x^2-1/4*b*c^2*d^2*arcsin(c*x)+1/2*I*c^2*d^2*b-1/2*b*c*d^2*(-c^2*x^2+1)^(1/2)/x-1/
2*d^2*b*arcsin(c*x)/x^2-2*c^2*d^2*b*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-2*c^2*d^2*b*arcsin(c*x)*ln(1-I*
c*x-(-c^2*x^2+1)^(1/2))+2*I*c^2*d^2*b*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))+2*I*c^2*d^2*b*polylog(2,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 \[ \frac {1}{2} \, a c^{4} d^{2} x^{2} - 2 \, a c^{2} d^{2} \log \relax (x) - \frac {1}{2} \, b d^{2} {\left (\frac {\sqrt {-c^{2} x^{2} + 1} c}{x} + \frac {\arcsin \left (c x\right )}{x^{2}}\right )} - \frac {a d^{2}}{2 \, x^{2}} + \int \frac {{\left (b c^{4} d^{2} x^{2} - 2 \, b c^{2} d^{2}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^2*(a+b*arcsin(c*x))/x^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^2}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*asin(c*x))*(d - c^2*d*x^2)^2)/x^3,x)

[Out]

int(((a + b*asin(c*x))*(d - c^2*d*x^2)^2)/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ d^{2} \left (\int \frac {a}{x^{3}}\, dx + \int \left (- \frac {2 a c^{2}}{x}\right )\, dx + \int a c^{4} x\, dx + \int \frac {b \operatorname {asin}{\left (c x \right )}}{x^{3}}\, dx + \int \left (- \frac {2 b c^{2} \operatorname {asin}{\left (c x \right )}}{x}\right )\, dx + \int b c^{4} x \operatorname {asin}{\left (c x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c**2*d*x**2+d)**2*(a+b*asin(c*x))/x**3,x)

[Out]

d**2*(Integral(a/x**3, x) + Integral(-2*a*c**2/x, x) + Integral(a*c**4*x, x) + Integral(b*asin(c*x)/x**3, x) +
Integral(-2*b*c**2*asin(c*x)/x, x) + Integral(b*c**4*x*asin(c*x), x))

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