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

Optimal. Leaf size=184 \[ -\frac{1}{2} i b d^2 \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )+\frac{1}{4} d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{2} d^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac{i d^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}+d^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{16} b c d^2 x \left (1-c^2 x^2\right )^{3/2}-\frac{11}{32} b c d^2 x \sqrt{1-c^2 x^2}-\frac{11}{32} b d^2 \sin ^{-1}(c x) \]

[Out]

(-11*b*c*d^2*x*Sqrt[1 - c^2*x^2])/32 - (b*c*d^2*x*(1 - c^2*x^2)^(3/2))/16 - (11*b*d^2*ArcSin[c*x])/32 + (d^2*(
1 - c^2*x^2)*(a + b*ArcSin[c*x]))/2 + (d^2*(1 - c^2*x^2)^2*(a + b*ArcSin[c*x]))/4 - ((I/2)*d^2*(a + b*ArcSin[c
*x])^2)/b + d^2*(a + b*ArcSin[c*x])*Log[1 - E^((2*I)*ArcSin[c*x])] - (I/2)*b*d^2*PolyLog[2, E^((2*I)*ArcSin[c*
x])]

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Rubi [A]  time = 0.201675, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {4683, 4625, 3717, 2190, 2279, 2391, 195, 216} \[ -\frac{1}{2} i b d^2 \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )+\frac{1}{4} d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{2} d^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac{i d^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}+d^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{16} b c d^2 x \left (1-c^2 x^2\right )^{3/2}-\frac{11}{32} b c d^2 x \sqrt{1-c^2 x^2}-\frac{11}{32} b d^2 \sin ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*b*c*d^2*x*Sqrt[1 - c^2*x^2])/32 - (b*c*d^2*x*(1 - c^2*x^2)^(3/2))/16 - (11*b*d^2*ArcSin[c*x])/32 + (d^2*(
1 - c^2*x^2)*(a + b*ArcSin[c*x]))/2 + (d^2*(1 - c^2*x^2)^2*(a + b*ArcSin[c*x]))/4 - ((I/2)*d^2*(a + b*ArcSin[c
*x])^2)/b + d^2*(a + b*ArcSin[c*x])*Log[1 - E^((2*I)*ArcSin[c*x])] - (I/2)*b*d^2*PolyLog[2, E^((2*I)*ArcSin[c*
x])]

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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{\left (d-c^2 d x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{x} \, dx &=\frac{1}{4} d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )+d \int \frac{\left (d-c^2 d x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{x} \, dx-\frac{1}{4} \left (b c d^2\right ) \int \left (1-c^2 x^2\right )^{3/2} \, dx\\ &=-\frac{1}{16} b c d^2 x \left (1-c^2 x^2\right )^{3/2}+\frac{1}{2} d^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{4} d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )+d^2 \int \frac{a+b \sin ^{-1}(c x)}{x} \, dx-\frac{1}{16} \left (3 b c d^2\right ) \int \sqrt{1-c^2 x^2} \, dx-\frac{1}{2} \left (b c d^2\right ) \int \sqrt{1-c^2 x^2} \, dx\\ &=-\frac{11}{32} b c d^2 x \sqrt{1-c^2 x^2}-\frac{1}{16} b c d^2 x \left (1-c^2 x^2\right )^{3/2}+\frac{1}{2} d^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{4} d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )+d^2 \operatorname{Subst}\left (\int (a+b x) \cot (x) \, dx,x,\sin ^{-1}(c x)\right )-\frac{1}{32} \left (3 b c d^2\right ) \int \frac{1}{\sqrt{1-c^2 x^2}} \, dx-\frac{1}{4} \left (b c d^2\right ) \int \frac{1}{\sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{11}{32} b c d^2 x \sqrt{1-c^2 x^2}-\frac{1}{16} b c d^2 x \left (1-c^2 x^2\right )^{3/2}-\frac{11}{32} b d^2 \sin ^{-1}(c x)+\frac{1}{2} d^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{4} d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )-\frac{i d^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}-\left (2 i 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{11}{32} b c d^2 x \sqrt{1-c^2 x^2}-\frac{1}{16} b c d^2 x \left (1-c^2 x^2\right )^{3/2}-\frac{11}{32} b d^2 \sin ^{-1}(c x)+\frac{1}{2} d^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{4} d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )-\frac{i d^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}+d^2 \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-\left (b d^2\right ) \operatorname{Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac{11}{32} b c d^2 x \sqrt{1-c^2 x^2}-\frac{1}{16} b c d^2 x \left (1-c^2 x^2\right )^{3/2}-\frac{11}{32} b d^2 \sin ^{-1}(c x)+\frac{1}{2} d^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{4} d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )-\frac{i d^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}+d^2 \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+\frac{1}{2} \left (i b d^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )\\ &=-\frac{11}{32} b c d^2 x \sqrt{1-c^2 x^2}-\frac{1}{16} b c d^2 x \left (1-c^2 x^2\right )^{3/2}-\frac{11}{32} b d^2 \sin ^{-1}(c x)+\frac{1}{2} d^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{4} d^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )-\frac{i d^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}+d^2 \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-\frac{1}{2} i b d^2 \text{Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.161432, size = 142, normalized size = 0.77 \[ \frac{1}{32} d^2 \left (-16 i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )+8 a c^4 x^4-32 a c^2 x^2+32 a \log (x)+2 b c^3 x^3 \sqrt{1-c^2 x^2}-13 b c x \sqrt{1-c^2 x^2}+b \sin ^{-1}(c x) \left (8 c^4 x^4-32 c^2 x^2+32 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+13\right )-16 i b \sin ^{-1}(c x)^2\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d^2*(-32*a*c^2*x^2 + 8*a*c^4*x^4 - 13*b*c*x*Sqrt[1 - c^2*x^2] + 2*b*c^3*x^3*Sqrt[1 - c^2*x^2] - (16*I)*b*ArcS
in[c*x]^2 + b*ArcSin[c*x]*(13 - 32*c^2*x^2 + 8*c^4*x^4 + 32*Log[1 - E^((2*I)*ArcSin[c*x])]) + 32*a*Log[x] - (1
6*I)*b*PolyLog[2, E^((2*I)*ArcSin[c*x])]))/32

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Maple [A]  time = 0.219, size = 250, normalized size = 1.4 \begin{align*}{\frac{{d}^{2}a{c}^{4}{x}^{4}}{4}}-{d}^{2}a{c}^{2}{x}^{2}+{d}^{2}a\ln \left ( cx \right ) +{\frac{{d}^{2}b\arcsin \left ( cx \right ){c}^{4}{x}^{4}}{4}}-{d}^{2}b\arcsin \left ( cx \right ){c}^{2}{x}^{2}+{\frac{13\,b{d}^{2}\arcsin \left ( cx \right ) }{32}}+{\frac{{d}^{2}b{c}^{3}{x}^{3}}{16}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{13\,{d}^{2}bcx}{32}\sqrt{-{c}^{2}{x}^{2}+1}}-i{d}^{2}b{\it polylog} \left ( 2,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) +{d}^{2}b\arcsin \left ( cx \right ) \ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) +{d}^{2}b\arcsin \left ( cx \right ) \ln \left ( 1-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) -i{d}^{2}b{\it polylog} \left ( 2,-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) -{\frac{i}{2}}b{d}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2} \end{align*}

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,x)

[Out]

1/4*d^2*a*c^4*x^4-d^2*a*c^2*x^2+d^2*a*ln(c*x)+1/4*d^2*b*arcsin(c*x)*c^4*x^4-d^2*b*arcsin(c*x)*c^2*x^2+13/32*b*
d^2*arcsin(c*x)+1/16*d^2*b*(-c^2*x^2+1)^(1/2)*c^3*x^3-13/32*b*c*d^2*x*(-c^2*x^2+1)^(1/2)-I*d^2*b*polylog(2,I*c
*x+(-c^2*x^2+1)^(1/2))+d^2*b*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))+d^2*b*arcsin(c*x)*ln(1-I*c*x-(-c^2*x^2
+1)^(1/2))-I*d^2*b*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))-1/2*I*b*d^2*arcsin(c*x)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{4} \, a c^{4} d^{2} x^{4} - a c^{2} d^{2} x^{2} + a d^{2} \log \left (x\right ) + \int \frac{{\left (b c^{4} d^{2} x^{4} - 2 \, b c^{2} d^{2} x^{2} + b d^{2}\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )}{x}\,{d x} \end{align*}

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,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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}, x\right ) \end{align*}

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,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,
x)

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

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,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c^{2} d x^{2} - d\right )}^{2}{\left (b \arcsin \left (c x\right ) + a\right )}}{x}\,{d x} \end{align*}

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,x, algorithm="giac")

[Out]

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