3.1.15 \(\int \frac {(d-c^2 d x^2)^2 (a+b \text {ArcSin}(c x))}{x} \, dx\) [15]

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

[Out]

-1/16*b*c*d^2*x*(-c^2*x^2+1)^(3/2)-11/32*b*d^2*arcsin(c*x)+1/2*d^2*(-c^2*x^2+1)*(a+b*arcsin(c*x))+1/4*d^2*(-c^
2*x^2+1)^2*(a+b*arcsin(c*x))-1/2*I*d^2*(a+b*arcsin(c*x))^2/b+d^2*(a+b*arcsin(c*x))*ln(1-(I*c*x+(-c^2*x^2+1)^(1
/2))^2)-1/2*I*b*d^2*polylog(2,(I*c*x+(-c^2*x^2+1)^(1/2))^2)-11/32*b*c*d^2*x*(-c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 184, 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 = {4773, 4721, 3798, 2221, 2317, 2438, 201, 222} \begin {gather*} \frac {1}{4} d^2 \left (1-c^2 x^2\right )^2 (a+b \text {ArcSin}(c x))+\frac {1}{2} d^2 \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))-\frac {i d^2 (a+b \text {ArcSin}(c x))^2}{2 b}+d^2 \log \left (1-e^{2 i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))-\frac {1}{2} i b d^2 \text {Li}_2\left (e^{2 i \text {ArcSin}(c x)}\right )-\frac {11}{32} b d^2 \text {ArcSin}(c x)-\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} \end {gather*}

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 201

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 222

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 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 3798

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 4721

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

Rule 4773

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]

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 \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.21, size = 166, normalized size = 0.90 \begin {gather*} \frac {1}{32} d^2 \left (-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 \text {ArcSin}(c x)^2+26 b \text {ArcTan}\left (\frac {c x}{-1+\sqrt {1-c^2 x^2}}\right )+8 b \text {ArcSin}(c x) \left (-4 c^2 x^2+c^4 x^4+4 \log \left (1-e^{2 i \text {ArcSin}(c x)}\right )\right )+32 a \log (x)-16 i b \text {PolyLog}\left (2,e^{2 i \text {ArcSin}(c x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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 + 26*b*ArcTan[(c*x)/(-1 + Sqrt[1 - c^2*x^2])] + 8*b*ArcSin[c*x]*(-4*c^2*x^2 + c^4*x^4 + 4*Log[1 - E^
((2*I)*ArcSin[c*x])]) + 32*a*Log[x] - (16*I)*b*PolyLog[2, E^((2*I)*ArcSin[c*x])]))/32

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Maple [A]
time = 0.21, size = 224, normalized size = 1.22

method	result	size
derivativedivides	\(\frac {d^{2} a \,c^{4} x^{4}}{4}-d^{2} a \,c^{2} x^{2}+d^{2} a \ln \left (c x \right )-\frac {i b \,d^{2} \arcsin \left (c x \right )^{2}}{2}+d^{2} b \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+d^{2} b \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-i d^{2} b \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-i d^{2} b \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )+\frac {d^{2} b \arcsin \left (c x \right ) \cos \left (4 \arcsin \left (c x \right )\right )}{32}-\frac {d^{2} b \sin \left (4 \arcsin \left (c x \right )\right )}{128}+\frac {3 d^{2} b \arcsin \left (c x \right ) \cos \left (2 \arcsin \left (c x \right )\right )}{8}-\frac {3 d^{2} b \sin \left (2 \arcsin \left (c x \right )\right )}{16}\)	\(224\)
default	\(\frac {d^{2} a \,c^{4} x^{4}}{4}-d^{2} a \,c^{2} x^{2}+d^{2} a \ln \left (c x \right )-\frac {i b \,d^{2} \arcsin \left (c x \right )^{2}}{2}+d^{2} b \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+d^{2} b \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-i d^{2} b \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-i d^{2} b \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )+\frac {d^{2} b \arcsin \left (c x \right ) \cos \left (4 \arcsin \left (c x \right )\right )}{32}-\frac {d^{2} b \sin \left (4 \arcsin \left (c x \right )\right )}{128}+\frac {3 d^{2} b \arcsin \left (c x \right ) \cos \left (2 \arcsin \left (c x \right )\right )}{8}-\frac {3 d^{2} b \sin \left (2 \arcsin \left (c x \right )\right )}{16}\)	\(224\)

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,method=_RETURNVERBOSE)

[Out]

1/4*d^2*a*c^4*x^4-d^2*a*c^2*x^2+d^2*a*ln(c*x)-1/2*I*b*d^2*arcsin(c*x)^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))-I*d^2
*b*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))+1/32*d^2*b*arcsin(c*x)*cos(4*arcsin(c*x))-1/128*d^2*b*sin(4*arcsin(c*x)
)+3/8*d^2*b*arcsin(c*x)*cos(2*arcsin(c*x))-3/16*d^2*b*sin(2*arcsin(c*x))

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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((-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 = 0.00, 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((-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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} d^{2} \left (\int \frac {a}{x}\, dx + \int \left (- 2 a c^{2} x\right )\, dx + \int a c^{4} x^{3}\, dx + \int \frac {b \operatorname {asin}{\left (c x \right )}}{x}\, dx + \int \left (- 2 b c^{2} x \operatorname {asin}{\left (c x \right )}\right )\, dx + \int b c^{4} x^{3} \operatorname {asin}{\left (c x \right )}\, dx\right ) \end {gather*}

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 = 0.00, 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((-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)

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

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,x)

[Out]

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

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