3.3.4 \(\int \frac {(a+b \text {ArcSin}(c+d x))^3}{(c e+d e x)^3} \, dx\) [204]

Optimal. Leaf size=167 \[ -\frac {3 i b (a+b \text {ArcSin}(c+d x))^2}{2 d e^3}-\frac {3 b \sqrt {1-(c+d x)^2} (a+b \text {ArcSin}(c+d x))^2}{2 d e^3 (c+d x)}-\frac {(a+b \text {ArcSin}(c+d x))^3}{2 d e^3 (c+d x)^2}+\frac {3 b^2 (a+b \text {ArcSin}(c+d x)) \log \left (1-e^{2 i \text {ArcSin}(c+d x)}\right )}{d e^3}-\frac {3 i b^3 \text {PolyLog}\left (2,e^{2 i \text {ArcSin}(c+d x)}\right )}{2 d e^3} \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {4889, 12, 4723, 4771, 4721, 3798, 2221, 2317, 2438} \begin {gather*} \frac {3 b^2 \log \left (1-e^{2 i \text {ArcSin}(c+d x)}\right ) (a+b \text {ArcSin}(c+d x))}{d e^3}-\frac {3 b \sqrt {1-(c+d x)^2} (a+b \text {ArcSin}(c+d x))^2}{2 d e^3 (c+d x)}-\frac {3 i b (a+b \text {ArcSin}(c+d x))^2}{2 d e^3}-\frac {(a+b \text {ArcSin}(c+d x))^3}{2 d e^3 (c+d x)^2}-\frac {3 i b^3 \text {Li}_2\left (e^{2 i \text {ArcSin}(c+d x)}\right )}{2 d e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 4723

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

Rule 4771

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[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] /;
FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]

Rule 4889

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + f*(x/d))^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rubi steps

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

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Mathematica [A]
time = 0.51, size = 248, normalized size = 1.49 \begin {gather*} -\frac {3 b^2 \left (a+b (c+d x) \left (i c+i d x+\sqrt {1-c^2-2 c d x-d^2 x^2}\right )\right ) \text {ArcSin}(c+d x)^2+b^3 \text {ArcSin}(c+d x)^3+3 b \text {ArcSin}(c+d x) \left (a \left (a+2 b (c+d x) \sqrt {1-c^2-2 c d x-d^2 x^2}\right )-2 b^2 (c+d x)^2 \log \left (1-e^{2 i \text {ArcSin}(c+d x)}\right )\right )+a \left (a \left (a+3 b (c+d x) \sqrt {1-c^2-2 c d x-d^2 x^2}\right )-6 b^2 (c+d x)^2 \log (c+d x)\right )+3 i b^3 (c+d x)^2 \text {PolyLog}\left (2,e^{2 i \text {ArcSin}(c+d x)}\right )}{2 d e^3 (c+d x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 363 vs. \(2 (179 ) = 358\).
time = 0.38, size = 364, normalized size = 2.18

method	result	size
derivativedivides	\(\frac {-\frac {a^{3}}{2 e^{3} \left (d x +c \right )^{2}}-\frac {3 i b^{3} \arcsin \left (d x +c \right )^{2}}{2 e^{3}}-\frac {3 b^{3} \arcsin \left (d x +c \right )^{2} \sqrt {1-\left (d x +c \right )^{2}}}{2 e^{3} \left (d x +c \right )}-\frac {b^{3} \arcsin \left (d x +c \right )^{3}}{2 e^{3} \left (d x +c \right )^{2}}+\frac {3 b^{3} \arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}+\frac {3 b^{3} \arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {3 i b^{3} \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {3 i b^{3} \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {3 a \,b^{2} \arcsin \left (d x +c \right )^{2}}{2 e^{3} \left (d x +c \right )^{2}}-\frac {3 a \,b^{2} \arcsin \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{e^{3} \left (d x +c \right )}+\frac {3 a \,b^{2} \ln \left (d x +c \right )}{e^{3}}+\frac {3 a^{2} b \left (-\frac {\arcsin \left (d x +c \right )}{2 \left (d x +c \right )^{2}}-\frac {\sqrt {1-\left (d x +c \right )^{2}}}{2 \left (d x +c \right )}\right )}{e^{3}}}{d}\)	\(364\)
default	\(\frac {-\frac {a^{3}}{2 e^{3} \left (d x +c \right )^{2}}-\frac {3 i b^{3} \arcsin \left (d x +c \right )^{2}}{2 e^{3}}-\frac {3 b^{3} \arcsin \left (d x +c \right )^{2} \sqrt {1-\left (d x +c \right )^{2}}}{2 e^{3} \left (d x +c \right )}-\frac {b^{3} \arcsin \left (d x +c \right )^{3}}{2 e^{3} \left (d x +c \right )^{2}}+\frac {3 b^{3} \arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}+\frac {3 b^{3} \arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {3 i b^{3} \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {3 i b^{3} \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {3 a \,b^{2} \arcsin \left (d x +c \right )^{2}}{2 e^{3} \left (d x +c \right )^{2}}-\frac {3 a \,b^{2} \arcsin \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{e^{3} \left (d x +c \right )}+\frac {3 a \,b^{2} \ln \left (d x +c \right )}{e^{3}}+\frac {3 a^{2} b \left (-\frac {\arcsin \left (d x +c \right )}{2 \left (d x +c \right )^{2}}-\frac {\sqrt {1-\left (d x +c \right )^{2}}}{2 \left (d x +c \right )}\right )}{e^{3}}}{d}\)	\(364\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(d*x+c))^3/(d*e*x+c*e)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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(d*x+c))^3/(d*e*x+c*e)^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{3}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {3 a b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {3 a^{2} b \operatorname {asin}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(d*x+c))^3/(d*e*x+c*e)^3,x, algorithm="giac")

[Out]

integrate((b*arcsin(d*x + c) + a)^3/(d*e*x + c*e)^3, 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+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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