\(\int x^3 (a+b \arcsin (c x)) \, dx\) [140]

Optimal result

Integrand size = 12, antiderivative size = 76 \[ \int x^3 (a+b \arcsin (c x)) \, dx=\frac {3 b x \sqrt {1-c^2 x^2}}{32 c^3}+\frac {b x^3 \sqrt {1-c^2 x^2}}{16 c}-\frac {3 b \arcsin (c x)}{32 c^4}+\frac {1}{4} x^4 (a+b \arcsin (c x)) \]

[Out]

-3/32*b*arcsin(c*x)/c^4+1/4*x^4*(a+b*arcsin(c*x))+3/32*b*x*(-c^2*x^2+1)^(1/2)/c^3+1/16*b*x^3*(-c^2*x^2+1)^(1/2
)/c

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4723, 327, 222} \[ \int x^3 (a+b \arcsin (c x)) \, dx=\frac {1}{4} x^4 (a+b \arcsin (c x))-\frac {3 b \arcsin (c x)}{32 c^4}+\frac {b x^3 \sqrt {1-c^2 x^2}}{16 c}+\frac {3 b x \sqrt {1-c^2 x^2}}{32 c^3} \]

[In]

Int[x^3*(a + b*ArcSin[c*x]),x]

[Out]

(3*b*x*Sqrt[1 - c^2*x^2])/(32*c^3) + (b*x^3*Sqrt[1 - c^2*x^2])/(16*c) - (3*b*ArcSin[c*x])/(32*c^4) + (x^4*(a +
b*ArcSin[c*x]))/4

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 327

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

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} x^4 (a+b \arcsin (c x))-\frac {1}{4} (b c) \int \frac {x^4}{\sqrt {1-c^2 x^2}} \, dx \\ & = \frac {b x^3 \sqrt {1-c^2 x^2}}{16 c}+\frac {1}{4} x^4 (a+b \arcsin (c x))-\frac {(3 b) \int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx}{16 c} \\ & = \frac {3 b x \sqrt {1-c^2 x^2}}{32 c^3}+\frac {b x^3 \sqrt {1-c^2 x^2}}{16 c}+\frac {1}{4} x^4 (a+b \arcsin (c x))-\frac {(3 b) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{32 c^3} \\ & = \frac {3 b x \sqrt {1-c^2 x^2}}{32 c^3}+\frac {b x^3 \sqrt {1-c^2 x^2}}{16 c}-\frac {3 b \arcsin (c x)}{32 c^4}+\frac {1}{4} x^4 (a+b \arcsin (c x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.07 \[ \int x^3 (a+b \arcsin (c x)) \, dx=\frac {a x^4}{4}+\frac {3 b x \sqrt {1-c^2 x^2}}{32 c^3}+\frac {b x^3 \sqrt {1-c^2 x^2}}{16 c}-\frac {3 b \arcsin (c x)}{32 c^4}+\frac {1}{4} b x^4 \arcsin (c x) \]

[In]

Integrate[x^3*(a + b*ArcSin[c*x]),x]

[Out]

(a*x^4)/4 + (3*b*x*Sqrt[1 - c^2*x^2])/(32*c^3) + (b*x^3*Sqrt[1 - c^2*x^2])/(16*c) - (3*b*ArcSin[c*x])/(32*c^4)
+ (b*x^4*ArcSin[c*x])/4

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.89

method	result	size
parts	\(\frac {a \,x^{4}}{4}+\frac {b \left (\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}+\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{16}+\frac {3 c x \sqrt {-c^{2} x^{2}+1}}{32}-\frac {3 \arcsin \left (c x \right )}{32}\right )}{c^{4}}\)	\(68\)
derivativedivides	\(\frac {\frac {a \,c^{4} x^{4}}{4}+b \left (\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}+\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{16}+\frac {3 c x \sqrt {-c^{2} x^{2}+1}}{32}-\frac {3 \arcsin \left (c x \right )}{32}\right )}{c^{4}}\)	\(72\)
default	\(\frac {\frac {a \,c^{4} x^{4}}{4}+b \left (\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}+\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{16}+\frac {3 c x \sqrt {-c^{2} x^{2}+1}}{32}-\frac {3 \arcsin \left (c x \right )}{32}\right )}{c^{4}}\)	\(72\)

[In]

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

[Out]

1/4*a*x^4+b/c^4*(1/4*c^4*x^4*arcsin(c*x)+1/16*c^3*x^3*(-c^2*x^2+1)^(1/2)+3/32*c*x*(-c^2*x^2+1)^(1/2)-3/32*arcs
in(c*x))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.80 \[ \int x^3 (a+b \arcsin (c x)) \, dx=\frac {8 \, a c^{4} x^{4} + {\left (8 \, b c^{4} x^{4} - 3 \, b\right )} \arcsin \left (c x\right ) + {\left (2 \, b c^{3} x^{3} + 3 \, b c x\right )} \sqrt {-c^{2} x^{2} + 1}}{32 \, c^{4}} \]

[In]

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

[Out]

1/32*(8*a*c^4*x^4 + (8*b*c^4*x^4 - 3*b)*arcsin(c*x) + (2*b*c^3*x^3 + 3*b*c*x)*sqrt(-c^2*x^2 + 1))/c^4

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.05 \[ \int x^3 (a+b \arcsin (c x)) \, dx=\begin {cases} \frac {a x^{4}}{4} + \frac {b x^{4} \operatorname {asin}{\left (c x \right )}}{4} + \frac {b x^{3} \sqrt {- c^{2} x^{2} + 1}}{16 c} + \frac {3 b x \sqrt {- c^{2} x^{2} + 1}}{32 c^{3}} - \frac {3 b \operatorname {asin}{\left (c x \right )}}{32 c^{4}} & \text {for}\: c \neq 0 \\\frac {a x^{4}}{4} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a*x**4/4 + b*x**4*asin(c*x)/4 + b*x**3*sqrt(-c**2*x**2 + 1)/(16*c) + 3*b*x*sqrt(-c**2*x**2 + 1)/(32
*c**3) - 3*b*asin(c*x)/(32*c**4), Ne(c, 0)), (a*x**4/4, True))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.92 \[ \int x^3 (a+b \arcsin (c x)) \, dx=\frac {1}{4} \, a x^{4} + \frac {1}{32} \, {\left (8 \, x^{4} \arcsin \left (c x\right ) + {\left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1} x^{3}}{c^{2}} + \frac {3 \, \sqrt {-c^{2} x^{2} + 1} x}{c^{4}} - \frac {3 \, \arcsin \left (c x\right )}{c^{5}}\right )} c\right )} b \]

[In]

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

[Out]

1/4*a*x^4 + 1/32*(8*x^4*arcsin(c*x) + (2*sqrt(-c^2*x^2 + 1)*x^3/c^2 + 3*sqrt(-c^2*x^2 + 1)*x/c^4 - 3*arcsin(c*
x)/c^5)*c)*b

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.25 \[ \int x^3 (a+b \arcsin (c x)) \, dx=\frac {1}{4} \, a x^{4} - \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b x}{16 \, c^{3}} + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b \arcsin \left (c x\right )}{4 \, c^{4}} + \frac {5 \, \sqrt {-c^{2} x^{2} + 1} b x}{32 \, c^{3}} + \frac {{\left (c^{2} x^{2} - 1\right )} b \arcsin \left (c x\right )}{2 \, c^{4}} + \frac {5 \, b \arcsin \left (c x\right )}{32 \, c^{4}} \]

[In]

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

[Out]

1/4*a*x^4 - 1/16*(-c^2*x^2 + 1)^(3/2)*b*x/c^3 + 1/4*(c^2*x^2 - 1)^2*b*arcsin(c*x)/c^4 + 5/32*sqrt(-c^2*x^2 + 1
)*b*x/c^3 + 1/2*(c^2*x^2 - 1)*b*arcsin(c*x)/c^4 + 5/32*b*arcsin(c*x)/c^4

Mupad [F(-1)]

Timed out. \[ \int x^3 (a+b \arcsin (c x)) \, dx=\int x^3\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right ) \,d x \]

[In]

int(x^3*(a + b*asin(c*x)),x)

[Out]

int(x^3*(a + b*asin(c*x)), x)