3.2.11 \(\int \frac {x^3 (a+b \text {ArcSin}(c x))}{\sqrt {d-c^2 d x^2}} \, dx\) [111]

Optimal. Leaf size=148 \[ \frac {2 b x \sqrt {1-c^2 x^2}}{3 c^3 \sqrt {d-c^2 d x^2}}+\frac {b x^3 \sqrt {1-c^2 x^2}}{9 c \sqrt {d-c^2 d x^2}}-\frac {2 \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))}{3 c^4 d}-\frac {x^2 \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))}{3 c^2 d} \]

[Out]

2/3*b*x*(-c^2*x^2+1)^(1/2)/c^3/(-c^2*d*x^2+d)^(1/2)+1/9*b*x^3*(-c^2*x^2+1)^(1/2)/c/(-c^2*d*x^2+d)^(1/2)-2/3*(a
+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/c^4/d-1/3*x^2*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/c^2/d

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Rubi [A]
time = 0.11, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {4795, 4767, 8, 30} \begin {gather*} -\frac {x^2 \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))}{3 c^2 d}-\frac {2 \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))}{3 c^4 d}+\frac {b x^3 \sqrt {1-c^2 x^2}}{9 c \sqrt {d-c^2 d x^2}}+\frac {2 b x \sqrt {1-c^2 x^2}}{3 c^3 \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*x*Sqrt[1 - c^2*x^2])/(3*c^3*Sqrt[d - c^2*d*x^2]) + (b*x^3*Sqrt[1 - c^2*x^2])/(9*c*Sqrt[d - c^2*d*x^2]) -
(2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(3*c^4*d) - (x^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(3*c^2*d
)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 4767

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

Rule 4795

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[f
*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])^n/(e*(m + 2*p + 1))), x] + (Dist[f^2*((m - 1)/(c^2*(m
+ 2*p + 1))), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] + Dist[b*f*(n/(c*(m + 2*p + 1)))*S
imp[(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, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && IGtQ[m, 1] && NeQ[m + 2*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {x^3 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}} \, dx &=-\frac {x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d}+\frac {2 \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}} \, dx}{3 c^2}+\frac {\left (b \sqrt {1-c^2 x^2}\right ) \int x^2 \, dx}{3 c \sqrt {d-c^2 d x^2}}\\ &=\frac {b x^3 \sqrt {1-c^2 x^2}}{9 c \sqrt {d-c^2 d x^2}}-\frac {2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d}-\frac {x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d}+\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \int 1 \, dx}{3 c^3 \sqrt {d-c^2 d x^2}}\\ &=\frac {2 b x \sqrt {1-c^2 x^2}}{3 c^3 \sqrt {d-c^2 d x^2}}+\frac {b x^3 \sqrt {1-c^2 x^2}}{9 c \sqrt {d-c^2 d x^2}}-\frac {2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^4 d}-\frac {x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^2 d}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 92, normalized size = 0.62 \begin {gather*} \frac {b c x \sqrt {1-c^2 x^2} \left (6+c^2 x^2\right )+3 a \left (-2+c^2 x^2+c^4 x^4\right )+3 b \left (-2+c^2 x^2+c^4 x^4\right ) \text {ArcSin}(c x)}{9 c^4 \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*x*Sqrt[1 - c^2*x^2]*(6 + c^2*x^2) + 3*a*(-2 + c^2*x^2 + c^4*x^4) + 3*b*(-2 + c^2*x^2 + c^4*x^4)*ArcSin[c*
x])/(9*c^4*Sqrt[d - c^2*d*x^2])

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Maple [C] Result contains complex when optimal does not.
time = 0.26, size = 407, normalized size = 2.75

method	result	size
default	\(a \left (-\frac {x^{2} \sqrt {-c^{2} d \,x^{2}+d}}{3 c^{2} d}-\frac {2 \sqrt {-c^{2} d \,x^{2}+d}}{3 d \,c^{4}}\right )+b \left (\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (2 c^{2} x^{2}-2 i \sqrt {-c^{2} x^{2}+1}\, x c -1\right ) \left (i+3 \arcsin \left (c x \right )\right )}{144 c^{4} d \left (c^{2} x^{2}-1\right )}-\frac {3 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (c^{2} x^{2}-i \sqrt {-c^{2} x^{2}+1}\, x c -1\right ) \left (\arcsin \left (c x \right )+i\right )}{8 c^{4} d \left (c^{2} x^{2}-1\right )}-\frac {3 \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) \left (\arcsin \left (c x \right )-i\right )}{8 c^{4} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (2 i \sqrt {-c^{2} x^{2}+1}\, x c +2 c^{2} x^{2}-1\right ) \left (-i+3 \arcsin \left (c x \right )\right )}{144 c^{4} d \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \cos \left (4 \arcsin \left (c x \right )\right )}{24 c^{4} d \left (c^{2} x^{2}-1\right )}+\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sin \left (4 \arcsin \left (c x \right )\right )}{72 c^{4} d \left (c^{2} x^{2}-1\right )}\right )\)	\(407\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(1/2),x,method=_RETURNVERBOSE)

[Out]

a*(-1/3*x^2/c^2/d*(-c^2*d*x^2+d)^(1/2)-2/3/d/c^4*(-c^2*d*x^2+d)^(1/2))+b*(1/144*(-d*(c^2*x^2-1))^(1/2)*(2*c^2*
x^2-2*I*(-c^2*x^2+1)^(1/2)*x*c-1)*(I+3*arcsin(c*x))/c^4/d/(c^2*x^2-1)-3/8*(-d*(c^2*x^2-1))^(1/2)*(c^2*x^2-I*(-
c^2*x^2+1)^(1/2)*x*c-1)*(arcsin(c*x)+I)/c^4/d/(c^2*x^2-1)-3/8*(-d*(c^2*x^2-1))^(1/2)*(I*(-c^2*x^2+1)^(1/2)*x*c
+c^2*x^2-1)*(arcsin(c*x)-I)/c^4/d/(c^2*x^2-1)+1/144*(-d*(c^2*x^2-1))^(1/2)*(2*I*(-c^2*x^2+1)^(1/2)*x*c+2*c^2*x
^2-1)*(-I+3*arcsin(c*x))/c^4/d/(c^2*x^2-1)-1/24*(-d*(c^2*x^2-1))^(1/2)/c^4/d/(c^2*x^2-1)*arcsin(c*x)*cos(4*arc
sin(c*x))+1/72*(-d*(c^2*x^2-1))^(1/2)/c^4/d/(c^2*x^2-1)*sin(4*arcsin(c*x)))

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Maxima [A]
time = 0.49, size = 121, normalized size = 0.82 \begin {gather*} -\frac {1}{3} \, b {\left (\frac {\sqrt {-c^{2} d x^{2} + d} x^{2}}{c^{2} d} + \frac {2 \, \sqrt {-c^{2} d x^{2} + d}}{c^{4} d}\right )} \arcsin \left (c x\right ) - \frac {1}{3} \, a {\left (\frac {\sqrt {-c^{2} d x^{2} + d} x^{2}}{c^{2} d} + \frac {2 \, \sqrt {-c^{2} d x^{2} + d}}{c^{4} d}\right )} + \frac {{\left (c^{2} x^{3} + 6 \, x\right )} b}{9 \, c^{3} \sqrt {d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

-1/3*b*(sqrt(-c^2*d*x^2 + d)*x^2/(c^2*d) + 2*sqrt(-c^2*d*x^2 + d)/(c^4*d))*arcsin(c*x) - 1/3*a*(sqrt(-c^2*d*x^
2 + d)*x^2/(c^2*d) + 2*sqrt(-c^2*d*x^2 + d)/(c^4*d)) + 1/9*(c^2*x^3 + 6*x)*b/(c^3*sqrt(d))

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Fricas [A]
time = 1.70, size = 120, normalized size = 0.81 \begin {gather*} -\frac {{\left (b c^{3} x^{3} + 6 \, b c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {-c^{2} x^{2} + 1} + 3 \, {\left (a c^{4} x^{4} + a c^{2} x^{2} + {\left (b c^{4} x^{4} + b c^{2} x^{2} - 2 \, b\right )} \arcsin \left (c x\right ) - 2 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{9 \, {\left (c^{6} d x^{2} - c^{4} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

-1/9*((b*c^3*x^3 + 6*b*c*x)*sqrt(-c^2*d*x^2 + d)*sqrt(-c^2*x^2 + 1) + 3*(a*c^4*x^4 + a*c^2*x^2 + (b*c^4*x^4 +
b*c^2*x^2 - 2*b)*arcsin(c*x) - 2*a)*sqrt(-c^2*d*x^2 + d))/(c^6*d*x^2 - c^4*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*asin(c*x))/(-c**2*d*x**2+d)**(1/2),x)

[Out]

Integral(x**3*(a + b*asin(c*x))/sqrt(-d*(c*x - 1)*(c*x + 1)), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{\sqrt {d-c^2\,d\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*asin(c*x)))/(d - c^2*d*x^2)^(1/2),x)

[Out]

int((x^3*(a + b*asin(c*x)))/(d - c^2*d*x^2)^(1/2), x)

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