3.2.60 \(\int (d-c^2 d x^2) (a+b \text {ArcSin}(c x))^2 \, dx\) [160]

Optimal. Leaf size=128 \[ -\frac {14}{9} b^2 d x+\frac {2}{27} b^2 c^2 d x^3+\frac {4 b d \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{3 c}+\frac {2 b d \left (1-c^2 x^2\right )^{3/2} (a+b \text {ArcSin}(c x))}{9 c}+\frac {2}{3} d x (a+b \text {ArcSin}(c x))^2+\frac {1}{3} d x \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))^2 \]

[Out]

-14/9*b^2*d*x+2/27*b^2*c^2*d*x^3+2/9*b*d*(-c^2*x^2+1)^(3/2)*(a+b*arcsin(c*x))/c+2/3*d*x*(a+b*arcsin(c*x))^2+1/
3*d*x*(-c^2*x^2+1)*(a+b*arcsin(c*x))^2+4/3*b*d*(a+b*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/c

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Rubi [A]
time = 0.10, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4743, 4715, 4767, 8} \begin {gather*} \frac {1}{3} d x \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))^2+\frac {2 b d \left (1-c^2 x^2\right )^{3/2} (a+b \text {ArcSin}(c x))}{9 c}+\frac {4 b d \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{3 c}+\frac {2}{3} d x (a+b \text {ArcSin}(c x))^2+\frac {2}{27} b^2 c^2 d x^3-\frac {14}{9} b^2 d x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-14*b^2*d*x)/9 + (2*b^2*c^2*d*x^3)/27 + (4*b*d*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(3*c) + (2*b*d*(1 - c^2
*x^2)^(3/2)*(a + b*ArcSin[c*x]))/(9*c) + (2*d*x*(a + b*ArcSin[c*x])^2)/3 + (d*x*(1 - c^2*x^2)*(a + b*ArcSin[c*
x])^2)/3

Rule 8

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

Rule 4715

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

Rule 4743

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

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]

Rubi steps

\begin {align*} \int \left (d-c^2 d x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2 \, dx &=\frac {1}{3} d x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{3} (2 d) \int \left (a+b \sin ^{-1}(c x)\right )^2 \, dx-\frac {1}{3} (2 b c d) \int x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx\\ &=\frac {2 b d \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{9 c}+\frac {2}{3} d x \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{3} d x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2-\frac {1}{9} \left (2 b^2 d\right ) \int \left (1-c^2 x^2\right ) \, dx-\frac {1}{3} (4 b c d) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {2}{9} b^2 d x+\frac {2}{27} b^2 c^2 d x^3+\frac {4 b d \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c}+\frac {2 b d \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{9 c}+\frac {2}{3} d x \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{3} d x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2-\frac {1}{3} \left (4 b^2 d\right ) \int 1 \, dx\\ &=-\frac {14}{9} b^2 d x+\frac {2}{27} b^2 c^2 d x^3+\frac {4 b d \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c}+\frac {2 b d \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{9 c}+\frac {2}{3} d x \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{3} d x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 137, normalized size = 1.07 \begin {gather*} -\frac {d \left (-2 b^2 c x \left (-21+c^2 x^2\right )+6 a b \sqrt {1-c^2 x^2} \left (-7+c^2 x^2\right )+9 a^2 c x \left (-3+c^2 x^2\right )+6 b \left (b \sqrt {1-c^2 x^2} \left (-7+c^2 x^2\right )+3 a c x \left (-3+c^2 x^2\right )\right ) \text {ArcSin}(c x)+9 b^2 c x \left (-3+c^2 x^2\right ) \text {ArcSin}(c x)^2\right )}{27 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/27*(d*(-2*b^2*c*x*(-21 + c^2*x^2) + 6*a*b*Sqrt[1 - c^2*x^2]*(-7 + c^2*x^2) + 9*a^2*c*x*(-3 + c^2*x^2) + 6*b
*(b*Sqrt[1 - c^2*x^2]*(-7 + c^2*x^2) + 3*a*c*x*(-3 + c^2*x^2))*ArcSin[c*x] + 9*b^2*c*x*(-3 + c^2*x^2)*ArcSin[c
*x]^2))/c

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Maple [A]
time = 0.04, size = 173, normalized size = 1.35

method	result	size
derivativedivides	\(\frac {-d \,a^{2} \left (\frac {1}{3} c^{3} x^{3}-c x \right )-d \,b^{2} \left (\frac {\arcsin \left (c x \right )^{2} \left (c^{2} x^{2}-3\right ) c x}{3}+\frac {4 c x}{3}-\frac {4 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{3}+\frac {2 \arcsin \left (c x \right ) \left (c^{2} x^{2}-1\right ) \sqrt {-c^{2} x^{2}+1}}{9}-\frac {2 \left (c^{2} x^{2}-3\right ) c x}{27}\right )-2 d a b \left (\frac {c^{3} x^{3} \arcsin \left (c x \right )}{3}-c x \arcsin \left (c x \right )+\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{9}-\frac {7 \sqrt {-c^{2} x^{2}+1}}{9}\right )}{c}\)	\(173\)
default	\(\frac {-d \,a^{2} \left (\frac {1}{3} c^{3} x^{3}-c x \right )-d \,b^{2} \left (\frac {\arcsin \left (c x \right )^{2} \left (c^{2} x^{2}-3\right ) c x}{3}+\frac {4 c x}{3}-\frac {4 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{3}+\frac {2 \arcsin \left (c x \right ) \left (c^{2} x^{2}-1\right ) \sqrt {-c^{2} x^{2}+1}}{9}-\frac {2 \left (c^{2} x^{2}-3\right ) c x}{27}\right )-2 d a b \left (\frac {c^{3} x^{3} \arcsin \left (c x \right )}{3}-c x \arcsin \left (c x \right )+\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{9}-\frac {7 \sqrt {-c^{2} x^{2}+1}}{9}\right )}{c}\)	\(173\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(-d*a^2*(1/3*c^3*x^3-c*x)-d*b^2*(1/3*arcsin(c*x)^2*(c^2*x^2-3)*c*x+4/3*c*x-4/3*arcsin(c*x)*(-c^2*x^2+1)^(1
/2)+2/9*arcsin(c*x)*(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)-2/27*(c^2*x^2-3)*c*x)-2*d*a*b*(1/3*c^3*x^3*arcsin(c*x)-c*x*
arcsin(c*x)+1/9*c^2*x^2*(-c^2*x^2+1)^(1/2)-7/9*(-c^2*x^2+1)^(1/2)))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (111) = 222\).
time = 0.49, size = 233, normalized size = 1.82 \begin {gather*} -\frac {1}{3} \, b^{2} c^{2} d x^{3} \arcsin \left (c x\right )^{2} - \frac {1}{3} \, a^{2} c^{2} d x^{3} - \frac {2}{9} \, {\left (3 \, x^{3} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} a b c^{2} d - \frac {2}{27} \, {\left (3 \, c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )} \arcsin \left (c x\right ) - \frac {c^{2} x^{3} + 6 \, x}{c^{2}}\right )} b^{2} c^{2} d + b^{2} d x \arcsin \left (c x\right )^{2} - 2 \, b^{2} d {\left (x - \frac {\sqrt {-c^{2} x^{2} + 1} \arcsin \left (c x\right )}{c}\right )} + a^{2} d x + \frac {2 \, {\left (c x \arcsin \left (c x\right ) + \sqrt {-c^{2} x^{2} + 1}\right )} a b d}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*b^2*c^2*d*x^3*arcsin(c*x)^2 - 1/3*a^2*c^2*d*x^3 - 2/9*(3*x^3*arcsin(c*x) + c*(sqrt(-c^2*x^2 + 1)*x^2/c^2
+ 2*sqrt(-c^2*x^2 + 1)/c^4))*a*b*c^2*d - 2/27*(3*c*(sqrt(-c^2*x^2 + 1)*x^2/c^2 + 2*sqrt(-c^2*x^2 + 1)/c^4)*arc
sin(c*x) - (c^2*x^3 + 6*x)/c^2)*b^2*c^2*d + b^2*d*x*arcsin(c*x)^2 - 2*b^2*d*(x - sqrt(-c^2*x^2 + 1)*arcsin(c*x
)/c) + a^2*d*x + 2*(c*x*arcsin(c*x) + sqrt(-c^2*x^2 + 1))*a*b*d/c

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Fricas [A]
time = 3.99, size = 146, normalized size = 1.14 \begin {gather*} -\frac {{\left (9 \, a^{2} - 2 \, b^{2}\right )} c^{3} d x^{3} - 3 \, {\left (9 \, a^{2} - 14 \, b^{2}\right )} c d x + 9 \, {\left (b^{2} c^{3} d x^{3} - 3 \, b^{2} c d x\right )} \arcsin \left (c x\right )^{2} + 18 \, {\left (a b c^{3} d x^{3} - 3 \, a b c d x\right )} \arcsin \left (c x\right ) + 6 \, {\left (a b c^{2} d x^{2} - 7 \, a b d + {\left (b^{2} c^{2} d x^{2} - 7 \, b^{2} d\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} x^{2} + 1}}{27 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/27*((9*a^2 - 2*b^2)*c^3*d*x^3 - 3*(9*a^2 - 14*b^2)*c*d*x + 9*(b^2*c^3*d*x^3 - 3*b^2*c*d*x)*arcsin(c*x)^2 +
18*(a*b*c^3*d*x^3 - 3*a*b*c*d*x)*arcsin(c*x) + 6*(a*b*c^2*d*x^2 - 7*a*b*d + (b^2*c^2*d*x^2 - 7*b^2*d)*arcsin(c
*x))*sqrt(-c^2*x^2 + 1))/c

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Sympy [A]
time = 0.31, size = 224, normalized size = 1.75 \begin {gather*} \begin {cases} - \frac {a^{2} c^{2} d x^{3}}{3} + a^{2} d x - \frac {2 a b c^{2} d x^{3} \operatorname {asin}{\left (c x \right )}}{3} - \frac {2 a b c d x^{2} \sqrt {- c^{2} x^{2} + 1}}{9} + 2 a b d x \operatorname {asin}{\left (c x \right )} + \frac {14 a b d \sqrt {- c^{2} x^{2} + 1}}{9 c} - \frac {b^{2} c^{2} d x^{3} \operatorname {asin}^{2}{\left (c x \right )}}{3} + \frac {2 b^{2} c^{2} d x^{3}}{27} - \frac {2 b^{2} c d x^{2} \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{9} + b^{2} d x \operatorname {asin}^{2}{\left (c x \right )} - \frac {14 b^{2} d x}{9} + \frac {14 b^{2} d \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{9 c} & \text {for}\: c \neq 0 \\a^{2} d x & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*c**2*d*x**3/3 + a**2*d*x - 2*a*b*c**2*d*x**3*asin(c*x)/3 - 2*a*b*c*d*x**2*sqrt(-c**2*x**2 + 1
)/9 + 2*a*b*d*x*asin(c*x) + 14*a*b*d*sqrt(-c**2*x**2 + 1)/(9*c) - b**2*c**2*d*x**3*asin(c*x)**2/3 + 2*b**2*c**
2*d*x**3/27 - 2*b**2*c*d*x**2*sqrt(-c**2*x**2 + 1)*asin(c*x)/9 + b**2*d*x*asin(c*x)**2 - 14*b**2*d*x/9 + 14*b*
*2*d*sqrt(-c**2*x**2 + 1)*asin(c*x)/(9*c), Ne(c, 0)), (a**2*d*x, True))

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Giac [A]
time = 0.45, size = 196, normalized size = 1.53 \begin {gather*} -\frac {1}{3} \, a^{2} c^{2} d x^{3} - \frac {1}{3} \, {\left (c^{2} x^{2} - 1\right )} b^{2} d x \arcsin \left (c x\right )^{2} - \frac {2}{3} \, {\left (c^{2} x^{2} - 1\right )} a b d x \arcsin \left (c x\right ) + \frac {2}{3} \, b^{2} d x \arcsin \left (c x\right )^{2} + \frac {2}{27} \, {\left (c^{2} x^{2} - 1\right )} b^{2} d x + \frac {4}{3} \, a b d x \arcsin \left (c x\right ) + \frac {2 \, {\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b^{2} d \arcsin \left (c x\right )}{9 \, c} + a^{2} d x - \frac {40}{27} \, b^{2} d x + \frac {2 \, {\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} a b d}{9 \, c} + \frac {4 \, \sqrt {-c^{2} x^{2} + 1} b^{2} d \arcsin \left (c x\right )}{3 \, c} + \frac {4 \, \sqrt {-c^{2} x^{2} + 1} a b d}{3 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a^2*c^2*d*x^3 - 1/3*(c^2*x^2 - 1)*b^2*d*x*arcsin(c*x)^2 - 2/3*(c^2*x^2 - 1)*a*b*d*x*arcsin(c*x) + 2/3*b^2
*d*x*arcsin(c*x)^2 + 2/27*(c^2*x^2 - 1)*b^2*d*x + 4/3*a*b*d*x*arcsin(c*x) + 2/9*(-c^2*x^2 + 1)^(3/2)*b^2*d*arc
sin(c*x)/c + a^2*d*x - 40/27*b^2*d*x + 2/9*(-c^2*x^2 + 1)^(3/2)*a*b*d/c + 4/3*sqrt(-c^2*x^2 + 1)*b^2*d*arcsin(
c*x)/c + 4/3*sqrt(-c^2*x^2 + 1)*a*b*d/c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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