3.7.61 \(\int (d+e x^2) (a+b \text {ArcSin}(c x))^2 \, dx\) [661]

Optimal. Leaf size=156 \[ -2 b^2 d x-\frac {4 b^2 e x}{9 c^2}-\frac {2}{27} b^2 e x^3+\frac {2 b d \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{c}+\frac {4 b e \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{9 c^3}+\frac {2 b e x^2 \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{9 c}+d x (a+b \text {ArcSin}(c x))^2+\frac {1}{3} e x^3 (a+b \text {ArcSin}(c x))^2 \]

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {4757, 4715, 4767, 8, 4723, 4795, 30} \begin {gather*} \frac {2 b d \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{c}+\frac {2 b e x^2 \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{9 c}+\frac {4 b e \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{9 c^3}+d x (a+b \text {ArcSin}(c x))^2+\frac {1}{3} e x^3 (a+b \text {ArcSin}(c x))^2-\frac {4 b^2 e x}{9 c^2}-2 b^2 d x-\frac {2}{27} b^2 e x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 4757

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

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

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Mathematica [A]
time = 0.13, size = 166, normalized size = 1.06 \begin {gather*} \frac {9 a^2 c^3 x \left (3 d+e x^2\right )+6 a b \sqrt {1-c^2 x^2} \left (2 e+c^2 \left (9 d+e x^2\right )\right )-2 b^2 c x \left (6 e+c^2 \left (27 d+e x^2\right )\right )+6 b \left (3 a c^3 x \left (3 d+e x^2\right )+b \sqrt {1-c^2 x^2} \left (2 e+c^2 \left (9 d+e x^2\right )\right )\right ) \text {ArcSin}(c x)+9 b^2 c^3 x \left (3 d+e x^2\right ) \text {ArcSin}(c x)^2}{27 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.09, size = 276, normalized size = 1.77

method	result	size
derivativedivides	\(\frac {\frac {a^{2} \left (d \,c^{3} x +\frac {1}{3} e \,c^{3} x^{3}\right )}{c^{2}}+\frac {b^{2} \left (\frac {e \left (9 c^{3} x^{3} \arcsin \left (c x \right )^{2}+6 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}-27 c x \arcsin \left (c x \right )^{2}-2 c^{3} x^{3}-42 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}+42 c x \right )}{27}+d \,c^{2} \left (c x \arcsin \left (c x \right )^{2}-2 c x +2 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\right )+e \left (c x \arcsin \left (c x \right )^{2}-2 c x +2 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{2}}+\frac {2 a b \left (\arcsin \left (c x \right ) d \,c^{3} x +\frac {\arcsin \left (c x \right ) e \,c^{3} x^{3}}{3}+d \,c^{2} \sqrt {-c^{2} x^{2}+1}-\frac {e \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{3}\right )}{c^{2}}}{c}\)	\(276\)
default	\(\frac {\frac {a^{2} \left (d \,c^{3} x +\frac {1}{3} e \,c^{3} x^{3}\right )}{c^{2}}+\frac {b^{2} \left (\frac {e \left (9 c^{3} x^{3} \arcsin \left (c x \right )^{2}+6 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}-27 c x \arcsin \left (c x \right )^{2}-2 c^{3} x^{3}-42 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}+42 c x \right )}{27}+d \,c^{2} \left (c x \arcsin \left (c x \right )^{2}-2 c x +2 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\right )+e \left (c x \arcsin \left (c x \right )^{2}-2 c x +2 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\right )\right )}{c^{2}}+\frac {2 a b \left (\arcsin \left (c x \right ) d \,c^{3} x +\frac {\arcsin \left (c x \right ) e \,c^{3} x^{3}}{3}+d \,c^{2} \sqrt {-c^{2} x^{2}+1}-\frac {e \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{3}\right )}{c^{2}}}{c}\)	\(276\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(a^2/c^2*(d*c^3*x+1/3*e*c^3*x^3)+b^2/c^2*(1/27*e*(9*c^3*x^3*arcsin(c*x)^2+6*arcsin(c*x)*(-c^2*x^2+1)^(1/2)
*c^2*x^2-27*c*x*arcsin(c*x)^2-2*c^3*x^3-42*arcsin(c*x)*(-c^2*x^2+1)^(1/2)+42*c*x)+d*c^2*(c*x*arcsin(c*x)^2-2*c
*x+2*arcsin(c*x)*(-c^2*x^2+1)^(1/2))+e*(c*x*arcsin(c*x)^2-2*c*x+2*arcsin(c*x)*(-c^2*x^2+1)^(1/2)))+2*a*b/c^2*(
arcsin(c*x)*d*c^3*x+1/3*arcsin(c*x)*e*c^3*x^3+d*c^2*(-c^2*x^2+1)^(1/2)-1/3*e*(-1/3*c^2*x^2*(-c^2*x^2+1)^(1/2)-
2/3*(-c^2*x^2+1)^(1/2))))

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Maxima [A]
time = 0.56, size = 225, normalized size = 1.44 \begin {gather*} \frac {1}{3} \, b^{2} x^{3} \arcsin \left (c x\right )^{2} e + b^{2} d x \arcsin \left (c x\right )^{2} + \frac {1}{3} \, a^{2} x^{3} e - 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}{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 e + \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} e + \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((e*x^2+d)*(a+b*arcsin(c*x))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.67, size = 183, normalized size = 1.17 \begin {gather*} \frac {27 \, {\left (a^{2} - 2 \, b^{2}\right )} c^{3} d x + 9 \, {\left (b^{2} c^{3} x^{3} e + 3 \, b^{2} c^{3} d x\right )} \arcsin \left (c x\right )^{2} + 18 \, {\left (a b c^{3} x^{3} e + 3 \, a b c^{3} d x\right )} \arcsin \left (c x\right ) + {\left ({\left (9 \, a^{2} - 2 \, b^{2}\right )} c^{3} x^{3} - 12 \, b^{2} c x\right )} e + 6 \, {\left (9 \, a b c^{2} d + {\left (9 \, b^{2} c^{2} d + {\left (b^{2} c^{2} x^{2} + 2 \, b^{2}\right )} e\right )} \arcsin \left (c x\right ) + {\left (a b c^{2} x^{2} + 2 \, a b\right )} e\right )} \sqrt {-c^{2} x^{2} + 1}}{27 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/27*(27*(a^2 - 2*b^2)*c^3*d*x + 9*(b^2*c^3*x^3*e + 3*b^2*c^3*d*x)*arcsin(c*x)^2 + 18*(a*b*c^3*x^3*e + 3*a*b*c
^3*d*x)*arcsin(c*x) + ((9*a^2 - 2*b^2)*c^3*x^3 - 12*b^2*c*x)*e + 6*(9*a*b*c^2*d + (9*b^2*c^2*d + (b^2*c^2*x^2
+ 2*b^2)*e)*arcsin(c*x) + (a*b*c^2*x^2 + 2*a*b)*e)*sqrt(-c^2*x^2 + 1))/c^3

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Sympy [A]
time = 0.36, size = 279, normalized size = 1.79 \begin {gather*} \begin {cases} a^{2} d x + \frac {a^{2} e x^{3}}{3} + 2 a b d x \operatorname {asin}{\left (c x \right )} + \frac {2 a b e x^{3} \operatorname {asin}{\left (c x \right )}}{3} + \frac {2 a b d \sqrt {- c^{2} x^{2} + 1}}{c} + \frac {2 a b e x^{2} \sqrt {- c^{2} x^{2} + 1}}{9 c} + \frac {4 a b e \sqrt {- c^{2} x^{2} + 1}}{9 c^{3}} + b^{2} d x \operatorname {asin}^{2}{\left (c x \right )} - 2 b^{2} d x + \frac {b^{2} e x^{3} \operatorname {asin}^{2}{\left (c x \right )}}{3} - \frac {2 b^{2} e x^{3}}{27} + \frac {2 b^{2} d \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{c} + \frac {2 b^{2} e x^{2} \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{9 c} - \frac {4 b^{2} e x}{9 c^{2}} + \frac {4 b^{2} e \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{9 c^{3}} & \text {for}\: c \neq 0 \\a^{2} \left (d x + \frac {e x^{3}}{3}\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*d*x + a**2*e*x**3/3 + 2*a*b*d*x*asin(c*x) + 2*a*b*e*x**3*asin(c*x)/3 + 2*a*b*d*sqrt(-c**2*x**2
+ 1)/c + 2*a*b*e*x**2*sqrt(-c**2*x**2 + 1)/(9*c) + 4*a*b*e*sqrt(-c**2*x**2 + 1)/(9*c**3) + b**2*d*x*asin(c*x)
**2 - 2*b**2*d*x + b**2*e*x**3*asin(c*x)**2/3 - 2*b**2*e*x**3/27 + 2*b**2*d*sqrt(-c**2*x**2 + 1)*asin(c*x)/c +
2*b**2*e*x**2*sqrt(-c**2*x**2 + 1)*asin(c*x)/(9*c) - 4*b**2*e*x/(9*c**2) + 4*b**2*e*sqrt(-c**2*x**2 + 1)*asin
(c*x)/(9*c**3), Ne(c, 0)), (a**2*(d*x + e*x**3/3), True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 285 vs. \(2 (140) = 280\).
time = 0.41, size = 285, normalized size = 1.83 \begin {gather*} \frac {1}{3} \, a^{2} e x^{3} + b^{2} d x \arcsin \left (c x\right )^{2} + 2 \, a b d x \arcsin \left (c x\right ) + \frac {{\left (c^{2} x^{2} - 1\right )} b^{2} e x \arcsin \left (c x\right )^{2}}{3 \, c^{2}} + a^{2} d x - 2 \, b^{2} d x + \frac {2 \, {\left (c^{2} x^{2} - 1\right )} a b e x \arcsin \left (c x\right )}{3 \, c^{2}} + \frac {b^{2} e x \arcsin \left (c x\right )^{2}}{3 \, c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1} b^{2} d \arcsin \left (c x\right )}{c} - \frac {2 \, {\left (c^{2} x^{2} - 1\right )} b^{2} e x}{27 \, c^{2}} + \frac {2 \, a b e x \arcsin \left (c x\right )}{3 \, c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1} a b d}{c} - \frac {2 \, {\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b^{2} e \arcsin \left (c x\right )}{9 \, c^{3}} - \frac {14 \, b^{2} e x}{27 \, c^{2}} - \frac {2 \, {\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} a b e}{9 \, c^{3}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1} b^{2} e \arcsin \left (c x\right )}{3 \, c^{3}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1} a b e}{3 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 (e\,x^2+d\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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