3.2.91 \(\int (c e+d e x) (a+b \text {ArcSin}(c+d x))^2 \, dx\) [191]

Optimal. Leaf size=105 \[ -\frac {b^2 e (c+d x)^2}{4 d}+\frac {b e (c+d x) \sqrt {1-(c+d x)^2} (a+b \text {ArcSin}(c+d x))}{2 d}-\frac {e (a+b \text {ArcSin}(c+d x))^2}{4 d}+\frac {e (c+d x)^2 (a+b \text {ArcSin}(c+d x))^2}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4889, 12, 4723, 4795, 4737, 30} \begin {gather*} \frac {e (c+d x)^2 (a+b \text {ArcSin}(c+d x))^2}{2 d}+\frac {b e \sqrt {1-(c+d x)^2} (c+d x) (a+b \text {ArcSin}(c+d x))}{2 d}-\frac {e (a+b \text {ArcSin}(c+d x))^2}{4 d}-\frac {b^2 e (c+d x)^2}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 30

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

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 4737

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

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

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Mathematica [A]
time = 0.05, size = 86, normalized size = 0.82 \begin {gather*} -\frac {e \left (b^2 (c+d x)^2-2 b (c+d x) \sqrt {1-(c+d x)^2} (a+b \text {ArcSin}(c+d x))+(a+b \text {ArcSin}(c+d x))^2-2 (c+d x)^2 (a+b \text {ArcSin}(c+d x))^2\right )}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.05, size = 146, normalized size = 1.39

method	result	size
derivativedivides	\(\frac {\frac {e \left (d x +c \right )^{2} a^{2}}{2}+e \,b^{2} \left (\frac {\left (\left (d x +c \right )^{2}-1\right ) \arcsin \left (d x +c \right )^{2}}{2}+\frac {\arcsin \left (d x +c \right ) \left (\left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}+\arcsin \left (d x +c \right )\right )}{2}-\frac {\arcsin \left (d x +c \right )^{2}}{4}-\frac {\left (d x +c \right )^{2}}{4}\right )+2 e a b \left (\frac {\left (d x +c \right )^{2} \arcsin \left (d x +c \right )}{2}+\frac {\left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{4}-\frac {\arcsin \left (d x +c \right )}{4}\right )}{d}\)	\(146\)
default	\(\frac {\frac {e \left (d x +c \right )^{2} a^{2}}{2}+e \,b^{2} \left (\frac {\left (\left (d x +c \right )^{2}-1\right ) \arcsin \left (d x +c \right )^{2}}{2}+\frac {\arcsin \left (d x +c \right ) \left (\left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}+\arcsin \left (d x +c \right )\right )}{2}-\frac {\arcsin \left (d x +c \right )^{2}}{4}-\frac {\left (d x +c \right )^{2}}{4}\right )+2 e a b \left (\frac {\left (d x +c \right )^{2} \arcsin \left (d x +c \right )}{2}+\frac {\left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{4}-\frac {\arcsin \left (d x +c \right )}{4}\right )}{d}\)	\(146\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*d*x^2*e + 1/2*(2*x^2*arcsin(d*x + c) + d*(3*c^2*arcsin(-(d^2*x + c*d)/sqrt(c^2*d^2 - (c^2 - 1)*d^2))/d
^3 + sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*x/d^2 - (c^2 - 1)*arcsin(-(d^2*x + c*d)/sqrt(c^2*d^2 - (c^2 - 1)*d^2))
/d^3 - 3*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*c/d^3))*a*b*d*e + a^2*c*x*e + 2*((d*x + c)*arcsin(d*x + c) + sqrt(
-(d*x + c)^2 + 1))*a*b*c*e/d + 1/2*(b^2*d*x^2*e + 2*b^2*c*x*e)*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x -
c + 1))^2 + integrate((b^2*d^2*x^2*e + 2*b^2*c*d*x*e)*sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)*arctan2(d*x + c, sq
rt(d*x + c + 1)*sqrt(-d*x - c + 1))/(d^2*x^2 + 2*c*d*x + c^2 - 1), x)

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Fricas [A]
time = 2.11, size = 186, normalized size = 1.77 \begin {gather*} \frac {{\left (2 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c d x + 2 \, b^{2} c^{2} - b^{2}\right )} \arcsin \left (d x + c\right )^{2} e + 2 \, {\left (2 \, a b d^{2} x^{2} + 4 \, a b c d x + 2 \, a b c^{2} - a b\right )} \arcsin \left (d x + c\right ) e + {\left ({\left (2 \, a^{2} - b^{2}\right )} d^{2} x^{2} + 2 \, {\left (2 \, a^{2} - b^{2}\right )} c d x\right )} e + 2 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} {\left ({\left (b^{2} d x + b^{2} c\right )} \arcsin \left (d x + c\right ) e + {\left (a b d x + a b c\right )} e\right )}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (88) = 176\).
time = 0.20, size = 335, normalized size = 3.19 \begin {gather*} \begin {cases} a^{2} c e x + \frac {a^{2} d e x^{2}}{2} + \frac {a b c^{2} e \operatorname {asin}{\left (c + d x \right )}}{d} + 2 a b c e x \operatorname {asin}{\left (c + d x \right )} + \frac {a b c e \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{2 d} + a b d e x^{2} \operatorname {asin}{\left (c + d x \right )} + \frac {a b e x \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{2} - \frac {a b e \operatorname {asin}{\left (c + d x \right )}}{2 d} + \frac {b^{2} c^{2} e \operatorname {asin}^{2}{\left (c + d x \right )}}{2 d} + b^{2} c e x \operatorname {asin}^{2}{\left (c + d x \right )} - \frac {b^{2} c e x}{2} + \frac {b^{2} c e \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1} \operatorname {asin}{\left (c + d x \right )}}{2 d} + \frac {b^{2} d e x^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}{2} - \frac {b^{2} d e x^{2}}{4} + \frac {b^{2} e x \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1} \operatorname {asin}{\left (c + d x \right )}}{2} - \frac {b^{2} e \operatorname {asin}^{2}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\c e x \left (a + b \operatorname {asin}{\left (c \right )}\right )^{2} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.43, size = 184, normalized size = 1.75 \begin {gather*} \frac {{\left ({\left (d x + c\right )}^{2} - 1\right )} b^{2} e \arcsin \left (d x + c\right )^{2}}{2 \, d} + \frac {\sqrt {-{\left (d x + c\right )}^{2} + 1} {\left (d x + c\right )} b^{2} e \arcsin \left (d x + c\right )}{2 \, d} + \frac {{\left ({\left (d x + c\right )}^{2} - 1\right )} a b e \arcsin \left (d x + c\right )}{d} + \frac {b^{2} e \arcsin \left (d x + c\right )^{2}}{4 \, d} + \frac {\sqrt {-{\left (d x + c\right )}^{2} + 1} {\left (d x + c\right )} a b e}{2 \, d} + \frac {{\left ({\left (d x + c\right )}^{2} - 1\right )} a^{2} e}{2 \, d} - \frac {{\left ({\left (d x + c\right )}^{2} - 1\right )} b^{2} e}{4 \, d} + \frac {a b e \arcsin \left (d x + c\right )}{2 \, d} - \frac {b^{2} e}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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