3.179 \(\int (c e+d e x)^2 (a+b \sin ^{-1}(c+d x)) \, dx\)

Optimal. Leaf size=80 \[ \frac {e^2 (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )}{3 d}-\frac {b e^2 \left (1-(c+d x)^2\right )^{3/2}}{9 d}+\frac {b e^2 \sqrt {1-(c+d x)^2}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {4805, 12, 4627, 266, 43} \[ \frac {e^2 (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )}{3 d}-\frac {b e^2 \left (1-(c+d x)^2\right )^{3/2}}{9 d}+\frac {b e^2 \sqrt {1-(c+d x)^2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*e^2*Sqrt[1 - (c + d*x)^2])/(3*d) - (b*e^2*(1 - (c + d*x)^2)^(3/2))/(9*d) + (e^2*(c + d*x)^3*(a + b*ArcSin[c
+ d*x]))/(3*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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 4627

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 4805

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)^2 \left (a+b \sin ^{-1}(c+d x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int e^2 x^2 \left (a+b \sin ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 \operatorname {Subst}\left (\int x^2 \left (a+b \sin ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )}{3 d}-\frac {\left (b e^2\right ) \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{3 d}\\ &=\frac {e^2 (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )}{3 d}-\frac {\left (b e^2\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt {1-x}} \, dx,x,(c+d x)^2\right )}{6 d}\\ &=\frac {e^2 (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )}{3 d}-\frac {\left (b e^2\right ) \operatorname {Subst}\left (\int \left (\frac {1}{\sqrt {1-x}}-\sqrt {1-x}\right ) \, dx,x,(c+d x)^2\right )}{6 d}\\ &=\frac {b e^2 \sqrt {1-(c+d x)^2}}{3 d}-\frac {b e^2 \left (1-(c+d x)^2\right )^{3/2}}{9 d}+\frac {e^2 (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 64, normalized size = 0.80 \[ \frac {e^2 \left (3 (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )+b \left (c^2+2 c d x+d^2 x^2+2\right ) \sqrt {1-(c+d x)^2}\right )}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.60, size = 151, normalized size = 1.89 \[ \frac {3 \, a d^{3} e^{2} x^{3} + 9 \, a c d^{2} e^{2} x^{2} + 9 \, a c^{2} d e^{2} x + 3 \, {\left (b d^{3} e^{2} x^{3} + 3 \, b c d^{2} e^{2} x^{2} + 3 \, b c^{2} d e^{2} x + b c^{3} e^{2}\right )} \arcsin \left (d x + c\right ) + {\left (b d^{2} e^{2} x^{2} + 2 \, b c d e^{2} x + {\left (b c^{2} + 2 \, b\right )} e^{2}\right )} \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}{9 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.76, size = 105, normalized size = 1.31 \[ \frac {{\left (d x + c\right )}^{3} a e^{2}}{3 \, d} + \frac {{\left ({\left (d x + c\right )}^{2} - 1\right )} {\left (d x + c\right )} b \arcsin \left (d x + c\right ) e^{2}}{3 \, d} + \frac {{\left (d x + c\right )} b \arcsin \left (d x + c\right ) e^{2}}{3 \, d} - \frac {{\left (-{\left (d x + c\right )}^{2} + 1\right )}^{\frac {3}{2}} b e^{2}}{9 \, d} + \frac {\sqrt {-{\left (d x + c\right )}^{2} + 1} b e^{2}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 77, normalized size = 0.96 \[ \frac {\frac {e^{2} \left (d x +c \right )^{3} a}{3}+e^{2} b \left (\frac {\arcsin \left (d x +c \right ) \left (d x +c \right )^{3}}{3}+\frac {\left (d x +c \right )^{2} \sqrt {1-\left (d x +c \right )^{2}}}{9}+\frac {2 \sqrt {1-\left (d x +c \right )^{2}}}{9}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.42, size = 457, normalized size = 5.71 \[ \frac {1}{3} \, a d^{2} e^{2} x^{3} + a c d e^{2} x^{2} + \frac {1}{2} \, {\left (2 \, x^{2} \arcsin \left (d x + c\right ) + d {\left (\frac {3 \, c^{2} \arcsin \left (-\frac {d^{2} x + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{3}} + \frac {\sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} x}{d^{2}} - \frac {{\left (c^{2} - 1\right )} \arcsin \left (-\frac {d^{2} x + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{3}} - \frac {3 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} c}{d^{3}}\right )}\right )} b c d e^{2} + \frac {1}{18} \, {\left (6 \, x^{3} \arcsin \left (d x + c\right ) + d {\left (\frac {2 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} x^{2}}{d^{2}} - \frac {15 \, c^{3} \arcsin \left (-\frac {d^{2} x + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{4}} - \frac {5 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} c x}{d^{3}} + \frac {9 \, {\left (c^{2} - 1\right )} c \arcsin \left (-\frac {d^{2} x + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{4}} + \frac {15 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} c^{2}}{d^{4}} - \frac {4 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} {\left (c^{2} - 1\right )}}{d^{4}}\right )}\right )} b d^{2} e^{2} + a c^{2} e^{2} x + \frac {{\left ({\left (d x + c\right )} \arcsin \left (d x + c\right ) + \sqrt {-{\left (d x + c\right )}^{2} + 1}\right )} b c^{2} e^{2}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*d^2*e^2*x^3 + a*c*d*e^2*x^2 + 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))*b*c*d*e^2 + 1/18*(6*x^3*arcsin(d*x + c)
+ d*(2*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*x^2/d^2 - 15*c^3*arcsin(-(d^2*x + c*d)/sqrt(c^2*d^2 - (c^2 - 1)*d^2)
)/d^4 - 5*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*c*x/d^3 + 9*(c^2 - 1)*c*arcsin(-(d^2*x + c*d)/sqrt(c^2*d^2 - (c^2
- 1)*d^2))/d^4 + 15*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*c^2/d^4 - 4*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*(c^2 -
1)/d^4))*b*d^2*e^2 + a*c^2*e^2*x + ((d*x + c)*arcsin(d*x + c) + sqrt(-(d*x + c)^2 + 1))*b*c^2*e^2/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.72, size = 258, normalized size = 3.22 \[ \begin {cases} a c^{2} e^{2} x + a c d e^{2} x^{2} + \frac {a d^{2} e^{2} x^{3}}{3} + \frac {b c^{3} e^{2} \operatorname {asin}{\left (c + d x \right )}}{3 d} + b c^{2} e^{2} x \operatorname {asin}{\left (c + d x \right )} + \frac {b c^{2} e^{2} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{9 d} + b c d e^{2} x^{2} \operatorname {asin}{\left (c + d x \right )} + \frac {2 b c e^{2} x \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{9} + \frac {b d^{2} e^{2} x^{3} \operatorname {asin}{\left (c + d x \right )}}{3} + \frac {b d e^{2} x^{2} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{9} + \frac {2 b e^{2} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{9 d} & \text {for}\: d \neq 0 \\c^{2} e^{2} x \left (a + b \operatorname {asin}{\relax (c )}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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