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

Optimal. Leaf size=124 \[ \frac {(d+e x)^3 \left (a+b \sin ^{-1}(c x)\right )}{3 e}-\frac {b d \left (\frac {3 e^2}{c^2}+2 d^2\right ) \sin ^{-1}(c x)}{6 e}+\frac {b \sqrt {1-c^2 x^2} (d+e x)^2}{9 c}+\frac {b \sqrt {1-c^2 x^2} \left (4 \left (4 c^2 d^2+e^2\right )+5 c^2 d e x\right )}{18 c^3} \]

[Out]

-1/6*b*d*(2*d^2+3*e^2/c^2)*arcsin(c*x)/e+1/3*(e*x+d)^3*(a+b*arcsin(c*x))/e+1/9*b*(e*x+d)^2*(-c^2*x^2+1)^(1/2)/
c+1/18*b*(5*c^2*d*e*x+16*c^2*d^2+4*e^2)*(-c^2*x^2+1)^(1/2)/c^3

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Rubi [A]  time = 0.10, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4743, 743, 780, 216} \[ \frac {(d+e x)^3 \left (a+b \sin ^{-1}(c x)\right )}{3 e}+\frac {b \sqrt {1-c^2 x^2} \left (4 \left (4 c^2 d^2+e^2\right )+5 c^2 d e x\right )}{18 c^3}-\frac {b d \left (\frac {3 e^2}{c^2}+2 d^2\right ) \sin ^{-1}(c x)}{6 e}+\frac {b \sqrt {1-c^2 x^2} (d+e x)^2}{9 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(d + e*x)^2*Sqrt[1 - c^2*x^2])/(9*c) + (b*(4*(4*c^2*d^2 + e^2) + 5*c^2*d*e*x)*Sqrt[1 - c^2*x^2])/(18*c^3) -
(b*d*(2*d^2 + (3*e^2)/c^2)*ArcSin[c*x])/(6*e) + ((d + e*x)^3*(a + b*ArcSin[c*x]))/(3*e)

Rule 216

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

Rule 743

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m
, p, x]

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 4743

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

Rubi steps

\begin {align*} \int (d+e x)^2 \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac {(d+e x)^3 \left (a+b \sin ^{-1}(c x)\right )}{3 e}-\frac {(b c) \int \frac {(d+e x)^3}{\sqrt {1-c^2 x^2}} \, dx}{3 e}\\ &=\frac {b (d+e x)^2 \sqrt {1-c^2 x^2}}{9 c}+\frac {(d+e x)^3 \left (a+b \sin ^{-1}(c x)\right )}{3 e}+\frac {b \int \frac {(d+e x) \left (-3 c^2 d^2-2 e^2-5 c^2 d e x\right )}{\sqrt {1-c^2 x^2}} \, dx}{9 c e}\\ &=\frac {b (d+e x)^2 \sqrt {1-c^2 x^2}}{9 c}+\frac {b \left (4 \left (4 c^2 d^2+e^2\right )+5 c^2 d e x\right ) \sqrt {1-c^2 x^2}}{18 c^3}+\frac {(d+e x)^3 \left (a+b \sin ^{-1}(c x)\right )}{3 e}-\frac {1}{6} \left (b d \left (\frac {2 c d^2}{e}+\frac {3 e}{c}\right )\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx\\ &=\frac {b (d+e x)^2 \sqrt {1-c^2 x^2}}{9 c}+\frac {b \left (4 \left (4 c^2 d^2+e^2\right )+5 c^2 d e x\right ) \sqrt {1-c^2 x^2}}{18 c^3}-\frac {b d \left (2 d^2+\frac {3 e^2}{c^2}\right ) \sin ^{-1}(c x)}{6 e}+\frac {(d+e x)^3 \left (a+b \sin ^{-1}(c x)\right )}{3 e}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 121, normalized size = 0.98 \[ \frac {6 a c^3 x \left (3 d^2+3 d e x+e^2 x^2\right )+b \sqrt {1-c^2 x^2} \left (c^2 \left (18 d^2+9 d e x+2 e^2 x^2\right )+4 e^2\right )+3 b c \sin ^{-1}(c x) \left (6 c^2 d^2 x+3 d e \left (2 c^2 x^2-1\right )+2 c^2 e^2 x^3\right )}{18 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a*c^3*x*(3*d^2 + 3*d*e*x + e^2*x^2) + b*Sqrt[1 - c^2*x^2]*(4*e^2 + c^2*(18*d^2 + 9*d*e*x + 2*e^2*x^2)) + 3*
b*c*(6*c^2*d^2*x + 2*c^2*e^2*x^3 + 3*d*e*(-1 + 2*c^2*x^2))*ArcSin[c*x])/(18*c^3)

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fricas [A]  time = 0.56, size = 135, normalized size = 1.09 \[ \frac {6 \, a c^{3} e^{2} x^{3} + 18 \, a c^{3} d e x^{2} + 18 \, a c^{3} d^{2} x + 3 \, {\left (2 \, b c^{3} e^{2} x^{3} + 6 \, b c^{3} d e x^{2} + 6 \, b c^{3} d^{2} x - 3 \, b c d e\right )} \arcsin \left (c x\right ) + {\left (2 \, b c^{2} e^{2} x^{2} + 9 \, b c^{2} d e x + 18 \, b c^{2} d^{2} + 4 \, b e^{2}\right )} \sqrt {-c^{2} x^{2} + 1}}{18 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.27, size = 193, normalized size = 1.56 \[ b d^{2} x \arcsin \left (c x\right ) + \frac {1}{3} \, a x^{3} e^{2} + a d^{2} x + \frac {\sqrt {-c^{2} x^{2} + 1} b d x e}{2 \, c} + \frac {{\left (c^{2} x^{2} - 1\right )} b x \arcsin \left (c x\right ) e^{2}}{3 \, c^{2}} + \frac {{\left (c^{2} x^{2} - 1\right )} b d \arcsin \left (c x\right ) e}{c^{2}} + \frac {\sqrt {-c^{2} x^{2} + 1} b d^{2}}{c} + \frac {b x \arcsin \left (c x\right ) e^{2}}{3 \, c^{2}} + \frac {{\left (c^{2} x^{2} - 1\right )} a d e}{c^{2}} + \frac {b d \arcsin \left (c x\right ) e}{2 \, c^{2}} - \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b e^{2}}{9 \, c^{3}} + \frac {\sqrt {-c^{2} x^{2} + 1} b e^{2}}{3 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 193, normalized size = 1.56 \[ \frac {\frac {\left (c e x +d c \right )^{3} a}{3 c^{2} e}+\frac {b \left (\frac {\arcsin \left (c x \right ) e^{2} c^{3} x^{3}}{3}+e \arcsin \left (c x \right ) c^{3} x^{2} d +\arcsin \left (c x \right ) c^{3} x \,d^{2}+\frac {\arcsin \left (c x \right ) c^{3} d^{3}}{3 e}-\frac {e^{3} \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )+3 d c \,e^{2} \left (-\frac {c x \sqrt {-c^{2} x^{2}+1}}{2}+\frac {\arcsin \left (c x \right )}{2}\right )-3 d^{2} c^{2} e \sqrt {-c^{2} x^{2}+1}+c^{3} d^{3} \arcsin \left (c x \right )}{3 e}\right )}{c^{2}}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 150, normalized size = 1.21 \[ \frac {1}{3} \, a e^{2} x^{3} + a d e x^{2} + \frac {1}{2} \, {\left (2 \, x^{2} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x}{c^{2}} - \frac {\arcsin \left (c x\right )}{c^{3}}\right )}\right )} b d e + \frac {1}{9} \, {\left (3 \, x^{3} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b e^{2} + a d^{2} x + \frac {{\left (c x \arcsin \left (c x\right ) + \sqrt {-c^{2} x^{2} + 1}\right )} b d^{2}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \left \{\begin {array}{cl} b\,e^2\,\left (\frac {\sqrt {\frac {1}{c^2}-x^2}\,\left (\frac {2}{c^2}+x^2\right )}{9}+\frac {x^3\,\mathrm {asin}\left (c\,x\right )}{3}\right )+\frac {a\,x\,\left (3\,d^2+3\,d\,e\,x+e^2\,x^2\right )}{3}+\frac {b\,d^2\,\left (\sqrt {1-c^2\,x^2}+c\,x\,\mathrm {asin}\left (c\,x\right )\right )}{c}+\frac {2\,b\,d\,e\,\left (\frac {\mathrm {asin}\left (c\,x\right )\,\left (2\,c^2\,x^2-1\right )}{4}+\frac {c\,x\,\sqrt {1-c^2\,x^2}}{4}\right )}{c^2} & \text {\ if\ \ }0<c\\ \int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d+e\,x\right )}^2 \,d x & \text {\ if\ \ }\neg 0<c \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))*(d + e*x)^2,x)

[Out]

piecewise(0 < c, b*e^2*(((1/c^2 - x^2)^(1/2)*(2/c^2 + x^2))/9 + (x^3*asin(c*x))/3) + (a*x*(3*d^2 + e^2*x^2 + 3
*d*e*x))/3 + (b*d^2*((- c^2*x^2 + 1)^(1/2) + c*x*asin(c*x)))/c + (2*b*d*e*((asin(c*x)*(2*c^2*x^2 - 1))/4 + (c*
x*(- c^2*x^2 + 1)^(1/2))/4))/c^2, ~0 < c, int((a + b*asin(c*x))*(d + e*x)^2, x))

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sympy [A]  time = 0.63, size = 190, normalized size = 1.53 \[ \begin {cases} a d^{2} x + a d e x^{2} + \frac {a e^{2} x^{3}}{3} + b d^{2} x \operatorname {asin}{\left (c x \right )} + b d e x^{2} \operatorname {asin}{\left (c x \right )} + \frac {b e^{2} x^{3} \operatorname {asin}{\left (c x \right )}}{3} + \frac {b d^{2} \sqrt {- c^{2} x^{2} + 1}}{c} + \frac {b d e x \sqrt {- c^{2} x^{2} + 1}}{2 c} + \frac {b e^{2} x^{2} \sqrt {- c^{2} x^{2} + 1}}{9 c} - \frac {b d e \operatorname {asin}{\left (c x \right )}}{2 c^{2}} + \frac {2 b e^{2} \sqrt {- c^{2} x^{2} + 1}}{9 c^{3}} & \text {for}\: c \neq 0 \\a \left (d^{2} x + d e x^{2} + \frac {e^{2} x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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