3.90 \(\int (d+e x) (f+g x) (a+b \sin ^{-1}(c x)) \, dx\)

Optimal. Leaf size=148 \[ \frac{1}{2} x^2 (d g+e f) \left (a+b \sin ^{-1}(c x)\right )+d f x \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} e g x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac{b \sqrt{1-c^2 x^2} \left (9 c^2 x (d g+e f)+4 \left (9 c^2 d f+2 e g\right )\right )}{36 c^3}-\frac{b \sin ^{-1}(c x) (d g+e f)}{4 c^2}+\frac{b e g x^2 \sqrt{1-c^2 x^2}}{9 c} \]

[Out]

(b*e*g*x^2*Sqrt[1 - c^2*x^2])/(9*c) + (b*(4*(9*c^2*d*f + 2*e*g) + 9*c^2*(e*f + d*g)*x)*Sqrt[1 - c^2*x^2])/(36*
c^3) - (b*(e*f + d*g)*ArcSin[c*x])/(4*c^2) + d*f*x*(a + b*ArcSin[c*x]) + ((e*f + d*g)*x^2*(a + b*ArcSin[c*x]))
/2 + (e*g*x^3*(a + b*ArcSin[c*x]))/3

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Rubi [A]  time = 0.192987, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {4749, 12, 1809, 780, 216} \[ \frac{1}{2} x^2 (d g+e f) \left (a+b \sin ^{-1}(c x)\right )+d f x \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} e g x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac{b \sqrt{1-c^2 x^2} \left (9 c^2 x (d g+e f)+4 \left (9 c^2 d f+2 e g\right )\right )}{36 c^3}-\frac{b \sin ^{-1}(c x) (d g+e f)}{4 c^2}+\frac{b e g x^2 \sqrt{1-c^2 x^2}}{9 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*(f + g*x)*(a + b*ArcSin[c*x]),x]

[Out]

(b*e*g*x^2*Sqrt[1 - c^2*x^2])/(9*c) + (b*(4*(9*c^2*d*f + 2*e*g) + 9*c^2*(e*f + d*g)*x)*Sqrt[1 - c^2*x^2])/(36*
c^3) - (b*(e*f + d*g)*ArcSin[c*x])/(4*c^2) + d*f*x*(a + b*ArcSin[c*x]) + ((e*f + d*g)*x^2*(a + b*ArcSin[c*x]))
/2 + (e*g*x^3*(a + b*ArcSin[c*x]))/3

Rule 4749

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(Px_), x_Symbol] :> With[{u = IntHide[ExpandExpression[Px, x], x]}, Dis
t[a + b*ArcSin[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 - c^2*x^2], x], x], x]] /; FreeQ[{a, b,
c}, x] && PolynomialQ[Px, x]

Rule 12

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

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

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 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]

Rubi steps

\begin{align*} \int (d+e x) (f+g x) \left (a+b \sin ^{-1}(c x)\right ) \, dx &=d f x \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{2} (e f+d g) x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} e g x^3 \left (a+b \sin ^{-1}(c x)\right )-(b c) \int \frac{x \left (6 d f+3 (e f+d g) x+2 e g x^2\right )}{6 \sqrt{1-c^2 x^2}} \, dx\\ &=d f x \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{2} (e f+d g) x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} e g x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{6} (b c) \int \frac{x \left (6 d f+3 (e f+d g) x+2 e g x^2\right )}{\sqrt{1-c^2 x^2}} \, dx\\ &=\frac{b e g x^2 \sqrt{1-c^2 x^2}}{9 c}+d f x \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{2} (e f+d g) x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} e g x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac{b \int \frac{x \left (-2 \left (9 c^2 d f+2 e g\right )-9 c^2 (e f+d g) x\right )}{\sqrt{1-c^2 x^2}} \, dx}{18 c}\\ &=\frac{b e g x^2 \sqrt{1-c^2 x^2}}{9 c}+\frac{b \left (4 \left (9 c^2 d f+2 e g\right )+9 c^2 (e f+d g) x\right ) \sqrt{1-c^2 x^2}}{36 c^3}+d f x \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{2} (e f+d g) x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} e g x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{(b (e f+d g)) \int \frac{1}{\sqrt{1-c^2 x^2}} \, dx}{4 c}\\ &=\frac{b e g x^2 \sqrt{1-c^2 x^2}}{9 c}+\frac{b \left (4 \left (9 c^2 d f+2 e g\right )+9 c^2 (e f+d g) x\right ) \sqrt{1-c^2 x^2}}{36 c^3}-\frac{b (e f+d g) \sin ^{-1}(c x)}{4 c^2}+d f x \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{2} (e f+d g) x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} e g x^3 \left (a+b \sin ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.187501, size = 138, normalized size = 0.93 \[ \frac{6 a c^3 x (3 d (2 f+g x)+e x (3 f+2 g x))+b \sqrt{1-c^2 x^2} \left (c^2 (9 d (4 f+g x)+e x (9 f+4 g x))+8 e g\right )+3 b c \sin ^{-1}(c x) \left (12 c^2 d f x+3 d g \left (2 c^2 x^2-1\right )+e f \left (6 c^2 x^2-3\right )+4 c^2 e g x^3\right )}{36 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)*(f + g*x)*(a + b*ArcSin[c*x]),x]

[Out]

(6*a*c^3*x*(3*d*(2*f + g*x) + e*x*(3*f + 2*g*x)) + b*Sqrt[1 - c^2*x^2]*(8*e*g + c^2*(9*d*(4*f + g*x) + e*x*(9*
f + 4*g*x))) + 3*b*c*(12*c^2*d*f*x + 4*c^2*e*g*x^3 + 3*d*g*(-1 + 2*c^2*x^2) + e*f*(-3 + 6*c^2*x^2))*ArcSin[c*x
])/(36*c^3)

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Maple [A]  time = 0.006, size = 198, normalized size = 1.3 \begin{align*}{\frac{1}{c} \left ({\frac{a}{{c}^{2}} \left ({\frac{eg{c}^{3}{x}^{3}}{3}}+{\frac{ \left ( dcg+ecf \right ){c}^{2}{x}^{2}}{2}}+{c}^{3}fdx \right ) }+{\frac{b}{{c}^{2}} \left ({\frac{\arcsin \left ( cx \right ) eg{c}^{3}{x}^{3}}{3}}+{\frac{\arcsin \left ( cx \right ){c}^{3}{x}^{2}dg}{2}}+{\frac{\arcsin \left ( cx \right ){c}^{3}{x}^{2}ef}{2}}+\arcsin \left ( cx \right ){c}^{3}fdx-{\frac{eg}{3} \left ( -{\frac{{c}^{2}{x}^{2}}{3}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{2}{3}\sqrt{-{c}^{2}{x}^{2}+1}} \right ) }-{\frac{3\,dcg+3\,ecf}{6} \left ( -{\frac{cx}{2}\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{\arcsin \left ( cx \right ) }{2}} \right ) }+{c}^{2}fd\sqrt{-{c}^{2}{x}^{2}+1} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(g*x+f)*(a+b*arcsin(c*x)),x)

[Out]

1/c*(a/c^2*(1/3*e*g*c^3*x^3+1/2*(c*d*g+c*e*f)*c^2*x^2+c^3*f*d*x)+b/c^2*(1/3*arcsin(c*x)*e*g*c^3*x^3+1/2*arcsin
(c*x)*c^3*x^2*d*g+1/2*arcsin(c*x)*c^3*x^2*e*f+arcsin(c*x)*c^3*f*d*x-1/3*e*g*(-1/3*c^2*x^2*(-c^2*x^2+1)^(1/2)-2
/3*(-c^2*x^2+1)^(1/2))-1/6*(3*c*d*g+3*c*e*f)*(-1/2*c*x*(-c^2*x^2+1)^(1/2)+1/2*arcsin(c*x))+c^2*f*d*(-c^2*x^2+1
)^(1/2)))

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Maxima [A]  time = 1.57911, size = 301, normalized size = 2.03 \begin{align*} \frac{1}{3} \, a e g x^{3} + \frac{1}{2} \, a e f x^{2} + \frac{1}{2} \, a d g x^{2} + \frac{1}{4} \,{\left (2 \, x^{2} \arcsin \left (c x\right ) + c{\left (\frac{\sqrt{-c^{2} x^{2} + 1} x}{c^{2}} - \frac{\arcsin \left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{2}}\right )}\right )} b e f + \frac{1}{4} \,{\left (2 \, x^{2} \arcsin \left (c x\right ) + c{\left (\frac{\sqrt{-c^{2} x^{2} + 1} x}{c^{2}} - \frac{\arcsin \left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{2}}\right )}\right )} b d g + \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 g + a d f x + \frac{{\left (c x \arcsin \left (c x\right ) + \sqrt{-c^{2} x^{2} + 1}\right )} b d f}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.76674, size = 378, normalized size = 2.55 \begin{align*} \frac{12 \, a c^{3} e g x^{3} + 36 \, a c^{3} d f x + 18 \,{\left (a c^{3} e f + a c^{3} d g\right )} x^{2} + 3 \,{\left (4 \, b c^{3} e g x^{3} + 12 \, b c^{3} d f x - 3 \, b c e f - 3 \, b c d g + 6 \,{\left (b c^{3} e f + b c^{3} d g\right )} x^{2}\right )} \arcsin \left (c x\right ) +{\left (4 \, b c^{2} e g x^{2} + 36 \, b c^{2} d f + 8 \, b e g + 9 \,{\left (b c^{2} e f + b c^{2} d g\right )} x\right )} \sqrt{-c^{2} x^{2} + 1}}{36 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/36*(12*a*c^3*e*g*x^3 + 36*a*c^3*d*f*x + 18*(a*c^3*e*f + a*c^3*d*g)*x^2 + 3*(4*b*c^3*e*g*x^3 + 12*b*c^3*d*f*x
- 3*b*c*e*f - 3*b*c*d*g + 6*(b*c^3*e*f + b*c^3*d*g)*x^2)*arcsin(c*x) + (4*b*c^2*e*g*x^2 + 36*b*c^2*d*f + 8*b*
e*g + 9*(b*c^2*e*f + b*c^2*d*g)*x)*sqrt(-c^2*x^2 + 1))/c^3

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Sympy [A]  time = 1.31798, size = 267, normalized size = 1.8 \begin{align*} \begin{cases} a d f x + \frac{a d g x^{2}}{2} + \frac{a e f x^{2}}{2} + \frac{a e g x^{3}}{3} + b d f x \operatorname{asin}{\left (c x \right )} + \frac{b d g x^{2} \operatorname{asin}{\left (c x \right )}}{2} + \frac{b e f x^{2} \operatorname{asin}{\left (c x \right )}}{2} + \frac{b e g x^{3} \operatorname{asin}{\left (c x \right )}}{3} + \frac{b d f \sqrt{- c^{2} x^{2} + 1}}{c} + \frac{b d g x \sqrt{- c^{2} x^{2} + 1}}{4 c} + \frac{b e f x \sqrt{- c^{2} x^{2} + 1}}{4 c} + \frac{b e g x^{2} \sqrt{- c^{2} x^{2} + 1}}{9 c} - \frac{b d g \operatorname{asin}{\left (c x \right )}}{4 c^{2}} - \frac{b e f \operatorname{asin}{\left (c x \right )}}{4 c^{2}} + \frac{2 b e g \sqrt{- c^{2} x^{2} + 1}}{9 c^{3}} & \text{for}\: c \neq 0 \\a \left (d f x + \frac{d g x^{2}}{2} + \frac{e f x^{2}}{2} + \frac{e g x^{3}}{3}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.31928, size = 362, normalized size = 2.45 \begin{align*} \frac{1}{3} \, a g x^{3} e + b d f x \arcsin \left (c x\right ) + a d f x + \frac{{\left (c^{2} x^{2} - 1\right )} b g x \arcsin \left (c x\right ) e}{3 \, c^{2}} + \frac{\sqrt{-c^{2} x^{2} + 1} b d g x}{4 \, c} + \frac{\sqrt{-c^{2} x^{2} + 1} b f x e}{4 \, c} + \frac{{\left (c^{2} x^{2} - 1\right )} b d g \arcsin \left (c x\right )}{2 \, c^{2}} + \frac{{\left (c^{2} x^{2} - 1\right )} b f \arcsin \left (c x\right ) e}{2 \, c^{2}} + \frac{b g x \arcsin \left (c x\right ) e}{3 \, c^{2}} + \frac{\sqrt{-c^{2} x^{2} + 1} b d f}{c} + \frac{{\left (c^{2} x^{2} - 1\right )} a d g}{2 \, c^{2}} + \frac{b d g \arcsin \left (c x\right )}{4 \, c^{2}} + \frac{{\left (c^{2} x^{2} - 1\right )} a f e}{2 \, c^{2}} + \frac{b f \arcsin \left (c x\right ) e}{4 \, c^{2}} - \frac{{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}} b g e}{9 \, c^{3}} + \frac{\sqrt{-c^{2} x^{2} + 1} b g e}{3 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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