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

Optimal. Leaf size=98 \[ \frac{(d+e x)^2 \left (a+b \sin ^{-1}(c x)\right )}{2 e}-\frac{b \left (\frac{e^2}{c^2}+2 d^2\right ) \sin ^{-1}(c x)}{4 e}+\frac{b \sqrt{1-c^2 x^2} (d+e x)}{4 c}+\frac{3 b d \sqrt{1-c^2 x^2}}{4 c} \]

[Out]

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

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Rubi [A]  time = 0.0510849, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4743, 743, 641, 216} \[ \frac{(d+e x)^2 \left (a+b \sin ^{-1}(c x)\right )}{2 e}-\frac{b \left (\frac{e^2}{c^2}+2 d^2\right ) \sin ^{-1}(c x)}{4 e}+\frac{b \sqrt{1-c^2 x^2} (d+e x)}{4 c}+\frac{3 b d \sqrt{1-c^2 x^2}}{4 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*(a + b*ArcSin[c*x]),x]

[Out]

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

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]

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 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[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) \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac{(d+e x)^2 \left (a+b \sin ^{-1}(c x)\right )}{2 e}-\frac{(b c) \int \frac{(d+e x)^2}{\sqrt{1-c^2 x^2}} \, dx}{2 e}\\ &=\frac{b (d+e x) \sqrt{1-c^2 x^2}}{4 c}+\frac{(d+e x)^2 \left (a+b \sin ^{-1}(c x)\right )}{2 e}+\frac{b \int \frac{-2 c^2 d^2-e^2-3 c^2 d e x}{\sqrt{1-c^2 x^2}} \, dx}{4 c e}\\ &=\frac{3 b d \sqrt{1-c^2 x^2}}{4 c}+\frac{b (d+e x) \sqrt{1-c^2 x^2}}{4 c}+\frac{(d+e x)^2 \left (a+b \sin ^{-1}(c x)\right )}{2 e}-\frac{\left (b \left (2 c^2 d^2+e^2\right )\right ) \int \frac{1}{\sqrt{1-c^2 x^2}} \, dx}{4 c e}\\ &=\frac{3 b d \sqrt{1-c^2 x^2}}{4 c}+\frac{b (d+e x) \sqrt{1-c^2 x^2}}{4 c}-\frac{b \left (2 d^2+\frac{e^2}{c^2}\right ) \sin ^{-1}(c x)}{4 e}+\frac{(d+e x)^2 \left (a+b \sin ^{-1}(c x)\right )}{2 e}\\ \end{align*}

Mathematica [A]  time = 0.0465409, size = 92, normalized size = 0.94 \[ a d x+\frac{1}{2} a e x^2+\frac{b d \sqrt{1-c^2 x^2}}{c}+\frac{b e x \sqrt{1-c^2 x^2}}{4 c}-\frac{b e \sin ^{-1}(c x)}{4 c^2}+b d x \sin ^{-1}(c x)+\frac{1}{2} b e x^2 \sin ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 97, normalized size = 1. \begin{align*}{\frac{1}{c} \left ({\frac{a}{c} \left ({\frac{{x}^{2}{c}^{2}e}{2}}+d{c}^{2}x \right ) }+{\frac{b}{c} \left ({\frac{\arcsin \left ( cx \right ){c}^{2}{x}^{2}e}{2}}+\arcsin \left ( cx \right ) d{c}^{2}x-{\frac{e}{2} \left ( -{\frac{cx}{2}\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{\arcsin \left ( cx \right ) }{2}} \right ) }+dc\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)*(a+b*arcsin(c*x)),x)

[Out]

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

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Maxima [A]  time = 1.44807, size = 126, normalized size = 1.29 \begin{align*} \frac{1}{2} \, a e 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 + a d x + \frac{{\left (c x \arcsin \left (c x\right ) + \sqrt{-c^{2} x^{2} + 1}\right )} b d}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.40443, size = 176, normalized size = 1.8 \begin{align*} \frac{2 \, a c^{2} e x^{2} + 4 \, a c^{2} d x +{\left (2 \, b c^{2} e x^{2} + 4 \, b c^{2} d x - b e\right )} \arcsin \left (c x\right ) +{\left (b c e x + 4 \, b c d\right )} \sqrt{-c^{2} x^{2} + 1}}{4 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.362445, size = 99, normalized size = 1.01 \begin{align*} \begin{cases} a d x + \frac{a e x^{2}}{2} + b d x \operatorname{asin}{\left (c x \right )} + \frac{b e x^{2} \operatorname{asin}{\left (c x \right )}}{2} + \frac{b d \sqrt{- c^{2} x^{2} + 1}}{c} + \frac{b e x \sqrt{- c^{2} x^{2} + 1}}{4 c} - \frac{b e \operatorname{asin}{\left (c x \right )}}{4 c^{2}} & \text{for}\: c \neq 0 \\a \left (d x + \frac{e x^{2}}{2}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.29413, size = 138, normalized size = 1.41 \begin{align*} b d x \arcsin \left (c x\right ) + a d x + \frac{\sqrt{-c^{2} x^{2} + 1} b x e}{4 \, c} + \frac{{\left (c^{2} x^{2} - 1\right )} b \arcsin \left (c x\right ) e}{2 \, c^{2}} + \frac{\sqrt{-c^{2} x^{2} + 1} b d}{c} + \frac{{\left (c^{2} x^{2} - 1\right )} a e}{2 \, c^{2}} + \frac{b \arcsin \left (c x\right ) e}{4 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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