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]

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

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Rubi [A]  time = 0.05, antiderivative size = 98, normalized size of antiderivative = 1.00, 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 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 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 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 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) \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*}

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Mathematica [A]  time = 0.05, 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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fricas [A]  time = 1.35, size = 76, normalized size = 0.78 \[ \frac {2 \, a c^{2} e x^{2} + 4 \, a c^{2} d x + {\left (2 \, b c^{2} e x^{2} + 4 \, b c^{2} d x - b e\right )} \arcsin \left (c x\right ) + {\left (b c e x + 4 \, b c d\right )} \sqrt {-c^{2} x^{2} + 1}}{4 \, c^{2}} \]

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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giac [A]  time = 0.37, size = 102, normalized size = 1.04 \[ b d x \arcsin \left (c x\right ) + a d x + \frac {\sqrt {-c^{2} x^{2} + 1} b x e}{4 \, c} + \frac {{\left (c^{2} x^{2} - 1\right )} b \arcsin \left (c x\right ) e}{2 \, c^{2}} + \frac {\sqrt {-c^{2} x^{2} + 1} b d}{c} + \frac {{\left (c^{2} x^{2} - 1\right )} a e}{2 \, c^{2}} + \frac {b \arcsin \left (c x\right ) e}{4 \, c^{2}} \]

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

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maple [A]  time = 0.01, size = 97, normalized size = 0.99 \[ \frac {\frac {a \left (\frac {1}{2} x^{2} c^{2} e +c^{2} d x \right )}{c}+\frac {b \left (\frac {\arcsin \left (c x \right ) c^{2} x^{2} e}{2}+\arcsin \left (c x \right ) d \,c^{2} x -\frac {e \left (-\frac {c x \sqrt {-c^{2} x^{2}+1}}{2}+\frac {\arcsin \left (c x \right )}{2}\right )}{2}+d c \sqrt {-c^{2} x^{2}+1}\right )}{c}}{c} \]

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+c^2*d*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 = 0.41, size = 81, normalized size = 0.83 \[ \frac {1}{2} \, a e x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x}{c^{2}} - \frac {\arcsin \left (c x\right )}{c^{3}}\right )}\right )} b e + a d x + \frac {{\left (c x \arcsin \left (c x\right ) + \sqrt {-c^{2} x^{2} + 1}\right )} b d}{c} \]

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

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mupad [B]  time = 0.41, size = 77, normalized size = 0.79 \[ \frac {a\,x\,\left (2\,d+e\,x\right )}{2}+\frac {b\,e\,\left (\frac {\mathrm {asin}\left (c\,x\right )\,\left (2\,c^2\,x^2-1\right )}{4}+\frac {c\,x\,\sqrt {1-c^2\,x^2}}{4}\right )}{c^2}+\frac {b\,d\,\left (\sqrt {1-c^2\,x^2}+c\,x\,\mathrm {asin}\left (c\,x\right )\right )}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))*(d + e*x),x)

[Out]

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

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sympy [A]  time = 0.28, size = 99, normalized size = 1.01 \[ \begin {cases} a d x + \frac {a e x^{2}}{2} + b d x \operatorname {asin}{\left (c x \right )} + \frac {b e x^{2} \operatorname {asin}{\left (c x \right )}}{2} + \frac {b d \sqrt {- c^{2} x^{2} + 1}}{c} + \frac {b e x \sqrt {- c^{2} x^{2} + 1}}{4 c} - \frac {b e \operatorname {asin}{\left (c x \right )}}{4 c^{2}} & \text {for}\: c \neq 0 \\a \left (d x + \frac {e x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

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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