3.33 \(\int (f+g x) \sqrt{d-c^2 d x^2} (a+b \sin ^{-1}(c x)) \, dx\)

Optimal. Leaf size=238 \[ \frac{1}{2} f x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{f \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c \sqrt{1-c^2 x^2}}-\frac{g \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^2}-\frac{b c f x^2 \sqrt{d-c^2 d x^2}}{4 \sqrt{1-c^2 x^2}}-\frac{b c g x^3 \sqrt{d-c^2 d x^2}}{9 \sqrt{1-c^2 x^2}}+\frac{b g x \sqrt{d-c^2 d x^2}}{3 c \sqrt{1-c^2 x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.24215, antiderivative size = 238, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {4777, 4763, 4647, 4641, 30, 4677} \[ \frac{1}{2} f x \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac{f \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c \sqrt{1-c^2 x^2}}-\frac{g \left (1-c^2 x^2\right ) \sqrt{d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^2}-\frac{b c f x^2 \sqrt{d-c^2 d x^2}}{4 \sqrt{1-c^2 x^2}}-\frac{b c g x^3 \sqrt{d-c^2 d x^2}}{9 \sqrt{1-c^2 x^2}}+\frac{b g x \sqrt{d-c^2 d x^2}}{3 c \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(f + g*x)*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]),x]

[Out]

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

Rule 4777

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :
> Dist[(d^IntPart[p]*(d + e*x^2)^FracPart[p])/(1 - c^2*x^2)^FracPart[p], Int[(f + g*x)^m*(1 - c^2*x^2)^p*(a +
b*ArcSin[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && IntegerQ
[p - 1/2] &&  !GtQ[d, 0]

Rule 4763

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g},
x] && EqQ[c^2*d + e, 0] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ[n, 0] && (m == 1 || p > 0 ||
(n == 1 && p > -1) || (m == 2 && p < -2))

Rule 4647

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

Rule 4641

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 4677

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

Rubi steps

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

Mathematica [A]  time = 0.249715, size = 132, normalized size = 0.55 \[ \frac{\sqrt{d-c^2 d x^2} \left (18 f x \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac{12 g \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{c^2}+\frac{9 f \left (a+b \sin ^{-1}(c x)\right )^2}{b c}-\frac{4 b g x \left (c^2 x^2-3\right )}{c}-9 b c f x^2\right )}{36 \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(f + g*x)*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]),x]

[Out]

(Sqrt[d - c^2*d*x^2]*(-9*b*c*f*x^2 - (4*b*g*x*(-3 + c^2*x^2))/c + 18*f*x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])
- (12*g*(1 - c^2*x^2)^(3/2)*(a + b*ArcSin[c*x]))/c^2 + (9*f*(a + b*ArcSin[c*x])^2)/(b*c)))/(36*Sqrt[1 - c^2*x
^2])

________________________________________________________________________________________

Maple [B]  time = 0.356, size = 491, normalized size = 2.1 \begin{align*} -{\frac{ag}{3\,{c}^{2}d} \left ( -{c}^{2}d{x}^{2}+d \right ) ^{{\frac{3}{2}}}}+{\frac{afx}{2}\sqrt{-{c}^{2}d{x}^{2}+d}}+{\frac{afd}{2}\arctan \left ({x\sqrt{{c}^{2}d}{\frac{1}{\sqrt{-{c}^{2}d{x}^{2}+d}}}} \right ){\frac{1}{\sqrt{{c}^{2}d}}}}+{\frac{bgc{x}^{3}}{9\,{c}^{2}{x}^{2}-9}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{bgx}{3\,c \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{bcf{x}^{2}}{4\,{c}^{2}{x}^{2}-4}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{bg{c}^{2}\arcsin \left ( cx \right ){x}^{4}}{3\,{c}^{2}{x}^{2}-3}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{2\,bg\arcsin \left ( cx \right ){x}^{2}}{3\,{c}^{2}{x}^{2}-3}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{bf{c}^{2}\arcsin \left ( cx \right ){x}^{3}}{2\,{c}^{2}{x}^{2}-2}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{bf\arcsin \left ( cx \right ) x}{2\,{c}^{2}{x}^{2}-2}\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{bf}{8\,c \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{b \left ( \arcsin \left ( cx \right ) \right ) ^{2}f}{4\,c \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{bg\arcsin \left ( cx \right ) }{3\,{c}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-d \left ({c}^{2}{x}^{2}-1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2),x)

[Out]

-1/3*a*g/c^2/d*(-c^2*d*x^2+d)^(3/2)+1/2*a*f*x*(-c^2*d*x^2+d)^(1/2)+1/2*a*f*d/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2
)*x/(-c^2*d*x^2+d)^(1/2))+1/9*b*(-d*(c^2*x^2-1))^(1/2)*g*c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^3-1/3*b*(-d*(c^2*x
^2-1))^(1/2)*g/c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x+1/4*b*(-d*(c^2*x^2-1))^(1/2)*f*c/(c^2*x^2-1)*(-c^2*x^2+1)^(1
/2)*x^2+1/3*b*(-d*(c^2*x^2-1))^(1/2)*g*c^2/(c^2*x^2-1)*arcsin(c*x)*x^4-2/3*b*(-d*(c^2*x^2-1))^(1/2)*g/(c^2*x^2
-1)*arcsin(c*x)*x^2+1/2*b*(-d*(c^2*x^2-1))^(1/2)*f*c^2/(c^2*x^2-1)*arcsin(c*x)*x^3-1/2*b*(-d*(c^2*x^2-1))^(1/2
)*f/(c^2*x^2-1)*arcsin(c*x)*x-1/8*b*(-d*(c^2*x^2-1))^(1/2)*f/c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)-1/4*b*(-d*(c^2*x
^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c/(c^2*x^2-1)*arcsin(c*x)^2*f+1/3*b*(-d*(c^2*x^2-1))^(1/2)*g/c^2/(c^2*x^2-1)*a
rcsin(c*x)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{-c^{2} d x^{2} + d}{\left (a g x + a f +{\left (b g x + b f\right )} \arcsin \left (c x\right )\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(-c^2*d*x^2 + d)*(a*g*x + a*f + (b*g*x + b*f)*arcsin(c*x)), x)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(a+b*asin(c*x))*(-c**2*d*x**2+d)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-c^{2} d x^{2} + d}{\left (g x + f\right )}{\left (b \arcsin \left (c x\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(-c^2*d*x^2 + d)*(g*x + f)*(b*arcsin(c*x) + a), x)