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

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

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 238, normalized size of antiderivative = 1.00, 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 30

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

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 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.30, 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])

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fricas [F]  time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {-c^{2} d x^{2} + d} {\left (a g x + a f + {\left (b g x + b f\right )} \arcsin \left (c x\right )\right )}, x\right ) \]

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)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:sym2poly/r2sym(co
nst gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.65, size = 491, normalized size = 2.06 \[ -\frac {a g \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{3 c^{2} d}+\frac {a f x \sqrt {-c^{2} d \,x^{2}+d}}{2}+\frac {a f d \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{2 \sqrt {c^{2} d}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, g \,c^{2} \arcsin \left (c x \right ) x^{4}}{3 c^{2} x^{2}-3}-\frac {2 b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, g \arcsin \left (c x \right ) x^{2}}{3 \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, f \,c^{2} \arcsin \left (c x \right ) x^{3}}{2 c^{2} x^{2}-2}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, f \sqrt {-c^{2} x^{2}+1}}{8 c \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, f \arcsin \left (c x \right ) x}{2 \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2} f}{4 c \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, g c \sqrt {-c^{2} x^{2}+1}\, x^{3}}{9 c^{2} x^{2}-9}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, g \sqrt {-c^{2} x^{2}+1}\, x}{3 c \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, f c \sqrt {-c^{2} x^{2}+1}\, x^{2}}{4 c^{2} x^{2}-4}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, g \arcsin \left (c x \right )}{3 c^{2} \left (c^{2} x^{2}-1\right )} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, {\left (\sqrt {-c^{2} d x^{2} + d} x + \frac {\sqrt {d} \arcsin \left (c x\right )}{c}\right )} a f + \sqrt {d} \int {\left (b g x + b f\right )} \sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\,{d x} - \frac {{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} a g}{3 \, c^{2} d} \]

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]

1/2*(sqrt(-c^2*d*x^2 + d)*x + sqrt(d)*arcsin(c*x)/c)*a*f + sqrt(d)*integrate((b*g*x + b*f)*sqrt(c*x + 1)*sqrt(
-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)), x) - 1/3*(-c^2*d*x^2 + d)^(3/2)*a*g/(c^2*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (f+g\,x\right )\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {d-c^2\,d\,x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right ) \left (f + g x\right )\, dx \]

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]

Integral(sqrt(-d*(c*x - 1)*(c*x + 1))*(a + b*asin(c*x))*(f + g*x), x)

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