3.31 \(\int (f+g x)^3 \sqrt {d-c^2 d x^2} (a+b \sin ^{-1}(c x)) \, dx\)

Optimal. Leaf size=669 \[ \frac {1}{2} f^3 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {f^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c \sqrt {1-c^2 x^2}}-\frac {f^2 g \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^2}-\frac {3 f g^2 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c^2}+\frac {3}{4} f g^2 x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {g^3 \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^4}-\frac {g^3 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^4}+\frac {3 f g^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c^3 \sqrt {1-c^2 x^2}}-\frac {b c f^3 x^2 \sqrt {d-c^2 d x^2}}{4 \sqrt {1-c^2 x^2}}+\frac {b f^2 g x \sqrt {d-c^2 d x^2}}{c \sqrt {1-c^2 x^2}}-\frac {b c f^2 g x^3 \sqrt {d-c^2 d x^2}}{3 \sqrt {1-c^2 x^2}}+\frac {3 b f g^2 x^2 \sqrt {d-c^2 d x^2}}{16 c \sqrt {1-c^2 x^2}}-\frac {3 b c f g^2 x^4 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}}-\frac {b c g^3 x^5 \sqrt {d-c^2 d x^2}}{25 \sqrt {1-c^2 x^2}}+\frac {b g^3 x^3 \sqrt {d-c^2 d x^2}}{45 c \sqrt {1-c^2 x^2}}+\frac {2 b g^3 x \sqrt {d-c^2 d x^2}}{15 c^3 \sqrt {1-c^2 x^2}} \]

[Out]

1/2*f^3*x*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)-3/8*f*g^2*x*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/c^2+3/4*f*
g^2*x^3*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)-f^2*g*(-c^2*x^2+1)*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/c^2-1
/3*g^3*(-c^2*x^2+1)*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/c^4+1/5*g^3*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))*(-c^2*
d*x^2+d)^(1/2)/c^4+b*f^2*g*x*(-c^2*d*x^2+d)^(1/2)/c/(-c^2*x^2+1)^(1/2)+2/15*b*g^3*x*(-c^2*d*x^2+d)^(1/2)/c^3/(
-c^2*x^2+1)^(1/2)-1/4*b*c*f^3*x^2*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+3/16*b*f*g^2*x^2*(-c^2*d*x^2+d)^(1/2
)/c/(-c^2*x^2+1)^(1/2)-1/3*b*c*f^2*g*x^3*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+1/45*b*g^3*x^3*(-c^2*d*x^2+d)
^(1/2)/c/(-c^2*x^2+1)^(1/2)-3/16*b*c*f*g^2*x^4*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)-1/25*b*c*g^3*x^5*(-c^2*
d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+1/4*f^3*(a+b*arcsin(c*x))^2*(-c^2*d*x^2+d)^(1/2)/b/c/(-c^2*x^2+1)^(1/2)+3/16
*f*g^2*(a+b*arcsin(c*x))^2*(-c^2*d*x^2+d)^(1/2)/b/c^3/(-c^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.69, antiderivative size = 669, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {4777, 4763, 4647, 4641, 30, 4677, 4697, 4707, 266, 43, 4689, 12} \[ -\frac {f^2 g \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^2}+\frac {1}{2} f^3 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {f^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c \sqrt {1-c^2 x^2}}+\frac {3}{4} f g^2 x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {3 f g^2 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c^2}+\frac {3 f g^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c^3 \sqrt {1-c^2 x^2}}+\frac {g^3 \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^4}-\frac {g^3 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^4}-\frac {b c f^2 g x^3 \sqrt {d-c^2 d x^2}}{3 \sqrt {1-c^2 x^2}}+\frac {b f^2 g x \sqrt {d-c^2 d x^2}}{c \sqrt {1-c^2 x^2}}-\frac {b c f^3 x^2 \sqrt {d-c^2 d x^2}}{4 \sqrt {1-c^2 x^2}}-\frac {3 b c f g^2 x^4 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}}+\frac {3 b f g^2 x^2 \sqrt {d-c^2 d x^2}}{16 c \sqrt {1-c^2 x^2}}-\frac {b c g^3 x^5 \sqrt {d-c^2 d x^2}}{25 \sqrt {1-c^2 x^2}}+\frac {b g^3 x^3 \sqrt {d-c^2 d x^2}}{45 c \sqrt {1-c^2 x^2}}+\frac {2 b g^3 x \sqrt {d-c^2 d x^2}}{15 c^3 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(f + g*x)^3*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]),x]

[Out]

(b*f^2*g*x*Sqrt[d - c^2*d*x^2])/(c*Sqrt[1 - c^2*x^2]) + (2*b*g^3*x*Sqrt[d - c^2*d*x^2])/(15*c^3*Sqrt[1 - c^2*x
^2]) - (b*c*f^3*x^2*Sqrt[d - c^2*d*x^2])/(4*Sqrt[1 - c^2*x^2]) + (3*b*f*g^2*x^2*Sqrt[d - c^2*d*x^2])/(16*c*Sqr
t[1 - c^2*x^2]) - (b*c*f^2*g*x^3*Sqrt[d - c^2*d*x^2])/(3*Sqrt[1 - c^2*x^2]) + (b*g^3*x^3*Sqrt[d - c^2*d*x^2])/
(45*c*Sqrt[1 - c^2*x^2]) - (3*b*c*f*g^2*x^4*Sqrt[d - c^2*d*x^2])/(16*Sqrt[1 - c^2*x^2]) - (b*c*g^3*x^5*Sqrt[d
- c^2*d*x^2])/(25*Sqrt[1 - c^2*x^2]) + (f^3*x*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/2 - (3*f*g^2*x*Sqrt[d -
c^2*d*x^2]*(a + b*ArcSin[c*x]))/(8*c^2) + (3*f*g^2*x^3*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/4 - (f^2*g*(1
- c^2*x^2)*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/c^2 - (g^3*(1 - c^2*x^2)*Sqrt[d - c^2*d*x^2]*(a + b*ArcSi
n[c*x]))/(3*c^4) + (g^3*(1 - c^2*x^2)^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(5*c^4) + (f^3*Sqrt[d - c^2*d
*x^2]*(a + b*ArcSin[c*x])^2)/(4*b*c*Sqrt[1 - c^2*x^2]) + (3*f*g^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^2)/(
16*b*c^3*Sqrt[1 - c^2*x^2])

Rule 12

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

Rule 30

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 4689

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x^
m*(1 - c^2*x^2)^p, x]}, Dist[d^p*(a + b*ArcSin[c*x]), u, x] - Dist[b*c*d^p, Int[SimplifyIntegrand[u/Sqrt[1 - c
^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IntegerQ[p - 1/2] && (IGtQ[(m + 1)/2
, 0] || ILtQ[(m + 2*p + 3)/2, 0]) && NeQ[p, -2^(-1)] && GtQ[d, 0]

Rule 4697

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

Rule 4707

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

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)^3 \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)^3 \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^3 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )+3 f^2 g x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )+3 f g^2 x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )+g^3 x^3 \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^3 \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 (3 f^2 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 {\left (3 f g^2 \sqrt {d-c^2 d x^2}\right ) \int x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt {1-c^2 x^2}}+\frac {\left (g^3 \sqrt {d-c^2 d x^2}\right ) \int x^3 \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^3 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {3}{4} f g^2 x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {f^2 g \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^2}-\frac {g^3 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^4}+\frac {g^3 \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^4}+\frac {\left (f^3 \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^3 \sqrt {d-c^2 d x^2}\right ) \int x \, dx}{2 \sqrt {1-c^2 x^2}}+\frac {\left (b f^2 g \sqrt {d-c^2 d x^2}\right ) \int \left (1-c^2 x^2\right ) \, dx}{c \sqrt {1-c^2 x^2}}+\frac {\left (3 f g^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{4 \sqrt {1-c^2 x^2}}-\frac {\left (3 b c f g^2 \sqrt {d-c^2 d x^2}\right ) \int x^3 \, dx}{4 \sqrt {1-c^2 x^2}}-\frac {\left (b c g^3 \sqrt {d-c^2 d x^2}\right ) \int \frac {-2-c^2 x^2+3 c^4 x^4}{15 c^4} \, dx}{\sqrt {1-c^2 x^2}}\\ &=\frac {b f^2 g x \sqrt {d-c^2 d x^2}}{c \sqrt {1-c^2 x^2}}-\frac {b c f^3 x^2 \sqrt {d-c^2 d x^2}}{4 \sqrt {1-c^2 x^2}}-\frac {b c f^2 g x^3 \sqrt {d-c^2 d x^2}}{3 \sqrt {1-c^2 x^2}}-\frac {3 b c f g^2 x^4 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}}+\frac {1}{2} f^3 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {3 f g^2 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c^2}+\frac {3}{4} f g^2 x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {f^2 g \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^2}-\frac {g^3 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^4}+\frac {g^3 \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^4}+\frac {f^3 \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 {\left (3 f g^2 \sqrt {d-c^2 d x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{8 c^2 \sqrt {1-c^2 x^2}}+\frac {\left (3 b f g^2 \sqrt {d-c^2 d x^2}\right ) \int x \, dx}{8 c \sqrt {1-c^2 x^2}}-\frac {\left (b g^3 \sqrt {d-c^2 d x^2}\right ) \int \left (-2-c^2 x^2+3 c^4 x^4\right ) \, dx}{15 c^3 \sqrt {1-c^2 x^2}}\\ &=\frac {b f^2 g x \sqrt {d-c^2 d x^2}}{c \sqrt {1-c^2 x^2}}+\frac {2 b g^3 x \sqrt {d-c^2 d x^2}}{15 c^3 \sqrt {1-c^2 x^2}}-\frac {b c f^3 x^2 \sqrt {d-c^2 d x^2}}{4 \sqrt {1-c^2 x^2}}+\frac {3 b f g^2 x^2 \sqrt {d-c^2 d x^2}}{16 c \sqrt {1-c^2 x^2}}-\frac {b c f^2 g x^3 \sqrt {d-c^2 d x^2}}{3 \sqrt {1-c^2 x^2}}+\frac {b g^3 x^3 \sqrt {d-c^2 d x^2}}{45 c \sqrt {1-c^2 x^2}}-\frac {3 b c f g^2 x^4 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}}-\frac {b c g^3 x^5 \sqrt {d-c^2 d x^2}}{25 \sqrt {1-c^2 x^2}}+\frac {1}{2} f^3 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {3 f g^2 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{8 c^2}+\frac {3}{4} f g^2 x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {f^2 g \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^2}-\frac {g^3 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c^4}+\frac {g^3 \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^4}+\frac {f^3 \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 {3 f g^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c^3 \sqrt {1-c^2 x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.47, size = 356, normalized size = 0.53 \[ \frac {\sqrt {d-c^2 d x^2} \left (225 a^2 \left (4 c^3 f^3+3 c f g^2\right )+30 a b \sqrt {1-c^2 x^2} \left (6 c^4 x \left (10 f^3+20 f^2 g x+15 f g^2 x^2+4 g^3 x^3\right )-c^2 g \left (120 f^2+45 f g x+8 g^2 x^2\right )-16 g^3\right )+30 b \sin ^{-1}(c x) \left (15 a \left (4 c^3 f^3+3 c f g^2\right )+b \sqrt {1-c^2 x^2} \left (6 c^4 x \left (10 f^3+20 f^2 g x+15 f g^2 x^2+4 g^3 x^3\right )-c^2 g \left (120 f^2+45 f g x+8 g^2 x^2\right )-16 g^3\right )\right )+225 b^2 c f \left (4 c^2 f^2+3 g^2\right ) \sin ^{-1}(c x)^2+b^2 c x \left (-3 c^4 x \left (300 f^3+400 f^2 g x+225 f g^2 x^2+48 g^3 x^3\right )+5 c^2 g \left (720 f^2+135 f g x+16 g^2 x^2\right )+480 g^3\right )\right )}{3600 b c^4 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(f + g*x)^3*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]),x]

[Out]

(Sqrt[d - c^2*d*x^2]*(225*a^2*(4*c^3*f^3 + 3*c*f*g^2) + 30*a*b*Sqrt[1 - c^2*x^2]*(-16*g^3 - c^2*g*(120*f^2 + 4
5*f*g*x + 8*g^2*x^2) + 6*c^4*x*(10*f^3 + 20*f^2*g*x + 15*f*g^2*x^2 + 4*g^3*x^3)) + b^2*c*x*(480*g^3 + 5*c^2*g*
(720*f^2 + 135*f*g*x + 16*g^2*x^2) - 3*c^4*x*(300*f^3 + 400*f^2*g*x + 225*f*g^2*x^2 + 48*g^3*x^3)) + 30*b*(15*
a*(4*c^3*f^3 + 3*c*f*g^2) + b*Sqrt[1 - c^2*x^2]*(-16*g^3 - c^2*g*(120*f^2 + 45*f*g*x + 8*g^2*x^2) + 6*c^4*x*(1
0*f^3 + 20*f^2*g*x + 15*f*g^2*x^2 + 4*g^3*x^3)))*ArcSin[c*x] + 225*b^2*c*f*(4*c^2*f^2 + 3*g^2)*ArcSin[c*x]^2))
/(3600*b*c^4*Sqrt[1 - c^2*x^2])

________________________________________________________________________________________

fricas [F]  time = 0.94, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a g^{3} x^{3} + 3 \, a f g^{2} x^{2} + 3 \, a f^{2} g x + a f^{3} + {\left (b g^{3} x^{3} + 3 \, b f g^{2} x^{2} + 3 \, b f^{2} g x + b f^{3}\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^3*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

integral((a*g^3*x^3 + 3*a*f*g^2*x^2 + 3*a*f^2*g*x + a*f^3 + (b*g^3*x^3 + 3*b*f*g^2*x^2 + 3*b*f^2*g*x + b*f^3)*
arcsin(c*x))*sqrt(-c^2*d*x^2 + d), x)

________________________________________________________________________________________

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)^3*(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

________________________________________________________________________________________

maple [B]  time = 1.56, size = 1286, normalized size = 1.92 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*f^3*d/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))-1/8*b*(-d*(c^2*x^2-1))^(1/2)*f^3/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^3/(c^2*x^2-1)*arcsin(c*x)*x+2/15*b*(-d*(c^2*x^2-1))^
(1/2)*g^3/c^4/(c^2*x^2-1)*arcsin(c*x)-4/15*b*(-d*(c^2*x^2-1))^(1/2)*g^3/(c^2*x^2-1)*arcsin(c*x)*x^4-1/5*a*g^3*
x^2*(-c^2*d*x^2+d)^(3/2)/c^2/d+3/8*a*f*g^2/c^2*x*(-c^2*d*x^2+d)^(1/2)-a*f^2*g/c^2/d*(-c^2*d*x^2+d)^(3/2)+1/2*a
*f^3*x*(-c^2*d*x^2+d)^(1/2)-3/4*a*f*g^2*x*(-c^2*d*x^2+d)^(3/2)/c^2/d+3/8*a*f*g^2/c^2*d/(c^2*d)^(1/2)*arctan((c
^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))+1/4*b*(-d*(c^2*x^2-1))^(1/2)*f^3*c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^2+b*(-
d*(c^2*x^2-1))^(1/2)*g/c^2/(c^2*x^2-1)*arcsin(c*x)*f^2+1/2*b*(-d*(c^2*x^2-1))^(1/2)*f^3*c^2/(c^2*x^2-1)*arcsin
(c*x)*x^3-1/4*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^3+1/5*b*(-d*(c^2*x^2-1
))^(1/2)*g^3*c^2/(c^2*x^2-1)*arcsin(c*x)*x^6-1/15*b*(-d*(c^2*x^2-1))^(1/2)*g^3/c^2/(c^2*x^2-1)*arcsin(c*x)*x^2
+1/25*b*(-d*(c^2*x^2-1))^(1/2)*g^3*c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^5-1/45*b*(-d*(c^2*x^2-1))^(1/2)*g^3/c/(c
^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^3-2/15*b*(-d*(c^2*x^2-1))^(1/2)*g^3/c^3/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x-9/8*b*
(-d*(c^2*x^2-1))^(1/2)*f*g^2/(c^2*x^2-1)*arcsin(c*x)*x^3+3/128*b*(-d*(c^2*x^2-1))^(1/2)*f*g^2/c^3/(c^2*x^2-1)*
(-c^2*x^2+1)^(1/2)-2*b*(-d*(c^2*x^2-1))^(1/2)*g/(c^2*x^2-1)*arcsin(c*x)*x^2*f^2-b*(-d*(c^2*x^2-1))^(1/2)*g/c/(
c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x*f^2+3/16*b*(-d*(c^2*x^2-1))^(1/2)*f*g^2*c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^4-3
/16*b*(-d*(c^2*x^2-1))^(1/2)*f*g^2/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/(c^2*
x^2-1)*(-c^2*x^2+1)^(1/2)*x^3*f^2+3/4*b*(-d*(c^2*x^2-1))^(1/2)*f*g^2*c^2/(c^2*x^2-1)*arcsin(c*x)*x^5+3/8*b*(-d
*(c^2*x^2-1))^(1/2)*f*g^2/c^2/(c^2*x^2-1)*arcsin(c*x)*x+b*(-d*(c^2*x^2-1))^(1/2)*g*c^2/(c^2*x^2-1)*arcsin(c*x)
*x^4*f^2-3/16*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^3/(c^2*x^2-1)*arcsin(c*x)^2*f*g^2-2/15*a*g^3/d/c^4
*(-c^2*d*x^2+d)^(3/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, {\left (\sqrt {-c^{2} d x^{2} + d} x + \frac {\sqrt {d} \arcsin \left (c x\right )}{c}\right )} a f^{3} - \frac {1}{15} \, a g^{3} {\left (\frac {3 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x^{2}}{c^{2} d} + \frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}{c^{4} d}\right )} + \frac {3}{8} \, a f g^{2} {\left (\frac {\sqrt {-c^{2} d x^{2} + d} x}{c^{2}} - \frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x}{c^{2} d} + \frac {\sqrt {d} \arcsin \left (c x\right )}{c^{3}}\right )} - \frac {{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} a f^{2} g}{c^{2} d} + \sqrt {d} \int {\left (b g^{3} x^{3} + 3 \, b f g^{2} x^{2} + 3 \, b f^{2} g x + b f^{3}\right )} \sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^3*(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^3 - 1/15*a*g^3*(3*(-c^2*d*x^2 + d)^(3/2)*x^2/(c^2*d)
+ 2*(-c^2*d*x^2 + d)^(3/2)/(c^4*d)) + 3/8*a*f*g^2*(sqrt(-c^2*d*x^2 + d)*x/c^2 - 2*(-c^2*d*x^2 + d)^(3/2)*x/(c^
2*d) + sqrt(d)*arcsin(c*x)/c^3) - (-c^2*d*x^2 + d)^(3/2)*a*f^2*g/(c^2*d) + sqrt(d)*integrate((b*g^3*x^3 + 3*b*
f*g^2*x^2 + 3*b*f^2*g*x + b*f^3)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (f+g\,x\right )}^3\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {d-c^2\,d\,x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f + g*x)^3*(a + b*asin(c*x))*(d - c^2*d*x^2)^(1/2),x)

[Out]

int((f + g*x)^3*(a + b*asin(c*x))*(d - c^2*d*x^2)^(1/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right ) \left (f + g x\right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)**3*(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)**3, x)

________________________________________________________________________________________