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

Optimal. Leaf size=959 \[ \frac {b c^3 d g^3 \sqrt {d-c^2 d x^2} x^7}{49 \sqrt {1-c^2 x^2}}+\frac {b c^3 d f g^2 \sqrt {d-c^2 d x^2} x^6}{12 \sqrt {1-c^2 x^2}}-\frac {8 b c d g^3 \sqrt {d-c^2 d x^2} x^5}{175 \sqrt {1-c^2 x^2}}+\frac {3 b c^3 d f^2 g \sqrt {d-c^2 d x^2} x^5}{25 \sqrt {1-c^2 x^2}}+\frac {b c^3 d f^3 \sqrt {d-c^2 d x^2} x^4}{16 \sqrt {1-c^2 x^2}}-\frac {7 b c d f g^2 \sqrt {d-c^2 d x^2} x^4}{32 \sqrt {1-c^2 x^2}}+\frac {3}{8} d f g^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) x^3+\frac {1}{2} d f g^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) x^3+\frac {b d g^3 \sqrt {d-c^2 d x^2} x^3}{105 c \sqrt {1-c^2 x^2}}-\frac {2 b c d f^2 g \sqrt {d-c^2 d x^2} x^3}{5 \sqrt {1-c^2 x^2}}-\frac {5 b c d f^3 \sqrt {d-c^2 d x^2} x^2}{16 \sqrt {1-c^2 x^2}}+\frac {3 b d f g^2 \sqrt {d-c^2 d x^2} x^2}{32 c \sqrt {1-c^2 x^2}}+\frac {3}{8} d f^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) x-\frac {3 d f g^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) x}{16 c^2}+\frac {1}{4} d f^3 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) x+\frac {2 b d g^3 \sqrt {d-c^2 d x^2} x}{35 c^3 \sqrt {1-c^2 x^2}}+\frac {3 b d f^2 g \sqrt {d-c^2 d x^2} x}{5 c \sqrt {1-c^2 x^2}}+\frac {3 d f^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \sqrt {1-c^2 x^2}}+\frac {3 d f g^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{32 b c^3 \sqrt {1-c^2 x^2}}+\frac {d g^3 \left (1-c^2 x^2\right )^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{7 c^4}-\frac {d g^3 \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^4}-\frac {3 d f^2 g \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^2} \]

[Out]

3/8*d*f^3*x*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)-3/16*d*f*g^2*x*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/c^2+3
/8*d*f*g^2*x^3*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)+1/4*d*f^3*x*(-c^2*x^2+1)*(a+b*arcsin(c*x))*(-c^2*d*x^2+d
)^(1/2)+1/2*d*f*g^2*x^3*(-c^2*x^2+1)*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)-3/5*d*f^2*g*(-c^2*x^2+1)^2*(a+b*ar
csin(c*x))*(-c^2*d*x^2+d)^(1/2)/c^2-1/5*d*g^3*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/c^4+1/7*d*
g^3*(-c^2*x^2+1)^3*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/c^4+3/5*b*d*f^2*g*x*(-c^2*d*x^2+d)^(1/2)/c/(-c^2*x^2
+1)^(1/2)+2/35*b*d*g^3*x*(-c^2*d*x^2+d)^(1/2)/c^3/(-c^2*x^2+1)^(1/2)-5/16*b*c*d*f^3*x^2*(-c^2*d*x^2+d)^(1/2)/(
-c^2*x^2+1)^(1/2)+3/32*b*d*f*g^2*x^2*(-c^2*d*x^2+d)^(1/2)/c/(-c^2*x^2+1)^(1/2)-2/5*b*c*d*f^2*g*x^3*(-c^2*d*x^2
+d)^(1/2)/(-c^2*x^2+1)^(1/2)+1/105*b*d*g^3*x^3*(-c^2*d*x^2+d)^(1/2)/c/(-c^2*x^2+1)^(1/2)+1/16*b*c^3*d*f^3*x^4*
(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)-7/32*b*c*d*f*g^2*x^4*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+3/25*b*c^
3*d*f^2*g*x^5*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)-8/175*b*c*d*g^3*x^5*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1
/2)+1/12*b*c^3*d*f*g^2*x^6*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+1/49*b*c^3*d*g^3*x^7*(-c^2*d*x^2+d)^(1/2)/(
-c^2*x^2+1)^(1/2)+3/16*d*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)+3/32*d*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 = 0.94, antiderivative size = 959, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 17, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.548, Rules used = {4777, 4763, 4649, 4647, 4641, 30, 14, 4677, 194, 4699, 4697, 4707, 266, 43, 4689, 12, 373} \[ \frac {b c^3 d g^3 \sqrt {d-c^2 d x^2} x^7}{49 \sqrt {1-c^2 x^2}}+\frac {b c^3 d f g^2 \sqrt {d-c^2 d x^2} x^6}{12 \sqrt {1-c^2 x^2}}-\frac {8 b c d g^3 \sqrt {d-c^2 d x^2} x^5}{175 \sqrt {1-c^2 x^2}}+\frac {3 b c^3 d f^2 g \sqrt {d-c^2 d x^2} x^5}{25 \sqrt {1-c^2 x^2}}+\frac {b c^3 d f^3 \sqrt {d-c^2 d x^2} x^4}{16 \sqrt {1-c^2 x^2}}-\frac {7 b c d f g^2 \sqrt {d-c^2 d x^2} x^4}{32 \sqrt {1-c^2 x^2}}+\frac {3}{8} d f g^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) x^3+\frac {1}{2} d f g^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) x^3+\frac {b d g^3 \sqrt {d-c^2 d x^2} x^3}{105 c \sqrt {1-c^2 x^2}}-\frac {2 b c d f^2 g \sqrt {d-c^2 d x^2} x^3}{5 \sqrt {1-c^2 x^2}}-\frac {5 b c d f^3 \sqrt {d-c^2 d x^2} x^2}{16 \sqrt {1-c^2 x^2}}+\frac {3 b d f g^2 \sqrt {d-c^2 d x^2} x^2}{32 c \sqrt {1-c^2 x^2}}+\frac {3}{8} d f^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) x-\frac {3 d f g^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) x}{16 c^2}+\frac {1}{4} d f^3 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right ) x+\frac {2 b d g^3 \sqrt {d-c^2 d x^2} x}{35 c^3 \sqrt {1-c^2 x^2}}+\frac {3 b d f^2 g \sqrt {d-c^2 d x^2} x}{5 c \sqrt {1-c^2 x^2}}+\frac {3 d f^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \sqrt {1-c^2 x^2}}+\frac {3 d f g^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{32 b c^3 \sqrt {1-c^2 x^2}}+\frac {d g^3 \left (1-c^2 x^2\right )^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{7 c^4}-\frac {d g^3 \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^4}-\frac {3 d f^2 g \left (1-c^2 x^2\right )^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{5 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*d*f^2*g*x*Sqrt[d - c^2*d*x^2])/(5*c*Sqrt[1 - c^2*x^2]) + (2*b*d*g^3*x*Sqrt[d - c^2*d*x^2])/(35*c^3*Sqrt[1
- c^2*x^2]) - (5*b*c*d*f^3*x^2*Sqrt[d - c^2*d*x^2])/(16*Sqrt[1 - c^2*x^2]) + (3*b*d*f*g^2*x^2*Sqrt[d - c^2*d*
x^2])/(32*c*Sqrt[1 - c^2*x^2]) - (2*b*c*d*f^2*g*x^3*Sqrt[d - c^2*d*x^2])/(5*Sqrt[1 - c^2*x^2]) + (b*d*g^3*x^3*
Sqrt[d - c^2*d*x^2])/(105*c*Sqrt[1 - c^2*x^2]) + (b*c^3*d*f^3*x^4*Sqrt[d - c^2*d*x^2])/(16*Sqrt[1 - c^2*x^2])
- (7*b*c*d*f*g^2*x^4*Sqrt[d - c^2*d*x^2])/(32*Sqrt[1 - c^2*x^2]) + (3*b*c^3*d*f^2*g*x^5*Sqrt[d - c^2*d*x^2])/(
25*Sqrt[1 - c^2*x^2]) - (8*b*c*d*g^3*x^5*Sqrt[d - c^2*d*x^2])/(175*Sqrt[1 - c^2*x^2]) + (b*c^3*d*f*g^2*x^6*Sqr
t[d - c^2*d*x^2])/(12*Sqrt[1 - c^2*x^2]) + (b*c^3*d*g^3*x^7*Sqrt[d - c^2*d*x^2])/(49*Sqrt[1 - c^2*x^2]) + (3*d
*f^3*x*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/8 - (3*d*f*g^2*x*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(16*
c^2) + (3*d*f*g^2*x^3*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/8 + (d*f^3*x*(1 - c^2*x^2)*Sqrt[d - c^2*d*x^2]*
(a + b*ArcSin[c*x]))/4 + (d*f*g^2*x^3*(1 - c^2*x^2)*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/2 - (3*d*f^2*g*(1
- c^2*x^2)^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(5*c^2) - (d*g^3*(1 - c^2*x^2)^2*Sqrt[d - c^2*d*x^2]*(a
+ b*ArcSin[c*x]))/(5*c^4) + (d*g^3*(1 - c^2*x^2)^3*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(7*c^4) + (3*d*f^
3*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^2)/(16*b*c*Sqrt[1 - c^2*x^2]) + (3*d*f*g^2*Sqrt[d - c^2*d*x^2]*(a +
b*ArcSin[c*x])^2)/(32*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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 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 373

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 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 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 4649

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(x*(d + e*x^2)^p*(
a + b*ArcSin[c*x])^n)/(2*p + 1), x] + (Dist[(2*d*p)/(2*p + 1), Int[(d + e*x^2)^(p - 1)*(a + b*ArcSin[c*x])^n,
x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/((2*p + 1)*(1 - c^2*x^2)^FracPart[p]), Int[x*(1 - c
^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && Gt
Q[n, 0] && GtQ[p, 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 4699

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[
((f*x)^(m + 1)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n)/(f*(m + 2*p + 1)), x] + (Dist[(2*d*p)/(m + 2*p + 1), Int[(
f*x)^m*(d + e*x^2)^(p - 1)*(a + b*ArcSin[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(
f*(m + 2*p + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p - 1/2)*(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] && GtQ[p, 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 \left (d-c^2 d x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac {\left (d \sqrt {d-c^2 d x^2}\right ) \int (f+g x)^3 \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt {1-c^2 x^2}}\\ &=\frac {\left (d \sqrt {d-c^2 d x^2}\right ) \int \left (f^3 \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )+3 f^2 g x \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )+3 f g^2 x^2 \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )+g^3 x^3 \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )\right ) \, dx}{\sqrt {1-c^2 x^2}}\\ &=\frac {\left (d f^3 \sqrt {d-c^2 d x^2}\right ) \int \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt {1-c^2 x^2}}+\frac {\left (3 d f^2 g \sqrt {d-c^2 d x^2}\right ) \int x \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt {1-c^2 x^2}}+\frac {\left (3 d f g^2 \sqrt {d-c^2 d x^2}\right ) \int x^2 \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt {1-c^2 x^2}}+\frac {\left (d g^3 \sqrt {d-c^2 d x^2}\right ) \int x^3 \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt {1-c^2 x^2}}\\ &=\frac {1}{4} d f^3 x \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{2} d f g^2 x^3 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {3 d f^2 g \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^2}-\frac {d 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 {d g^3 \left (1-c^2 x^2\right )^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{7 c^4}+\frac {\left (3 d 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}{4 \sqrt {1-c^2 x^2}}-\frac {\left (b c d f^3 \sqrt {d-c^2 d x^2}\right ) \int x \left (1-c^2 x^2\right ) \, dx}{4 \sqrt {1-c^2 x^2}}+\frac {\left (3 b d f^2 g \sqrt {d-c^2 d x^2}\right ) \int \left (1-c^2 x^2\right )^2 \, dx}{5 c \sqrt {1-c^2 x^2}}+\frac {\left (3 d 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}{2 \sqrt {1-c^2 x^2}}-\frac {\left (b c d f g^2 \sqrt {d-c^2 d x^2}\right ) \int x^3 \left (1-c^2 x^2\right ) \, dx}{2 \sqrt {1-c^2 x^2}}-\frac {\left (b c d g^3 \sqrt {d-c^2 d x^2}\right ) \int \frac {\left (-2-5 c^2 x^2\right ) \left (1-c^2 x^2\right )^2}{35 c^4} \, dx}{\sqrt {1-c^2 x^2}}\\ &=\frac {3}{8} d f^3 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {3}{8} d f g^2 x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} d f^3 x \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{2} d f g^2 x^3 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {3 d f^2 g \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^2}-\frac {d 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 {d g^3 \left (1-c^2 x^2\right )^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{7 c^4}+\frac {\left (3 d 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}{8 \sqrt {1-c^2 x^2}}-\frac {\left (b c d f^3 \sqrt {d-c^2 d x^2}\right ) \int \left (x-c^2 x^3\right ) \, dx}{4 \sqrt {1-c^2 x^2}}-\frac {\left (3 b c d f^3 \sqrt {d-c^2 d x^2}\right ) \int x \, dx}{8 \sqrt {1-c^2 x^2}}+\frac {\left (3 b d f^2 g \sqrt {d-c^2 d x^2}\right ) \int \left (1-2 c^2 x^2+c^4 x^4\right ) \, dx}{5 c \sqrt {1-c^2 x^2}}+\frac {\left (3 d 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}{8 \sqrt {1-c^2 x^2}}-\frac {\left (3 b c d f g^2 \sqrt {d-c^2 d x^2}\right ) \int x^3 \, dx}{8 \sqrt {1-c^2 x^2}}-\frac {\left (b c d f g^2 \sqrt {d-c^2 d x^2}\right ) \int \left (x^3-c^2 x^5\right ) \, dx}{2 \sqrt {1-c^2 x^2}}-\frac {\left (b d g^3 \sqrt {d-c^2 d x^2}\right ) \int \left (-2-5 c^2 x^2\right ) \left (1-c^2 x^2\right )^2 \, dx}{35 c^3 \sqrt {1-c^2 x^2}}\\ &=\frac {3 b d f^2 g x \sqrt {d-c^2 d x^2}}{5 c \sqrt {1-c^2 x^2}}-\frac {5 b c d f^3 x^2 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}}-\frac {2 b c d f^2 g x^3 \sqrt {d-c^2 d x^2}}{5 \sqrt {1-c^2 x^2}}+\frac {b c^3 d f^3 x^4 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}}-\frac {7 b c d f g^2 x^4 \sqrt {d-c^2 d x^2}}{32 \sqrt {1-c^2 x^2}}+\frac {3 b c^3 d f^2 g x^5 \sqrt {d-c^2 d x^2}}{25 \sqrt {1-c^2 x^2}}+\frac {b c^3 d f g^2 x^6 \sqrt {d-c^2 d x^2}}{12 \sqrt {1-c^2 x^2}}+\frac {3}{8} d f^3 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {3 d f g^2 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{16 c^2}+\frac {3}{8} d f g^2 x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} d f^3 x \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{2} d f g^2 x^3 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {3 d f^2 g \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^2}-\frac {d 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 {d g^3 \left (1-c^2 x^2\right )^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{7 c^4}+\frac {3 d f^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \sqrt {1-c^2 x^2}}+\frac {\left (3 d 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}{16 c^2 \sqrt {1-c^2 x^2}}+\frac {\left (3 b d f g^2 \sqrt {d-c^2 d x^2}\right ) \int x \, dx}{16 c \sqrt {1-c^2 x^2}}-\frac {\left (b d g^3 \sqrt {d-c^2 d x^2}\right ) \int \left (-2-c^2 x^2+8 c^4 x^4-5 c^6 x^6\right ) \, dx}{35 c^3 \sqrt {1-c^2 x^2}}\\ &=\frac {3 b d f^2 g x \sqrt {d-c^2 d x^2}}{5 c \sqrt {1-c^2 x^2}}+\frac {2 b d g^3 x \sqrt {d-c^2 d x^2}}{35 c^3 \sqrt {1-c^2 x^2}}-\frac {5 b c d f^3 x^2 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}}+\frac {3 b d f g^2 x^2 \sqrt {d-c^2 d x^2}}{32 c \sqrt {1-c^2 x^2}}-\frac {2 b c d f^2 g x^3 \sqrt {d-c^2 d x^2}}{5 \sqrt {1-c^2 x^2}}+\frac {b d g^3 x^3 \sqrt {d-c^2 d x^2}}{105 c \sqrt {1-c^2 x^2}}+\frac {b c^3 d f^3 x^4 \sqrt {d-c^2 d x^2}}{16 \sqrt {1-c^2 x^2}}-\frac {7 b c d f g^2 x^4 \sqrt {d-c^2 d x^2}}{32 \sqrt {1-c^2 x^2}}+\frac {3 b c^3 d f^2 g x^5 \sqrt {d-c^2 d x^2}}{25 \sqrt {1-c^2 x^2}}-\frac {8 b c d g^3 x^5 \sqrt {d-c^2 d x^2}}{175 \sqrt {1-c^2 x^2}}+\frac {b c^3 d f g^2 x^6 \sqrt {d-c^2 d x^2}}{12 \sqrt {1-c^2 x^2}}+\frac {b c^3 d g^3 x^7 \sqrt {d-c^2 d x^2}}{49 \sqrt {1-c^2 x^2}}+\frac {3}{8} d f^3 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {3 d f g^2 x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{16 c^2}+\frac {3}{8} d f g^2 x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} d f^3 x \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{2} d f g^2 x^3 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {3 d f^2 g \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^2}-\frac {d 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 {d g^3 \left (1-c^2 x^2\right )^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{7 c^4}+\frac {3 d f^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \sqrt {1-c^2 x^2}}+\frac {3 d f g^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{32 b c^3 \sqrt {1-c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 1.22, size = 463, normalized size = 0.48 \[ \frac {d \sqrt {d-c^2 d x^2} \left (11025 a^2 c f \left (2 c^2 f^2+g^2\right )-210 a b \sqrt {1-c^2 x^2} \left (4 c^6 x^3 \left (35 f^3+84 f^2 g x+70 f g^2 x^2+20 g^3 x^3\right )-2 c^4 x \left (175 f^3+336 f^2 g x+245 f g^2 x^2+64 g^3 x^3\right )+c^2 g \left (336 f^2+105 f g x+16 g^2 x^2\right )+32 g^3\right )-210 b \sin ^{-1}(c x) \left (b \sqrt {1-c^2 x^2} \left (4 c^6 x^3 \left (35 f^3+84 f^2 g x+70 f g^2 x^2+20 g^3 x^3\right )-2 c^4 x \left (175 f^3+336 f^2 g x+245 f g^2 x^2+64 g^3 x^3\right )+c^2 g \left (336 f^2+105 f g x+16 g^2 x^2\right )+32 g^3\right )-105 a c f \left (2 c^2 f^2+g^2\right )\right )+11025 b^2 c f \left (2 c^2 f^2+g^2\right ) \sin ^{-1}(c x)^2+b^2 c x \left (2 c^6 x^3 \left (3675 f^3+7056 f^2 g x+4900 f g^2 x^2+1200 g^3 x^3\right )-21 c^4 x \left (1750 f^3+2240 f^2 g x+1225 f g^2 x^2+256 g^3 x^3\right )+35 c^2 g \left (2016 f^2+315 f g x+32 g^2 x^2\right )+6720 g^3\right )\right )}{117600 b c^4 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*Sqrt[d - c^2*d*x^2]*(11025*a^2*c*f*(2*c^2*f^2 + g^2) - 210*a*b*Sqrt[1 - c^2*x^2]*(32*g^3 + c^2*g*(336*f^2 +
105*f*g*x + 16*g^2*x^2) + 4*c^6*x^3*(35*f^3 + 84*f^2*g*x + 70*f*g^2*x^2 + 20*g^3*x^3) - 2*c^4*x*(175*f^3 + 33
6*f^2*g*x + 245*f*g^2*x^2 + 64*g^3*x^3)) + b^2*c*x*(6720*g^3 + 35*c^2*g*(2016*f^2 + 315*f*g*x + 32*g^2*x^2) -
21*c^4*x*(1750*f^3 + 2240*f^2*g*x + 1225*f*g^2*x^2 + 256*g^3*x^3) + 2*c^6*x^3*(3675*f^3 + 7056*f^2*g*x + 4900*
f*g^2*x^2 + 1200*g^3*x^3)) - 210*b*(-105*a*c*f*(2*c^2*f^2 + g^2) + b*Sqrt[1 - c^2*x^2]*(32*g^3 + c^2*g*(336*f^
2 + 105*f*g*x + 16*g^2*x^2) + 4*c^6*x^3*(35*f^3 + 84*f^2*g*x + 70*f*g^2*x^2 + 20*g^3*x^3) - 2*c^4*x*(175*f^3 +
336*f^2*g*x + 245*f*g^2*x^2 + 64*g^3*x^3)))*ArcSin[c*x] + 11025*b^2*c*f*(2*c^2*f^2 + g^2)*ArcSin[c*x]^2))/(11
7600*b*c^4*Sqrt[1 - c^2*x^2])

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a c^{2} d g^{3} x^{5} + 3 \, a c^{2} d f g^{2} x^{4} - 3 \, a d f^{2} g x - a d f^{3} + {\left (3 \, a c^{2} d f^{2} g - a d g^{3}\right )} x^{3} + {\left (a c^{2} d f^{3} - 3 \, a d f g^{2}\right )} x^{2} + {\left (b c^{2} d g^{3} x^{5} + 3 \, b c^{2} d f g^{2} x^{4} - 3 \, b d f^{2} g x - b d f^{3} + {\left (3 \, b c^{2} d f^{2} g - b d g^{3}\right )} x^{3} + {\left (b c^{2} d f^{3} - 3 \, b d f g^{2}\right )} x^{2}\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} d x^{2} + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(a*c^2*d*g^3*x^5 + 3*a*c^2*d*f*g^2*x^4 - 3*a*d*f^2*g*x - a*d*f^3 + (3*a*c^2*d*f^2*g - a*d*g^3)*x^3 +
(a*c^2*d*f^3 - 3*a*d*f*g^2)*x^2 + (b*c^2*d*g^3*x^5 + 3*b*c^2*d*f*g^2*x^4 - 3*b*d*f^2*g*x - b*d*f^3 + (3*b*c^2
*d*f^2*g - b*d*g^3)*x^3 + (b*c^2*d*f^3 - 3*b*d*f*g^2)*x^2)*arcsin(c*x))*sqrt(-c^2*d*x^2 + d), 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)^3*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x)),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 [C]  time = 1.37, size = 5085, normalized size = 5.30 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{8} \, {\left (2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x + 3 \, \sqrt {-c^{2} d x^{2} + d} d x + \frac {3 \, d^{\frac {3}{2}} \arcsin \left (c x\right )}{c}\right )} a f^{3} - \frac {1}{35} \, {\left (\frac {5 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x^{2}}{c^{2} d} + \frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}}{c^{4} d}\right )} a g^{3} + \frac {1}{16} \, a f g^{2} {\left (\frac {2 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x}{c^{2}} - \frac {8 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} x}{c^{2} d} + \frac {3 \, \sqrt {-c^{2} d x^{2} + d} d x}{c^{2}} + \frac {3 \, d^{\frac {3}{2}} \arcsin \left (c x\right )}{c^{3}}\right )} - \frac {3 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}} a f^{2} g}{5 \, c^{2} d} + \sqrt {d} \int -{\left (b c^{2} d g^{3} x^{5} + 3 \, b c^{2} d f g^{2} x^{4} - 3 \, b d f^{2} g x - b d f^{3} + {\left (3 \, b c^{2} d f^{2} g - b d g^{3}\right )} x^{3} + {\left (b c^{2} d f^{3} - 3 \, b d f g^{2}\right )} x^{2}\right )} \sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*(-c^2*d*x^2 + d)^(3/2)*x + 3*sqrt(-c^2*d*x^2 + d)*d*x + 3*d^(3/2)*arcsin(c*x)/c)*a*f^3 - 1/35*(5*(-c^2*
d*x^2 + d)^(5/2)*x^2/(c^2*d) + 2*(-c^2*d*x^2 + d)^(5/2)/(c^4*d))*a*g^3 + 1/16*a*f*g^2*(2*(-c^2*d*x^2 + d)^(3/2
)*x/c^2 - 8*(-c^2*d*x^2 + d)^(5/2)*x/(c^2*d) + 3*sqrt(-c^2*d*x^2 + d)*d*x/c^2 + 3*d^(3/2)*arcsin(c*x)/c^3) - 3
/5*(-c^2*d*x^2 + d)^(5/2)*a*f^2*g/(c^2*d) + sqrt(d)*integrate(-(b*c^2*d*g^3*x^5 + 3*b*c^2*d*f*g^2*x^4 - 3*b*d*
f^2*g*x - b*d*f^3 + (3*b*c^2*d*f^2*g - b*d*g^3)*x^3 + (b*c^2*d*f^3 - 3*b*d*f*g^2)*x^2)*sqrt(c*x + 1)*sqrt(-c*x
+ 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)), x)

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

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)^(3/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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