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

Optimal. Leaf size=396 \[ -\frac {b c f x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 \sqrt {1-c^2 x^2}}+\frac {1}{2} f x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {f \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{6 b c \sqrt {1-c^2 x^2}}+\frac {2 b g x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c \sqrt {1-c^2 x^2}}-\frac {g \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2}-\frac {2 b c g x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 \sqrt {1-c^2 x^2}}-\frac {1}{4} b^2 f x \sqrt {d-c^2 d x^2}+\frac {b^2 f \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{4 c \sqrt {1-c^2 x^2}}+\frac {4 b^2 g \sqrt {d-c^2 d x^2}}{9 c^2}+\frac {2 b^2 g \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}{27 c^2} \]

[Out]

4/9*b^2*g*(-c^2*d*x^2+d)^(1/2)/c^2-1/4*b^2*f*x*(-c^2*d*x^2+d)^(1/2)+2/27*b^2*g*(-c^2*x^2+1)*(-c^2*d*x^2+d)^(1/
2)/c^2+1/2*f*x*(a+b*arcsin(c*x))^2*(-c^2*d*x^2+d)^(1/2)-1/3*g*(-c^2*x^2+1)*(a+b*arcsin(c*x))^2*(-c^2*d*x^2+d)^
(1/2)/c^2+1/4*b^2*f*arcsin(c*x)*(-c^2*d*x^2+d)^(1/2)/c/(-c^2*x^2+1)^(1/2)+2/3*b*g*x*(a+b*arcsin(c*x))*(-c^2*d*
x^2+d)^(1/2)/c/(-c^2*x^2+1)^(1/2)-1/2*b*c*f*x^2*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)-2/9*
b*c*g*x^3*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2)/(-c^2*x^2+1)^(1/2)+1/6*f*(a+b*arcsin(c*x))^3*(-c^2*d*x^2+d)^(
1/2)/b/c/(-c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.50, antiderivative size = 396, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {4777, 4763, 4647, 4641, 4627, 321, 216, 4677, 4645, 444, 43} \[ -\frac {b c f x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 \sqrt {1-c^2 x^2}}+\frac {1}{2} f x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2+\frac {f \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{6 b c \sqrt {1-c^2 x^2}}-\frac {2 b c g x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{9 \sqrt {1-c^2 x^2}}+\frac {2 b g x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c \sqrt {1-c^2 x^2}}-\frac {g \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{3 c^2}-\frac {1}{4} b^2 f x \sqrt {d-c^2 d x^2}+\frac {b^2 f \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{4 c \sqrt {1-c^2 x^2}}+\frac {4 b^2 g \sqrt {d-c^2 d x^2}}{9 c^2}+\frac {2 b^2 g \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}{27 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b^2*g*Sqrt[d - c^2*d*x^2])/(9*c^2) - (b^2*f*x*Sqrt[d - c^2*d*x^2])/4 + (2*b^2*g*(1 - c^2*x^2)*Sqrt[d - c^2*
d*x^2])/(27*c^2) + (b^2*f*Sqrt[d - c^2*d*x^2]*ArcSin[c*x])/(4*c*Sqrt[1 - c^2*x^2]) + (2*b*g*x*Sqrt[d - c^2*d*x
^2]*(a + b*ArcSin[c*x]))/(3*c*Sqrt[1 - c^2*x^2]) - (b*c*f*x^2*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(2*Sqrt
[1 - c^2*x^2]) - (2*b*c*g*x^3*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x]))/(9*Sqrt[1 - c^2*x^2]) + (f*x*Sqrt[d - c
^2*d*x^2]*(a + b*ArcSin[c*x])^2)/2 - (g*(1 - c^2*x^2)*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^2)/(3*c^2) + (f*
Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^3)/(6*b*c*Sqrt[1 - c^2*x^2])

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 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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && 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 4645

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

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 )^2 \, 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 )^2 \, 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 )^2+g x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2\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 )^2 \, 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 )^2 \, 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 )^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 )^2}{3 c^2}+\frac {\left (f \sqrt {d-c^2 d x^2}\right ) \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\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 \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt {1-c^2 x^2}}+\frac {\left (2 b g \sqrt {d-c^2 d x^2}\right ) \int \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right ) \, dx}{3 c \sqrt {1-c^2 x^2}}\\ &=\frac {2 b g x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c \sqrt {1-c^2 x^2}}-\frac {b c f x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 \sqrt {1-c^2 x^2}}-\frac {2 b c g x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{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 )^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 )^2}{3 c^2}+\frac {f \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{6 b c \sqrt {1-c^2 x^2}}+\frac {\left (b^2 c^2 f \sqrt {d-c^2 d x^2}\right ) \int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx}{2 \sqrt {1-c^2 x^2}}-\frac {\left (2 b^2 g \sqrt {d-c^2 d x^2}\right ) \int \frac {x \left (1-\frac {c^2 x^2}{3}\right )}{\sqrt {1-c^2 x^2}} \, dx}{3 \sqrt {1-c^2 x^2}}\\ &=-\frac {1}{4} b^2 f x \sqrt {d-c^2 d x^2}+\frac {2 b g x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c \sqrt {1-c^2 x^2}}-\frac {b c f x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 \sqrt {1-c^2 x^2}}-\frac {2 b c g x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{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 )^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 )^2}{3 c^2}+\frac {f \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{6 b c \sqrt {1-c^2 x^2}}+\frac {\left (b^2 f \sqrt {d-c^2 d x^2}\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{4 \sqrt {1-c^2 x^2}}-\frac {\left (b^2 g \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int \frac {1-\frac {c^2 x}{3}}{\sqrt {1-c^2 x}} \, dx,x,x^2\right )}{3 \sqrt {1-c^2 x^2}}\\ &=-\frac {1}{4} b^2 f x \sqrt {d-c^2 d x^2}+\frac {b^2 f \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{4 c \sqrt {1-c^2 x^2}}+\frac {2 b g x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c \sqrt {1-c^2 x^2}}-\frac {b c f x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 \sqrt {1-c^2 x^2}}-\frac {2 b c g x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{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 )^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 )^2}{3 c^2}+\frac {f \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{6 b c \sqrt {1-c^2 x^2}}-\frac {\left (b^2 g \sqrt {d-c^2 d x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {2}{3 \sqrt {1-c^2 x}}+\frac {1}{3} \sqrt {1-c^2 x}\right ) \, dx,x,x^2\right )}{3 \sqrt {1-c^2 x^2}}\\ &=\frac {4 b^2 g \sqrt {d-c^2 d x^2}}{9 c^2}-\frac {1}{4} b^2 f x \sqrt {d-c^2 d x^2}+\frac {2 b^2 g \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}{27 c^2}+\frac {b^2 f \sqrt {d-c^2 d x^2} \sin ^{-1}(c x)}{4 c \sqrt {1-c^2 x^2}}+\frac {2 b g x \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 c \sqrt {1-c^2 x^2}}-\frac {b c f x^2 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 \sqrt {1-c^2 x^2}}-\frac {2 b c g x^3 \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )}{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 )^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 )^2}{3 c^2}+\frac {f \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{6 b c \sqrt {1-c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 224, normalized size = 0.57 \[ \frac {\sqrt {d-c^2 d x^2} \left (-27 b^2 c f \left (c x \left (2 a c x+b \sqrt {1-c^2 x^2}\right )+b \left (2 c^2 x^2-1\right ) \sin ^{-1}(c x)\right )+8 b^2 g \left (-3 c^3 x^3 \left (a+b \sin ^{-1}(c x)\right )+9 c x \left (a+b \sin ^{-1}(c x)\right )-b \sqrt {1-c^2 x^2} \left (c^2 x^2-7\right )\right )+54 b c^2 f x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2-36 b g \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2+18 c f \left (a+b \sin ^{-1}(c x)\right )^3\right )}{108 b c^2 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d - c^2*d*x^2]*(54*b*c^2*f*x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2 - 36*b*g*(1 - c^2*x^2)^(3/2)*(a + b
*ArcSin[c*x])^2 + 18*c*f*(a + b*ArcSin[c*x])^3 - 27*b^2*c*f*(c*x*(2*a*c*x + b*Sqrt[1 - c^2*x^2]) + b*(-1 + 2*c
^2*x^2)*ArcSin[c*x]) + 8*b^2*g*(-(b*Sqrt[1 - c^2*x^2]*(-7 + c^2*x^2)) + 9*c*x*(a + b*ArcSin[c*x]) - 3*c^3*x^3*
(a + b*ArcSin[c*x]))))/(108*b*c^2*Sqrt[1 - c^2*x^2])

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {-c^{2} d x^{2} + d} {\left (a^{2} g x + a^{2} f + {\left (b^{2} g x + b^{2} f\right )} \arcsin \left (c x\right )^{2} + 2 \, {\left (a b g x + a b f\right )} \arcsin \left (c x\right )\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-c^2*d*x^2 + d)*(a^2*g*x + a^2*f + (b^2*g*x + b^2*f)*arcsin(c*x)^2 + 2*(a*b*g*x + a*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))^2*(-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 = 0.79, size = 1139, normalized size = 2.88 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3*a*b*(-d*(c^2*x^2-1))^(1/2)*g/(c^2*x^2-1)*arcsin(c*x)*x^2+a*b*(-d*(c^2*x^2-1))^(1/2)*f*c^2/(c^2*x^2-1)*arc
sin(c*x)*x^3+2/9*a*b*(-d*(c^2*x^2-1))^(1/2)*g*c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^3-2/3*a*b*(-d*(c^2*x^2-1))^(1
/2)*g/c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x+1/2*a*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
-1/2*a*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+2/3*a*b*(-d*(c^2*x^2-1))^(1/2
)*g*c^2/(c^2*x^2-1)*arcsin(c*x)*x^4+2/9*b^2*(-d*(c^2*x^2-1))^(1/2)*g*c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*arcsin(c
*x)*x^3-2/3*b^2*(-d*(c^2*x^2-1))^(1/2)*g/c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*x+1/2*b^2*(-d*(c^2*x^2-1
))^(1/2)*f*c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)*x^2-2/27*b^2*(-d*(c^2*x^2-1))^(1/2)*g*c^2/(c^2*x^2-1)*
x^4+1/3*b^2*(-d*(c^2*x^2-1))^(1/2)*g/c^2/(c^2*x^2-1)*arcsin(c*x)^2-1/4*b^2*(-d*(c^2*x^2-1))^(1/2)*f*c^2/(c^2*x
^2-1)*x^3-2/3*b^2*(-d*(c^2*x^2-1))^(1/2)*g/(c^2*x^2-1)*arcsin(c*x)^2*x^2-1/2*b^2*(-d*(c^2*x^2-1))^(1/2)*f/(c^2
*x^2-1)*arcsin(c*x)^2*x+1/4*b^2*(-d*(c^2*x^2-1))^(1/2)*f/(c^2*x^2-1)*x+1/2*a^2*f*d/(c^2*d)^(1/2)*arctan((c^2*d
)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))-1/3*a^2*g/c^2/d*(-c^2*d*x^2+d)^(3/2)-1/4*a*b*(-d*(c^2*x^2-1))^(1/2)*f/c/(c^2*x
^2-1)*(-c^2*x^2+1)^(1/2)-a*b*(-d*(c^2*x^2-1))^(1/2)*f/(c^2*x^2-1)*arcsin(c*x)*x+2/3*a*b*(-d*(c^2*x^2-1))^(1/2)
*g/c^2/(c^2*x^2-1)*arcsin(c*x)-1/4*b^2*(-d*(c^2*x^2-1))^(1/2)*f/c/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*arcsin(c*x)-1
/6*b^2*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c/(c^2*x^2-1)*arcsin(c*x)^3*f+1/3*b^2*(-d*(c^2*x^2-1))^(1/2)*
g*c^2/(c^2*x^2-1)*arcsin(c*x)^2*x^4+1/2*b^2*(-d*(c^2*x^2-1))^(1/2)*f*c^2/(c^2*x^2-1)*arcsin(c*x)^2*x^3-14/27*b
^2*(-d*(c^2*x^2-1))^(1/2)*g/c^2/(c^2*x^2-1)+16/27*b^2*(-d*(c^2*x^2-1))^(1/2)*g/(c^2*x^2-1)*x^2+1/2*a^2*f*x*(-c
^2*d*x^2+d)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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