3.6.5 \(\int (d+c d x)^{3/2} \sqrt {f-c f x} (a+b \text {ArcSin}(c x)) \, dx\) [505]

Optimal. Leaf size=273 \[ \frac {b d x \sqrt {d+c d x} \sqrt {f-c f x}}{3 \sqrt {1-c^2 x^2}}-\frac {b c d x^2 \sqrt {d+c d x} \sqrt {f-c f x}}{4 \sqrt {1-c^2 x^2}}-\frac {b c^2 d x^3 \sqrt {d+c d x} \sqrt {f-c f x}}{9 \sqrt {1-c^2 x^2}}+\frac {1}{2} d x \sqrt {d+c d x} \sqrt {f-c f x} (a+b \text {ArcSin}(c x))-\frac {d \sqrt {d+c d x} \sqrt {f-c f x} \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{3 c}+\frac {d \sqrt {d+c d x} \sqrt {f-c f x} (a+b \text {ArcSin}(c x))^2}{4 b c \sqrt {1-c^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.21, antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4763, 4847, 4741, 4737, 30, 4767} \begin {gather*} \frac {d \sqrt {c d x+d} \sqrt {f-c f x} (a+b \text {ArcSin}(c x))^2}{4 b c \sqrt {1-c^2 x^2}}-\frac {d \left (1-c^2 x^2\right ) \sqrt {c d x+d} \sqrt {f-c f x} (a+b \text {ArcSin}(c x))}{3 c}+\frac {1}{2} d x \sqrt {c d x+d} \sqrt {f-c f x} (a+b \text {ArcSin}(c x))-\frac {b c d x^2 \sqrt {c d x+d} \sqrt {f-c f x}}{4 \sqrt {1-c^2 x^2}}+\frac {b d x \sqrt {c d x+d} \sqrt {f-c f x}}{3 \sqrt {1-c^2 x^2}}-\frac {b c^2 d x^3 \sqrt {c d x+d} \sqrt {f-c f x}}{9 \sqrt {1-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(3*Sqrt[1 - c^2*x^2]) - (b*c*d*x^2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(4
*Sqrt[1 - c^2*x^2]) - (b*c^2*d*x^3*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(9*Sqrt[1 - c^2*x^2]) + (d*x*Sqrt[d + c*d*
x]*Sqrt[f - c*f*x]*(a + b*ArcSin[c*x]))/2 - (d*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(1 - c^2*x^2)*(a + b*ArcSin[c*x
]))/(3*c) + (d*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(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 4737

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

Rule 4741

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[(1/2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]], Int[(a + b*ArcSin[c*x])^n/S
qrt[1 - c^2*x^2], x], x] - Dist[b*c*(n/2)*Simp[Sqrt[d + e*x^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 4763

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :> D
ist[(d + e*x)^q*((f + g*x)^q/(1 - c^2*x^2)^q), Int[(d + e*x)^(p - q)*(1 - c^2*x^2)^q*(a + b*ArcSin[c*x])^n, x]
, x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q]
&& GeQ[p - q, 0]

Rule 4767

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^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 4847

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))

Rubi steps

\begin {align*} \int (d+c d x)^{3/2} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac {\left (\sqrt {d+c d x} \sqrt {f-c f x}\right ) \int (d+c d x) \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt {1-c^2 x^2}}\\ &=\frac {\left (\sqrt {d+c d x} \sqrt {f-c f x}\right ) \int \left (d \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )+c d 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 (d \sqrt {d+c d x} \sqrt {f-c f x}\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 (c d \sqrt {d+c d x} \sqrt {f-c f x}\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} d x \sqrt {d+c d x} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )-\frac {d \sqrt {d+c d x} \sqrt {f-c f x} \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c}+\frac {\left (d \sqrt {d+c d x} \sqrt {f-c f x}\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 d \sqrt {d+c d x} \sqrt {f-c f x}\right ) \int \left (1-c^2 x^2\right ) \, dx}{3 \sqrt {1-c^2 x^2}}-\frac {\left (b c d \sqrt {d+c d x} \sqrt {f-c f x}\right ) \int x \, dx}{2 \sqrt {1-c^2 x^2}}\\ &=\frac {b d x \sqrt {d+c d x} \sqrt {f-c f x}}{3 \sqrt {1-c^2 x^2}}-\frac {b c d x^2 \sqrt {d+c d x} \sqrt {f-c f x}}{4 \sqrt {1-c^2 x^2}}-\frac {b c^2 d x^3 \sqrt {d+c d x} \sqrt {f-c f x}}{9 \sqrt {1-c^2 x^2}}+\frac {1}{2} d x \sqrt {d+c d x} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )-\frac {d \sqrt {d+c d x} \sqrt {f-c f x} \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c}+\frac {d \sqrt {d+c d x} \sqrt {f-c f x} \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 = 0.55, size = 260, normalized size = 0.95 \begin {gather*} \frac {18 b d \sqrt {d+c d x} \sqrt {f-c f x} \text {ArcSin}(c x)^2-36 a d^{3/2} \sqrt {f} \sqrt {1-c^2 x^2} \text {ArcTan}\left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )+d \sqrt {d+c d x} \sqrt {f-c f x} \left (-8 b c x \left (-3+c^2 x^2\right )+12 a \sqrt {1-c^2 x^2} \left (-2+3 c x+2 c^2 x^2\right )+9 b \cos (2 \text {ArcSin}(c x))\right )+6 b d \sqrt {d+c d x} \sqrt {f-c f x} \text {ArcSin}(c x) \left (-4 \left (1-c^2 x^2\right )^{3/2}+3 \sin (2 \text {ArcSin}(c x))\right )}{72 c \sqrt {1-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(18*b*d*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2 - 36*a*d^(3/2)*Sqrt[f]*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqr
t[d + c*d*x]*Sqrt[f - c*f*x])/(Sqrt[d]*Sqrt[f]*(-1 + c^2*x^2))] + d*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(-8*b*c*x*
(-3 + c^2*x^2) + 12*a*Sqrt[1 - c^2*x^2]*(-2 + 3*c*x + 2*c^2*x^2) + 9*b*Cos[2*ArcSin[c*x]]) + 6*b*d*Sqrt[d + c*
d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]*(-4*(1 - c^2*x^2)^(3/2) + 3*Sin[2*ArcSin[c*x]]))/(72*c*Sqrt[1 - c^2*x^2])

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Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \left (c d x +d \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right ) \sqrt {-c f x +f}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*sqrt(d)*sqrt(f)*integrate((c*d*x + d)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)
), x) + 1/6*(3*sqrt(-c^2*d*f*x^2 + d*f)*d*x + 3*d^2*f*arcsin(c*x)/(sqrt(d*f)*c) - 2*(-c^2*d*f*x^2 + d*f)^(3/2)
/(c*f))*a

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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