3.6.23 \(\int \frac {(d+c d x)^{3/2} (a+b \text {ArcSin}(c x))}{\sqrt {f-c f x}} \, dx\) [523]

Optimal. Leaf size=242 \[ \frac {2 b d^2 x \sqrt {1-c^2 x^2}}{\sqrt {d+c d x} \sqrt {f-c f x}}+\frac {b c d^2 x^2 \sqrt {1-c^2 x^2}}{4 \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {2 d^2 \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{c \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {d^2 x \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{2 \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {3 d^2 \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^2}{4 b c \sqrt {d+c d x} \sqrt {f-c f x}} \]

[Out]

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

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Rubi [A]
time = 0.31, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {4763, 4847, 4737, 4767, 8, 4795, 30} \begin {gather*} \frac {3 d^2 \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^2}{4 b c \sqrt {c d x+d} \sqrt {f-c f x}}-\frac {d^2 x \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{2 \sqrt {c d x+d} \sqrt {f-c f x}}-\frac {2 d^2 \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{c \sqrt {c d x+d} \sqrt {f-c f x}}+\frac {b c d^2 x^2 \sqrt {1-c^2 x^2}}{4 \sqrt {c d x+d} \sqrt {f-c f x}}+\frac {2 b d^2 x \sqrt {1-c^2 x^2}}{\sqrt {c d x+d} \sqrt {f-c f x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 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 4795

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[f
*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])^n/(e*(m + 2*p + 1))), x] + (Dist[f^2*((m - 1)/(c^2*(m
+ 2*p + 1))), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] + Dist[b*f*(n/(c*(m + 2*p + 1)))*S
imp[(d + e*x^2)^p/(1 - c^2*x^2)^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, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && IGtQ[m, 1] && NeQ[m + 2*p + 1, 0]

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

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Mathematica [A]
time = 0.63, size = 238, normalized size = 0.98 \begin {gather*} \frac {-4 b d (4+c x) \sqrt {d+c d x} \sqrt {f-c f x} \sqrt {1-c^2 x^2} \text {ArcSin}(c x)+6 b d \sqrt {d+c d x} \sqrt {f-c f x} \text {ArcSin}(c x)^2-12 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 (16 b c x-4 a (4+c x) \sqrt {1-c^2 x^2}-b \cos (2 \text {ArcSin}(c x))\right )}{8 c f \sqrt {1-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b*d*(4 + c*x)*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*Sqrt[1 - c^2*x^2]*ArcSin[c*x] + 6*b*d*Sqrt[d + c*d*x]*Sqrt[f
- c*f*x]*ArcSin[c*x]^2 - 12*a*d^(3/2)*Sqrt[f]*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[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]*(16*b*c*x - 4*a*(4 + c*x)*Sqrt[1 - c^2*x
^2] - b*Cos[2*ArcSin[c*x]]))/(8*c*f*Sqrt[1 - c^2*x^2])

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Maple [F]
time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {\left (c d x +d \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )}{\sqrt {-c f x +f}}\, dx\]

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]

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

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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)/(c*f*x - f), x)

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\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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