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

Optimal. Leaf size=162 \[ -\frac{f^2 \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c (c d x+d)^{3/2} (f-c f x)^{3/2}}-\frac{2 f^2 (1-c x) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{3/2} (f-c f x)^{3/2}}+\frac{2 b f^2 \left (1-c^2 x^2\right )^{3/2} \log (c x+1)}{c (c d x+d)^{3/2} (f-c f x)^{3/2}} \]

[Out]

(-2*f^2*(1 - c*x)*(1 - c^2*x^2)*(a + b*ArcSin[c*x]))/(c*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)) - (f^2*(1 - c^2*x
^2)^(3/2)*(a + b*ArcSin[c*x])^2)/(2*b*c*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)) + (2*b*f^2*(1 - c^2*x^2)^(3/2)*Lo
g[1 + c*x])/(c*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2))

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Rubi [A]  time = 0.360423, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {4673, 4775, 637, 4761, 12, 627, 31, 4641} \[ -\frac{f^2 \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c (c d x+d)^{3/2} (f-c f x)^{3/2}}-\frac{2 f^2 (1-c x) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{3/2} (f-c f x)^{3/2}}+\frac{2 b f^2 \left (1-c^2 x^2\right )^{3/2} \log (c x+1)}{c (c d x+d)^{3/2} (f-c f x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*f^2*(1 - c*x)*(1 - c^2*x^2)*(a + b*ArcSin[c*x]))/(c*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)) - (f^2*(1 - c^2*x
^2)^(3/2)*(a + b*ArcSin[c*x])^2)/(2*b*c*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)) + (2*b*f^2*(1 - c^2*x^2)^(3/2)*Lo
g[1 + c*x])/(c*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2))

Rule 4673

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 4775

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :
> Int[ExpandIntegrand[(a + b*ArcSin[c*x])^n/Sqrt[d + e*x^2], (f + g*x)^m*(d + e*x^2)^(p + 1/2), x], x] /; Free
Q[{a, b, c, d, e, f, g}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && ILtQ[p + 1/2, 0] && GtQ[d, 0] && IGtQ[n, 0]

Rule 637

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-(a*e) + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

Rule 4761

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

Rule 12

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

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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]

Rubi steps

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

Mathematica [A]  time = 1.32478, size = 248, normalized size = 1.53 \[ -\frac{-2 a \sqrt{d} \sqrt{f} \tan ^{-1}\left (\frac{c x \sqrt{c d x+d} \sqrt{f-c f x}}{\sqrt{d} \sqrt{f} \left (c^2 x^2-1\right )}\right )+\frac{4 a \sqrt{c d x+d} \sqrt{f-c f x}}{c x+1}+\frac{b \sqrt{c d x+d} \sqrt{f-c f x} \left (\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right ) \left (\sin ^{-1}(c x) \left (\sin ^{-1}(c x)+4\right )-8 \log \left (\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )\right )+\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right ) \left (\left (\sin ^{-1}(c x)-4\right ) \sin ^{-1}(c x)-8 \log \left (\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )\right )\right )}{\sqrt{1-c^2 x^2} \left (\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )}}{2 c d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((4*a*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(1 + c*x) - 2*a*Sqrt[d]*Sqrt[f]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[f - c
*f*x])/(Sqrt[d]*Sqrt[f]*(-1 + c^2*x^2))] + (b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(Cos[ArcSin[c*x]/2]*(ArcSin[c*x]
*(4 + ArcSin[c*x]) - 8*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]]) + ((-4 + ArcSin[c*x])*ArcSin[c*x] - 8*Log
[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]])*Sin[ArcSin[c*x]/2]))/(Sqrt[1 - c^2*x^2]*(Cos[ArcSin[c*x]/2] + Sin[A
rcSin[c*x]/2])))/(2*c*d^2)

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Maple [F]  time = 0.285, size = 0, normalized size = 0. \begin{align*} \int{(a+b\arcsin \left ( cx \right ) )\sqrt{-cfx+f} \left ( cdx+d \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c d x + d} \sqrt{-c f x + f}{\left (b \arcsin \left (c x\right ) + a\right )}}{c^{2} d^{2} x^{2} + 2 \, c d^{2} x + d^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-c f x + f}{\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (c d x + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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