3.6.8 \(\int \frac {\sqrt {f-c f x} (a+b \text {ArcSin}(c x))}{(d+c d x)^{3/2}} \, dx\) [508]

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

[Out]

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

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Rubi [A]
time = 0.26, antiderivative size = 162, normalized size of antiderivative = 1.00, 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 = {4763, 4859, 651, 4845, 12, 641, 31, 4737} \begin {gather*} -\frac {f^2 \left (1-c^2 x^2\right )^{3/2} (a+b \text {ArcSin}(c x))^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 ) (a+b \text {ArcSin}(c x))}{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}} \end {gather*}

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 12

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

Rule 31

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

Rule 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
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 651

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 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 4845

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 4859

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]

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*}

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Mathematica [A]
time = 0.74, size = 248, normalized size = 1.53 \begin {gather*} -\frac {\frac {4 a \sqrt {d+c d x} \sqrt {f-c f x}}{1+c x}-2 a \sqrt {d} \sqrt {f} \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 )+\frac {b \sqrt {d+c d x} \sqrt {f-c f x} \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right ) \left (\text {ArcSin}(c x) (4+\text {ArcSin}(c x))-8 \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )+\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )\right )+\left ((-4+\text {ArcSin}(c x)) \text {ArcSin}(c x)-8 \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )+\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )\right ) \sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )}{\sqrt {1-c^2 x^2} \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )+\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )}}{2 c d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((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] + S
in[ArcSin[c*x]/2])))/(c*d^2)

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Maple [F]
time = 0.34, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arcsin \left (c x \right )\right ) \sqrt {-c f x +f}}{\left (c d x +d \right )^{\frac {3}{2}}}\, dx\]

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

[Out]

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

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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((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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, 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((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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {f-c\,f\,x}}{{\left (d+c\,d\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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