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

Optimal. Leaf size=98 \[ \frac{d (c x+1) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{3/2} (f-c f x)^{3/2}}+\frac{b d \left (1-c^2 x^2\right )^{3/2} \log (1-c x)}{c (c d x+d)^{3/2} (f-c f x)^{3/2}} \]

[Out]

(d*(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)) + (b*d*(1 - c^2*x^2)^(
3/2)*Log[1 - c*x])/(c*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2))

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Rubi [A]  time = 0.210481, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 6, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4673, 637, 4761, 12, 627, 31} \[ \frac{d (c x+1) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{3/2} (f-c f x)^{3/2}}+\frac{b d \left (1-c^2 x^2\right )^{3/2} \log (1-c x)}{c (c d x+d)^{3/2} (f-c f x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(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)) + (b*d*(1 - c^2*x^2)^(
3/2)*Log[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 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]

Rubi steps

\begin{align*} \int \frac{a+b \sin ^{-1}(c x)}{\sqrt{d+c d x} (f-c f x)^{3/2}} \, dx &=\frac{\left (1-c^2 x^2\right )^{3/2} \int \frac{(d+c d x) \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{d (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{\left (b c \left (1-c^2 x^2\right )^{3/2}\right ) \int \frac{d (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{d (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{\left (b d \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{d (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{\left (b d \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{d (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{b d \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 = 0.428999, size = 106, normalized size = 1.08 \[ \frac{\sqrt{c d x+d} \sqrt{f-c f x} \left (a \left (-\sqrt{1-c^2 x^2}\right )-b \sqrt{1-c^2 x^2} \sin ^{-1}(c x)+b (c x-1) \log (f-c f x)\right )}{c d f^2 (c x-1) \sqrt{1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(-(a*Sqrt[1 - c^2*x^2]) - b*Sqrt[1 - c^2*x^2]*ArcSin[c*x] + b*(-1 + c*x)*Log[
f - c*f*x]))/(c*d*f^2*(-1 + c*x)*Sqrt[1 - c^2*x^2])

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Maple [F]  time = 0.231, size = 0, normalized size = 0. \begin{align*} \int{(a+b\arcsin \left ( cx \right ) ){\frac{1}{\sqrt{cdx+d}}} \left ( -cfx+f \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*d*x+d)^(1/2)/(-c*f*x+f)^(3/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.5051, size = 803, normalized size = 8.19 \begin{align*} \left [\frac{{\left (b c x - b\right )} \sqrt{d f} \log \left (\frac{c^{6} d f x^{6} - 4 \, c^{5} d f x^{5} + 5 \, c^{4} d f x^{4} - 4 \, c^{2} d f x^{2} + 4 \, c d f x -{\left (c^{4} x^{4} - 4 \, c^{3} x^{3} + 6 \, c^{2} x^{2} - 4 \, c x\right )} \sqrt{-c^{2} x^{2} + 1} \sqrt{c d x + d} \sqrt{-c f x + f} \sqrt{d f} - 2 \, d f}{c^{4} x^{4} - 2 \, c^{3} x^{3} + 2 \, c x - 1}\right ) - 2 \, \sqrt{c d x + d} \sqrt{-c f x + f}{\left (b \arcsin \left (c x\right ) + a\right )}}{2 \,{\left (c^{2} d f^{2} x - c d f^{2}\right )}}, \frac{{\left (b c x - b\right )} \sqrt{-d f} \arctan \left (\frac{{\left (c^{2} x^{2} - 2 \, c x + 2\right )} \sqrt{-c^{2} x^{2} + 1} \sqrt{c d x + d} \sqrt{-c f x + f} \sqrt{-d f}}{c^{4} d f x^{4} - 2 \, c^{3} d f x^{3} - c^{2} d f x^{2} + 2 \, c d f x}\right ) - \sqrt{c d x + d} \sqrt{-c f x + f}{\left (b \arcsin \left (c x\right ) + a\right )}}{c^{2} d f^{2} x - c d f^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((b*c*x - b)*sqrt(d*f)*log((c^6*d*f*x^6 - 4*c^5*d*f*x^5 + 5*c^4*d*f*x^4 - 4*c^2*d*f*x^2 + 4*c*d*f*x - (c^
4*x^4 - 4*c^3*x^3 + 6*c^2*x^2 - 4*c*x)*sqrt(-c^2*x^2 + 1)*sqrt(c*d*x + d)*sqrt(-c*f*x + f)*sqrt(d*f) - 2*d*f)/
(c^4*x^4 - 2*c^3*x^3 + 2*c*x - 1)) - 2*sqrt(c*d*x + d)*sqrt(-c*f*x + f)*(b*arcsin(c*x) + a))/(c^2*d*f^2*x - c*
d*f^2), ((b*c*x - b)*sqrt(-d*f)*arctan((c^2*x^2 - 2*c*x + 2)*sqrt(-c^2*x^2 + 1)*sqrt(c*d*x + d)*sqrt(-c*f*x +
f)*sqrt(-d*f)/(c^4*d*f*x^4 - 2*c^3*d*f*x^3 - c^2*d*f*x^2 + 2*c*d*f*x)) - sqrt(c*d*x + d)*sqrt(-c*f*x + f)*(b*a
rcsin(c*x) + a))/(c^2*d*f^2*x - c*d*f^2)]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \arcsin \left (c x\right ) + a}{\sqrt{c d x + d}{\left (-c f x + f\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*d*x+d)^(1/2)/(-c*f*x+f)^(3/2),x, algorithm="giac")

[Out]

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