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

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

[Out]

2*d^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*d^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*d^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 = 0.35, 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 = {4673, 4775, 637, 4761, 12, 627, 31, 4641} \[ -\frac {d^2 \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c (c d x+d)^{3/2} (f-c f x)^{3/2}}+\frac {2 d^2 (c x+1) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{3/2} (f-c f x)^{3/2}}+\frac {2 b d^2 \left (1-c^2 x^2\right )^{3/2} \log (1-c x)}{c (c d x+d)^{3/2} (f-c f x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d^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)) - (d^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*d^2*(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 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 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 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 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]

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 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 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]

Rubi steps

\begin {align*} \int \frac {\sqrt {d+c d x} \left (a+b \sin ^{-1}(c x)\right )}{(f-c f x)^{3/2}} \, dx &=\frac {\left (1-c^2 x^2\right )^{3/2} \int \frac {(d+c d 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 (d^2+c d^2 x\right ) \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{3/2}}-\frac {d^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 (d^2+c d^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 (d^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 d^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 {d^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 {d^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 d^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 {d^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 d^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 d^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 {d^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 d^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 d^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 {d^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 d^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 = 1.72, size = 281, normalized size = 1.73 \[ -\frac {-2 a \sqrt {d} \sqrt {f} \tan ^{-1}\left (\frac {c x \sqrt {c d x+d} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (c^2 x^2-1\right )}\right )+\frac {4 a \sqrt {c d x+d} \sqrt {f-c f x}}{c x-1}+\frac {b (c x+1) \sqrt {c d x+d} \sqrt {f-c f x} \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right ) \left (\left (\sin ^{-1}(c x)-4\right ) \sin ^{-1}(c x)-8 \log \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right ) \left (\sin ^{-1}(c x) \left (\sin ^{-1}(c x)+4\right )-8 \log \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )\right )\right )}{\sqrt {1-c^2 x^2} \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right ) \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )^2}}{2 c f^2} \]

Antiderivative was successfully verified.

[In]

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

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c d x + d} \sqrt {-c f x + f} {\left (b \arcsin \left (c x\right ) + a\right )}}{c^{2} f^{2} x^{2} - 2 \, c f^{2} x + f^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.35, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c d x +d}\, \left (a +b \arcsin \left (c x \right )\right )}{\left (-c f x +f \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -a {\left (\frac {2 \, \sqrt {-c^{2} d f x^{2} + d f}}{c^{2} f^{2} x - c f^{2}} + \frac {d \arcsin \left (c x\right )}{c f^{2} \sqrt {\frac {d}{f}}}\right )} - \frac {\frac {b \sqrt {d} \int \frac {\sqrt {c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{{\left (c x - 1\right )} \sqrt {-c x + 1}}\,{d x}}{f}}{\sqrt {f}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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