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

Optimal. Leaf size=252 \[ -\frac {3 f^3 \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c (c d x+d)^{3/2} (f-c f x)^{3/2}}-\frac {f^3 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{3/2} (f-c f x)^{3/2}}-\frac {4 f^3 (1-c x) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{3/2} (f-c f x)^{3/2}}+\frac {b f^3 x \left (1-c^2 x^2\right )^{3/2}}{(c d x+d)^{3/2} (f-c f x)^{3/2}}+\frac {4 b f^3 \left (1-c^2 x^2\right )^{3/2} \log (c x+1)}{c (c d x+d)^{3/2} (f-c f x)^{3/2}} \]

[Out]

b*f^3*x*(-c^2*x^2+1)^(3/2)/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2)-4*f^3*(-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)-f^3*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))/c/(c*d*x+d)^(3/2)/(-c*f*x+f)^(3/2)-3/2*f^3
*(-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)+4*b*f^3*(-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 = 0.44, antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4673, 4775, 637, 4761, 12, 627, 31, 4641, 4677, 8} \[ -\frac {3 f^3 \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c (c d x+d)^{3/2} (f-c f x)^{3/2}}-\frac {f^3 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{3/2} (f-c f x)^{3/2}}-\frac {4 f^3 (1-c x) \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{3/2} (f-c f x)^{3/2}}+\frac {b f^3 x \left (1-c^2 x^2\right )^{3/2}}{(c d x+d)^{3/2} (f-c f x)^{3/2}}+\frac {4 b f^3 \left (1-c^2 x^2\right )^{3/2} \log (c x+1)}{c (c d x+d)^{3/2} (f-c f x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*f^3*x*(1 - c^2*x^2)^(3/2))/((d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)) - (4*f^3*(1 - c*x)*(1 - c^2*x^2)*(a + b*Ar
cSin[c*x]))/(c*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)) - (f^3*(1 - c^2*x^2)^2*(a + b*ArcSin[c*x]))/(c*(d + c*d*x)
^(3/2)*(f - c*f*x)^(3/2)) - (3*f^3*(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)) + (4*b*f^3*(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 8

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

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 4677

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p + 1
)*(1 - c^2*x^2)^FracPart[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 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 {(f-c f x)^{3/2} \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)^3 \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 {4 \left (f^3-c f^3 x\right ) \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{3/2}}-\frac {3 f^3 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}+\frac {c f^3 x \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 (4 \left (1-c^2 x^2\right )^{3/2}\right ) \int \frac {\left (f^3-c f^3 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 (3 f^3 \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 {\left (c f^3 \left (1-c^2 x^2\right )^{3/2}\right ) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{(d+c d x)^{3/2} (f-c f x)^{3/2}}\\ &=-\frac {4 f^3 (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^3 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {3 f^3 \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 (4 b c \left (1-c^2 x^2\right )^{3/2}\right ) \int \frac {f^3 (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 {\left (b f^3 \left (1-c^2 x^2\right )^{3/2}\right ) \int 1 \, dx}{(d+c d x)^{3/2} (f-c f x)^{3/2}}\\ &=\frac {b f^3 x \left (1-c^2 x^2\right )^{3/2}}{(d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {4 f^3 (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^3 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {3 f^3 \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 (4 b f^3 \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 {b f^3 x \left (1-c^2 x^2\right )^{3/2}}{(d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {4 f^3 (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^3 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {3 f^3 \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 (4 b f^3 \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 {b f^3 x \left (1-c^2 x^2\right )^{3/2}}{(d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {4 f^3 (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^3 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{3/2} (f-c f x)^{3/2}}-\frac {3 f^3 \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 {4 b f^3 \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 = 3.55, size = 291, normalized size = 1.15 \[ \frac {f \left (6 a \sqrt {d} \sqrt {f} \tan ^{-1}\left (\frac {c x \sqrt {c d x+d} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (c^2 x^2-1\right )}\right )-\frac {\sqrt {c d x+d} \sqrt {f-c f x} \csc ^2\left (\frac {1}{2} \sin ^{-1}(c x)\right ) \left (2 \left (a (c x+5) \left (\sqrt {1-c^2 x^2}+c x-1\right )+b c x \left (\sqrt {1-c^2 x^2}-c x-1\right )+8 b \left (\sqrt {1-c^2 x^2}-c x-1\right ) \log \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac {1}{2} \sin ^{-1}(c x)\right )\right )\right )-3 b \left (\sqrt {1-c^2 x^2}-c x-1\right ) \sin ^{-1}(c x)^2+2 b (c x+5) \left (\sqrt {1-c^2 x^2}+c x-1\right ) \sin ^{-1}(c x)\right )}{2 \sqrt {1-c^2 x^2} \left (\cot \left (\frac {1}{2} \sin ^{-1}(c x)\right )+1\right )}\right )}{2 c d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f*(6*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))] - (Sqrt
[d + c*d*x]*Sqrt[f - c*f*x]*Csc[ArcSin[c*x]/2]^2*(2*b*(5 + c*x)*(-1 + c*x + Sqrt[1 - c^2*x^2])*ArcSin[c*x] - 3
*b*(-1 - c*x + Sqrt[1 - c^2*x^2])*ArcSin[c*x]^2 + 2*(b*c*x*(-1 - c*x + Sqrt[1 - c^2*x^2]) + a*(5 + c*x)*(-1 +
c*x + Sqrt[1 - c^2*x^2]) + 8*b*(-1 - c*x + Sqrt[1 - c^2*x^2])*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]])))/
(2*Sqrt[1 - c^2*x^2]*(1 + Cot[ArcSin[c*x]/2]))))/(2*c*d^2)

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a c f x - a f + {\left (b c f x - b f\right )} \arcsin \left (c x\right )\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{c^{2} d^{2} x^{2} + 2 \, c d^{2} x + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -b \sqrt {d} \sqrt {f} \int \frac {{\left (c f x - f\right )} \sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{c^{2} d^{2} x^{2} + 2 \, c d^{2} x + d^{2}}\,{d x} + a {\left (\frac {{\left (-c^{2} d f x^{2} + d f\right )}^{\frac {3}{2}}}{c^{3} d^{3} x^{2} + 2 \, c^{2} d^{3} x + c d^{3}} - \frac {6 \, \sqrt {-c^{2} d f x^{2} + d f} f}{c^{2} d^{2} x + c d^{2}} - \frac {3 \, f^{2} \arcsin \left (c x\right )}{c d^{2} \sqrt {\frac {f}{d}}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*sqrt(d)*sqrt(f)*integrate((c*f*x - f)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1
))/(c^2*d^2*x^2 + 2*c*d^2*x + d^2), x) + a*((-c^2*d*f*x^2 + d*f)^(3/2)/(c^3*d^3*x^2 + 2*c^2*d^3*x + c*d^3) - 6
*sqrt(-c^2*d*f*x^2 + d*f)*f/(c^2*d^2*x + c*d^2) - 3*f^2*arcsin(c*x)/(c*d^2*sqrt(f/d)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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