3.6.14 \(\int \frac {(f-c f x)^{3/2} (a+b \text {ArcSin}(c x))}{(d+c d x)^{3/2}} \, dx\) [514]

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

[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 = 0.30, 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 = {4763, 4859, 651, 4845, 12, 641, 31, 4737, 4767, 8} \begin {gather*} -\frac {3 f^3 \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 {f^3 \left (1-c^2 x^2\right )^2 (a+b \text {ArcSin}(c x))}{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 ) (a+b \text {ArcSin}(c x))}{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}} \end {gather*}

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 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 4767

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^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 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 {(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 = 2.02, size = 291, normalized size = 1.15 \begin {gather*} \frac {f \left (6 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 {\sqrt {d+c d x} \sqrt {f-c f x} \csc ^2\left (\frac {1}{2} \text {ArcSin}(c x)\right ) \left (2 b (5+c x) \left (-1+c x+\sqrt {1-c^2 x^2}\right ) \text {ArcSin}(c x)-3 b \left (-1-c x+\sqrt {1-c^2 x^2}\right ) \text {ArcSin}(c x)^2+2 \left (b c x \left (-1-c x+\sqrt {1-c^2 x^2}\right )+a (5+c x) \left (-1+c x+\sqrt {1-c^2 x^2}\right )+8 b \left (-1-c x+\sqrt {1-c^2 x^2}\right ) \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )+\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )\right )\right )}{2 \sqrt {1-c^2 x^2} \left (1+\cot \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )}\right )}{2 c d^2} \end {gather*}

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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Maple [F]
time = 0.28, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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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((-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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- f \left (c x - 1\right )\right )^{\frac {3}{2}} \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((-c*f*x+f)**(3/2)*(a+b*asin(c*x))/(c*d*x+d)**(3/2),x)

[Out]

Integral((-f*(c*x - 1))**(3/2)*(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((-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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 \end {gather*}

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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