3.515 \(\int \frac{(f-c f x)^{3/2} (a+b \sin ^{-1}(c x))}{(d+c d x)^{5/2}} \, dx\)

Optimal. Leaf size=324 \[ \frac{2 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{2 f^4 (1-c x)^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac{f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{4 b f^4 \left (1-c^2 x^2\right )^{5/2}}{3 c (c x+1) (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{8 b f^4 \left (1-c^2 x^2\right )^{5/2} \log (c x+1)}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{b f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]

[Out]

(-4*b*f^4*(1 - c^2*x^2)^(5/2))/(3*c*(1 + c*x)*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)) - (b*f^4*(1 - c^2*x^2)^(5/2
)*ArcSin[c*x]^2)/(2*c*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)) - (2*f^4*(1 - c*x)^3*(1 - c^2*x^2)*(a + b*ArcSin[c*
x]))/(3*c*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)) + (2*f^4*(1 - c*x)*(1 - c^2*x^2)^2*(a + b*ArcSin[c*x]))/(c*(d +
c*d*x)^(5/2)*(f - c*f*x)^(5/2)) + (f^4*(1 - c^2*x^2)^(5/2)*ArcSin[c*x]*(a + b*ArcSin[c*x]))/(c*(d + c*d*x)^(5
/2)*(f - c*f*x)^(5/2)) - (8*b*f^4*(1 - c^2*x^2)^(5/2)*Log[1 + c*x])/(3*c*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.353152, antiderivative size = 324, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {4673, 669, 653, 216, 4761, 627, 43, 31, 4641} \[ \frac{2 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{2 f^4 (1-c x)^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac{f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{4 b f^4 \left (1-c^2 x^2\right )^{5/2}}{3 c (c x+1) (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{8 b f^4 \left (1-c^2 x^2\right )^{5/2} \log (c x+1)}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{b f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[((f - c*f*x)^(3/2)*(a + b*ArcSin[c*x]))/(d + c*d*x)^(5/2),x]

[Out]

(-4*b*f^4*(1 - c^2*x^2)^(5/2))/(3*c*(1 + c*x)*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)) - (b*f^4*(1 - c^2*x^2)^(5/2
)*ArcSin[c*x]^2)/(2*c*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)) - (2*f^4*(1 - c*x)^3*(1 - c^2*x^2)*(a + b*ArcSin[c*
x]))/(3*c*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)) + (2*f^4*(1 - c*x)*(1 - c^2*x^2)^2*(a + b*ArcSin[c*x]))/(c*(d +
c*d*x)^(5/2)*(f - c*f*x)^(5/2)) + (f^4*(1 - c^2*x^2)^(5/2)*ArcSin[c*x]*(a + b*ArcSin[c*x]))/(c*(d + c*d*x)^(5
/2)*(f - c*f*x)^(5/2)) - (8*b*f^4*(1 - c^2*x^2)^(5/2)*Log[1 + c*x])/(3*c*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2))

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 669

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(p + 1)), x] - Dist[(e^2*(m + p))/(c*(p + 1)), Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rule 653

Int[((d_) + (e_.)*(x_))^2*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)*(a + c*x^2)^(p + 1))/(c*(
p + 1)), x] - Dist[(e^2*(p + 2))/(c*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, p}, x] &&
EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && LtQ[p, -1]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

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 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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 31

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

Rubi steps

\begin{align*} \int \frac{(f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{(d+c d x)^{5/2}} \, dx &=\frac{\left (1-c^2 x^2\right )^{5/2} \int \frac{(f-c f x)^4 \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{(d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac{2 f^4 (1-c x)^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{2 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac{\left (b c \left (1-c^2 x^2\right )^{5/2}\right ) \int \left (-\frac{2 f^4 (1-c x)^3}{3 c \left (1-c^2 x^2\right )^2}+\frac{2 f^4 (1-c x)}{c \left (1-c^2 x^2\right )}+\frac{f^4 \sin ^{-1}(c x)}{c \sqrt{1-c^2 x^2}}\right ) \, dx}{(d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac{2 f^4 (1-c x)^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{2 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{\left (2 b f^4 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac{(1-c x)^3}{\left (1-c^2 x^2\right )^2} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac{\left (b f^4 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac{\sin ^{-1}(c x)}{\sqrt{1-c^2 x^2}} \, dx}{(d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac{\left (2 b f^4 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac{1-c x}{1-c^2 x^2} \, dx}{(d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac{b f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac{2 f^4 (1-c x)^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{2 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{\left (2 b f^4 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac{1-c x}{(1+c x)^2} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac{\left (2 b f^4 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac{1}{1+c x} \, dx}{(d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac{b f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac{2 f^4 (1-c x)^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{2 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac{2 b f^4 \left (1-c^2 x^2\right )^{5/2} \log (1+c x)}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{\left (2 b f^4 \left (1-c^2 x^2\right )^{5/2}\right ) \int \left (\frac{1}{-1-c x}+\frac{2}{(1+c x)^2}\right ) \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac{4 b f^4 \left (1-c^2 x^2\right )^{5/2}}{3 c (1+c x) (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac{b f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac{2 f^4 (1-c x)^3 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{2 f^4 (1-c x) \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{f^4 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac{8 b f^4 \left (1-c^2 x^2\right )^{5/2} \log (1+c x)}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ \end{align*}

Mathematica [A]  time = 5.10115, size = 599, normalized size = 1.85 \[ \frac{f \left (-12 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{16 a (2 c x+1) \sqrt{c d x+d} \sqrt{f-c f x}}{(c x+1)^2}-\frac{b \sqrt{c d x+d} \sqrt{f-c f x} \left (\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right ) \left (2 \sin \left (\frac{1}{2} \sin ^{-1}(c x)\right ) \left (3 \left (\sqrt{1-c^2 x^2}+2\right ) \sin ^{-1}(c x)^2+2 \left (7 \sqrt{1-c^2 x^2}+2\right ) \sin ^{-1}(c x)-28 \left (\sqrt{1-c^2 x^2}+2\right ) \log \left (\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )-4\right )+\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right ) \left (9 \sin ^{-1}(c x)^2+6 \sin ^{-1}(c x)-84 \log \left (\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )-8\right )+\cos \left (\frac{3}{2} \sin ^{-1}(c x)\right ) \left (\left (14-3 \sin ^{-1}(c x)\right ) \sin ^{-1}(c x)+28 \log \left (\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )\right )\right )}{(c x-1) \left (\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )^4}-\frac{2 b \sqrt{c d x+d} \sqrt{f-c f x} \left (\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right ) \left (2 \sin \left (\frac{1}{2} \sin ^{-1}(c x)\right ) \left (\left (\sqrt{1-c^2 x^2}+2\right ) \sin ^{-1}(c x)-2 \left (\sqrt{1-c^2 x^2}+2\right ) \log \left (\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )-2\right )+\cos \left (\frac{3}{2} \sin ^{-1}(c x)\right ) \left (\sin ^{-1}(c x)+2 \log \left (\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )\right )-\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right ) \left (3 \sin ^{-1}(c x)+6 \log \left (\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )+4\right )\right )}{(c x-1) \left (\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )^4}\right )}{12 c d^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((f - c*f*x)^(3/2)*(a + b*ArcSin[c*x]))/(d + c*d*x)^(5/2),x]

[Out]

(f*((16*a*(1 + 2*c*x)*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(1 + c*x)^2 - 12*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] - Sin[ArcSin[c*x]/2])*(Cos[ArcSin[c*x]/2]*(-8 + 6*ArcSin[c*x] + 9*ArcSin[c*x]^2 - 84*Log[Cos[ArcSin[c*x]
/2] + Sin[ArcSin[c*x]/2]]) + Cos[(3*ArcSin[c*x])/2]*((14 - 3*ArcSin[c*x])*ArcSin[c*x] + 28*Log[Cos[ArcSin[c*x]
/2] + Sin[ArcSin[c*x]/2]]) + 2*(-4 + 2*(2 + 7*Sqrt[1 - c^2*x^2])*ArcSin[c*x] + 3*(2 + Sqrt[1 - c^2*x^2])*ArcSi
n[c*x]^2 - 28*(2 + Sqrt[1 - c^2*x^2])*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]])*Sin[ArcSin[c*x]/2]))/((-1
+ c*x)*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^4) - (2*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(Cos[ArcSin[c*x]/2]
- Sin[ArcSin[c*x]/2])*(Cos[(3*ArcSin[c*x])/2]*(ArcSin[c*x] + 2*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]])
- Cos[ArcSin[c*x]/2]*(4 + 3*ArcSin[c*x] + 6*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]]) + 2*(-2 + (2 + Sqrt[
1 - c^2*x^2])*ArcSin[c*x] - 2*(2 + Sqrt[1 - c^2*x^2])*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]])*Sin[ArcSin
[c*x]/2]))/((-1 + c*x)*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2])^4)))/(12*c*d^3)

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Maple [F]  time = 0.233, size = 0, normalized size = 0. \begin{align*} \int{(a+b\arcsin \left ( cx \right ) ) \left ( -cfx+f \right ) ^{{\frac{3}{2}}} \left ( cdx+d \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

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)^(5/2),x)

[Out]

int((-c*f*x+f)^(3/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2),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((-c*f*x+f)^(3/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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^{3} d^{3} x^{3} + 3 \, c^{2} d^{3} x^{2} + 3 \, c d^{3} x + d^{3}}, x\right ) \end{align*}

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)^(5/2),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^3*d^3*x^3 + 3*c^2*
d^3*x^2 + 3*c*d^3*x + d^3), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{5}{2}}}\,{d x} \end{align*}

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)^(5/2),x, algorithm="giac")

[Out]

integrate((-c*f*x + f)^(3/2)*(b*arcsin(c*x) + a)/(c*d*x + d)^(5/2), x)