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

Optimal. Leaf size=420 \[ \frac{5 f^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac{10 f^5 (1-c x)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{2 f^5 (1-c x)^4 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac{5 f^5 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{b f^5 x \left (1-c^2 x^2\right )^{5/2}}{(c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{8 b f^5 \left (1-c^2 x^2\right )^{5/2}}{3 c (c x+1) (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{28 b f^5 \left (1-c^2 x^2\right )^{5/2} \log (c x+1)}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{5 b f^5 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]

[Out]

-((b*f^5*x*(1 - c^2*x^2)^(5/2))/((d + c*d*x)^(5/2)*(f - c*f*x)^(5/2))) - (8*b*f^5*(1 - c^2*x^2)^(5/2))/(3*c*(1
+ c*x)*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)) - (5*b*f^5*(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^5*(1 - c*x)^4*(1 - c^2*x^2)*(a + b*ArcSin[c*x]))/(3*c*(d + c*d*x)^(5/2)*(f - c*f
*x)^(5/2)) + (10*f^5*(1 - c*x)^2*(1 - c^2*x^2)^2*(a + b*ArcSin[c*x]))/(3*c*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)
) + (5*f^5*(1 - c^2*x^2)^3*(a + b*ArcSin[c*x]))/(c*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)) + (5*f^5*(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)) - (28*b*f^5*(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 = 0.394031, antiderivative size = 420, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {4673, 669, 641, 216, 4761, 627, 43, 4641} \[ \frac{5 f^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac{10 f^5 (1-c x)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{2 f^5 (1-c x)^4 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}+\frac{5 f^5 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x) \left (a+b \sin ^{-1}(c x)\right )}{c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{b f^5 x \left (1-c^2 x^2\right )^{5/2}}{(c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{8 b f^5 \left (1-c^2 x^2\right )^{5/2}}{3 c (c x+1) (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{28 b f^5 \left (1-c^2 x^2\right )^{5/2} \log (c x+1)}{3 c (c d x+d)^{5/2} (f-c f x)^{5/2}}-\frac{5 b f^5 \left (1-c^2 x^2\right )^{5/2} \sin ^{-1}(c x)^2}{2 c (c d x+d)^{5/2} (f-c f x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*f^5*x*(1 - c^2*x^2)^(5/2))/((d + c*d*x)^(5/2)*(f - c*f*x)^(5/2))) - (8*b*f^5*(1 - c^2*x^2)^(5/2))/(3*c*(1
+ c*x)*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)) - (5*b*f^5*(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^5*(1 - c*x)^4*(1 - c^2*x^2)*(a + b*ArcSin[c*x]))/(3*c*(d + c*d*x)^(5/2)*(f - c*f
*x)^(5/2)) + (10*f^5*(1 - c*x)^2*(1 - c^2*x^2)^2*(a + b*ArcSin[c*x]))/(3*c*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)
) + (5*f^5*(1 - c^2*x^2)^3*(a + b*ArcSin[c*x]))/(c*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)) + (5*f^5*(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)) - (28*b*f^5*(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 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[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 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)^{5/2} \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)^5 \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^5 (1-c x)^4 \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{10 f^5 (1-c x)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{5 f^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{5 f^5 \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{5 f^5}{c}-\frac{2 f^5 (1-c x)^4}{3 c \left (1-c^2 x^2\right )^2}+\frac{10 f^5 (1-c x)^2}{3 c \left (1-c^2 x^2\right )}+\frac{5 f^5 \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{5 b f^5 x \left (1-c^2 x^2\right )^{5/2}}{(d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac{2 f^5 (1-c x)^4 \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{10 f^5 (1-c x)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{5 f^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{5 f^5 \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^5 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac{(1-c x)^4}{\left (1-c^2 x^2\right )^2} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac{\left (10 b f^5 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac{(1-c x)^2}{1-c^2 x^2} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac{\left (5 b f^5 \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{5 b f^5 x \left (1-c^2 x^2\right )^{5/2}}{(d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac{5 b f^5 \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^5 (1-c x)^4 \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{10 f^5 (1-c x)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{5 f^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{5 f^5 \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^5 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac{(1-c x)^2}{(1+c x)^2} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac{\left (10 b f^5 \left (1-c^2 x^2\right )^{5/2}\right ) \int \frac{1-c x}{1+c x} \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac{5 b f^5 x \left (1-c^2 x^2\right )^{5/2}}{(d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac{5 b f^5 \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^5 (1-c x)^4 \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{10 f^5 (1-c x)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{5 f^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{5 f^5 \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^5 \left (1-c^2 x^2\right )^{5/2}\right ) \int \left (1+\frac{4}{(1+c x)^2}-\frac{4}{1+c x}\right ) \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac{\left (10 b f^5 \left (1-c^2 x^2\right )^{5/2}\right ) \int \left (-1+\frac{2}{1+c x}\right ) \, dx}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}\\ &=-\frac{b f^5 x \left (1-c^2 x^2\right )^{5/2}}{(d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac{8 b f^5 \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{5 b f^5 \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^5 (1-c x)^4 \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{10 f^5 (1-c x)^2 \left (1-c^2 x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{5 f^5 \left (1-c^2 x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac{5 f^5 \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{28 b f^5 \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 [B]  time = 6.72092, size = 847, normalized size = 2.02 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f^2*((4*a*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(23 + 34*c*x + 3*c^2*x^2))/(1 + c*x)^2 - 60*a*Sqrt[d]*Sqrt[f]*ArcTa
n[(c*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(Sqrt[d]*Sqrt[f]*(-1 + c^2*x^2))] + (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*Lo
g[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*Lo
g[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])*ArcSin[c*x]^2 - 28*(2 + Sqrt[1 - c^2*x^2])*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]])*Sin[ArcSi
n[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]*(Co
s[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[A
rcSin[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) + (b*Sqrt[d + c*d*x]*Sqrt[f
- c*f*x]*(Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2])*(2*(4 + 6*c*x + 6*c^2*x^2 + 52*(1 + c*x)*Log[Cos[ArcSin[c*
x]/2] + Sin[ArcSin[c*x]/2]])*(Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]) - 18*ArcSin[c*x]^2*(Cos[ArcSin[c*x]/2]
+ Sin[ArcSin[c*x]/2])^3 + ArcSin[c*x]*(-24*Cos[ArcSin[c*x]/2] - 35*Cos[(3*ArcSin[c*x])/2] + 3*Cos[(5*ArcSin[c*
x])/2] + 24*Sin[ArcSin[c*x]/2] - 35*Sin[(3*ArcSin[c*x])/2] - 3*Sin[(5*ArcSin[c*x])/2])))/((-1 + c*x)*(Cos[ArcS
in[c*x]/2] + Sin[ArcSin[c*x]/2])^4)))/(12*c*d^3)

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Maple [F]  time = 0.242, size = 0, normalized size = 0. \begin{align*} \int{(a+b\arcsin \left ( cx \right ) ) \left ( -cfx+f \right ) ^{{\frac{5}{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)^(5/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2),x)

[Out]

int((-c*f*x+f)^(5/2)*(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)^(5/2)*(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^{2} f^{2} x^{2} - 2 \, a c f^{2} x + a f^{2} +{\left (b c^{2} f^{2} x^{2} - 2 \, b c f^{2} x + b f^{2}\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)^(5/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2),x, algorithm="fricas")

[Out]

integral((a*c^2*f^2*x^2 - 2*a*c*f^2*x + a*f^2 + (b*c^2*f^2*x^2 - 2*b*c*f^2*x + b*f^2)*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)**(5/2)*(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{5}{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)^(5/2)*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2),x, algorithm="giac")

[Out]

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