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

Optimal. Leaf size=414 \[ \frac {3 d x (c d x+d)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{8 \left (1-c^2 x^2\right )}+\frac {3 d (c d x+d)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \left (1-c^2 x^2\right )^{3/2}}-\frac {d \left (1-c^2 x^2\right ) (c d x+d)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{5 c}+\frac {1}{4} d x (c d x+d)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )-\frac {5 b c d x^2 (c d x+d)^{3/2} (f-c f x)^{3/2}}{16 \left (1-c^2 x^2\right )^{3/2}}+\frac {b d x (c d x+d)^{3/2} (f-c f x)^{3/2}}{5 \left (1-c^2 x^2\right )^{3/2}}-\frac {2 b c^2 d x^3 (c d x+d)^{3/2} (f-c f x)^{3/2}}{15 \left (1-c^2 x^2\right )^{3/2}}+\frac {b c^4 d x^5 (c d x+d)^{3/2} (f-c f x)^{3/2}}{25 \left (1-c^2 x^2\right )^{3/2}}+\frac {b c^3 d x^4 (c d x+d)^{3/2} (f-c f x)^{3/2}}{16 \left (1-c^2 x^2\right )^{3/2}} \]

[Out]

1/5*b*d*x*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)/(-c^2*x^2+1)^(3/2)-5/16*b*c*d*x^2*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)/
(-c^2*x^2+1)^(3/2)-2/15*b*c^2*d*x^3*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)/(-c^2*x^2+1)^(3/2)+1/16*b*c^3*d*x^4*(c*d*
x+d)^(3/2)*(-c*f*x+f)^(3/2)/(-c^2*x^2+1)^(3/2)+1/25*b*c^4*d*x^5*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)/(-c^2*x^2+1)^
(3/2)+1/4*d*x*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)*(a+b*arcsin(c*x))+3/8*d*x*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)*(a+b
*arcsin(c*x))/(-c^2*x^2+1)-1/5*d*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)*(-c^2*x^2+1)*(a+b*arcsin(c*x))/c+3/16*d*(c*d
*x+d)^(3/2)*(-c*f*x+f)^(3/2)*(a+b*arcsin(c*x))^2/b/c/(-c^2*x^2+1)^(3/2)

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Rubi [A]  time = 0.39, antiderivative size = 414, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4673, 4763, 4649, 4647, 4641, 30, 14, 4677, 194} \[ \frac {3 d x (c d x+d)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{8 \left (1-c^2 x^2\right )}+\frac {3 d (c d x+d)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \left (1-c^2 x^2\right )^{3/2}}-\frac {d \left (1-c^2 x^2\right ) (c d x+d)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{5 c}+\frac {1}{4} d x (c d x+d)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {b c^4 d x^5 (c d x+d)^{3/2} (f-c f x)^{3/2}}{25 \left (1-c^2 x^2\right )^{3/2}}+\frac {b c^3 d x^4 (c d x+d)^{3/2} (f-c f x)^{3/2}}{16 \left (1-c^2 x^2\right )^{3/2}}-\frac {2 b c^2 d x^3 (c d x+d)^{3/2} (f-c f x)^{3/2}}{15 \left (1-c^2 x^2\right )^{3/2}}-\frac {5 b c d x^2 (c d x+d)^{3/2} (f-c f x)^{3/2}}{16 \left (1-c^2 x^2\right )^{3/2}}+\frac {b d x (c d x+d)^{3/2} (f-c f x)^{3/2}}{5 \left (1-c^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*x*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2))/(5*(1 - c^2*x^2)^(3/2)) - (5*b*c*d*x^2*(d + c*d*x)^(3/2)*(f - c*f*
x)^(3/2))/(16*(1 - c^2*x^2)^(3/2)) - (2*b*c^2*d*x^3*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2))/(15*(1 - c^2*x^2)^(3/
2)) + (b*c^3*d*x^4*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2))/(16*(1 - c^2*x^2)^(3/2)) + (b*c^4*d*x^5*(d + c*d*x)^(3
/2)*(f - c*f*x)^(3/2))/(25*(1 - c^2*x^2)^(3/2)) + (d*x*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)*(a + b*ArcSin[c*x])
)/4 + (3*d*x*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)*(a + b*ArcSin[c*x]))/(8*(1 - c^2*x^2)) - (d*(d + c*d*x)^(3/2)
*(f - c*f*x)^(3/2)*(1 - c^2*x^2)*(a + b*ArcSin[c*x]))/(5*c) + (3*d*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)*(a + b*
ArcSin[c*x])^2)/(16*b*c*(1 - c^2*x^2)^(3/2))

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 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]

Rule 4647

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(x*Sqrt[d + e*x^2]*(
a + b*ArcSin[c*x])^n)/2, x] + (Dist[Sqrt[d + e*x^2]/(2*Sqrt[1 - c^2*x^2]), Int[(a + b*ArcSin[c*x])^n/Sqrt[1 -
c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(2*Sqrt[1 - c^2*x^2]), Int[x*(a + b*ArcSin[c*x])^(n - 1), x],
x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0]

Rule 4649

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(x*(d + e*x^2)^p*(
a + b*ArcSin[c*x])^n)/(2*p + 1), x] + (Dist[(2*d*p)/(2*p + 1), Int[(d + e*x^2)^(p - 1)*(a + b*ArcSin[c*x])^n,
x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/((2*p + 1)*(1 - c^2*x^2)^FracPart[p]), Int[x*(1 - c
^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && Gt
Q[n, 0] && GtQ[p, 0]

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 4763

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g},
x] && EqQ[c^2*d + e, 0] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ[n, 0] && (m == 1 || p > 0 ||
(n == 1 && p > -1) || (m == 2 && p < -2))

Rubi steps

\begin {align*} \int (d+c d x)^{5/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac {\left ((d+c d x)^{3/2} (f-c f x)^{3/2}\right ) \int (d+c d x) \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\left (1-c^2 x^2\right )^{3/2}}\\ &=\frac {\left ((d+c d x)^{3/2} (f-c f x)^{3/2}\right ) \int \left (d \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )+c d x \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )\right ) \, dx}{\left (1-c^2 x^2\right )^{3/2}}\\ &=\frac {\left (d (d+c d x)^{3/2} (f-c f x)^{3/2}\right ) \int \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\left (1-c^2 x^2\right )^{3/2}}+\frac {\left (c d (d+c d x)^{3/2} (f-c f x)^{3/2}\right ) \int x \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\left (1-c^2 x^2\right )^{3/2}}\\ &=\frac {1}{4} d x (d+c d x)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )-\frac {d (d+c d x)^{3/2} (f-c f x)^{3/2} \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{5 c}+\frac {\left (3 d (d+c d x)^{3/2} (f-c f x)^{3/2}\right ) \int \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{4 \left (1-c^2 x^2\right )^{3/2}}+\frac {\left (b d (d+c d x)^{3/2} (f-c f x)^{3/2}\right ) \int \left (1-c^2 x^2\right )^2 \, dx}{5 \left (1-c^2 x^2\right )^{3/2}}-\frac {\left (b c d (d+c d x)^{3/2} (f-c f x)^{3/2}\right ) \int x \left (1-c^2 x^2\right ) \, dx}{4 \left (1-c^2 x^2\right )^{3/2}}\\ &=\frac {1}{4} d x (d+c d x)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {3 d x (d+c d x)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{8 \left (1-c^2 x^2\right )}-\frac {d (d+c d x)^{3/2} (f-c f x)^{3/2} \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{5 c}+\frac {\left (3 d (d+c d x)^{3/2} (f-c f x)^{3/2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{8 \left (1-c^2 x^2\right )^{3/2}}+\frac {\left (b d (d+c d x)^{3/2} (f-c f x)^{3/2}\right ) \int \left (1-2 c^2 x^2+c^4 x^4\right ) \, dx}{5 \left (1-c^2 x^2\right )^{3/2}}-\frac {\left (b c d (d+c d x)^{3/2} (f-c f x)^{3/2}\right ) \int \left (x-c^2 x^3\right ) \, dx}{4 \left (1-c^2 x^2\right )^{3/2}}-\frac {\left (3 b c d (d+c d x)^{3/2} (f-c f x)^{3/2}\right ) \int x \, dx}{8 \left (1-c^2 x^2\right )^{3/2}}\\ &=\frac {b d x (d+c d x)^{3/2} (f-c f x)^{3/2}}{5 \left (1-c^2 x^2\right )^{3/2}}-\frac {5 b c d x^2 (d+c d x)^{3/2} (f-c f x)^{3/2}}{16 \left (1-c^2 x^2\right )^{3/2}}-\frac {2 b c^2 d x^3 (d+c d x)^{3/2} (f-c f x)^{3/2}}{15 \left (1-c^2 x^2\right )^{3/2}}+\frac {b c^3 d x^4 (d+c d x)^{3/2} (f-c f x)^{3/2}}{16 \left (1-c^2 x^2\right )^{3/2}}+\frac {b c^4 d x^5 (d+c d x)^{3/2} (f-c f x)^{3/2}}{25 \left (1-c^2 x^2\right )^{3/2}}+\frac {1}{4} d x (d+c d x)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {3 d x (d+c d x)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )}{8 \left (1-c^2 x^2\right )}-\frac {d (d+c d x)^{3/2} (f-c f x)^{3/2} \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{5 c}+\frac {3 d (d+c d x)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \left (1-c^2 x^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 1.74, size = 305, normalized size = 0.74 \[ \frac {d^2 f \left (\sqrt {c d x+d} \sqrt {f-c f x} \left (-240 a \sqrt {1-c^2 x^2} \left (8 c^4 x^4+10 c^3 x^3-16 c^2 x^2-25 c x+8\right )+128 b c x \left (3 c^4 x^4-10 c^2 x^2+15\right )+1200 b \cos \left (2 \sin ^{-1}(c x)\right )+75 b \cos \left (4 \sin ^{-1}(c x)\right )\right )-3600 a \sqrt {d} \sqrt {f} \sqrt {1-c^2 x^2} \tan ^{-1}\left (\frac {c x \sqrt {c d x+d} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (c^2 x^2-1\right )}\right )-60 b \sqrt {c d x+d} \sqrt {f-c f x} \left (32 \left (1-c^2 x^2\right )^{5/2}-40 \sin \left (2 \sin ^{-1}(c x)\right )-5 \sin \left (4 \sin ^{-1}(c x)\right )\right ) \sin ^{-1}(c x)+1800 b \sqrt {c d x+d} \sqrt {f-c f x} \sin ^{-1}(c x)^2\right )}{9600 c \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*f*(1800*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2 - 3600*a*Sqrt[d]*Sqrt[f]*Sqrt[1 - c^2*x^2]*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]*(12
8*b*c*x*(15 - 10*c^2*x^2 + 3*c^4*x^4) - 240*a*Sqrt[1 - c^2*x^2]*(8 - 25*c*x - 16*c^2*x^2 + 10*c^3*x^3 + 8*c^4*
x^4) + 1200*b*Cos[2*ArcSin[c*x]] + 75*b*Cos[4*ArcSin[c*x]]) - 60*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]
*(32*(1 - c^2*x^2)^(5/2) - 40*Sin[2*ArcSin[c*x]] - 5*Sin[4*ArcSin[c*x]])))/(9600*c*Sqrt[1 - c^2*x^2])

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a c^{3} d^{2} f x^{3} + a c^{2} d^{2} f x^{2} - a c d^{2} f x - a d^{2} f + {\left (b c^{3} d^{2} f x^{3} + b c^{2} d^{2} f x^{2} - b c d^{2} f x - b d^{2} f\right )} \arcsin \left (c x\right )\right )} \sqrt {c d x + d} \sqrt {-c f x + f}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c d x + d\right )}^{\frac {5}{2}} {\left (-c f x + f\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.28, size = 0, normalized size = 0.00 \[ \int \left (c d x +d \right )^{\frac {5}{2}} \left (-c f x +f \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b \sqrt {d} \sqrt {f} \int -{\left (c^{3} d^{2} f x^{3} + c^{2} d^{2} f x^{2} - c d^{2} f x - d^{2} f\right )} \sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\,{d x} + \frac {1}{40} \, {\left (15 \, \sqrt {-c^{2} d f x^{2} + d f} d^{2} f x + \frac {15 \, d^{3} f^{2} \arcsin \left (c x\right )}{\sqrt {d f} c} + 10 \, {\left (-c^{2} d f x^{2} + d f\right )}^{\frac {3}{2}} d x - \frac {8 \, {\left (-c^{2} d f x^{2} + d f\right )}^{\frac {5}{2}}}{c f}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*sqrt(d)*sqrt(f)*integrate(-(c^3*d^2*f*x^3 + c^2*d^2*f*x^2 - c*d^2*f*x - d^2*f)*sqrt(c*x + 1)*sqrt(-c*x + 1)*
arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)), x) + 1/40*(15*sqrt(-c^2*d*f*x^2 + d*f)*d^2*f*x + 15*d^3*f^2*arcsin
(c*x)/(sqrt(d*f)*c) + 10*(-c^2*d*f*x^2 + d*f)^(3/2)*d*x - 8*(-c^2*d*f*x^2 + d*f)^(5/2)/(c*f))*a

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\right )}^{5/2}\,{\left (f-c\,f\,x\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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