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

Optimal. Leaf size=376 \[ \frac {1}{4} c^2 f^2 x^3 \sqrt {c d x+d} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )+\frac {5 f^2 \sqrt {c d x+d} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \sqrt {1-c^2 x^2}}+\frac {2 f^2 \left (1-c^2 x^2\right ) \sqrt {c d x+d} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )}{3 c}+\frac {3}{8} f^2 x \sqrt {c d x+d} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )-\frac {3 b c f^2 x^2 \sqrt {c d x+d} \sqrt {f-c f x}}{16 \sqrt {1-c^2 x^2}}-\frac {2 b f^2 x \sqrt {c d x+d} \sqrt {f-c f x}}{3 \sqrt {1-c^2 x^2}}+\frac {2 b c^2 f^2 x^3 \sqrt {c d x+d} \sqrt {f-c f x}}{9 \sqrt {1-c^2 x^2}}-\frac {b c^3 f^2 x^4 \sqrt {c d x+d} \sqrt {f-c f x}}{16 \sqrt {1-c^2 x^2}} \]

[Out]

3/8*f^2*x*(a+b*arcsin(c*x))*(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)+1/4*c^2*f^2*x^3*(a+b*arcsin(c*x))*(c*d*x+d)^(1/2)
*(-c*f*x+f)^(1/2)+2/3*f^2*(-c^2*x^2+1)*(a+b*arcsin(c*x))*(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)/c-2/3*b*f^2*x*(c*d*x
+d)^(1/2)*(-c*f*x+f)^(1/2)/(-c^2*x^2+1)^(1/2)-3/16*b*c*f^2*x^2*(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)/(-c^2*x^2+1)^(
1/2)+2/9*b*c^2*f^2*x^3*(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)/(-c^2*x^2+1)^(1/2)-1/16*b*c^3*f^2*x^4*(c*d*x+d)^(1/2)*
(-c*f*x+f)^(1/2)/(-c^2*x^2+1)^(1/2)+5/16*f^2*(a+b*arcsin(c*x))^2*(c*d*x+d)^(1/2)*(-c*f*x+f)^(1/2)/b/c/(-c^2*x^
2+1)^(1/2)

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Rubi [A]  time = 0.54, antiderivative size = 376, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {4673, 4763, 4647, 4641, 30, 4677, 4697, 4707} \[ \frac {1}{4} c^2 f^2 x^3 \sqrt {c d x+d} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )+\frac {5 f^2 \sqrt {c d x+d} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \sqrt {1-c^2 x^2}}+\frac {2 f^2 \left (1-c^2 x^2\right ) \sqrt {c d x+d} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )}{3 c}+\frac {3}{8} f^2 x \sqrt {c d x+d} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )-\frac {b c^3 f^2 x^4 \sqrt {c d x+d} \sqrt {f-c f x}}{16 \sqrt {1-c^2 x^2}}+\frac {2 b c^2 f^2 x^3 \sqrt {c d x+d} \sqrt {f-c f x}}{9 \sqrt {1-c^2 x^2}}-\frac {3 b c f^2 x^2 \sqrt {c d x+d} \sqrt {f-c f x}}{16 \sqrt {1-c^2 x^2}}-\frac {2 b f^2 x \sqrt {c d x+d} \sqrt {f-c f x}}{3 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*f^2*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(3*Sqrt[1 - c^2*x^2]) - (3*b*c*f^2*x^2*Sqrt[d + c*d*x]*Sqrt[f - c
*f*x])/(16*Sqrt[1 - c^2*x^2]) + (2*b*c^2*f^2*x^3*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(9*Sqrt[1 - c^2*x^2]) - (b*c
^3*f^2*x^4*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(16*Sqrt[1 - c^2*x^2]) + (3*f^2*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*
(a + b*ArcSin[c*x]))/8 + (c^2*f^2*x^3*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(a + b*ArcSin[c*x]))/4 + (2*f^2*Sqrt[d +
c*d*x]*Sqrt[f - c*f*x]*(1 - c^2*x^2)*(a + b*ArcSin[c*x]))/(3*c) + (5*f^2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(a +
b*ArcSin[c*x])^2)/(16*b*c*Sqrt[1 - c^2*x^2])

Rule 30

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

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 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 4697

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

Rule 4707

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

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 \sqrt {d+c d x} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac {\left (\sqrt {d+c d x} \sqrt {f-c f x}\right ) \int (f-c f x)^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt {1-c^2 x^2}}\\ &=\frac {\left (\sqrt {d+c d x} \sqrt {f-c f x}\right ) \int \left (f^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )-2 c f^2 x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )+c^2 f^2 x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )\right ) \, dx}{\sqrt {1-c^2 x^2}}\\ &=\frac {\left (f^2 \sqrt {d+c d x} \sqrt {f-c f x}\right ) \int \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt {1-c^2 x^2}}-\frac {\left (2 c f^2 \sqrt {d+c d x} \sqrt {f-c f x}\right ) \int x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt {1-c^2 x^2}}+\frac {\left (c^2 f^2 \sqrt {d+c d x} \sqrt {f-c f x}\right ) \int x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\sqrt {1-c^2 x^2}}\\ &=\frac {1}{2} f^2 x \sqrt {d+c d x} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} c^2 f^2 x^3 \sqrt {d+c d x} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )+\frac {2 f^2 \sqrt {d+c d x} \sqrt {f-c f x} \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c}+\frac {\left (f^2 \sqrt {d+c d x} \sqrt {f-c f x}\right ) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{2 \sqrt {1-c^2 x^2}}-\frac {\left (2 b f^2 \sqrt {d+c d x} \sqrt {f-c f x}\right ) \int \left (1-c^2 x^2\right ) \, dx}{3 \sqrt {1-c^2 x^2}}-\frac {\left (b c f^2 \sqrt {d+c d x} \sqrt {f-c f x}\right ) \int x \, dx}{2 \sqrt {1-c^2 x^2}}+\frac {\left (c^2 f^2 \sqrt {d+c d x} \sqrt {f-c f x}\right ) \int \frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{4 \sqrt {1-c^2 x^2}}-\frac {\left (b c^3 f^2 \sqrt {d+c d x} \sqrt {f-c f x}\right ) \int x^3 \, dx}{4 \sqrt {1-c^2 x^2}}\\ &=-\frac {2 b f^2 x \sqrt {d+c d x} \sqrt {f-c f x}}{3 \sqrt {1-c^2 x^2}}-\frac {b c f^2 x^2 \sqrt {d+c d x} \sqrt {f-c f x}}{4 \sqrt {1-c^2 x^2}}+\frac {2 b c^2 f^2 x^3 \sqrt {d+c d x} \sqrt {f-c f x}}{9 \sqrt {1-c^2 x^2}}-\frac {b c^3 f^2 x^4 \sqrt {d+c d x} \sqrt {f-c f x}}{16 \sqrt {1-c^2 x^2}}+\frac {3}{8} f^2 x \sqrt {d+c d x} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} c^2 f^2 x^3 \sqrt {d+c d x} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )+\frac {2 f^2 \sqrt {d+c d x} \sqrt {f-c f x} \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c}+\frac {f^2 \sqrt {d+c d x} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )^2}{4 b c \sqrt {1-c^2 x^2}}+\frac {\left (f^2 \sqrt {d+c d x} \sqrt {f-c f x}\right ) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{8 \sqrt {1-c^2 x^2}}+\frac {\left (b c f^2 \sqrt {d+c d x} \sqrt {f-c f x}\right ) \int x \, dx}{8 \sqrt {1-c^2 x^2}}\\ &=-\frac {2 b f^2 x \sqrt {d+c d x} \sqrt {f-c f x}}{3 \sqrt {1-c^2 x^2}}-\frac {3 b c f^2 x^2 \sqrt {d+c d x} \sqrt {f-c f x}}{16 \sqrt {1-c^2 x^2}}+\frac {2 b c^2 f^2 x^3 \sqrt {d+c d x} \sqrt {f-c f x}}{9 \sqrt {1-c^2 x^2}}-\frac {b c^3 f^2 x^4 \sqrt {d+c d x} \sqrt {f-c f x}}{16 \sqrt {1-c^2 x^2}}+\frac {3}{8} f^2 x \sqrt {d+c d x} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} c^2 f^2 x^3 \sqrt {d+c d x} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )+\frac {2 f^2 \sqrt {d+c d x} \sqrt {f-c f x} \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 c}+\frac {5 f^2 \sqrt {d+c d x} \sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )^2}{16 b c \sqrt {1-c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 1.35, size = 293, normalized size = 0.78 \[ \frac {f^2 \sqrt {c d x+d} \sqrt {f-c f x} \left (48 a \sqrt {1-c^2 x^2} \left (6 c^3 x^3-16 c^2 x^2+9 c x+16\right )+256 b c x \left (c^2 x^2-3\right )+144 b \cos \left (2 \sin ^{-1}(c x)\right )-9 b \cos \left (4 \sin ^{-1}(c x)\right )\right )-720 a \sqrt {d} f^{5/2} \sqrt {1-c^2 x^2} \tan ^{-1}\left (\frac {c x \sqrt {c d x+d} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (c^2 x^2-1\right )}\right )-12 b f^2 \sqrt {c d x+d} \sqrt {f-c f x} \sin ^{-1}(c x) \left (-64 \left (1-c^2 x^2\right )^{3/2}-24 \sin \left (2 \sin ^{-1}(c x)\right )+3 \sin \left (4 \sin ^{-1}(c x)\right )\right )+360 b f^2 \sqrt {c d x+d} \sqrt {f-c f x} \sin ^{-1}(c x)^2}{1152 c \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(360*b*f^2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2 - 720*a*Sqrt[d]*f^(5/2)*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))] + f^2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(256
*b*c*x*(-3 + c^2*x^2) + 48*a*Sqrt[1 - c^2*x^2]*(16 + 9*c*x - 16*c^2*x^2 + 6*c^3*x^3) + 144*b*Cos[2*ArcSin[c*x]
] - 9*b*Cos[4*ArcSin[c*x]]) - 12*b*f^2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]*(-64*(1 - c^2*x^2)^(3/2) -
24*Sin[2*ArcSin[c*x]] + 3*Sin[4*ArcSin[c*x]]))/(1152*c*Sqrt[1 - c^2*x^2])

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a c^{2} f^{2} x^{2} - 2 \, a c f^{2} x + a f^{2} + {\left (b c^{2} f^{2} x^{2} - 2 \, b c f^{2} x + b f^{2}\right )} \arcsin \left (c x\right )\right )} \sqrt {c d x + d} \sqrt {-c f x + f}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^(1/2)*(-c*f*x+f)^(5/2)*(a+b*arcsin(c*x)),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), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^(1/2)*(-c*f*x+f)^(5/2)*(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*x+d)^(1/2)*(-c*f*x+f)^(5/2)*(a+b*arcsin(c*x)),x)

[Out]

int((c*d*x+d)^(1/2)*(-c*f*x+f)^(5/2)*(a+b*arcsin(c*x)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b \sqrt {d} \sqrt {f} \int {\left (c^{2} f^{2} x^{2} - 2 \, c f^{2} x + f^{2}\right )} \sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\,{d x} + \frac {1}{24} \, {\left (15 \, \sqrt {-c^{2} d f x^{2} + d f} f^{2} x + \frac {15 \, d f^{3} \arcsin \left (c x\right )}{\sqrt {d f} c} - \frac {6 \, {\left (-c^{2} d f x^{2} + d f\right )}^{\frac {3}{2}} f x}{d} + \frac {16 \, {\left (-c^{2} d f x^{2} + d f\right )}^{\frac {3}{2}} f}{c d}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)^(1/2)*(-c*f*x+f)^(5/2)*(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

b*sqrt(d)*sqrt(f)*integrate((c^2*f^2*x^2 - 2*c*f^2*x + f^2)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x
+ 1)*sqrt(-c*x + 1)), x) + 1/24*(15*sqrt(-c^2*d*f*x^2 + d*f)*f^2*x + 15*d*f^3*arcsin(c*x)/(sqrt(d*f)*c) - 6*(
-c^2*d*f*x^2 + d*f)^(3/2)*f*x/d + 16*(-c^2*d*f*x^2 + d*f)^(3/2)*f/(c*d))*a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))*(d + c*d*x)^(1/2)*(f - c*f*x)^(5/2),x)

[Out]

int((a + b*asin(c*x))*(d + c*d*x)^(1/2)*(f - c*f*x)^(5/2), 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)**(1/2)*(-c*f*x+f)**(5/2)*(a+b*asin(c*x)),x)

[Out]

Timed out

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