3.6.17 \(\int (d+c d x)^{3/2} (f-c f x)^{5/2} (a+b \text {ArcSin}(c x)) \, dx\) [517]

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

[Out]

-1/5*b*f*x*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)/(-c^2*x^2+1)^(3/2)-5/16*b*c*f*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*f*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*f*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*f*x^5*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)/(-c^2*x^2+1)
^(3/2)+1/4*f*x*(c*d*x+d)^(3/2)*(-c*f*x+f)^(3/2)*(a+b*arcsin(c*x))+3/8*f*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*f*(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*f*(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 = 0.28, 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 = {4763, 4847, 4743, 4741, 4737, 30, 14, 4767, 200} \begin {gather*} \frac {3 f x (c d x+d)^{3/2} (f-c f x)^{3/2} (a+b \text {ArcSin}(c x))}{8 \left (1-c^2 x^2\right )}+\frac {3 f (c d x+d)^{3/2} (f-c f x)^{3/2} (a+b \text {ArcSin}(c x))^2}{16 b c \left (1-c^2 x^2\right )^{3/2}}+\frac {f \left (1-c^2 x^2\right ) (c d x+d)^{3/2} (f-c f x)^{3/2} (a+b \text {ArcSin}(c x))}{5 c}+\frac {1}{4} f x (c d x+d)^{3/2} (f-c f x)^{3/2} (a+b \text {ArcSin}(c x))-\frac {5 b c f 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 f 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 f 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 f 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 f x^4 (c d x+d)^{3/2} (f-c f x)^{3/2}}{16 \left (1-c^2 x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(b*f*x*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2))/(1 - c^2*x^2)^(3/2) - (5*b*c*f*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*f*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*f*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*f*x^5*(d + c*d*x)^(
3/2)*(f - c*f*x)^(3/2))/(25*(1 - c^2*x^2)^(3/2)) + (f*x*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)*(a + b*ArcSin[c*x]
))/4 + (3*f*x*(d + c*d*x)^(3/2)*(f - c*f*x)^(3/2)*(a + b*ArcSin[c*x]))/(8*(1 - c^2*x^2)) + (f*(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*f*(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 200

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 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 4741

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[(1/2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]], Int[(a + b*ArcSin[c*x])^n/S
qrt[1 - c^2*x^2], x], x] - Dist[b*c*(n/2)*Simp[Sqrt[d + e*x^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 4743

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/(2*p + 1))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p], Int[x*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcS
in[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0]

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 4847

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)^{3/2} (f-c f x)^{5/2} \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 (f-c f 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 (f \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right )-c f 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 (f (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 f (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} f x (d+c d x)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {f (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 f (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 f (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 f (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} f x (d+c d x)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {3 f 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 {f (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 f (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 f (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 f (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 f (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 f 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 f 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 f 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 f 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 f 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} f x (d+c d x)^{3/2} (f-c f x)^{3/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {3 f 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 {f (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 f (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 = 0.89, size = 305, normalized size = 0.74 \begin {gather*} \frac {d f^2 \left (1800 b \sqrt {d+c d x} \sqrt {f-c f x} \text {ArcSin}(c x)^2-3600 a \sqrt {d} \sqrt {f} \sqrt {1-c^2 x^2} \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 )+\sqrt {d+c d x} \sqrt {f-c f x} \left (-128 b c x \left (15-10 c^2 x^2+3 c^4 x^4\right )+240 a \sqrt {1-c^2 x^2} \left (8+25 c x-16 c^2 x^2-10 c^3 x^3+8 c^4 x^4\right )+1200 b \cos (2 \text {ArcSin}(c x))+75 b \cos (4 \text {ArcSin}(c x))\right )+60 b \sqrt {d+c d x} \sqrt {f-c f x} \text {ArcSin}(c x) \left (32 \left (1-c^2 x^2\right )^{5/2}+40 \sin (2 \text {ArcSin}(c x))+5 \sin (4 \text {ArcSin}(c x))\right )\right )}{9600 c \sqrt {1-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*f^2*(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]*(-1
28*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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Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \left (c d x +d \right )^{\frac {3}{2}} \left (-c f x +f \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((c*d*x+d)^(3/2)*(-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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*sqrt(d)*sqrt(f)*integrate((c^3*d*f^2*x^3 - c^2*d*f^2*x^2 - c*d*f^2*x + d*f^2)*sqrt(c*x + 1)*sqrt(-c*x + 1)*a
rctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)), x) + 1/40*(15*sqrt(-c^2*d*f*x^2 + d*f)*d*f^2*x + 15*d^2*f^3*arcsin(
c*x)/(sqrt(d*f)*c) + 10*(-c^2*d*f*x^2 + d*f)^(3/2)*f*x + 8*(-c^2*d*f*x^2 + d*f)^(5/2)/(c*d))*a

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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