3.6.16 \(\int (d+c d x)^{5/2} (f-c f x)^{5/2} (a+b \text {ArcSin}(c x)) \, dx\) [516]

Optimal. Leaf size=315 \[ -\frac {25 b c x^2 (d+c d x)^{5/2} (f-c f x)^{5/2}}{96 \left (1-c^2 x^2\right )^{5/2}}+\frac {5 b c^3 x^4 (d+c d x)^{5/2} (f-c f x)^{5/2}}{96 \left (1-c^2 x^2\right )^{5/2}}+\frac {b (d+c d x)^{5/2} (f-c f x)^{5/2} \sqrt {1-c^2 x^2}}{36 c}+\frac {1}{6} x (d+c d x)^{5/2} (f-c f x)^{5/2} (a+b \text {ArcSin}(c x))+\frac {5 x (d+c d x)^{5/2} (f-c f x)^{5/2} (a+b \text {ArcSin}(c x))}{16 \left (1-c^2 x^2\right )^2}+\frac {5 x (d+c d x)^{5/2} (f-c f x)^{5/2} (a+b \text {ArcSin}(c x))}{24 \left (1-c^2 x^2\right )}+\frac {5 (d+c d x)^{5/2} (f-c f x)^{5/2} (a+b \text {ArcSin}(c x))^2}{32 b c \left (1-c^2 x^2\right )^{5/2}} \]

[Out]

-25/96*b*c*x^2*(c*d*x+d)^(5/2)*(-c*f*x+f)^(5/2)/(-c^2*x^2+1)^(5/2)+5/96*b*c^3*x^4*(c*d*x+d)^(5/2)*(-c*f*x+f)^(
5/2)/(-c^2*x^2+1)^(5/2)+1/6*x*(c*d*x+d)^(5/2)*(-c*f*x+f)^(5/2)*(a+b*arcsin(c*x))+5/16*x*(c*d*x+d)^(5/2)*(-c*f*
x+f)^(5/2)*(a+b*arcsin(c*x))/(-c^2*x^2+1)^2+5/24*x*(c*d*x+d)^(5/2)*(-c*f*x+f)^(5/2)*(a+b*arcsin(c*x))/(-c^2*x^
2+1)+5/32*(c*d*x+d)^(5/2)*(-c*f*x+f)^(5/2)*(a+b*arcsin(c*x))^2/b/c/(-c^2*x^2+1)^(5/2)+1/36*b*(c*d*x+d)^(5/2)*(
-c*f*x+f)^(5/2)*(-c^2*x^2+1)^(1/2)/c

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Rubi [A]
time = 0.19, antiderivative size = 315, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {4763, 4743, 4741, 4737, 30, 14, 267} \begin {gather*} \frac {5 x (c d x+d)^{5/2} (f-c f x)^{5/2} (a+b \text {ArcSin}(c x))}{24 \left (1-c^2 x^2\right )}+\frac {5 x (c d x+d)^{5/2} (f-c f x)^{5/2} (a+b \text {ArcSin}(c x))}{16 \left (1-c^2 x^2\right )^2}+\frac {5 (c d x+d)^{5/2} (f-c f x)^{5/2} (a+b \text {ArcSin}(c x))^2}{32 b c \left (1-c^2 x^2\right )^{5/2}}+\frac {1}{6} x (c d x+d)^{5/2} (f-c f x)^{5/2} (a+b \text {ArcSin}(c x))-\frac {25 b c x^2 (c d x+d)^{5/2} (f-c f x)^{5/2}}{96 \left (1-c^2 x^2\right )^{5/2}}+\frac {b \sqrt {1-c^2 x^2} (c d x+d)^{5/2} (f-c f x)^{5/2}}{36 c}+\frac {5 b c^3 x^4 (c d x+d)^{5/2} (f-c f x)^{5/2}}{96 \left (1-c^2 x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-25*b*c*x^2*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2))/(96*(1 - c^2*x^2)^(5/2)) + (5*b*c^3*x^4*(d + c*d*x)^(5/2)*(f
- c*f*x)^(5/2))/(96*(1 - c^2*x^2)^(5/2)) + (b*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)*Sqrt[1 - c^2*x^2])/(36*c) +
(x*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)*(a + b*ArcSin[c*x]))/6 + (5*x*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)*(a +
b*ArcSin[c*x]))/(16*(1 - c^2*x^2)^2) + (5*x*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)*(a + b*ArcSin[c*x]))/(24*(1 -
c^2*x^2)) + (5*(d + c*d*x)^(5/2)*(f - c*f*x)^(5/2)*(a + b*ArcSin[c*x])^2)/(32*b*c*(1 - c^2*x^2)^(5/2))

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 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

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]

Rubi steps

\begin {align*} \int (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac {\left ((d+c d x)^{5/2} (f-c f x)^{5/2}\right ) \int \left (1-c^2 x^2\right )^{5/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{\left (1-c^2 x^2\right )^{5/2}}\\ &=\frac {1}{6} x (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {\left (5 (d+c d x)^{5/2} (f-c f x)^{5/2}\right ) \int \left (1-c^2 x^2\right )^{3/2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{6 \left (1-c^2 x^2\right )^{5/2}}-\frac {\left (b c (d+c d x)^{5/2} (f-c f x)^{5/2}\right ) \int x \left (1-c^2 x^2\right )^2 \, dx}{6 \left (1-c^2 x^2\right )^{5/2}}\\ &=\frac {b (d+c d x)^{5/2} (f-c f x)^{5/2} \sqrt {1-c^2 x^2}}{36 c}+\frac {1}{6} x (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {5 x (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{24 \left (1-c^2 x^2\right )}+\frac {\left (5 (d+c d x)^{5/2} (f-c f x)^{5/2}\right ) \int \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \, dx}{8 \left (1-c^2 x^2\right )^{5/2}}-\frac {\left (5 b c (d+c d x)^{5/2} (f-c f x)^{5/2}\right ) \int x \left (1-c^2 x^2\right ) \, dx}{24 \left (1-c^2 x^2\right )^{5/2}}\\ &=\frac {b (d+c d x)^{5/2} (f-c f x)^{5/2} \sqrt {1-c^2 x^2}}{36 c}+\frac {1}{6} x (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {5 x (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{16 \left (1-c^2 x^2\right )^2}+\frac {5 x (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{24 \left (1-c^2 x^2\right )}+\frac {\left (5 (d+c d x)^{5/2} (f-c f x)^{5/2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{16 \left (1-c^2 x^2\right )^{5/2}}-\frac {\left (5 b c (d+c d x)^{5/2} (f-c f x)^{5/2}\right ) \int \left (x-c^2 x^3\right ) \, dx}{24 \left (1-c^2 x^2\right )^{5/2}}-\frac {\left (5 b c (d+c d x)^{5/2} (f-c f x)^{5/2}\right ) \int x \, dx}{16 \left (1-c^2 x^2\right )^{5/2}}\\ &=-\frac {25 b c x^2 (d+c d x)^{5/2} (f-c f x)^{5/2}}{96 \left (1-c^2 x^2\right )^{5/2}}+\frac {5 b c^3 x^4 (d+c d x)^{5/2} (f-c f x)^{5/2}}{96 \left (1-c^2 x^2\right )^{5/2}}+\frac {b (d+c d x)^{5/2} (f-c f x)^{5/2} \sqrt {1-c^2 x^2}}{36 c}+\frac {1}{6} x (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )+\frac {5 x (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{16 \left (1-c^2 x^2\right )^2}+\frac {5 x (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )}{24 \left (1-c^2 x^2\right )}+\frac {5 (d+c d x)^{5/2} (f-c f x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )^2}{32 b c \left (1-c^2 x^2\right )^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.98, size = 303, normalized size = 0.96 \begin {gather*} \frac {d^2 f^2 \left (360 b \sqrt {d+c d x} \sqrt {f-c f x} \text {ArcSin}(c x)^2-720 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 (1584 a c x \sqrt {1-c^2 x^2}-1248 a c^3 x^3 \sqrt {1-c^2 x^2}+384 a c^5 x^5 \sqrt {1-c^2 x^2}+270 b \cos (2 \text {ArcSin}(c x))+27 b \cos (4 \text {ArcSin}(c x))+2 b \cos (6 \text {ArcSin}(c x))\right )+12 b \sqrt {d+c d x} \sqrt {f-c f x} \text {ArcSin}(c x) (45 \sin (2 \text {ArcSin}(c x))+9 \sin (4 \text {ArcSin}(c x))+\sin (6 \text {ArcSin}(c x)))\right )}{2304 c \sqrt {1-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*f^2*(360*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2 - 720*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]*(15
84*a*c*x*Sqrt[1 - c^2*x^2] - 1248*a*c^3*x^3*Sqrt[1 - c^2*x^2] + 384*a*c^5*x^5*Sqrt[1 - c^2*x^2] + 270*b*Cos[2*
ArcSin[c*x]] + 27*b*Cos[4*ArcSin[c*x]] + 2*b*Cos[6*ArcSin[c*x]]) + 12*b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin
[c*x]*(45*Sin[2*ArcSin[c*x]] + 9*Sin[4*ArcSin[c*x]] + Sin[6*ArcSin[c*x]])))/(2304*c*Sqrt[1 - c^2*x^2])

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Maple [F]
time = 0.13, size = 0, normalized size = 0.00 \[\int \left (c d x +d \right )^{\frac {5}{2}} \left (-c f x +f \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

b*sqrt(d)*sqrt(f)*integrate((c^4*d^2*f^2*x^4 - 2*c^2*d^2*f^2*x^2 + d^2*f^2)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arcta
n2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)), x) + 1/48*(15*sqrt(-c^2*d*f*x^2 + d*f)*d^2*f^2*x + 15*d^3*f^3*arcsin(c*
x)/(sqrt(d*f)*c) + 10*(-c^2*d*f*x^2 + d*f)^(3/2)*d*f*x + 8*(-c^2*d*f*x^2 + d*f)^(5/2)*x)*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)^(5/2)*(-c*f*x+f)^(5/2)*(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

integral((a*c^4*d^2*f^2*x^4 - 2*a*c^2*d^2*f^2*x^2 + a*d^2*f^2 + (b*c^4*d^2*f^2*x^4 - 2*b*c^2*d^2*f^2*x^2 + b*d
^2*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)**(5/2)*(-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)^(5/2)*(-c*f*x+f)^(5/2)*(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

integrate((c*d*x + d)^(5/2)*(-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 )}^{5/2}\,{\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)^(5/2)*(f - c*f*x)^(5/2),x)

[Out]

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

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