3.596 \(\int x^4 (d+e x^2) (a+b \sin ^{-1}(c x)) \, dx\)

Optimal. Leaf size=152 \[ \frac{1}{5} d x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{7} e x^7 \left (a+b \sin ^{-1}(c x)\right )+\frac{b \left (1-c^2 x^2\right )^{5/2} \left (7 c^2 d+15 e\right )}{175 c^7}-\frac{b \left (1-c^2 x^2\right )^{3/2} \left (14 c^2 d+15 e\right )}{105 c^7}+\frac{b \sqrt{1-c^2 x^2} \left (7 c^2 d+5 e\right )}{35 c^7}-\frac{b e \left (1-c^2 x^2\right )^{7/2}}{49 c^7} \]

[Out]

(b*(7*c^2*d + 5*e)*Sqrt[1 - c^2*x^2])/(35*c^7) - (b*(14*c^2*d + 15*e)*(1 - c^2*x^2)^(3/2))/(105*c^7) + (b*(7*c
^2*d + 15*e)*(1 - c^2*x^2)^(5/2))/(175*c^7) - (b*e*(1 - c^2*x^2)^(7/2))/(49*c^7) + (d*x^5*(a + b*ArcSin[c*x]))
/5 + (e*x^7*(a + b*ArcSin[c*x]))/7

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Rubi [A]  time = 0.150382, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {14, 4731, 12, 446, 77} \[ \frac{1}{5} d x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{7} e x^7 \left (a+b \sin ^{-1}(c x)\right )+\frac{b \left (1-c^2 x^2\right )^{5/2} \left (7 c^2 d+15 e\right )}{175 c^7}-\frac{b \left (1-c^2 x^2\right )^{3/2} \left (14 c^2 d+15 e\right )}{105 c^7}+\frac{b \sqrt{1-c^2 x^2} \left (7 c^2 d+5 e\right )}{35 c^7}-\frac{b e \left (1-c^2 x^2\right )^{7/2}}{49 c^7} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(d + e*x^2)*(a + b*ArcSin[c*x]),x]

[Out]

(b*(7*c^2*d + 5*e)*Sqrt[1 - c^2*x^2])/(35*c^7) - (b*(14*c^2*d + 15*e)*(1 - c^2*x^2)^(3/2))/(105*c^7) + (b*(7*c
^2*d + 15*e)*(1 - c^2*x^2)^(5/2))/(175*c^7) - (b*e*(1 - c^2*x^2)^(7/2))/(49*c^7) + (d*x^5*(a + b*ArcSin[c*x]))
/5 + (e*x^7*(a + b*ArcSin[c*x]))/7

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 4731

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSin[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 -
 c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[c^2*d + e, 0] && IntegerQ[p] && (GtQ[p, 0] ||
 (IGtQ[(m - 1)/2, 0] && LeQ[m + p, 0]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
 NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int x^4 \left (d+e x^2\right ) \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac{1}{5} d x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{7} e x^7 \left (a+b \sin ^{-1}(c x)\right )-(b c) \int \frac{x^5 \left (7 d+5 e x^2\right )}{35 \sqrt{1-c^2 x^2}} \, dx\\ &=\frac{1}{5} d x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{7} e x^7 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{35} (b c) \int \frac{x^5 \left (7 d+5 e x^2\right )}{\sqrt{1-c^2 x^2}} \, dx\\ &=\frac{1}{5} d x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{7} e x^7 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{70} (b c) \operatorname{Subst}\left (\int \frac{x^2 (7 d+5 e x)}{\sqrt{1-c^2 x}} \, dx,x,x^2\right )\\ &=\frac{1}{5} d x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{7} e x^7 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{70} (b c) \operatorname{Subst}\left (\int \left (\frac{7 c^2 d+5 e}{c^6 \sqrt{1-c^2 x}}+\frac{\left (-14 c^2 d-15 e\right ) \sqrt{1-c^2 x}}{c^6}+\frac{\left (7 c^2 d+15 e\right ) \left (1-c^2 x\right )^{3/2}}{c^6}-\frac{5 e \left (1-c^2 x\right )^{5/2}}{c^6}\right ) \, dx,x,x^2\right )\\ &=\frac{b \left (7 c^2 d+5 e\right ) \sqrt{1-c^2 x^2}}{35 c^7}-\frac{b \left (14 c^2 d+15 e\right ) \left (1-c^2 x^2\right )^{3/2}}{105 c^7}+\frac{b \left (7 c^2 d+15 e\right ) \left (1-c^2 x^2\right )^{5/2}}{175 c^7}-\frac{b e \left (1-c^2 x^2\right )^{7/2}}{49 c^7}+\frac{1}{5} d x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{7} e x^7 \left (a+b \sin ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.114022, size = 115, normalized size = 0.76 \[ \frac{105 a x^5 \left (7 d+5 e x^2\right )+\frac{b \sqrt{1-c^2 x^2} \left (3 c^6 \left (49 d x^4+25 e x^6\right )+2 c^4 \left (98 d x^2+45 e x^4\right )+8 c^2 \left (49 d+15 e x^2\right )+240 e\right )}{c^7}+105 b x^5 \sin ^{-1}(c x) \left (7 d+5 e x^2\right )}{3675} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(d + e*x^2)*(a + b*ArcSin[c*x]),x]

[Out]

(105*a*x^5*(7*d + 5*e*x^2) + (b*Sqrt[1 - c^2*x^2]*(240*e + 8*c^2*(49*d + 15*e*x^2) + 2*c^4*(98*d*x^2 + 45*e*x^
4) + 3*c^6*(49*d*x^4 + 25*e*x^6)))/c^7 + 105*b*x^5*(7*d + 5*e*x^2)*ArcSin[c*x])/3675

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Maple [A]  time = 0.005, size = 201, normalized size = 1.3 \begin{align*}{\frac{1}{{c}^{5}} \left ({\frac{a}{{c}^{2}} \left ({\frac{e{c}^{7}{x}^{7}}{7}}+{\frac{{c}^{7}{x}^{5}d}{5}} \right ) }+{\frac{b}{{c}^{2}} \left ({\frac{\arcsin \left ( cx \right ) e{c}^{7}{x}^{7}}{7}}+{\frac{\arcsin \left ( cx \right ){c}^{7}{x}^{5}d}{5}}-{\frac{e}{7} \left ( -{\frac{{c}^{6}{x}^{6}}{7}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{6\,{c}^{4}{x}^{4}}{35}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{8\,{c}^{2}{x}^{2}}{35}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{16}{35}\sqrt{-{c}^{2}{x}^{2}+1}} \right ) }-{\frac{{c}^{2}d}{5} \left ( -{\frac{{c}^{4}{x}^{4}}{5}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{4\,{c}^{2}{x}^{2}}{15}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{8}{15}\sqrt{-{c}^{2}{x}^{2}+1}} \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(e*x^2+d)*(a+b*arcsin(c*x)),x)

[Out]

1/c^5*(a/c^2*(1/7*e*c^7*x^7+1/5*c^7*x^5*d)+b/c^2*(1/7*arcsin(c*x)*e*c^7*x^7+1/5*arcsin(c*x)*c^7*x^5*d-1/7*e*(-
1/7*c^6*x^6*(-c^2*x^2+1)^(1/2)-6/35*c^4*x^4*(-c^2*x^2+1)^(1/2)-8/35*c^2*x^2*(-c^2*x^2+1)^(1/2)-16/35*(-c^2*x^2
+1)^(1/2))-1/5*c^2*d*(-1/5*c^4*x^4*(-c^2*x^2+1)^(1/2)-4/15*c^2*x^2*(-c^2*x^2+1)^(1/2)-8/15*(-c^2*x^2+1)^(1/2))
))

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Maxima [A]  time = 1.46336, size = 247, normalized size = 1.62 \begin{align*} \frac{1}{7} \, a e x^{7} + \frac{1}{5} \, a d x^{5} + \frac{1}{75} \,{\left (15 \, x^{5} \arcsin \left (c x\right ) +{\left (\frac{3 \, \sqrt{-c^{2} x^{2} + 1} x^{4}}{c^{2}} + \frac{4 \, \sqrt{-c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac{8 \, \sqrt{-c^{2} x^{2} + 1}}{c^{6}}\right )} c\right )} b d + \frac{1}{245} \,{\left (35 \, x^{7} \arcsin \left (c x\right ) +{\left (\frac{5 \, \sqrt{-c^{2} x^{2} + 1} x^{6}}{c^{2}} + \frac{6 \, \sqrt{-c^{2} x^{2} + 1} x^{4}}{c^{4}} + \frac{8 \, \sqrt{-c^{2} x^{2} + 1} x^{2}}{c^{6}} + \frac{16 \, \sqrt{-c^{2} x^{2} + 1}}{c^{8}}\right )} c\right )} b e \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

1/7*a*e*x^7 + 1/5*a*d*x^5 + 1/75*(15*x^5*arcsin(c*x) + (3*sqrt(-c^2*x^2 + 1)*x^4/c^2 + 4*sqrt(-c^2*x^2 + 1)*x^
2/c^4 + 8*sqrt(-c^2*x^2 + 1)/c^6)*c)*b*d + 1/245*(35*x^7*arcsin(c*x) + (5*sqrt(-c^2*x^2 + 1)*x^6/c^2 + 6*sqrt(
-c^2*x^2 + 1)*x^4/c^4 + 8*sqrt(-c^2*x^2 + 1)*x^2/c^6 + 16*sqrt(-c^2*x^2 + 1)/c^8)*c)*b*e

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Fricas [A]  time = 2.27816, size = 308, normalized size = 2.03 \begin{align*} \frac{525 \, a c^{7} e x^{7} + 735 \, a c^{7} d x^{5} + 105 \,{\left (5 \, b c^{7} e x^{7} + 7 \, b c^{7} d x^{5}\right )} \arcsin \left (c x\right ) +{\left (75 \, b c^{6} e x^{6} + 3 \,{\left (49 \, b c^{6} d + 30 \, b c^{4} e\right )} x^{4} + 392 \, b c^{2} d + 4 \,{\left (49 \, b c^{4} d + 30 \, b c^{2} e\right )} x^{2} + 240 \, b e\right )} \sqrt{-c^{2} x^{2} + 1}}{3675 \, c^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

1/3675*(525*a*c^7*e*x^7 + 735*a*c^7*d*x^5 + 105*(5*b*c^7*e*x^7 + 7*b*c^7*d*x^5)*arcsin(c*x) + (75*b*c^6*e*x^6
+ 3*(49*b*c^6*d + 30*b*c^4*e)*x^4 + 392*b*c^2*d + 4*(49*b*c^4*d + 30*b*c^2*e)*x^2 + 240*b*e)*sqrt(-c^2*x^2 + 1
))/c^7

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Sympy [A]  time = 7.81841, size = 223, normalized size = 1.47 \begin{align*} \begin{cases} \frac{a d x^{5}}{5} + \frac{a e x^{7}}{7} + \frac{b d x^{5} \operatorname{asin}{\left (c x \right )}}{5} + \frac{b e x^{7} \operatorname{asin}{\left (c x \right )}}{7} + \frac{b d x^{4} \sqrt{- c^{2} x^{2} + 1}}{25 c} + \frac{b e x^{6} \sqrt{- c^{2} x^{2} + 1}}{49 c} + \frac{4 b d x^{2} \sqrt{- c^{2} x^{2} + 1}}{75 c^{3}} + \frac{6 b e x^{4} \sqrt{- c^{2} x^{2} + 1}}{245 c^{3}} + \frac{8 b d \sqrt{- c^{2} x^{2} + 1}}{75 c^{5}} + \frac{8 b e x^{2} \sqrt{- c^{2} x^{2} + 1}}{245 c^{5}} + \frac{16 b e \sqrt{- c^{2} x^{2} + 1}}{245 c^{7}} & \text{for}\: c \neq 0 \\a \left (\frac{d x^{5}}{5} + \frac{e x^{7}}{7}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(e*x**2+d)*(a+b*asin(c*x)),x)

[Out]

Piecewise((a*d*x**5/5 + a*e*x**7/7 + b*d*x**5*asin(c*x)/5 + b*e*x**7*asin(c*x)/7 + b*d*x**4*sqrt(-c**2*x**2 +
1)/(25*c) + b*e*x**6*sqrt(-c**2*x**2 + 1)/(49*c) + 4*b*d*x**2*sqrt(-c**2*x**2 + 1)/(75*c**3) + 6*b*e*x**4*sqrt
(-c**2*x**2 + 1)/(245*c**3) + 8*b*d*sqrt(-c**2*x**2 + 1)/(75*c**5) + 8*b*e*x**2*sqrt(-c**2*x**2 + 1)/(245*c**5
) + 16*b*e*sqrt(-c**2*x**2 + 1)/(245*c**7), Ne(c, 0)), (a*(d*x**5/5 + e*x**7/7), True))

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Giac [B]  time = 1.2158, size = 439, normalized size = 2.89 \begin{align*} \frac{1}{7} \, a x^{7} e + \frac{1}{5} \, a d x^{5} + \frac{{\left (c^{2} x^{2} - 1\right )}^{2} b d x \arcsin \left (c x\right )}{5 \, c^{4}} + \frac{2 \,{\left (c^{2} x^{2} - 1\right )} b d x \arcsin \left (c x\right )}{5 \, c^{4}} + \frac{{\left (c^{2} x^{2} - 1\right )}^{3} b x \arcsin \left (c x\right ) e}{7 \, c^{6}} + \frac{b d x \arcsin \left (c x\right )}{5 \, c^{4}} + \frac{3 \,{\left (c^{2} x^{2} - 1\right )}^{2} b x \arcsin \left (c x\right ) e}{7 \, c^{6}} + \frac{{\left (c^{2} x^{2} - 1\right )}^{2} \sqrt{-c^{2} x^{2} + 1} b d}{25 \, c^{5}} + \frac{3 \,{\left (c^{2} x^{2} - 1\right )} b x \arcsin \left (c x\right ) e}{7 \, c^{6}} - \frac{2 \,{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}} b d}{15 \, c^{5}} + \frac{{\left (c^{2} x^{2} - 1\right )}^{3} \sqrt{-c^{2} x^{2} + 1} b e}{49 \, c^{7}} + \frac{b x \arcsin \left (c x\right ) e}{7 \, c^{6}} + \frac{\sqrt{-c^{2} x^{2} + 1} b d}{5 \, c^{5}} + \frac{3 \,{\left (c^{2} x^{2} - 1\right )}^{2} \sqrt{-c^{2} x^{2} + 1} b e}{35 \, c^{7}} - \frac{{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}} b e}{7 \, c^{7}} + \frac{\sqrt{-c^{2} x^{2} + 1} b e}{7 \, c^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(e*x^2+d)*(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

1/7*a*x^7*e + 1/5*a*d*x^5 + 1/5*(c^2*x^2 - 1)^2*b*d*x*arcsin(c*x)/c^4 + 2/5*(c^2*x^2 - 1)*b*d*x*arcsin(c*x)/c^
4 + 1/7*(c^2*x^2 - 1)^3*b*x*arcsin(c*x)*e/c^6 + 1/5*b*d*x*arcsin(c*x)/c^4 + 3/7*(c^2*x^2 - 1)^2*b*x*arcsin(c*x
)*e/c^6 + 1/25*(c^2*x^2 - 1)^2*sqrt(-c^2*x^2 + 1)*b*d/c^5 + 3/7*(c^2*x^2 - 1)*b*x*arcsin(c*x)*e/c^6 - 2/15*(-c
^2*x^2 + 1)^(3/2)*b*d/c^5 + 1/49*(c^2*x^2 - 1)^3*sqrt(-c^2*x^2 + 1)*b*e/c^7 + 1/7*b*x*arcsin(c*x)*e/c^6 + 1/5*
sqrt(-c^2*x^2 + 1)*b*d/c^5 + 3/35*(c^2*x^2 - 1)^2*sqrt(-c^2*x^2 + 1)*b*e/c^7 - 1/7*(-c^2*x^2 + 1)^(3/2)*b*e/c^
7 + 1/7*sqrt(-c^2*x^2 + 1)*b*e/c^7