3.1.1 \(\int x^4 (d-c^2 d x^2) (a+b \text {ArcSin}(c x)) \, dx\) [1]

Optimal. Leaf size=128 \[ \frac {2 b d \sqrt {1-c^2 x^2}}{35 c^5}+\frac {b d \left (1-c^2 x^2\right )^{3/2}}{105 c^5}-\frac {8 b d \left (1-c^2 x^2\right )^{5/2}}{175 c^5}+\frac {b d \left (1-c^2 x^2\right )^{7/2}}{49 c^5}+\frac {1}{5} d x^5 (a+b \text {ArcSin}(c x))-\frac {1}{7} c^2 d x^7 (a+b \text {ArcSin}(c x)) \]

[Out]

1/105*b*d*(-c^2*x^2+1)^(3/2)/c^5-8/175*b*d*(-c^2*x^2+1)^(5/2)/c^5+1/49*b*d*(-c^2*x^2+1)^(7/2)/c^5+1/5*d*x^5*(a
+b*arcsin(c*x))-1/7*c^2*d*x^7*(a+b*arcsin(c*x))+2/35*b*d*(-c^2*x^2+1)^(1/2)/c^5

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Rubi [A]
time = 0.08, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {14, 4777, 12, 457, 78} \begin {gather*} -\frac {1}{7} c^2 d x^7 (a+b \text {ArcSin}(c x))+\frac {1}{5} d x^5 (a+b \text {ArcSin}(c x))+\frac {b d \left (1-c^2 x^2\right )^{7/2}}{49 c^5}-\frac {8 b d \left (1-c^2 x^2\right )^{5/2}}{175 c^5}+\frac {b d \left (1-c^2 x^2\right )^{3/2}}{105 c^5}+\frac {2 b d \sqrt {1-c^2 x^2}}{35 c^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*d*Sqrt[1 - c^2*x^2])/(35*c^5) + (b*d*(1 - c^2*x^2)^(3/2))/(105*c^5) - (8*b*d*(1 - c^2*x^2)^(5/2))/(175*c^
5) + (b*d*(1 - c^2*x^2)^(7/2))/(49*c^5) + (d*x^5*(a + b*ArcSin[c*x]))/5 - (c^2*d*x^7*(a + b*ArcSin[c*x]))/7

Rule 12

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

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 78

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]))))

Rule 457

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 4777

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = I
ntHide[(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] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int x^4 \left (d-c^2 d 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} c^2 d x^7 \left (a+b \sin ^{-1}(c x)\right )-(b c) \int \frac {d x^5 \left (7-5 c^2 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} c^2 d x^7 \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{35} (b c d) \int \frac {x^5 \left (7-5 c^2 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} c^2 d x^7 \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{70} (b c d) \text {Subst}\left (\int \frac {x^2 \left (7-5 c^2 x\right )}{\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} c^2 d x^7 \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{70} (b c d) \text {Subst}\left (\int \left (\frac {2}{c^4 \sqrt {1-c^2 x}}+\frac {\sqrt {1-c^2 x}}{c^4}-\frac {8 \left (1-c^2 x\right )^{3/2}}{c^4}+\frac {5 \left (1-c^2 x\right )^{5/2}}{c^4}\right ) \, dx,x,x^2\right )\\ &=\frac {2 b d \sqrt {1-c^2 x^2}}{35 c^5}+\frac {b d \left (1-c^2 x^2\right )^{3/2}}{105 c^5}-\frac {8 b d \left (1-c^2 x^2\right )^{5/2}}{175 c^5}+\frac {b d \left (1-c^2 x^2\right )^{7/2}}{49 c^5}+\frac {1}{5} d x^5 \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{7} c^2 d x^7 \left (a+b \sin ^{-1}(c x)\right )\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 87, normalized size = 0.68 \begin {gather*} \frac {d \left (-105 a x^5 \left (-7+5 c^2 x^2\right )+\frac {b \sqrt {1-c^2 x^2} \left (152+76 c^2 x^2+57 c^4 x^4-75 c^6 x^6\right )}{c^5}-105 b x^5 \left (-7+5 c^2 x^2\right ) \text {ArcSin}(c x)\right )}{3675} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(-105*a*x^5*(-7 + 5*c^2*x^2) + (b*Sqrt[1 - c^2*x^2]*(152 + 76*c^2*x^2 + 57*c^4*x^4 - 75*c^6*x^6))/c^5 - 105
*b*x^5*(-7 + 5*c^2*x^2)*ArcSin[c*x]))/3675

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Maple [A]
time = 0.06, size = 130, normalized size = 1.02

method	result	size
derivativedivides	\(\frac {-d a \left (\frac {1}{7} c^{7} x^{7}-\frac {1}{5} c^{5} x^{5}\right )-d b \left (\frac {\arcsin \left (c x \right ) c^{7} x^{7}}{7}-\frac {\arcsin \left (c x \right ) c^{5} x^{5}}{5}+\frac {c^{6} x^{6} \sqrt {-c^{2} x^{2}+1}}{49}-\frac {19 c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{1225}-\frac {76 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3675}-\frac {152 \sqrt {-c^{2} x^{2}+1}}{3675}\right )}{c^{5}}\)	\(130\)
default	\(\frac {-d a \left (\frac {1}{7} c^{7} x^{7}-\frac {1}{5} c^{5} x^{5}\right )-d b \left (\frac {\arcsin \left (c x \right ) c^{7} x^{7}}{7}-\frac {\arcsin \left (c x \right ) c^{5} x^{5}}{5}+\frac {c^{6} x^{6} \sqrt {-c^{2} x^{2}+1}}{49}-\frac {19 c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{1225}-\frac {76 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3675}-\frac {152 \sqrt {-c^{2} x^{2}+1}}{3675}\right )}{c^{5}}\)	\(130\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(-c^2*d*x^2+d)*(a+b*arcsin(c*x)),x,method=_RETURNVERBOSE)

[Out]

1/c^5*(-d*a*(1/7*c^7*x^7-1/5*c^5*x^5)-d*b*(1/7*arcsin(c*x)*c^7*x^7-1/5*arcsin(c*x)*c^5*x^5+1/49*c^6*x^6*(-c^2*
x^2+1)^(1/2)-19/1225*c^4*x^4*(-c^2*x^2+1)^(1/2)-76/3675*c^2*x^2*(-c^2*x^2+1)^(1/2)-152/3675*(-c^2*x^2+1)^(1/2)
))

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Maxima [A]
time = 0.48, size = 189, normalized size = 1.48 \begin {gather*} -\frac {1}{7} \, a c^{2} d x^{7} + \frac {1}{5} \, a d x^{5} - \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 c^{2} d + \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*a*c^2*d*x^7 + 1/5*a*d*x^5 - 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*c^2*d + 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

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Fricas [A]
time = 2.88, size = 101, normalized size = 0.79 \begin {gather*} -\frac {525 \, a c^{7} d x^{7} - 735 \, a c^{5} d x^{5} + 105 \, {\left (5 \, b c^{7} d x^{7} - 7 \, b c^{5} d x^{5}\right )} \arcsin \left (c x\right ) + {\left (75 \, b c^{6} d x^{6} - 57 \, b c^{4} d x^{4} - 76 \, b c^{2} d x^{2} - 152 \, b d\right )} \sqrt {-c^{2} x^{2} + 1}}{3675 \, c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3675*(525*a*c^7*d*x^7 - 735*a*c^5*d*x^5 + 105*(5*b*c^7*d*x^7 - 7*b*c^5*d*x^5)*arcsin(c*x) + (75*b*c^6*d*x^6
- 57*b*c^4*d*x^4 - 76*b*c^2*d*x^2 - 152*b*d)*sqrt(-c^2*x^2 + 1))/c^5

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Sympy [A]
time = 0.74, size = 151, normalized size = 1.18 \begin {gather*} \begin {cases} - \frac {a c^{2} d x^{7}}{7} + \frac {a d x^{5}}{5} - \frac {b c^{2} d x^{7} \operatorname {asin}{\left (c x \right )}}{7} - \frac {b c d x^{6} \sqrt {- c^{2} x^{2} + 1}}{49} + \frac {b d x^{5} \operatorname {asin}{\left (c x \right )}}{5} + \frac {19 b d x^{4} \sqrt {- c^{2} x^{2} + 1}}{1225 c} + \frac {76 b d x^{2} \sqrt {- c^{2} x^{2} + 1}}{3675 c^{3}} + \frac {152 b d \sqrt {- c^{2} x^{2} + 1}}{3675 c^{5}} & \text {for}\: c \neq 0 \\\frac {a d x^{5}}{5} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a*c**2*d*x**7/7 + a*d*x**5/5 - b*c**2*d*x**7*asin(c*x)/7 - b*c*d*x**6*sqrt(-c**2*x**2 + 1)/49 + b*
d*x**5*asin(c*x)/5 + 19*b*d*x**4*sqrt(-c**2*x**2 + 1)/(1225*c) + 76*b*d*x**2*sqrt(-c**2*x**2 + 1)/(3675*c**3)
+ 152*b*d*sqrt(-c**2*x**2 + 1)/(3675*c**5), Ne(c, 0)), (a*d*x**5/5, True))

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Giac [A]
time = 0.42, size = 195, normalized size = 1.52 \begin {gather*} -\frac {1}{7} \, a c^{2} d x^{7} + \frac {1}{5} \, a d x^{5} - \frac {{\left (c^{2} x^{2} - 1\right )}^{3} b d x \arcsin \left (c x\right )}{7 \, c^{4}} - \frac {8 \, {\left (c^{2} x^{2} - 1\right )}^{2} b d x \arcsin \left (c x\right )}{35 \, c^{4}} - \frac {{\left (c^{2} x^{2} - 1\right )} b d x \arcsin \left (c x\right )}{35 \, c^{4}} - \frac {{\left (c^{2} x^{2} - 1\right )}^{3} \sqrt {-c^{2} x^{2} + 1} b d}{49 \, c^{5}} + \frac {2 \, b d x \arcsin \left (c x\right )}{35 \, c^{4}} - \frac {8 \, {\left (c^{2} x^{2} - 1\right )}^{2} \sqrt {-c^{2} x^{2} + 1} b d}{175 \, c^{5}} + \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b d}{105 \, c^{5}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1} b d}{35 \, c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*a*c^2*d*x^7 + 1/5*a*d*x^5 - 1/7*(c^2*x^2 - 1)^3*b*d*x*arcsin(c*x)/c^4 - 8/35*(c^2*x^2 - 1)^2*b*d*x*arcsin
(c*x)/c^4 - 1/35*(c^2*x^2 - 1)*b*d*x*arcsin(c*x)/c^4 - 1/49*(c^2*x^2 - 1)^3*sqrt(-c^2*x^2 + 1)*b*d/c^5 + 2/35*
b*d*x*arcsin(c*x)/c^4 - 8/175*(c^2*x^2 - 1)^2*sqrt(-c^2*x^2 + 1)*b*d/c^5 + 1/105*(-c^2*x^2 + 1)^(3/2)*b*d/c^5
+ 2/35*sqrt(-c^2*x^2 + 1)*b*d/c^5

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^4\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\left (d-c^2\,d\,x^2\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + b*asin(c*x))*(d - c^2*d*x^2),x)

[Out]

int(x^4*(a + b*asin(c*x))*(d - c^2*d*x^2), x)

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