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

Optimal. Leaf size=161 \[ \frac {1}{7} c^4 d^2 x^7 \left (a+b \sin ^{-1}(c x)\right )-\frac {2}{5} c^2 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{3} d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac {b d^2 \left (1-c^2 x^2\right )^{7/2}}{49 c^3}+\frac {b d^2 \left (1-c^2 x^2\right )^{5/2}}{175 c^3}+\frac {4 b d^2 \left (1-c^2 x^2\right )^{3/2}}{315 c^3}+\frac {8 b d^2 \sqrt {1-c^2 x^2}}{105 c^3} \]

[Out]

4/315*b*d^2*(-c^2*x^2+1)^(3/2)/c^3+1/175*b*d^2*(-c^2*x^2+1)^(5/2)/c^3-1/49*b*d^2*(-c^2*x^2+1)^(7/2)/c^3+1/3*d^
2*x^3*(a+b*arcsin(c*x))-2/5*c^2*d^2*x^5*(a+b*arcsin(c*x))+1/7*c^4*d^2*x^7*(a+b*arcsin(c*x))+8/105*b*d^2*(-c^2*
x^2+1)^(1/2)/c^3

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Rubi [A]  time = 0.17, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {270, 4687, 12, 1251, 771} \[ \frac {1}{7} c^4 d^2 x^7 \left (a+b \sin ^{-1}(c x)\right )-\frac {2}{5} c^2 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{3} d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac {b d^2 \left (1-c^2 x^2\right )^{7/2}}{49 c^3}+\frac {b d^2 \left (1-c^2 x^2\right )^{5/2}}{175 c^3}+\frac {4 b d^2 \left (1-c^2 x^2\right )^{3/2}}{315 c^3}+\frac {8 b d^2 \sqrt {1-c^2 x^2}}{105 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b*d^2*Sqrt[1 - c^2*x^2])/(105*c^3) + (4*b*d^2*(1 - c^2*x^2)^(3/2))/(315*c^3) + (b*d^2*(1 - c^2*x^2)^(5/2))/
(175*c^3) - (b*d^2*(1 - c^2*x^2)^(7/2))/(49*c^3) + (d^2*x^3*(a + b*ArcSin[c*x]))/3 - (2*c^2*d^2*x^5*(a + b*Arc
Sin[c*x]))/5 + (c^4*d^2*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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 771

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 4687

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^2 \left (d-c^2 d x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac {1}{3} d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac {2}{5} c^2 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{7} c^4 d^2 x^7 \left (a+b \sin ^{-1}(c x)\right )-(b c) \int \frac {d^2 x^3 \left (35-42 c^2 x^2+15 c^4 x^4\right )}{105 \sqrt {1-c^2 x^2}} \, dx\\ &=\frac {1}{3} d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac {2}{5} c^2 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{7} c^4 d^2 x^7 \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{105} \left (b c d^2\right ) \int \frac {x^3 \left (35-42 c^2 x^2+15 c^4 x^4\right )}{\sqrt {1-c^2 x^2}} \, dx\\ &=\frac {1}{3} d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac {2}{5} c^2 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{7} c^4 d^2 x^7 \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{210} \left (b c d^2\right ) \operatorname {Subst}\left (\int \frac {x \left (35-42 c^2 x+15 c^4 x^2\right )}{\sqrt {1-c^2 x}} \, dx,x,x^2\right )\\ &=\frac {1}{3} d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac {2}{5} c^2 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{7} c^4 d^2 x^7 \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{210} \left (b c d^2\right ) \operatorname {Subst}\left (\int \left (\frac {8}{c^2 \sqrt {1-c^2 x}}+\frac {4 \sqrt {1-c^2 x}}{c^2}+\frac {3 \left (1-c^2 x\right )^{3/2}}{c^2}-\frac {15 \left (1-c^2 x\right )^{5/2}}{c^2}\right ) \, dx,x,x^2\right )\\ &=\frac {8 b d^2 \sqrt {1-c^2 x^2}}{105 c^3}+\frac {4 b d^2 \left (1-c^2 x^2\right )^{3/2}}{315 c^3}+\frac {b d^2 \left (1-c^2 x^2\right )^{5/2}}{175 c^3}-\frac {b d^2 \left (1-c^2 x^2\right )^{7/2}}{49 c^3}+\frac {1}{3} d^2 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac {2}{5} c^2 d^2 x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{7} c^4 d^2 x^7 \left (a+b \sin ^{-1}(c x)\right )\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 111, normalized size = 0.69 \[ \frac {d^2 \left (105 a c^3 x^3 \left (15 c^4 x^4-42 c^2 x^2+35\right )+b \sqrt {1-c^2 x^2} \left (225 c^6 x^6-612 c^4 x^4+409 c^2 x^2+818\right )+105 b c^3 x^3 \left (15 c^4 x^4-42 c^2 x^2+35\right ) \sin ^{-1}(c x)\right )}{11025 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(105*a*c^3*x^3*(35 - 42*c^2*x^2 + 15*c^4*x^4) + b*Sqrt[1 - c^2*x^2]*(818 + 409*c^2*x^2 - 612*c^4*x^4 + 22
5*c^6*x^6) + 105*b*c^3*x^3*(35 - 42*c^2*x^2 + 15*c^4*x^4)*ArcSin[c*x]))/(11025*c^3)

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fricas [A]  time = 0.50, size = 141, normalized size = 0.88 \[ \frac {1575 \, a c^{7} d^{2} x^{7} - 4410 \, a c^{5} d^{2} x^{5} + 3675 \, a c^{3} d^{2} x^{3} + 105 \, {\left (15 \, b c^{7} d^{2} x^{7} - 42 \, b c^{5} d^{2} x^{5} + 35 \, b c^{3} d^{2} x^{3}\right )} \arcsin \left (c x\right ) + {\left (225 \, b c^{6} d^{2} x^{6} - 612 \, b c^{4} d^{2} x^{4} + 409 \, b c^{2} d^{2} x^{2} + 818 \, b d^{2}\right )} \sqrt {-c^{2} x^{2} + 1}}{11025 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/11025*(1575*a*c^7*d^2*x^7 - 4410*a*c^5*d^2*x^5 + 3675*a*c^3*d^2*x^3 + 105*(15*b*c^7*d^2*x^7 - 42*b*c^5*d^2*x
^5 + 35*b*c^3*d^2*x^3)*arcsin(c*x) + (225*b*c^6*d^2*x^6 - 612*b*c^4*d^2*x^4 + 409*b*c^2*d^2*x^2 + 818*b*d^2)*s
qrt(-c^2*x^2 + 1))/c^3

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giac [A]  time = 0.60, size = 227, normalized size = 1.41 \[ \frac {1}{7} \, a c^{4} d^{2} x^{7} - \frac {2}{5} \, a c^{2} d^{2} x^{5} + \frac {1}{3} \, a d^{2} x^{3} + \frac {{\left (c^{2} x^{2} - 1\right )}^{3} b d^{2} x \arcsin \left (c x\right )}{7 \, c^{2}} + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b d^{2} x \arcsin \left (c x\right )}{35 \, c^{2}} - \frac {4 \, {\left (c^{2} x^{2} - 1\right )} b d^{2} x \arcsin \left (c x\right )}{105 \, c^{2}} + \frac {{\left (c^{2} x^{2} - 1\right )}^{3} \sqrt {-c^{2} x^{2} + 1} b d^{2}}{49 \, c^{3}} + \frac {8 \, b d^{2} x \arcsin \left (c x\right )}{105 \, c^{2}} + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} \sqrt {-c^{2} x^{2} + 1} b d^{2}}{175 \, c^{3}} + \frac {4 \, {\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b d^{2}}{315 \, c^{3}} + \frac {8 \, \sqrt {-c^{2} x^{2} + 1} b d^{2}}{105 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*a*c^4*d^2*x^7 - 2/5*a*c^2*d^2*x^5 + 1/3*a*d^2*x^3 + 1/7*(c^2*x^2 - 1)^3*b*d^2*x*arcsin(c*x)/c^2 + 1/35*(c^
2*x^2 - 1)^2*b*d^2*x*arcsin(c*x)/c^2 - 4/105*(c^2*x^2 - 1)*b*d^2*x*arcsin(c*x)/c^2 + 1/49*(c^2*x^2 - 1)^3*sqrt
(-c^2*x^2 + 1)*b*d^2/c^3 + 8/105*b*d^2*x*arcsin(c*x)/c^2 + 1/175*(c^2*x^2 - 1)^2*sqrt(-c^2*x^2 + 1)*b*d^2/c^3
+ 4/315*(-c^2*x^2 + 1)^(3/2)*b*d^2/c^3 + 8/105*sqrt(-c^2*x^2 + 1)*b*d^2/c^3

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maple [A]  time = 0.01, size = 152, normalized size = 0.94 \[ \frac {d^{2} a \left (\frac {1}{7} c^{7} x^{7}-\frac {2}{5} c^{5} x^{5}+\frac {1}{3} c^{3} x^{3}\right )+d^{2} b \left (\frac {\arcsin \left (c x \right ) c^{7} x^{7}}{7}-\frac {2 \arcsin \left (c x \right ) c^{5} x^{5}}{5}+\frac {c^{3} x^{3} \arcsin \left (c x \right )}{3}+\frac {c^{6} x^{6} \sqrt {-c^{2} x^{2}+1}}{49}-\frac {68 c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{1225}+\frac {409 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{11025}+\frac {818 \sqrt {-c^{2} x^{2}+1}}{11025}\right )}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^3*(d^2*a*(1/7*c^7*x^7-2/5*c^5*x^5+1/3*c^3*x^3)+d^2*b*(1/7*arcsin(c*x)*c^7*x^7-2/5*arcsin(c*x)*c^5*x^5+1/3*
c^3*x^3*arcsin(c*x)+1/49*c^6*x^6*(-c^2*x^2+1)^(1/2)-68/1225*c^4*x^4*(-c^2*x^2+1)^(1/2)+409/11025*c^2*x^2*(-c^2
*x^2+1)^(1/2)+818/11025*(-c^2*x^2+1)^(1/2)))

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maxima [A]  time = 0.49, size = 267, normalized size = 1.66 \[ \frac {1}{7} \, a c^{4} d^{2} x^{7} - \frac {2}{5} \, a c^{2} d^{2} x^{5} + \frac {1}{245} \, {\left (35 \, x^{7} \arcsin \left (c x\right ) + {\left (\frac {5 \, \sqrt {-c^{2} x^{2} + 1} x^{6}}{c^{2}} + \frac {6 \, \sqrt {-c^{2} x^{2} + 1} x^{4}}{c^{4}} + \frac {8 \, \sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{6}} + \frac {16 \, \sqrt {-c^{2} x^{2} + 1}}{c^{8}}\right )} c\right )} b c^{4} d^{2} - \frac {2}{75} \, {\left (15 \, x^{5} \arcsin \left (c x\right ) + {\left (\frac {3 \, \sqrt {-c^{2} x^{2} + 1} x^{4}}{c^{2}} + \frac {4 \, \sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac {8 \, \sqrt {-c^{2} x^{2} + 1}}{c^{6}}\right )} c\right )} b c^{2} d^{2} + \frac {1}{3} \, a d^{2} x^{3} + \frac {1}{9} \, {\left (3 \, x^{3} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b d^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*a*c^4*d^2*x^7 - 2/5*a*c^2*d^2*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^4*d^2 - 2/75*(15*x^5*arc
sin(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*c^2*d
^2 + 1/3*a*d^2*x^3 + 1/9*(3*x^3*arcsin(c*x) + c*(sqrt(-c^2*x^2 + 1)*x^2/c^2 + 2*sqrt(-c^2*x^2 + 1)/c^4))*b*d^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 5.93, size = 202, normalized size = 1.25 \[ \begin {cases} \frac {a c^{4} d^{2} x^{7}}{7} - \frac {2 a c^{2} d^{2} x^{5}}{5} + \frac {a d^{2} x^{3}}{3} + \frac {b c^{4} d^{2} x^{7} \operatorname {asin}{\left (c x \right )}}{7} + \frac {b c^{3} d^{2} x^{6} \sqrt {- c^{2} x^{2} + 1}}{49} - \frac {2 b c^{2} d^{2} x^{5} \operatorname {asin}{\left (c x \right )}}{5} - \frac {68 b c d^{2} x^{4} \sqrt {- c^{2} x^{2} + 1}}{1225} + \frac {b d^{2} x^{3} \operatorname {asin}{\left (c x \right )}}{3} + \frac {409 b d^{2} x^{2} \sqrt {- c^{2} x^{2} + 1}}{11025 c} + \frac {818 b d^{2} \sqrt {- c^{2} x^{2} + 1}}{11025 c^{3}} & \text {for}\: c \neq 0 \\\frac {a d^{2} x^{3}}{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*c**4*d**2*x**7/7 - 2*a*c**2*d**2*x**5/5 + a*d**2*x**3/3 + b*c**4*d**2*x**7*asin(c*x)/7 + b*c**3*d
**2*x**6*sqrt(-c**2*x**2 + 1)/49 - 2*b*c**2*d**2*x**5*asin(c*x)/5 - 68*b*c*d**2*x**4*sqrt(-c**2*x**2 + 1)/1225
+ b*d**2*x**3*asin(c*x)/3 + 409*b*d**2*x**2*sqrt(-c**2*x**2 + 1)/(11025*c) + 818*b*d**2*sqrt(-c**2*x**2 + 1)/
(11025*c**3), Ne(c, 0)), (a*d**2*x**3/3, True))

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