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

Optimal. Leaf size=129 \[ -\frac {c^2 d x^{m+3} \left (a+b \sin ^{-1}(c x)\right )}{m+3}+\frac {d x^{m+1} \left (a+b \sin ^{-1}(c x)\right )}{m+1}-\frac {b c d (3 m+7) x^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{(m+1) (m+2) (m+3)^2}-\frac {b c d \sqrt {1-c^2 x^2} x^{m+2}}{(m+3)^2} \]

[Out]

d*x^(1+m)*(a+b*arcsin(c*x))/(1+m)-c^2*d*x^(3+m)*(a+b*arcsin(c*x))/(3+m)-b*c*d*(7+3*m)*x^(2+m)*hypergeom([1/2,
1+1/2*m],[2+1/2*m],c^2*x^2)/(3+m)^2/(m^2+3*m+2)-b*c*d*x^(2+m)*(-c^2*x^2+1)^(1/2)/(3+m)^2

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Rubi [A]  time = 0.14, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {14, 4687, 12, 459, 364} \[ -\frac {c^2 d x^{m+3} \left (a+b \sin ^{-1}(c x)\right )}{m+3}+\frac {d x^{m+1} \left (a+b \sin ^{-1}(c x)\right )}{m+1}-\frac {b c d (3 m+7) x^{m+2} \, _2F_1\left (\frac {1}{2},\frac {m+2}{2};\frac {m+4}{2};c^2 x^2\right )}{(m+1) (m+2) (m+3)^2}-\frac {b c d \sqrt {1-c^2 x^2} x^{m+2}}{(m+3)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c*d*x^(2 + m)*Sqrt[1 - c^2*x^2])/(3 + m)^2) + (d*x^(1 + m)*(a + b*ArcSin[c*x]))/(1 + m) - (c^2*d*x^(3 + m
)*(a + b*ArcSin[c*x]))/(3 + m) - (b*c*d*(7 + 3*m)*x^(2 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, c^2*x
^2])/((1 + m)*(2 + m)*(3 + m)^2)

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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 459

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

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^m \left (d-c^2 d x^2\right ) \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac {d x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{1+m}-\frac {c^2 d x^{3+m} \left (a+b \sin ^{-1}(c x)\right )}{3+m}-(b c) \int \frac {d x^{1+m} \left (\frac {1}{1+m}-\frac {c^2 x^2}{3+m}\right )}{\sqrt {1-c^2 x^2}} \, dx\\ &=\frac {d x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{1+m}-\frac {c^2 d x^{3+m} \left (a+b \sin ^{-1}(c x)\right )}{3+m}-(b c d) \int \frac {x^{1+m} \left (\frac {1}{1+m}-\frac {c^2 x^2}{3+m}\right )}{\sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {b c d x^{2+m} \sqrt {1-c^2 x^2}}{(3+m)^2}+\frac {d x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{1+m}-\frac {c^2 d x^{3+m} \left (a+b \sin ^{-1}(c x)\right )}{3+m}-\frac {(b c d (7+3 m)) \int \frac {x^{1+m}}{\sqrt {1-c^2 x^2}} \, dx}{(1+m) (3+m)^2}\\ &=-\frac {b c d x^{2+m} \sqrt {1-c^2 x^2}}{(3+m)^2}+\frac {d x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{1+m}-\frac {c^2 d x^{3+m} \left (a+b \sin ^{-1}(c x)\right )}{3+m}-\frac {b c d (7+3 m) x^{2+m} \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};c^2 x^2\right )}{(1+m) (2+m) (3+m)^2}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 118, normalized size = 0.91 \[ -\frac {d x^{m+1} \left ((m+2) \left (m \left (c^2 x^2-1\right )+c^2 x^2-3\right ) \left (a+b \sin ^{-1}(c x)\right )+b c (m+1) x \, _2F_1\left (-\frac {1}{2},\frac {m}{2}+1;\frac {m}{2}+2;c^2 x^2\right )+2 b c x \, _2F_1\left (\frac {1}{2},\frac {m}{2}+1;\frac {m}{2}+2;c^2 x^2\right )\right )}{(m+1) (m+2) (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*x^(1 + m)*((2 + m)*(-3 + c^2*x^2 + m*(-1 + c^2*x^2))*(a + b*ArcSin[c*x]) + b*c*(1 + m)*x*Hypergeometric2F
1[-1/2, 1 + m/2, 2 + m/2, c^2*x^2] + 2*b*c*x*Hypergeometric2F1[1/2, 1 + m/2, 2 + m/2, c^2*x^2]))/((1 + m)*(2 +
m)*(3 + m)))

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (a c^{2} d x^{2} - a d + {\left (b c^{2} d x^{2} - b d\right )} \arcsin \left (c x\right )\right )} x^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -{\left (c^{2} d x^{2} - d\right )} {\left (b \arcsin \left (c x\right ) + a\right )} x^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 4.43, size = 0, normalized size = 0.00 \[ \int x^{m} \left (-c^{2} d \,x^{2}+d \right ) \left (a +b \arcsin \left (c x \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - d \left (\int \left (- a x^{m}\right )\, dx + \int \left (- b x^{m} \operatorname {asin}{\left (c x \right )}\right )\, dx + \int a c^{2} x^{2} x^{m}\, dx + \int b c^{2} x^{2} x^{m} \operatorname {asin}{\left (c x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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