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

Optimal. Leaf size=129 \[ -\frac{b c d (3 m+7) x^{m+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},c^2 x^2\right )}{(m+1) (m+2) (m+3)^2}-\frac{c^2 d x^{m+3} \left (a+b \sin ^{-1}(c x)\right )}{m+3}+\frac{d x^{m+1} \left (a+b \sin ^{-1}(c x)\right )}{m+1}-\frac{b c d \sqrt{1-c^2 x^2} x^{m+2}}{(m+3)^2} \]

[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)

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Rubi [A]  time = 0.141078, antiderivative size = 129, normalized size of antiderivative = 1., 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 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 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]

Rule 12

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

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 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])

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*}

Mathematica [A]  time = 0.0820227, size = 118, normalized size = 0.91 \[ -\frac{d x^{m+1} \left (b c (m+1) x \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{m}{2}+1,\frac{m}{2}+2,c^2 x^2\right )+2 b c x \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m}{2}+1,\frac{m}{2}+2,c^2 x^2\right )+(m+2) \left (m \left (c^2 x^2-1\right )+c^2 x^2-3\right ) \left (a+b \sin ^{-1}(c x)\right )\right )}{(m+1) (m+2) (m+3)} \]

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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Maple [F]  time = 2.898, size = 0, normalized size = 0. \begin{align*} \int{x}^{m} \left ( -{c}^{2}d{x}^{2}+d \right ) \left ( a+b\arcsin \left ( cx \right ) \right ) \, dx \end{align*}

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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - d \left (\int - a x^{m}\, dx + \int - b x^{m} \operatorname{asin}{\left (c x \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 ) \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -{\left (c^{2} d x^{2} - d\right )}{\left (b \arcsin \left (c x\right ) + a\right )} x^{m}\,{d x} \end{align*}

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)