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

Optimal. Leaf size=217 \[ -\frac{b c d^2 \left (15 m^2+100 m+149\right ) x^{m+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},c^2 x^2\right )}{(m+1) (m+2) (m+3)^2 (m+5)^2}-\frac{2 c^2 d^2 x^{m+3} \left (a+b \sin ^{-1}(c x)\right )}{m+3}+\frac{c^4 d^2 x^{m+5} \left (a+b \sin ^{-1}(c x)\right )}{m+5}+\frac{d^2 x^{m+1} \left (a+b \sin ^{-1}(c x)\right )}{m+1}-\frac{b c d^2 \left (m^2+13 m+38\right ) \sqrt{1-c^2 x^2} x^{m+2}}{(m+3)^2 (m+5)^2}+\frac{b c^3 d^2 \sqrt{1-c^2 x^2} x^{m+4}}{(m+5)^2} \]

[Out]

-((b*c*d^2*(38 + 13*m + m^2)*x^(2 + m)*Sqrt[1 - c^2*x^2])/((3 + m)^2*(5 + m)^2)) + (b*c^3*d^2*x^(4 + m)*Sqrt[1
- c^2*x^2])/(5 + m)^2 + (d^2*x^(1 + m)*(a + b*ArcSin[c*x]))/(1 + m) - (2*c^2*d^2*x^(3 + m)*(a + b*ArcSin[c*x]
))/(3 + m) + (c^4*d^2*x^(5 + m)*(a + b*ArcSin[c*x]))/(5 + m) - (b*c*d^2*(149 + 100*m + 15*m^2)*x^(2 + m)*Hyper
geometric2F1[1/2, (2 + m)/2, (4 + m)/2, c^2*x^2])/((1 + m)*(2 + m)*(3 + m)^2*(5 + m)^2)

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Rubi [A]  time = 0.30633, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {270, 4687, 12, 1267, 459, 364} \[ -\frac{2 c^2 d^2 x^{m+3} \left (a+b \sin ^{-1}(c x)\right )}{m+3}+\frac{c^4 d^2 x^{m+5} \left (a+b \sin ^{-1}(c x)\right )}{m+5}+\frac{d^2 x^{m+1} \left (a+b \sin ^{-1}(c x)\right )}{m+1}-\frac{b c d^2 \left (15 m^2+100 m+149\right ) x^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};c^2 x^2\right )}{(m+1) (m+2) (m+3)^2 (m+5)^2}-\frac{b c d^2 \left (m^2+13 m+38\right ) \sqrt{1-c^2 x^2} x^{m+2}}{(m+3)^2 (m+5)^2}+\frac{b c^3 d^2 \sqrt{1-c^2 x^2} x^{m+4}}{(m+5)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c*d^2*(38 + 13*m + m^2)*x^(2 + m)*Sqrt[1 - c^2*x^2])/((3 + m)^2*(5 + m)^2)) + (b*c^3*d^2*x^(4 + m)*Sqrt[1
- c^2*x^2])/(5 + m)^2 + (d^2*x^(1 + m)*(a + b*ArcSin[c*x]))/(1 + m) - (2*c^2*d^2*x^(3 + m)*(a + b*ArcSin[c*x]
))/(3 + m) + (c^4*d^2*x^(5 + m)*(a + b*ArcSin[c*x]))/(5 + m) - (b*c*d^2*(149 + 100*m + 15*m^2)*x^(2 + m)*Hyper
geometric2F1[1/2, (2 + m)/2, (4 + m)/2, c^2*x^2])/((1 + m)*(2 + m)*(3 + m)^2*(5 + m)^2)

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 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 1267

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Si
mp[(c^p*(f*x)^(m + 4*p - 1)*(d + e*x^2)^(q + 1))/(e*f^(4*p - 1)*(m + 4*p + 2*q + 1)), x] + Dist[1/(e*(m + 4*p
+ 2*q + 1)), Int[(f*x)^m*(d + e*x^2)^q*ExpandToSum[e*(m + 4*p + 2*q + 1)*((a + b*x^2 + c*x^4)^p - c^p*x^(4*p))
- d*c^p*(m + 4*p - 1)*x^(4*p - 2), x], x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[b^2 - 4*a*c, 0] &&
IGtQ[p, 0] &&  !IntegerQ[q] && NeQ[m + 4*p + 2*q + 1, 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 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 )^2 \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac{d^2 x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{1+m}-\frac{2 c^2 d^2 x^{3+m} \left (a+b \sin ^{-1}(c x)\right )}{3+m}+\frac{c^4 d^2 x^{5+m} \left (a+b \sin ^{-1}(c x)\right )}{5+m}-(b c) \int \frac{d^2 x^{1+m} \left (\frac{1}{1+m}-\frac{2 c^2 x^2}{3+m}+\frac{c^4 x^4}{5+m}\right )}{\sqrt{1-c^2 x^2}} \, dx\\ &=\frac{d^2 x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{1+m}-\frac{2 c^2 d^2 x^{3+m} \left (a+b \sin ^{-1}(c x)\right )}{3+m}+\frac{c^4 d^2 x^{5+m} \left (a+b \sin ^{-1}(c x)\right )}{5+m}-\left (b c d^2\right ) \int \frac{x^{1+m} \left (\frac{1}{1+m}-\frac{2 c^2 x^2}{3+m}+\frac{c^4 x^4}{5+m}\right )}{\sqrt{1-c^2 x^2}} \, dx\\ &=\frac{b c^3 d^2 x^{4+m} \sqrt{1-c^2 x^2}}{(5+m)^2}+\frac{d^2 x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{1+m}-\frac{2 c^2 d^2 x^{3+m} \left (a+b \sin ^{-1}(c x)\right )}{3+m}+\frac{c^4 d^2 x^{5+m} \left (a+b \sin ^{-1}(c x)\right )}{5+m}+\frac{\left (b d^2\right ) \int \frac{x^{1+m} \left (-\frac{c^2 (5+m)}{1+m}+\frac{c^4 \left (38+13 m+m^2\right ) x^2}{(3+m) (5+m)}\right )}{\sqrt{1-c^2 x^2}} \, dx}{c (5+m)}\\ &=-\frac{b c d^2 \left (38+13 m+m^2\right ) x^{2+m} \sqrt{1-c^2 x^2}}{(3+m)^2 (5+m)^2}+\frac{b c^3 d^2 x^{4+m} \sqrt{1-c^2 x^2}}{(5+m)^2}+\frac{d^2 x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{1+m}-\frac{2 c^2 d^2 x^{3+m} \left (a+b \sin ^{-1}(c x)\right )}{3+m}+\frac{c^4 d^2 x^{5+m} \left (a+b \sin ^{-1}(c x)\right )}{5+m}-\frac{\left (b c d^2 \left (149+100 m+15 m^2\right )\right ) \int \frac{x^{1+m}}{\sqrt{1-c^2 x^2}} \, dx}{(1+m) (3+m)^2 (5+m)^2}\\ &=-\frac{b c d^2 \left (38+13 m+m^2\right ) x^{2+m} \sqrt{1-c^2 x^2}}{(3+m)^2 (5+m)^2}+\frac{b c^3 d^2 x^{4+m} \sqrt{1-c^2 x^2}}{(5+m)^2}+\frac{d^2 x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{1+m}-\frac{2 c^2 d^2 x^{3+m} \left (a+b \sin ^{-1}(c x)\right )}{3+m}+\frac{c^4 d^2 x^{5+m} \left (a+b \sin ^{-1}(c x)\right )}{5+m}-\frac{b c d^2 \left (149+100 m+15 m^2\right ) 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 (5+m)^2}\\ \end{align*}

Mathematica [A]  time = 0.015439, size = 187, normalized size = 0.86 \[ \frac{x^{m+1} \left (-\frac{4 d^2 \left (b c (m+1) x \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{m}{2}+1,\frac{m}{2}+2,c^2 x^2\right )+2 b c x \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m}{2}+1,\frac{m}{2}+2,c^2 x^2\right )+(m+2) \left (m \left (c^2 x^2-1\right )+c^2 x^2-3\right ) \left (a+b \sin ^{-1}(c x)\right )\right )}{(m+1) (m+2) (m+3)}-\frac{b c d^2 x \text{Hypergeometric2F1}\left (-\frac{3}{2},\frac{m}{2}+1,\frac{m}{2}+2,c^2 x^2\right )}{m+2}+\left (d-c^2 d x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right )\right )}{m+5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*((d - c^2*d*x^2)^2*(a + b*ArcSin[c*x]) - (b*c*d^2*x*Hypergeometric2F1[-3/2, 1 + m/2, 2 + m/2, c^2*x
^2])/(2 + m) - (4*d^2*((2 + m)*(-3 + c^2*x^2 + m*(-1 + c^2*x^2))*(a + b*ArcSin[c*x]) + b*c*(1 + m)*x*Hypergeom
etric2F1[-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))))/(5 + m)

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Maple [F]  time = 5.034, size = 0, normalized size = 0. \begin{align*} \int{x}^{m} \left ( -{c}^{2}d{x}^{2}+d \right ) ^{2} \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)^2*(a+b*arcsin(c*x)),x)

[Out]

int(x^m*(-c^2*d*x^2+d)^2*(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)^2*(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^{4} d^{2} x^{4} - 2 \, a c^{2} d^{2} x^{2} + a d^{2} +{\left (b c^{4} d^{2} x^{4} - 2 \, b c^{2} d^{2} x^{2} + b d^{2}\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)^2*(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

integral((a*c^4*d^2*x^4 - 2*a*c^2*d^2*x^2 + a*d^2 + (b*c^4*d^2*x^4 - 2*b*c^2*d^2*x^2 + b*d^2)*arcsin(c*x))*x^m
, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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