3.386 \(\int x^5 (a+b \sin ^{-1}(c+d x^2)) \, dx\)

Optimal. Leaf size=129 \[ \frac{1}{6} x^6 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )+\frac{b x^4 \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}{18 d}+\frac{b \left (11 c^2-5 c d x^2+4\right ) \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}{36 d^3}+\frac{b c \left (2 c^2+3\right ) \sin ^{-1}\left (c+d x^2\right )}{12 d^3} \]

[Out]

(b*x^4*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])/(18*d) + (b*(4 + 11*c^2 - 5*c*d*x^2)*Sqrt[1 - c^2 - 2*c*d*x^2 - d^
2*x^4])/(36*d^3) + (b*c*(3 + 2*c^2)*ArcSin[c + d*x^2])/(12*d^3) + (x^6*(a + b*ArcSin[c + d*x^2]))/6

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Rubi [A]  time = 0.155358, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {4842, 12, 1114, 742, 779, 619, 216} \[ \frac{1}{6} x^6 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )+\frac{b x^4 \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}{18 d}+\frac{b \left (11 c^2-5 c d x^2+4\right ) \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}{36 d^3}+\frac{b c \left (2 c^2+3\right ) \sin ^{-1}\left (c+d x^2\right )}{12 d^3} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*ArcSin[c + d*x^2]),x]

[Out]

(b*x^4*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])/(18*d) + (b*(4 + 11*c^2 - 5*c*d*x^2)*Sqrt[1 - c^2 - 2*c*d*x^2 - d^
2*x^4])/(36*d^3) + (b*c*(3 + 2*c^2)*ArcSin[c + d*x^2])/(12*d^3) + (x^6*(a + b*ArcSin[c + d*x^2]))/6

Rule 4842

Int[((a_.) + ArcSin[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcSin[
u]))/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/Sqrt[1 - u^2], x]
, x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(
m + 1), u, x] &&  !FunctionOfExponentialQ[u, x]

Rule 12

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

Rule 1114

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

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int x^5 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right ) \, dx &=\frac{1}{6} x^6 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )-\frac{1}{6} b \int \frac{2 d x^7}{\sqrt{1-c^2-2 c d x^2-d^2 x^4}} \, dx\\ &=\frac{1}{6} x^6 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )-\frac{1}{3} (b d) \int \frac{x^7}{\sqrt{1-c^2-2 c d x^2-d^2 x^4}} \, dx\\ &=\frac{1}{6} x^6 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )-\frac{1}{6} (b d) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{1-c^2-2 c d x-d^2 x^2}} \, dx,x,x^2\right )\\ &=\frac{b x^4 \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{18 d}+\frac{1}{6} x^6 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )+\frac{b \operatorname{Subst}\left (\int \frac{x \left (-2 \left (1-c^2\right )+5 c d x\right )}{\sqrt{1-c^2-2 c d x-d^2 x^2}} \, dx,x,x^2\right )}{18 d}\\ &=\frac{b x^4 \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{18 d}+\frac{b \left (4+11 c^2-5 c d x^2\right ) \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{36 d^3}+\frac{1}{6} x^6 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )+\frac{\left (b c \left (3+2 c^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-c^2-2 c d x-d^2 x^2}} \, dx,x,x^2\right )}{12 d^2}\\ &=\frac{b x^4 \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{18 d}+\frac{b \left (4+11 c^2-5 c d x^2\right ) \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{36 d^3}+\frac{1}{6} x^6 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )-\frac{\left (b c \left (3+2 c^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{4 d^2}}} \, dx,x,-2 d \left (c+d x^2\right )\right )}{24 d^4}\\ &=\frac{b x^4 \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{18 d}+\frac{b \left (4+11 c^2-5 c d x^2\right ) \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{36 d^3}+\frac{b c \left (3+2 c^2\right ) \sin ^{-1}\left (c+d x^2\right )}{12 d^3}+\frac{1}{6} x^6 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.111501, size = 116, normalized size = 0.9 \[ \frac{a x^6}{6}+\frac{1}{2} b \left (\frac{11 c^2+4}{18 d^3}-\frac{5 c x^2}{18 d^2}+\frac{x^4}{9 d}\right ) \sqrt{-c^2-2 c d x^2-d^2 x^4+1}+\frac{b c \left (2 c^2+3\right ) \sin ^{-1}\left (c+d x^2\right )}{12 d^3}+\frac{1}{6} b x^6 \sin ^{-1}\left (c+d x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*ArcSin[c + d*x^2]),x]

[Out]

(a*x^6)/6 + (b*((4 + 11*c^2)/(18*d^3) - (5*c*x^2)/(18*d^2) + x^4/(9*d))*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])/2
+ (b*c*(3 + 2*c^2)*ArcSin[c + d*x^2])/(12*d^3) + (b*x^6*ArcSin[c + d*x^2])/6

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Maple [B]  time = 0.054, size = 258, normalized size = 2. \begin{align*}{\frac{{x}^{6}a}{6}}+{\frac{b{x}^{6}\arcsin \left ( d{x}^{2}+c \right ) }{6}}+{\frac{b{x}^{4}}{18\,d}\sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1}}-{\frac{5\,bc{x}^{2}}{36\,{d}^{2}}\sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1}}+{\frac{11\,b{c}^{2}}{36\,{d}^{3}}\sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1}}+{\frac{b{c}^{3}}{6\,{d}^{2}}\arctan \left ({\sqrt{{d}^{2}} \left ({x}^{2}+{\frac{c}{d}} \right ){\frac{1}{\sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1}}}} \right ){\frac{1}{\sqrt{{d}^{2}}}}}+{\frac{bc}{4\,{d}^{2}}\arctan \left ({\sqrt{{d}^{2}} \left ({x}^{2}+{\frac{c}{d}} \right ){\frac{1}{\sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1}}}} \right ){\frac{1}{\sqrt{{d}^{2}}}}}+{\frac{b}{9\,{d}^{3}}\sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*arcsin(d*x^2+c)),x)

[Out]

1/6*x^6*a+1/6*b*x^6*arcsin(d*x^2+c)+1/18*b*x^4*(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)/d-5/36*b*c/d^2*x^2*(-d^2*x^4-2
*c*d*x^2-c^2+1)^(1/2)+11/36*b*c^2/d^3*(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)+1/6*b*c^3/d^2/(d^2)^(1/2)*arctan((d^2)^
(1/2)*(x^2+c/d)/(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2))+1/4*b*c/d^2/(d^2)^(1/2)*arctan((d^2)^(1/2)*(x^2+c/d)/(-d^2*x
^4-2*c*d*x^2-c^2+1)^(1/2))+1/9*b/d^3*(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)

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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^5*(a+b*arcsin(d*x^2+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.29689, size = 219, normalized size = 1.7 \begin{align*} \frac{6 \, a d^{3} x^{6} + 3 \,{\left (2 \, b d^{3} x^{6} + 2 \, b c^{3} + 3 \, b c\right )} \arcsin \left (d x^{2} + c\right ) +{\left (2 \, b d^{2} x^{4} - 5 \, b c d x^{2} + 11 \, b c^{2} + 4 \, b\right )} \sqrt{-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1}}{36 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsin(d*x^2+c)),x, algorithm="fricas")

[Out]

1/36*(6*a*d^3*x^6 + 3*(2*b*d^3*x^6 + 2*b*c^3 + 3*b*c)*arcsin(d*x^2 + c) + (2*b*d^2*x^4 - 5*b*c*d*x^2 + 11*b*c^
2 + 4*b)*sqrt(-d^2*x^4 - 2*c*d*x^2 - c^2 + 1))/d^3

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Sympy [A]  time = 4.71836, size = 204, normalized size = 1.58 \begin{align*} \begin{cases} \frac{a x^{6}}{6} + \frac{b c^{3} \operatorname{asin}{\left (c + d x^{2} \right )}}{6 d^{3}} + \frac{11 b c^{2} \sqrt{- c^{2} - 2 c d x^{2} - d^{2} x^{4} + 1}}{36 d^{3}} - \frac{5 b c x^{2} \sqrt{- c^{2} - 2 c d x^{2} - d^{2} x^{4} + 1}}{36 d^{2}} + \frac{b c \operatorname{asin}{\left (c + d x^{2} \right )}}{4 d^{3}} + \frac{b x^{6} \operatorname{asin}{\left (c + d x^{2} \right )}}{6} + \frac{b x^{4} \sqrt{- c^{2} - 2 c d x^{2} - d^{2} x^{4} + 1}}{18 d} + \frac{b \sqrt{- c^{2} - 2 c d x^{2} - d^{2} x^{4} + 1}}{9 d^{3}} & \text{for}\: d \neq 0 \\\frac{x^{6} \left (a + b \operatorname{asin}{\left (c \right )}\right )}{6} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*asin(d*x**2+c)),x)

[Out]

Piecewise((a*x**6/6 + b*c**3*asin(c + d*x**2)/(6*d**3) + 11*b*c**2*sqrt(-c**2 - 2*c*d*x**2 - d**2*x**4 + 1)/(3
6*d**3) - 5*b*c*x**2*sqrt(-c**2 - 2*c*d*x**2 - d**2*x**4 + 1)/(36*d**2) + b*c*asin(c + d*x**2)/(4*d**3) + b*x*
*6*asin(c + d*x**2)/6 + b*x**4*sqrt(-c**2 - 2*c*d*x**2 - d**2*x**4 + 1)/(18*d) + b*sqrt(-c**2 - 2*c*d*x**2 - d
**2*x**4 + 1)/(9*d**3), Ne(d, 0)), (x**6*(a + b*asin(c))/6, True))

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Giac [A]  time = 1.16537, size = 297, normalized size = 2.3 \begin{align*} \frac{6 \, a d x^{6} +{\left (\frac{18 \,{\left (d x^{2} + c\right )} c^{2} \arcsin \left (d x^{2} + c\right )}{d^{2}} + \frac{6 \,{\left (d x^{2} + c\right )}{\left ({\left (d x^{2} + c\right )}^{2} - 1\right )} \arcsin \left (d x^{2} + c\right )}{d^{2}} - \frac{18 \,{\left ({\left (d x^{2} + c\right )}^{2} - 1\right )} c \arcsin \left (d x^{2} + c\right )}{d^{2}} - \frac{9 \,{\left (d x^{2} + c\right )} \sqrt{-{\left (d x^{2} + c\right )}^{2} + 1} c}{d^{2}} + \frac{18 \, \sqrt{-{\left (d x^{2} + c\right )}^{2} + 1} c^{2}}{d^{2}} + \frac{6 \,{\left (d x^{2} + c\right )} \arcsin \left (d x^{2} + c\right )}{d^{2}} - \frac{9 \, c \arcsin \left (d x^{2} + c\right )}{d^{2}} - \frac{2 \,{\left (-{\left (d x^{2} + c\right )}^{2} + 1\right )}^{\frac{3}{2}}}{d^{2}} + \frac{6 \, \sqrt{-{\left (d x^{2} + c\right )}^{2} + 1}}{d^{2}}\right )} b}{36 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsin(d*x^2+c)),x, algorithm="giac")

[Out]

1/36*(6*a*d*x^6 + (18*(d*x^2 + c)*c^2*arcsin(d*x^2 + c)/d^2 + 6*(d*x^2 + c)*((d*x^2 + c)^2 - 1)*arcsin(d*x^2 +
c)/d^2 - 18*((d*x^2 + c)^2 - 1)*c*arcsin(d*x^2 + c)/d^2 - 9*(d*x^2 + c)*sqrt(-(d*x^2 + c)^2 + 1)*c/d^2 + 18*s
qrt(-(d*x^2 + c)^2 + 1)*c^2/d^2 + 6*(d*x^2 + c)*arcsin(d*x^2 + c)/d^2 - 9*c*arcsin(d*x^2 + c)/d^2 - 2*(-(d*x^2
+ c)^2 + 1)^(3/2)/d^2 + 6*sqrt(-(d*x^2 + c)^2 + 1)/d^2)*b)/d