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

Optimal. Leaf size=150 \[ d^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac {2}{3} d e x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{5} e^2 x^5 \left (a+b \sin ^{-1}(c x)\right )-\frac {2 b e \left (1-c^2 x^2\right )^{3/2} \left (5 c^2 d+3 e\right )}{45 c^5}+\frac {b e^2 \left (1-c^2 x^2\right )^{5/2}}{25 c^5}+\frac {b \sqrt {1-c^2 x^2} \left (15 c^4 d^2+10 c^2 d e+3 e^2\right )}{15 c^5} \]

[Out]

-2/45*b*e*(5*c^2*d+3*e)*(-c^2*x^2+1)^(3/2)/c^5+1/25*b*e^2*(-c^2*x^2+1)^(5/2)/c^5+d^2*x*(a+b*arcsin(c*x))+2/3*d
*e*x^3*(a+b*arcsin(c*x))+1/5*e^2*x^5*(a+b*arcsin(c*x))+1/15*b*(15*c^4*d^2+10*c^2*d*e+3*e^2)*(-c^2*x^2+1)^(1/2)
/c^5

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Rubi [A]  time = 0.14, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {194, 4665, 12, 1247, 698} \[ d^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac {2}{3} d e x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{5} e^2 x^5 \left (a+b \sin ^{-1}(c x)\right )+\frac {b \sqrt {1-c^2 x^2} \left (15 c^4 d^2+10 c^2 d e+3 e^2\right )}{15 c^5}-\frac {2 b e \left (1-c^2 x^2\right )^{3/2} \left (5 c^2 d+3 e\right )}{45 c^5}+\frac {b e^2 \left (1-c^2 x^2\right )^{5/2}}{25 c^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(15*c^4*d^2 + 10*c^2*d*e + 3*e^2)*Sqrt[1 - c^2*x^2])/(15*c^5) - (2*b*e*(5*c^2*d + 3*e)*(1 - c^2*x^2)^(3/2))
/(45*c^5) + (b*e^2*(1 - c^2*x^2)^(5/2))/(25*c^5) + d^2*x*(a + b*ArcSin[c*x]) + (2*d*e*x^3*(a + b*ArcSin[c*x]))
/3 + (e^2*x^5*(a + b*ArcSin[c*x]))/5

Rule 12

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

Rule 194

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

Rule 698

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

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(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]

Rule 4665

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(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]] /; F
reeQ[{a, b, c, d, e}, x] && NeQ[c^2*d + e, 0] && (IGtQ[p, 0] || ILtQ[p + 1/2, 0])

Rubi steps

\begin {align*} \int \left (d+e x^2\right )^2 \left (a+b \sin ^{-1}(c x)\right ) \, dx &=d^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac {2}{3} d e x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{5} e^2 x^5 \left (a+b \sin ^{-1}(c x)\right )-(b c) \int \frac {x \left (15 d^2+10 d e x^2+3 e^2 x^4\right )}{15 \sqrt {1-c^2 x^2}} \, dx\\ &=d^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac {2}{3} d e x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{5} e^2 x^5 \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{15} (b c) \int \frac {x \left (15 d^2+10 d e x^2+3 e^2 x^4\right )}{\sqrt {1-c^2 x^2}} \, dx\\ &=d^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac {2}{3} d e x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{5} e^2 x^5 \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{30} (b c) \operatorname {Subst}\left (\int \frac {15 d^2+10 d e x+3 e^2 x^2}{\sqrt {1-c^2 x}} \, dx,x,x^2\right )\\ &=d^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac {2}{3} d e x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{5} e^2 x^5 \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{30} (b c) \operatorname {Subst}\left (\int \left (\frac {15 c^4 d^2+10 c^2 d e+3 e^2}{c^4 \sqrt {1-c^2 x}}-\frac {2 e \left (5 c^2 d+3 e\right ) \sqrt {1-c^2 x}}{c^4}+\frac {3 e^2 \left (1-c^2 x\right )^{3/2}}{c^4}\right ) \, dx,x,x^2\right )\\ &=\frac {b \left (15 c^4 d^2+10 c^2 d e+3 e^2\right ) \sqrt {1-c^2 x^2}}{15 c^5}-\frac {2 b e \left (5 c^2 d+3 e\right ) \left (1-c^2 x^2\right )^{3/2}}{45 c^5}+\frac {b e^2 \left (1-c^2 x^2\right )^{5/2}}{25 c^5}+d^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac {2}{3} d e x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{5} e^2 x^5 \left (a+b \sin ^{-1}(c x)\right )\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 125, normalized size = 0.83 \[ \frac {1}{225} \left (15 a x \left (15 d^2+10 d e x^2+3 e^2 x^4\right )+\frac {b \sqrt {1-c^2 x^2} \left (c^4 \left (225 d^2+50 d e x^2+9 e^2 x^4\right )+4 c^2 e \left (25 d+3 e x^2\right )+24 e^2\right )}{c^5}+15 b x \sin ^{-1}(c x) \left (15 d^2+10 d e x^2+3 e^2 x^4\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*a*x*(15*d^2 + 10*d*e*x^2 + 3*e^2*x^4) + (b*Sqrt[1 - c^2*x^2]*(24*e^2 + 4*c^2*e*(25*d + 3*e*x^2) + c^4*(225
*d^2 + 50*d*e*x^2 + 9*e^2*x^4)))/c^5 + 15*b*x*(15*d^2 + 10*d*e*x^2 + 3*e^2*x^4)*ArcSin[c*x])/225

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fricas [A]  time = 0.66, size = 151, normalized size = 1.01 \[ \frac {45 \, a c^{5} e^{2} x^{5} + 150 \, a c^{5} d e x^{3} + 225 \, a c^{5} d^{2} x + 15 \, {\left (3 \, b c^{5} e^{2} x^{5} + 10 \, b c^{5} d e x^{3} + 15 \, b c^{5} d^{2} x\right )} \arcsin \left (c x\right ) + {\left (9 \, b c^{4} e^{2} x^{4} + 225 \, b c^{4} d^{2} + 100 \, b c^{2} d e + 24 \, b e^{2} + 2 \, {\left (25 \, b c^{4} d e + 6 \, b c^{2} e^{2}\right )} x^{2}\right )} \sqrt {-c^{2} x^{2} + 1}}{225 \, c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/225*(45*a*c^5*e^2*x^5 + 150*a*c^5*d*e*x^3 + 225*a*c^5*d^2*x + 15*(3*b*c^5*e^2*x^5 + 10*b*c^5*d*e*x^3 + 15*b*
c^5*d^2*x)*arcsin(c*x) + (9*b*c^4*e^2*x^4 + 225*b*c^4*d^2 + 100*b*c^2*d*e + 24*b*e^2 + 2*(25*b*c^4*d*e + 6*b*c
^2*e^2)*x^2)*sqrt(-c^2*x^2 + 1))/c^5

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giac [A]  time = 0.31, size = 263, normalized size = 1.75 \[ \frac {1}{5} \, a x^{5} e^{2} + \frac {2}{3} \, a d x^{3} e + b d^{2} x \arcsin \left (c x\right ) + a d^{2} x + \frac {2 \, {\left (c^{2} x^{2} - 1\right )} b d x \arcsin \left (c x\right ) e}{3 \, c^{2}} + \frac {2 \, b d x \arcsin \left (c x\right ) e}{3 \, c^{2}} + \frac {\sqrt {-c^{2} x^{2} + 1} b d^{2}}{c} + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b x \arcsin \left (c x\right ) e^{2}}{5 \, c^{4}} - \frac {2 \, {\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b d e}{9 \, c^{3}} + \frac {2 \, {\left (c^{2} x^{2} - 1\right )} b x \arcsin \left (c x\right ) e^{2}}{5 \, c^{4}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1} b d e}{3 \, c^{3}} + \frac {b x \arcsin \left (c x\right ) e^{2}}{5 \, c^{4}} + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} \sqrt {-c^{2} x^{2} + 1} b e^{2}}{25 \, c^{5}} - \frac {2 \, {\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b e^{2}}{15 \, c^{5}} + \frac {\sqrt {-c^{2} x^{2} + 1} b e^{2}}{5 \, c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*x^5*e^2 + 2/3*a*d*x^3*e + b*d^2*x*arcsin(c*x) + a*d^2*x + 2/3*(c^2*x^2 - 1)*b*d*x*arcsin(c*x)*e/c^2 + 2/
3*b*d*x*arcsin(c*x)*e/c^2 + sqrt(-c^2*x^2 + 1)*b*d^2/c + 1/5*(c^2*x^2 - 1)^2*b*x*arcsin(c*x)*e^2/c^4 - 2/9*(-c
^2*x^2 + 1)^(3/2)*b*d*e/c^3 + 2/5*(c^2*x^2 - 1)*b*x*arcsin(c*x)*e^2/c^4 + 2/3*sqrt(-c^2*x^2 + 1)*b*d*e/c^3 + 1
/5*b*x*arcsin(c*x)*e^2/c^4 + 1/25*(c^2*x^2 - 1)^2*sqrt(-c^2*x^2 + 1)*b*e^2/c^5 - 2/15*(-c^2*x^2 + 1)^(3/2)*b*e
^2/c^5 + 1/5*sqrt(-c^2*x^2 + 1)*b*e^2/c^5

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maple [A]  time = 0.00, size = 209, normalized size = 1.39 \[ \frac {\frac {a \left (\frac {1}{5} e^{2} c^{5} x^{5}+\frac {2}{3} c^{5} e d \,x^{3}+d^{2} c^{5} x \right )}{c^{4}}+\frac {b \left (\frac {\arcsin \left (c x \right ) e^{2} c^{5} x^{5}}{5}+\frac {2 \arcsin \left (c x \right ) c^{5} e d \,x^{3}}{3}+\arcsin \left (c x \right ) d^{2} c^{5} x -\frac {e^{2} \left (-\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{5}-\frac {4 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{15}-\frac {8 \sqrt {-c^{2} x^{2}+1}}{15}\right )}{5}-\frac {2 c^{2} e d \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )}{3}+d^{2} c^{4} \sqrt {-c^{2} x^{2}+1}\right )}{c^{4}}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(a/c^4*(1/5*e^2*c^5*x^5+2/3*c^5*e*d*x^3+d^2*c^5*x)+b/c^4*(1/5*arcsin(c*x)*e^2*c^5*x^5+2/3*arcsin(c*x)*c^5*
e*d*x^3+arcsin(c*x)*d^2*c^5*x-1/5*e^2*(-1/5*c^4*x^4*(-c^2*x^2+1)^(1/2)-4/15*c^2*x^2*(-c^2*x^2+1)^(1/2)-8/15*(-
c^2*x^2+1)^(1/2))-2/3*c^2*e*d*(-1/3*c^2*x^2*(-c^2*x^2+1)^(1/2)-2/3*(-c^2*x^2+1)^(1/2))+d^2*c^4*(-c^2*x^2+1)^(1
/2)))

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maxima [A]  time = 0.48, size = 182, normalized size = 1.21 \[ \frac {1}{5} \, a e^{2} x^{5} + \frac {2}{3} \, a d e x^{3} + \frac {2}{9} \, {\left (3 \, x^{3} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b d e + \frac {1}{75} \, {\left (15 \, x^{5} \arcsin \left (c x\right ) + {\left (\frac {3 \, \sqrt {-c^{2} x^{2} + 1} x^{4}}{c^{2}} + \frac {4 \, \sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac {8 \, \sqrt {-c^{2} x^{2} + 1}}{c^{6}}\right )} c\right )} b e^{2} + a d^{2} x + \frac {{\left (c x \arcsin \left (c x\right ) + \sqrt {-c^{2} x^{2} + 1}\right )} b d^{2}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.27, size = 240, normalized size = 1.60 \[ \begin {cases} a d^{2} x + \frac {2 a d e x^{3}}{3} + \frac {a e^{2} x^{5}}{5} + b d^{2} x \operatorname {asin}{\left (c x \right )} + \frac {2 b d e x^{3} \operatorname {asin}{\left (c x \right )}}{3} + \frac {b e^{2} x^{5} \operatorname {asin}{\left (c x \right )}}{5} + \frac {b d^{2} \sqrt {- c^{2} x^{2} + 1}}{c} + \frac {2 b d e x^{2} \sqrt {- c^{2} x^{2} + 1}}{9 c} + \frac {b e^{2} x^{4} \sqrt {- c^{2} x^{2} + 1}}{25 c} + \frac {4 b d e \sqrt {- c^{2} x^{2} + 1}}{9 c^{3}} + \frac {4 b e^{2} x^{2} \sqrt {- c^{2} x^{2} + 1}}{75 c^{3}} + \frac {8 b e^{2} \sqrt {- c^{2} x^{2} + 1}}{75 c^{5}} & \text {for}\: c \neq 0 \\a \left (d^{2} x + \frac {2 d e x^{3}}{3} + \frac {e^{2} x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*d**2*x + 2*a*d*e*x**3/3 + a*e**2*x**5/5 + b*d**2*x*asin(c*x) + 2*b*d*e*x**3*asin(c*x)/3 + b*e**2*
x**5*asin(c*x)/5 + b*d**2*sqrt(-c**2*x**2 + 1)/c + 2*b*d*e*x**2*sqrt(-c**2*x**2 + 1)/(9*c) + b*e**2*x**4*sqrt(
-c**2*x**2 + 1)/(25*c) + 4*b*d*e*sqrt(-c**2*x**2 + 1)/(9*c**3) + 4*b*e**2*x**2*sqrt(-c**2*x**2 + 1)/(75*c**3)
+ 8*b*e**2*sqrt(-c**2*x**2 + 1)/(75*c**5), Ne(c, 0)), (a*(d**2*x + 2*d*e*x**3/3 + e**2*x**5/5), True))

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