3.1 \(\int (d+e x)^3 (a+b \sin ^{-1}(c x)) \, dx\)

Optimal. Leaf size=179 \[ \frac {(d+e x)^4 \left (a+b \sin ^{-1}(c x)\right )}{4 e}+\frac {b \sqrt {1-c^2 x^2} (d+e x)^3}{16 c}+\frac {7 b d \sqrt {1-c^2 x^2} (d+e x)^2}{48 c}-\frac {b \left (8 c^4 d^4+24 c^2 d^2 e^2+3 e^4\right ) \sin ^{-1}(c x)}{32 c^4 e}+\frac {b \sqrt {1-c^2 x^2} \left (e x \left (26 c^2 d^2+9 e^2\right )+4 d \left (19 c^2 d^2+16 e^2\right )\right )}{96 c^3} \]

[Out]

-1/32*b*(8*c^4*d^4+24*c^2*d^2*e^2+3*e^4)*arcsin(c*x)/c^4/e+1/4*(e*x+d)^4*(a+b*arcsin(c*x))/e+7/48*b*d*(e*x+d)^
2*(-c^2*x^2+1)^(1/2)/c+1/16*b*(e*x+d)^3*(-c^2*x^2+1)^(1/2)/c+1/96*b*(4*d*(19*c^2*d^2+16*e^2)+e*(26*c^2*d^2+9*e
^2)*x)*(-c^2*x^2+1)^(1/2)/c^3

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Rubi [A]  time = 0.18, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {4743, 743, 833, 780, 216} \[ \frac {(d+e x)^4 \left (a+b \sin ^{-1}(c x)\right )}{4 e}+\frac {b \sqrt {1-c^2 x^2} \left (e x \left (26 c^2 d^2+9 e^2\right )+4 d \left (19 c^2 d^2+16 e^2\right )\right )}{96 c^3}-\frac {b \left (24 c^2 d^2 e^2+8 c^4 d^4+3 e^4\right ) \sin ^{-1}(c x)}{32 c^4 e}+\frac {b \sqrt {1-c^2 x^2} (d+e x)^3}{16 c}+\frac {7 b d \sqrt {1-c^2 x^2} (d+e x)^2}{48 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3*(a + b*ArcSin[c*x]),x]

[Out]

(7*b*d*(d + e*x)^2*Sqrt[1 - c^2*x^2])/(48*c) + (b*(d + e*x)^3*Sqrt[1 - c^2*x^2])/(16*c) + (b*(4*d*(19*c^2*d^2
+ 16*e^2) + e*(26*c^2*d^2 + 9*e^2)*x)*Sqrt[1 - c^2*x^2])/(96*c^3) - (b*(8*c^4*d^4 + 24*c^2*d^2*e^2 + 3*e^4)*Ar
cSin[c*x])/(32*c^4*e) + ((d + e*x)^4*(a + b*ArcSin[c*x]))/(4*e)

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]

Rule 743

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

Rule 780

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

Rule 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(d + e*x)
^m*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 0])

Rule 4743

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*ArcSin[c*x])^n)/(e*(m + 1)), x] - Dist[(b*c*n)/(e*(m + 1)), Int[((d + e*x)^(m + 1)*(a + b*ArcSin[c*x])^(
n - 1))/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int (d+e x)^3 \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac {(d+e x)^4 \left (a+b \sin ^{-1}(c x)\right )}{4 e}-\frac {(b c) \int \frac {(d+e x)^4}{\sqrt {1-c^2 x^2}} \, dx}{4 e}\\ &=\frac {b (d+e x)^3 \sqrt {1-c^2 x^2}}{16 c}+\frac {(d+e x)^4 \left (a+b \sin ^{-1}(c x)\right )}{4 e}+\frac {b \int \frac {(d+e x)^2 \left (-4 c^2 d^2-3 e^2-7 c^2 d e x\right )}{\sqrt {1-c^2 x^2}} \, dx}{16 c e}\\ &=\frac {7 b d (d+e x)^2 \sqrt {1-c^2 x^2}}{48 c}+\frac {b (d+e x)^3 \sqrt {1-c^2 x^2}}{16 c}+\frac {(d+e x)^4 \left (a+b \sin ^{-1}(c x)\right )}{4 e}-\frac {b \int \frac {(d+e x) \left (c^2 d \left (12 c^2 d^2+23 e^2\right )+c^2 e \left (26 c^2 d^2+9 e^2\right ) x\right )}{\sqrt {1-c^2 x^2}} \, dx}{48 c^3 e}\\ &=\frac {7 b d (d+e x)^2 \sqrt {1-c^2 x^2}}{48 c}+\frac {b (d+e x)^3 \sqrt {1-c^2 x^2}}{16 c}+\frac {b \left (4 d \left (19 c^2 d^2+16 e^2\right )+e \left (26 c^2 d^2+9 e^2\right ) x\right ) \sqrt {1-c^2 x^2}}{96 c^3}+\frac {(d+e x)^4 \left (a+b \sin ^{-1}(c x)\right )}{4 e}-\frac {\left (b \left (8 c^4 d^4+24 c^2 d^2 e^2+3 e^4\right )\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{32 c^3 e}\\ &=\frac {7 b d (d+e x)^2 \sqrt {1-c^2 x^2}}{48 c}+\frac {b (d+e x)^3 \sqrt {1-c^2 x^2}}{16 c}+\frac {b \left (4 d \left (19 c^2 d^2+16 e^2\right )+e \left (26 c^2 d^2+9 e^2\right ) x\right ) \sqrt {1-c^2 x^2}}{96 c^3}-\frac {b \left (8 c^4 d^4+24 c^2 d^2 e^2+3 e^4\right ) \sin ^{-1}(c x)}{32 c^4 e}+\frac {(d+e x)^4 \left (a+b \sin ^{-1}(c x)\right )}{4 e}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 165, normalized size = 0.92 \[ \frac {24 a c^4 x \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )+b c \sqrt {1-c^2 x^2} \left (c^2 \left (96 d^3+72 d^2 e x+32 d e^2 x^2+6 e^3 x^3\right )+e^2 (64 d+9 e x)\right )+3 b \sin ^{-1}(c x) \left (8 c^4 x \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )-24 c^2 d^2 e-3 e^3\right )}{96 c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3*(a + b*ArcSin[c*x]),x]

[Out]

(24*a*c^4*x*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) + b*c*Sqrt[1 - c^2*x^2]*(e^2*(64*d + 9*e*x) + c^2*(96*
d^3 + 72*d^2*e*x + 32*d*e^2*x^2 + 6*e^3*x^3)) + 3*b*(-24*c^2*d^2*e - 3*e^3 + 8*c^4*x*(4*d^3 + 6*d^2*e*x + 4*d*
e^2*x^2 + e^3*x^3))*ArcSin[c*x])/(96*c^4)

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fricas [A]  time = 0.62, size = 201, normalized size = 1.12 \[ \frac {24 \, a c^{4} e^{3} x^{4} + 96 \, a c^{4} d e^{2} x^{3} + 144 \, a c^{4} d^{2} e x^{2} + 96 \, a c^{4} d^{3} x + 3 \, {\left (8 \, b c^{4} e^{3} x^{4} + 32 \, b c^{4} d e^{2} x^{3} + 48 \, b c^{4} d^{2} e x^{2} + 32 \, b c^{4} d^{3} x - 24 \, b c^{2} d^{2} e - 3 \, b e^{3}\right )} \arcsin \left (c x\right ) + {\left (6 \, b c^{3} e^{3} x^{3} + 32 \, b c^{3} d e^{2} x^{2} + 96 \, b c^{3} d^{3} + 64 \, b c d e^{2} + 9 \, {\left (8 \, b c^{3} d^{2} e + b c e^{3}\right )} x\right )} \sqrt {-c^{2} x^{2} + 1}}{96 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

1/96*(24*a*c^4*e^3*x^4 + 96*a*c^4*d*e^2*x^3 + 144*a*c^4*d^2*e*x^2 + 96*a*c^4*d^3*x + 3*(8*b*c^4*e^3*x^4 + 32*b
*c^4*d*e^2*x^3 + 48*b*c^4*d^2*e*x^2 + 32*b*c^4*d^3*x - 24*b*c^2*d^2*e - 3*b*e^3)*arcsin(c*x) + (6*b*c^3*e^3*x^
3 + 32*b*c^3*d*e^2*x^2 + 96*b*c^3*d^3 + 64*b*c*d*e^2 + 9*(8*b*c^3*d^2*e + b*c*e^3)*x)*sqrt(-c^2*x^2 + 1))/c^4

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giac [A]  time = 0.34, size = 310, normalized size = 1.73 \[ b d^{3} x \arcsin \left (c x\right ) + \frac {1}{4} \, a x^{4} e^{3} + a d x^{3} e^{2} + a d^{3} x + \frac {3 \, \sqrt {-c^{2} x^{2} + 1} b d^{2} x e}{4 \, c} + \frac {{\left (c^{2} x^{2} - 1\right )} b d x \arcsin \left (c x\right ) e^{2}}{c^{2}} + \frac {3 \, {\left (c^{2} x^{2} - 1\right )} b d^{2} \arcsin \left (c x\right ) e}{2 \, c^{2}} + \frac {\sqrt {-c^{2} x^{2} + 1} b d^{3}}{c} + \frac {b d x \arcsin \left (c x\right ) e^{2}}{c^{2}} + \frac {3 \, {\left (c^{2} x^{2} - 1\right )} a d^{2} e}{2 \, c^{2}} + \frac {3 \, b d^{2} \arcsin \left (c x\right ) e}{4 \, c^{2}} - \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b x e^{3}}{16 \, c^{3}} - \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b d e^{2}}{3 \, c^{3}} + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b \arcsin \left (c x\right ) e^{3}}{4 \, c^{4}} + \frac {5 \, \sqrt {-c^{2} x^{2} + 1} b x e^{3}}{32 \, c^{3}} + \frac {\sqrt {-c^{2} x^{2} + 1} b d e^{2}}{c^{3}} + \frac {{\left (c^{2} x^{2} - 1\right )} b \arcsin \left (c x\right ) e^{3}}{2 \, c^{4}} + \frac {5 \, b \arcsin \left (c x\right ) e^{3}}{32 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

b*d^3*x*arcsin(c*x) + 1/4*a*x^4*e^3 + a*d*x^3*e^2 + a*d^3*x + 3/4*sqrt(-c^2*x^2 + 1)*b*d^2*x*e/c + (c^2*x^2 -
1)*b*d*x*arcsin(c*x)*e^2/c^2 + 3/2*(c^2*x^2 - 1)*b*d^2*arcsin(c*x)*e/c^2 + sqrt(-c^2*x^2 + 1)*b*d^3/c + b*d*x*
arcsin(c*x)*e^2/c^2 + 3/2*(c^2*x^2 - 1)*a*d^2*e/c^2 + 3/4*b*d^2*arcsin(c*x)*e/c^2 - 1/16*(-c^2*x^2 + 1)^(3/2)*
b*x*e^3/c^3 - 1/3*(-c^2*x^2 + 1)^(3/2)*b*d*e^2/c^3 + 1/4*(c^2*x^2 - 1)^2*b*arcsin(c*x)*e^3/c^4 + 5/32*sqrt(-c^
2*x^2 + 1)*b*x*e^3/c^3 + sqrt(-c^2*x^2 + 1)*b*d*e^2/c^3 + 1/2*(c^2*x^2 - 1)*b*arcsin(c*x)*e^3/c^4 + 5/32*b*arc
sin(c*x)*e^3/c^4

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maple [A]  time = 0.01, size = 265, normalized size = 1.48 \[ \frac {\frac {\left (c e x +d c \right )^{4} a}{4 c^{3} e}+\frac {b \left (\frac {e^{3} \arcsin \left (c x \right ) c^{4} x^{4}}{4}+e^{2} \arcsin \left (c x \right ) c^{4} x^{3} d +\frac {3 e \arcsin \left (c x \right ) c^{4} x^{2} d^{2}}{2}+\arcsin \left (c x \right ) c^{4} x \,d^{3}+\frac {\arcsin \left (c x \right ) c^{4} d^{4}}{4 e}-\frac {e^{4} \left (-\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{4}-\frac {3 c x \sqrt {-c^{2} x^{2}+1}}{8}+\frac {3 \arcsin \left (c x \right )}{8}\right )+4 d c \,e^{3} \left (-\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-c^{2} x^{2}+1}}{3}\right )+6 c^{2} d^{2} e^{2} \left (-\frac {c x \sqrt {-c^{2} x^{2}+1}}{2}+\frac {\arcsin \left (c x \right )}{2}\right )-4 c^{3} d^{3} e \sqrt {-c^{2} x^{2}+1}+c^{4} d^{4} \arcsin \left (c x \right )}{4 e}\right )}{c^{3}}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(a+b*arcsin(c*x)),x)

[Out]

1/c*(1/4*(c*e*x+c*d)^4*a/c^3/e+b/c^3*(1/4*e^3*arcsin(c*x)*c^4*x^4+e^2*arcsin(c*x)*c^4*x^3*d+3/2*e*arcsin(c*x)*
c^4*x^2*d^2+arcsin(c*x)*c^4*x*d^3+1/4/e*arcsin(c*x)*c^4*d^4-1/4/e*(e^4*(-1/4*c^3*x^3*(-c^2*x^2+1)^(1/2)-3/8*c*
x*(-c^2*x^2+1)^(1/2)+3/8*arcsin(c*x))+4*d*c*e^3*(-1/3*c^2*x^2*(-c^2*x^2+1)^(1/2)-2/3*(-c^2*x^2+1)^(1/2))+6*c^2
*d^2*e^2*(-1/2*c*x*(-c^2*x^2+1)^(1/2)+1/2*arcsin(c*x))-4*c^3*d^3*e*(-c^2*x^2+1)^(1/2)+c^4*d^4*arcsin(c*x))))

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maxima [A]  time = 0.42, size = 231, normalized size = 1.29 \[ \frac {1}{4} \, a e^{3} x^{4} + a d e^{2} x^{3} + \frac {3}{2} \, a d^{2} e x^{2} + \frac {3}{4} \, {\left (2 \, x^{2} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x}{c^{2}} - \frac {\arcsin \left (c x\right )}{c^{3}}\right )}\right )} b d^{2} e + \frac {1}{3} \, {\left (3 \, x^{3} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b d e^{2} + \frac {1}{32} \, {\left (8 \, x^{4} \arcsin \left (c x\right ) + {\left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1} x^{3}}{c^{2}} + \frac {3 \, \sqrt {-c^{2} x^{2} + 1} x}{c^{4}} - \frac {3 \, \arcsin \left (c x\right )}{c^{5}}\right )} c\right )} b e^{3} + a d^{3} x + \frac {{\left (c x \arcsin \left (c x\right ) + \sqrt {-c^{2} x^{2} + 1}\right )} b d^{3}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

1/4*a*e^3*x^4 + a*d*e^2*x^3 + 3/2*a*d^2*e*x^2 + 3/4*(2*x^2*arcsin(c*x) + c*(sqrt(-c^2*x^2 + 1)*x/c^2 - arcsin(
c*x)/c^3))*b*d^2*e + 1/3*(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
^2 + 1/32*(8*x^4*arcsin(c*x) + (2*sqrt(-c^2*x^2 + 1)*x^3/c^2 + 3*sqrt(-c^2*x^2 + 1)*x/c^4 - 3*arcsin(c*x)/c^5)
*c)*b*e^3 + a*d^3*x + (c*x*arcsin(c*x) + sqrt(-c^2*x^2 + 1))*b*d^3/c

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d+e\,x\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))*(d + e*x)^3,x)

[Out]

int((a + b*asin(c*x))*(d + e*x)^3, x)

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sympy [A]  time = 1.37, size = 316, normalized size = 1.77 \[ \begin {cases} a d^{3} x + \frac {3 a d^{2} e x^{2}}{2} + a d e^{2} x^{3} + \frac {a e^{3} x^{4}}{4} + b d^{3} x \operatorname {asin}{\left (c x \right )} + \frac {3 b d^{2} e x^{2} \operatorname {asin}{\left (c x \right )}}{2} + b d e^{2} x^{3} \operatorname {asin}{\left (c x \right )} + \frac {b e^{3} x^{4} \operatorname {asin}{\left (c x \right )}}{4} + \frac {b d^{3} \sqrt {- c^{2} x^{2} + 1}}{c} + \frac {3 b d^{2} e x \sqrt {- c^{2} x^{2} + 1}}{4 c} + \frac {b d e^{2} x^{2} \sqrt {- c^{2} x^{2} + 1}}{3 c} + \frac {b e^{3} x^{3} \sqrt {- c^{2} x^{2} + 1}}{16 c} - \frac {3 b d^{2} e \operatorname {asin}{\left (c x \right )}}{4 c^{2}} + \frac {2 b d e^{2} \sqrt {- c^{2} x^{2} + 1}}{3 c^{3}} + \frac {3 b e^{3} x \sqrt {- c^{2} x^{2} + 1}}{32 c^{3}} - \frac {3 b e^{3} \operatorname {asin}{\left (c x \right )}}{32 c^{4}} & \text {for}\: c \neq 0 \\a \left (d^{3} x + \frac {3 d^{2} e x^{2}}{2} + d e^{2} x^{3} + \frac {e^{3} x^{4}}{4}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3*(a+b*asin(c*x)),x)

[Out]

Piecewise((a*d**3*x + 3*a*d**2*e*x**2/2 + a*d*e**2*x**3 + a*e**3*x**4/4 + b*d**3*x*asin(c*x) + 3*b*d**2*e*x**2
*asin(c*x)/2 + b*d*e**2*x**3*asin(c*x) + b*e**3*x**4*asin(c*x)/4 + b*d**3*sqrt(-c**2*x**2 + 1)/c + 3*b*d**2*e*
x*sqrt(-c**2*x**2 + 1)/(4*c) + b*d*e**2*x**2*sqrt(-c**2*x**2 + 1)/(3*c) + b*e**3*x**3*sqrt(-c**2*x**2 + 1)/(16
*c) - 3*b*d**2*e*asin(c*x)/(4*c**2) + 2*b*d*e**2*sqrt(-c**2*x**2 + 1)/(3*c**3) + 3*b*e**3*x*sqrt(-c**2*x**2 +
1)/(32*c**3) - 3*b*e**3*asin(c*x)/(32*c**4), Ne(c, 0)), (a*(d**3*x + 3*d**2*e*x**2/2 + d*e**2*x**3 + e**3*x**4
/4), True))

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