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

Optimal. Leaf size=109 \[ \frac {e^3 (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )}{4 d}+\frac {b e^3 \sqrt {1-(c+d x)^2} (c+d x)^3}{16 d}+\frac {3 b e^3 \sqrt {1-(c+d x)^2} (c+d x)}{32 d}-\frac {3 b e^3 \sin ^{-1}(c+d x)}{32 d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {4805, 12, 4627, 321, 216} \[ \frac {e^3 (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )}{4 d}+\frac {b e^3 \sqrt {1-(c+d x)^2} (c+d x)^3}{16 d}+\frac {3 b e^3 \sqrt {1-(c+d x)^2} (c+d x)}{32 d}-\frac {3 b e^3 \sin ^{-1}(c+d x)}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*e^3*(c + d*x)*Sqrt[1 - (c + d*x)^2])/(32*d) + (b*e^3*(c + d*x)^3*Sqrt[1 - (c + d*x)^2])/(16*d) - (3*b*e^3
*ArcSin[c + d*x])/(32*d) + (e^3*(c + d*x)^4*(a + b*ArcSin[c + d*x]))/(4*d)

Rule 12

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

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 4627

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

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rubi steps

\begin {align*} \int (c e+d e x)^3 \left (a+b \sin ^{-1}(c+d x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int e^3 x^3 \left (a+b \sin ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 \operatorname {Subst}\left (\int x^3 \left (a+b \sin ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )}{4 d}-\frac {\left (b e^3\right ) \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{4 d}\\ &=\frac {b e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{16 d}+\frac {e^3 (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )}{4 d}-\frac {\left (3 b e^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{16 d}\\ &=\frac {3 b e^3 (c+d x) \sqrt {1-(c+d x)^2}}{32 d}+\frac {b e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{16 d}+\frac {e^3 (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )}{4 d}-\frac {\left (3 b e^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{32 d}\\ &=\frac {3 b e^3 (c+d x) \sqrt {1-(c+d x)^2}}{32 d}+\frac {b e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{16 d}-\frac {3 b e^3 \sin ^{-1}(c+d x)}{32 d}+\frac {e^3 (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 87, normalized size = 0.80 \[ \frac {e^3 \left (8 (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )+2 b \sqrt {1-(c+d x)^2} (c+d x)^3+3 b \sqrt {1-(c+d x)^2} (c+d x)-3 b \sin ^{-1}(c+d x)\right )}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^3*(3*b*(c + d*x)*Sqrt[1 - (c + d*x)^2] + 2*b*(c + d*x)^3*Sqrt[1 - (c + d*x)^2] - 3*b*ArcSin[c + d*x] + 8*(c
+ d*x)^4*(a + b*ArcSin[c + d*x])))/(32*d)

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fricas [B]  time = 0.59, size = 209, normalized size = 1.92 \[ \frac {8 \, a d^{4} e^{3} x^{4} + 32 \, a c d^{3} e^{3} x^{3} + 48 \, a c^{2} d^{2} e^{3} x^{2} + 32 \, a c^{3} d e^{3} x + {\left (8 \, b d^{4} e^{3} x^{4} + 32 \, b c d^{3} e^{3} x^{3} + 48 \, b c^{2} d^{2} e^{3} x^{2} + 32 \, b c^{3} d e^{3} x + {\left (8 \, b c^{4} - 3 \, b\right )} e^{3}\right )} \arcsin \left (d x + c\right ) + {\left (2 \, b d^{3} e^{3} x^{3} + 6 \, b c d^{2} e^{3} x^{2} + 3 \, {\left (2 \, b c^{2} + b\right )} d e^{3} x + {\left (2 \, b c^{3} + 3 \, b c\right )} e^{3}\right )} \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.41, size = 130, normalized size = 1.19 \[ \frac {{\left (d x + c\right )}^{4} a e^{3}}{4 \, d} + \frac {{\left ({\left (d x + c\right )}^{2} - 1\right )}^{2} b \arcsin \left (d x + c\right ) e^{3}}{4 \, d} - \frac {{\left (-{\left (d x + c\right )}^{2} + 1\right )}^{\frac {3}{2}} {\left (d x + c\right )} b e^{3}}{16 \, d} + \frac {{\left ({\left (d x + c\right )}^{2} - 1\right )} b \arcsin \left (d x + c\right ) e^{3}}{2 \, d} + \frac {5 \, \sqrt {-{\left (d x + c\right )}^{2} + 1} {\left (d x + c\right )} b e^{3}}{32 \, d} + \frac {5 \, b \arcsin \left (d x + c\right ) e^{3}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 90, normalized size = 0.83 \[ \frac {\frac {e^{3} \left (d x +c \right )^{4} a}{4}+e^{3} b \left (\frac {\left (d x +c \right )^{4} \arcsin \left (d x +c \right )}{4}+\frac {\left (d x +c \right )^{3} \sqrt {1-\left (d x +c \right )^{2}}}{16}+\frac {3 \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{32}-\frac {3 \arcsin \left (d x +c \right )}{32}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.43, size = 816, normalized size = 7.49 \[ \frac {1}{4} \, a d^{3} e^{3} x^{4} + a c d^{2} e^{3} x^{3} + \frac {3}{2} \, a c^{2} d e^{3} x^{2} + \frac {3}{4} \, {\left (2 \, x^{2} \arcsin \left (d x + c\right ) + d {\left (\frac {3 \, c^{2} \arcsin \left (-\frac {d^{2} x + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{3}} + \frac {\sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} x}{d^{2}} - \frac {{\left (c^{2} - 1\right )} \arcsin \left (-\frac {d^{2} x + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{3}} - \frac {3 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} c}{d^{3}}\right )}\right )} b c^{2} d e^{3} + \frac {1}{6} \, {\left (6 \, x^{3} \arcsin \left (d x + c\right ) + d {\left (\frac {2 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} x^{2}}{d^{2}} - \frac {15 \, c^{3} \arcsin \left (-\frac {d^{2} x + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{4}} - \frac {5 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} c x}{d^{3}} + \frac {9 \, {\left (c^{2} - 1\right )} c \arcsin \left (-\frac {d^{2} x + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{4}} + \frac {15 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} c^{2}}{d^{4}} - \frac {4 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} {\left (c^{2} - 1\right )}}{d^{4}}\right )}\right )} b c d^{2} e^{3} + \frac {1}{96} \, {\left (24 \, x^{4} \arcsin \left (d x + c\right ) + {\left (\frac {6 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} x^{3}}{d^{2}} - \frac {14 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} c x^{2}}{d^{3}} + \frac {105 \, c^{4} \arcsin \left (-\frac {d^{2} x + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{5}} + \frac {35 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} c^{2} x}{d^{4}} - \frac {90 \, {\left (c^{2} - 1\right )} c^{2} \arcsin \left (-\frac {d^{2} x + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{5}} - \frac {105 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} c^{3}}{d^{5}} - \frac {9 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} {\left (c^{2} - 1\right )} x}{d^{4}} + \frac {9 \, {\left (c^{2} - 1\right )}^{2} \arcsin \left (-\frac {d^{2} x + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{5}} + \frac {55 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} {\left (c^{2} - 1\right )} c}{d^{5}}\right )} d\right )} b d^{3} e^{3} + a c^{3} e^{3} x + \frac {{\left ({\left (d x + c\right )} \arcsin \left (d x + c\right ) + \sqrt {-{\left (d x + c\right )}^{2} + 1}\right )} b c^{3} e^{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*d^3*e^3*x^4 + a*c*d^2*e^3*x^3 + 3/2*a*c^2*d*e^3*x^2 + 3/4*(2*x^2*arcsin(d*x + c) + d*(3*c^2*arcsin(-(d^2
*x + c*d)/sqrt(c^2*d^2 - (c^2 - 1)*d^2))/d^3 + sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*x/d^2 - (c^2 - 1)*arcsin(-(d
^2*x + c*d)/sqrt(c^2*d^2 - (c^2 - 1)*d^2))/d^3 - 3*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*c/d^3))*b*c^2*d*e^3 + 1/
6*(6*x^3*arcsin(d*x + c) + d*(2*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*x^2/d^2 - 15*c^3*arcsin(-(d^2*x + c*d)/sqrt
(c^2*d^2 - (c^2 - 1)*d^2))/d^4 - 5*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*c*x/d^3 + 9*(c^2 - 1)*c*arcsin(-(d^2*x +
c*d)/sqrt(c^2*d^2 - (c^2 - 1)*d^2))/d^4 + 15*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*c^2/d^4 - 4*sqrt(-d^2*x^2 - 2
*c*d*x - c^2 + 1)*(c^2 - 1)/d^4))*b*c*d^2*e^3 + 1/96*(24*x^4*arcsin(d*x + c) + (6*sqrt(-d^2*x^2 - 2*c*d*x - c^
2 + 1)*x^3/d^2 - 14*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*c*x^2/d^3 + 105*c^4*arcsin(-(d^2*x + c*d)/sqrt(c^2*d^2
- (c^2 - 1)*d^2))/d^5 + 35*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*c^2*x/d^4 - 90*(c^2 - 1)*c^2*arcsin(-(d^2*x + c*
d)/sqrt(c^2*d^2 - (c^2 - 1)*d^2))/d^5 - 105*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*c^3/d^5 - 9*sqrt(-d^2*x^2 - 2*c
*d*x - c^2 + 1)*(c^2 - 1)*x/d^4 + 9*(c^2 - 1)^2*arcsin(-(d^2*x + c*d)/sqrt(c^2*d^2 - (c^2 - 1)*d^2))/d^5 + 55*
sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*(c^2 - 1)*c/d^5)*d)*b*d^3*e^3 + a*c^3*e^3*x + ((d*x + c)*arcsin(d*x + c) +
sqrt(-(d*x + c)^2 + 1))*b*c^3*e^3/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.58, size = 394, normalized size = 3.61 \[ \begin {cases} a c^{3} e^{3} x + \frac {3 a c^{2} d e^{3} x^{2}}{2} + a c d^{2} e^{3} x^{3} + \frac {a d^{3} e^{3} x^{4}}{4} + \frac {b c^{4} e^{3} \operatorname {asin}{\left (c + d x \right )}}{4 d} + b c^{3} e^{3} x \operatorname {asin}{\left (c + d x \right )} + \frac {b c^{3} e^{3} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{16 d} + \frac {3 b c^{2} d e^{3} x^{2} \operatorname {asin}{\left (c + d x \right )}}{2} + \frac {3 b c^{2} e^{3} x \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{16} + b c d^{2} e^{3} x^{3} \operatorname {asin}{\left (c + d x \right )} + \frac {3 b c d e^{3} x^{2} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{16} + \frac {3 b c e^{3} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{32 d} + \frac {b d^{3} e^{3} x^{4} \operatorname {asin}{\left (c + d x \right )}}{4} + \frac {b d^{2} e^{3} x^{3} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{16} + \frac {3 b e^{3} x \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{32} - \frac {3 b e^{3} \operatorname {asin}{\left (c + d x \right )}}{32 d} & \text {for}\: d \neq 0 \\c^{3} e^{3} x \left (a + b \operatorname {asin}{\relax (c )}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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