3.177 \(\int (c e+d e x)^4 (a+b \sin ^{-1}(c+d x)) \, dx\)

Optimal. Leaf size=106 \[ \frac{e^4 (c+d x)^5 \left (a+b \sin ^{-1}(c+d x)\right )}{5 d}+\frac{b e^4 \left (1-(c+d x)^2\right )^{5/2}}{25 d}-\frac{2 b e^4 \left (1-(c+d x)^2\right )^{3/2}}{15 d}+\frac{b e^4 \sqrt{1-(c+d x)^2}}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.0926155, antiderivative size = 106, normalized size of antiderivative = 1., 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, 266, 43} \[ \frac{e^4 (c+d x)^5 \left (a+b \sin ^{-1}(c+d x)\right )}{5 d}+\frac{b e^4 \left (1-(c+d x)^2\right )^{5/2}}{25 d}-\frac{2 b e^4 \left (1-(c+d x)^2\right )^{3/2}}{15 d}+\frac{b e^4 \sqrt{1-(c+d x)^2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^4*(a + b*ArcSin[c + d*x]),x]

[Out]

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

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]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 266

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.111426, size = 77, normalized size = 0.73 \[ \frac{e^4 \left (\frac{1}{5} (c+d x)^5 \left (a+b \sin ^{-1}(c+d x)\right )-\frac{1}{75} b \sqrt{1-(c+d x)^2} \left (-3 \left ((c+d x)^2-1\right )^2+10 \left (1-(c+d x)^2\right )-15\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^4*(a + b*ArcSin[c + d*x]),x]

[Out]

(e^4*(-(b*Sqrt[1 - (c + d*x)^2]*(-15 + 10*(1 - (c + d*x)^2) - 3*(-1 + (c + d*x)^2)^2))/75 + ((c + d*x)^5*(a +
b*ArcSin[c + d*x]))/5))/d

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Maple [A]  time = 0.011, size = 99, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{{e}^{4} \left ( dx+c \right ) ^{5}a}{5}}+{e}^{4}b \left ({\frac{ \left ( dx+c \right ) ^{5}\arcsin \left ( dx+c \right ) }{5}}+{\frac{ \left ( dx+c \right ) ^{4}}{25}\sqrt{1- \left ( dx+c \right ) ^{2}}}+{\frac{4\, \left ( dx+c \right ) ^{2}}{75}\sqrt{1- \left ( dx+c \right ) ^{2}}}+{\frac{8}{75}\sqrt{1- \left ( dx+c \right ) ^{2}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^4*(a+b*arcsin(d*x+c)),x)

[Out]

1/d*(1/5*e^4*(d*x+c)^5*a+e^4*b*(1/5*(d*x+c)^5*arcsin(d*x+c)+1/25*(d*x+c)^4*(1-(d*x+c)^2)^(1/2)+4/75*(d*x+c)^2*
(1-(d*x+c)^2)^(1/2)+8/75*(1-(d*x+c)^2)^(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((d*e*x+c*e)^4*(a+b*arcsin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.41032, size = 564, normalized size = 5.32 \begin{align*} \frac{15 \, a d^{5} e^{4} x^{5} + 75 \, a c d^{4} e^{4} x^{4} + 150 \, a c^{2} d^{3} e^{4} x^{3} + 150 \, a c^{3} d^{2} e^{4} x^{2} + 75 \, a c^{4} d e^{4} x + 15 \,{\left (b d^{5} e^{4} x^{5} + 5 \, b c d^{4} e^{4} x^{4} + 10 \, b c^{2} d^{3} e^{4} x^{3} + 10 \, b c^{3} d^{2} e^{4} x^{2} + 5 \, b c^{4} d e^{4} x + b c^{5} e^{4}\right )} \arcsin \left (d x + c\right ) +{\left (3 \, b d^{4} e^{4} x^{4} + 12 \, b c d^{3} e^{4} x^{3} + 2 \,{\left (9 \, b c^{2} + 2 \, b\right )} d^{2} e^{4} x^{2} + 4 \,{\left (3 \, b c^{3} + 2 \, b c\right )} d e^{4} x +{\left (3 \, b c^{4} + 4 \, b c^{2} + 8 \, b\right )} e^{4}\right )} \sqrt{-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}{75 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4*(a+b*arcsin(d*x+c)),x, algorithm="fricas")

[Out]

1/75*(15*a*d^5*e^4*x^5 + 75*a*c*d^4*e^4*x^4 + 150*a*c^2*d^3*e^4*x^3 + 150*a*c^3*d^2*e^4*x^2 + 75*a*c^4*d*e^4*x
+ 15*(b*d^5*e^4*x^5 + 5*b*c*d^4*e^4*x^4 + 10*b*c^2*d^3*e^4*x^3 + 10*b*c^3*d^2*e^4*x^2 + 5*b*c^4*d*e^4*x + b*c
^5*e^4)*arcsin(d*x + c) + (3*b*d^4*e^4*x^4 + 12*b*c*d^3*e^4*x^3 + 2*(9*b*c^2 + 2*b)*d^2*e^4*x^2 + 4*(3*b*c^3 +
2*b*c)*d*e^4*x + (3*b*c^4 + 4*b*c^2 + 8*b)*e^4)*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1))/d

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Sympy [A]  time = 6.2101, size = 527, normalized size = 4.97 \begin{align*} \begin{cases} a c^{4} e^{4} x + 2 a c^{3} d e^{4} x^{2} + 2 a c^{2} d^{2} e^{4} x^{3} + a c d^{3} e^{4} x^{4} + \frac{a d^{4} e^{4} x^{5}}{5} + \frac{b c^{5} e^{4} \operatorname{asin}{\left (c + d x \right )}}{5 d} + b c^{4} e^{4} x \operatorname{asin}{\left (c + d x \right )} + \frac{b c^{4} e^{4} \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{25 d} + 2 b c^{3} d e^{4} x^{2} \operatorname{asin}{\left (c + d x \right )} + \frac{4 b c^{3} e^{4} x \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{25} + 2 b c^{2} d^{2} e^{4} x^{3} \operatorname{asin}{\left (c + d x \right )} + \frac{6 b c^{2} d e^{4} x^{2} \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{25} + \frac{4 b c^{2} e^{4} \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{75 d} + b c d^{3} e^{4} x^{4} \operatorname{asin}{\left (c + d x \right )} + \frac{4 b c d^{2} e^{4} x^{3} \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{25} + \frac{8 b c e^{4} x \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{75} + \frac{b d^{4} e^{4} x^{5} \operatorname{asin}{\left (c + d x \right )}}{5} + \frac{b d^{3} e^{4} x^{4} \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{25} + \frac{4 b d e^{4} x^{2} \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{75} + \frac{8 b e^{4} \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{75 d} & \text{for}\: d \neq 0 \\c^{4} e^{4} x \left (a + b \operatorname{asin}{\left (c \right )}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**4*(a+b*asin(d*x+c)),x)

[Out]

Piecewise((a*c**4*e**4*x + 2*a*c**3*d*e**4*x**2 + 2*a*c**2*d**2*e**4*x**3 + a*c*d**3*e**4*x**4 + a*d**4*e**4*x
**5/5 + b*c**5*e**4*asin(c + d*x)/(5*d) + b*c**4*e**4*x*asin(c + d*x) + b*c**4*e**4*sqrt(-c**2 - 2*c*d*x - d**
2*x**2 + 1)/(25*d) + 2*b*c**3*d*e**4*x**2*asin(c + d*x) + 4*b*c**3*e**4*x*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1
)/25 + 2*b*c**2*d**2*e**4*x**3*asin(c + d*x) + 6*b*c**2*d*e**4*x**2*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/25 +
4*b*c**2*e**4*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/(75*d) + b*c*d**3*e**4*x**4*asin(c + d*x) + 4*b*c*d**2*e*
*4*x**3*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/25 + 8*b*c*e**4*x*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/75 + b*d
**4*e**4*x**5*asin(c + d*x)/5 + b*d**3*e**4*x**4*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/25 + 4*b*d*e**4*x**2*sq
rt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/75 + 8*b*e**4*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/(75*d), Ne(d, 0)), (c*
*4*e**4*x*(a + b*asin(c)), True))

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Giac [A]  time = 1.35057, size = 225, normalized size = 2.12 \begin{align*} \frac{{\left (d x + c\right )}^{5} a e^{4}}{5 \, d} + \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )}^{2}{\left (d x + c\right )} b \arcsin \left (d x + c\right ) e^{4}}{5 \, d} + \frac{2 \,{\left ({\left (d x + c\right )}^{2} - 1\right )}{\left (d x + c\right )} b \arcsin \left (d x + c\right ) e^{4}}{5 \, d} + \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )}^{2} \sqrt{-{\left (d x + c\right )}^{2} + 1} b e^{4}}{25 \, d} + \frac{{\left (d x + c\right )} b \arcsin \left (d x + c\right ) e^{4}}{5 \, d} - \frac{2 \,{\left (-{\left (d x + c\right )}^{2} + 1\right )}^{\frac{3}{2}} b e^{4}}{15 \, d} + \frac{\sqrt{-{\left (d x + c\right )}^{2} + 1} b e^{4}}{5 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4*(a+b*arcsin(d*x+c)),x, algorithm="giac")

[Out]

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