3.186 \(\int \frac{a+b \sin ^{-1}(c+d x)}{(c e+d e x)^5} \, dx\)

Optimal. Leaf size=94 \[ -\frac{a+b \sin ^{-1}(c+d x)}{4 d e^5 (c+d x)^4}-\frac{b \sqrt{1-(c+d x)^2}}{6 d e^5 (c+d x)}-\frac{b \sqrt{1-(c+d x)^2}}{12 d e^5 (c+d x)^3} \]

[Out]

-(b*Sqrt[1 - (c + d*x)^2])/(12*d*e^5*(c + d*x)^3) - (b*Sqrt[1 - (c + d*x)^2])/(6*d*e^5*(c + d*x)) - (a + b*Arc
Sin[c + d*x])/(4*d*e^5*(c + d*x)^4)

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Rubi [A]  time = 0.0681326, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {4805, 12, 4627, 271, 264} \[ -\frac{a+b \sin ^{-1}(c+d x)}{4 d e^5 (c+d x)^4}-\frac{b \sqrt{1-(c+d x)^2}}{6 d e^5 (c+d x)}-\frac{b \sqrt{1-(c+d x)^2}}{12 d e^5 (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])/(c*e + d*e*x)^5,x]

[Out]

-(b*Sqrt[1 - (c + d*x)^2])/(12*d*e^5*(c + d*x)^3) - (b*Sqrt[1 - (c + d*x)^2])/(6*d*e^5*(c + d*x)) - (a + b*Arc
Sin[c + d*x])/(4*d*e^5*(c + d*x)^4)

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 271

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

Rule 264

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

Rubi steps

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

Mathematica [A]  time = 0.0610868, size = 63, normalized size = 0.67 \[ -\frac{3 \left (a+b \sin ^{-1}(c+d x)\right )+b (c+d x) \sqrt{1-(c+d x)^2} \left (2 (c+d x)^2+1\right )}{12 d e^5 (c+d x)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 84, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ( -{\frac{a}{4\,{e}^{5} \left ( dx+c \right ) ^{4}}}+{\frac{b}{{e}^{5}} \left ( -{\frac{\arcsin \left ( dx+c \right ) }{4\, \left ( dx+c \right ) ^{4}}}-{\frac{1}{12\, \left ( dx+c \right ) ^{3}}\sqrt{1- \left ( dx+c \right ) ^{2}}}-{\frac{1}{6\,dx+6\,c}\sqrt{1- \left ( dx+c \right ) ^{2}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-1/4*a/e^5/(d*x+c)^4+b/e^5*(-1/4/(d*x+c)^4*arcsin(d*x+c)-1/12/(d*x+c)^3*(1-(d*x+c)^2)^(1/2)-1/6/(d*x+c)*(
1-(d*x+c)^2)^(1/2)))

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Maxima [B]  time = 1.61405, size = 355, normalized size = 3.78 \begin{align*} \frac{1}{12} \, b{\left (\frac{{\left (2 \, d^{4} x^{4} + 8 \, c d^{3} x^{3} + 2 \, c^{4} +{\left (12 \, c^{2} d^{2} - d^{2}\right )} x^{2} - c^{2} + 2 \,{\left (4 \, c^{3} d - c d\right )} x - 1\right )} d}{{\left (d^{5} e^{5} x^{3} + 3 \, c d^{4} e^{5} x^{2} + 3 \, c^{2} d^{3} e^{5} x + c^{3} d^{2} e^{5}\right )} \sqrt{d x + c + 1} \sqrt{-d x - c + 1}} - \frac{3 \, \arcsin \left (d x + c\right )}{d^{5} e^{5} x^{4} + 4 \, c d^{4} e^{5} x^{3} + 6 \, c^{2} d^{3} e^{5} x^{2} + 4 \, c^{3} d^{2} e^{5} x + c^{4} d e^{5}}\right )} - \frac{a}{4 \,{\left (d^{5} e^{5} x^{4} + 4 \, c d^{4} e^{5} x^{3} + 6 \, c^{2} d^{3} e^{5} x^{2} + 4 \, c^{3} d^{2} e^{5} x + c^{4} d e^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*b*((2*d^4*x^4 + 8*c*d^3*x^3 + 2*c^4 + (12*c^2*d^2 - d^2)*x^2 - c^2 + 2*(4*c^3*d - c*d)*x - 1)*d/((d^5*e^5
*x^3 + 3*c*d^4*e^5*x^2 + 3*c^2*d^3*e^5*x + c^3*d^2*e^5)*sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) - 3*arcsin(d*x +
c)/(d^5*e^5*x^4 + 4*c*d^4*e^5*x^3 + 6*c^2*d^3*e^5*x^2 + 4*c^3*d^2*e^5*x + c^4*d*e^5)) - 1/4*a/(d^5*e^5*x^4 +
4*c*d^4*e^5*x^3 + 6*c^2*d^3*e^5*x^2 + 4*c^3*d^2*e^5*x + c^4*d*e^5)

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Fricas [B]  time = 2.29779, size = 405, normalized size = 4.31 \begin{align*} \frac{3 \, a d^{4} x^{4} + 12 \, a c d^{3} x^{3} + 18 \, a c^{2} d^{2} x^{2} + 12 \, a c^{3} d x - 3 \, b c^{4} \arcsin \left (d x + c\right ) -{\left (2 \, b c^{4} d^{3} x^{3} + 6 \, b c^{5} d^{2} x^{2} + 2 \, b c^{7} + b c^{5} +{\left (6 \, b c^{6} + b c^{4}\right )} d x\right )} \sqrt{-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}{12 \,{\left (c^{4} d^{5} e^{5} x^{4} + 4 \, c^{5} d^{4} e^{5} x^{3} + 6 \, c^{6} d^{3} e^{5} x^{2} + 4 \, c^{7} d^{2} e^{5} x + c^{8} d e^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a*d^4*x^4 + 12*a*c*d^3*x^3 + 18*a*c^2*d^2*x^2 + 12*a*c^3*d*x - 3*b*c^4*arcsin(d*x + c) - (2*b*c^4*d^3*
x^3 + 6*b*c^5*d^2*x^2 + 2*b*c^7 + b*c^5 + (6*b*c^6 + b*c^4)*d*x)*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1))/(c^4*d^5*
e^5*x^4 + 4*c^5*d^4*e^5*x^3 + 6*c^6*d^3*e^5*x^2 + 4*c^7*d^2*e^5*x + c^8*d*e^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a}{c^{5} + 5 c^{4} d x + 10 c^{3} d^{2} x^{2} + 10 c^{2} d^{3} x^{3} + 5 c d^{4} x^{4} + d^{5} x^{5}}\, dx + \int \frac{b \operatorname{asin}{\left (c + d x \right )}}{c^{5} + 5 c^{4} d x + 10 c^{3} d^{2} x^{2} + 10 c^{2} d^{3} x^{3} + 5 c d^{4} x^{4} + d^{5} x^{5}}\, dx}{e^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c**5 + 5*c**4*d*x + 10*c**3*d**2*x**2 + 10*c**2*d**3*x**3 + 5*c*d**4*x**4 + d**5*x**5), x) + Inte
gral(b*asin(c + d*x)/(c**5 + 5*c**4*d*x + 10*c**3*d**2*x**2 + 10*c**2*d**3*x**3 + 5*c*d**4*x**4 + d**5*x**5),
x))/e**5

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Giac [B]  time = 1.68219, size = 576, normalized size = 6.13 \begin{align*} -\frac{3 \, b \arcsin \left (d x + c\right ) e^{\left (-5\right )}}{32 \, d} - \frac{{\left (d x + c\right )}^{4} b \arcsin \left (d x + c\right ) e^{\left (-5\right )}}{64 \, d{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{4}} - \frac{{\left (d x + c\right )}^{2} b \arcsin \left (d x + c\right ) e^{\left (-5\right )}}{16 \, d{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{2}} - \frac{b{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{2} \arcsin \left (d x + c\right ) e^{\left (-5\right )}}{16 \,{\left (d x + c\right )}^{2} d} - \frac{b{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{4} \arcsin \left (d x + c\right ) e^{\left (-5\right )}}{64 \,{\left (d x + c\right )}^{4} d} - \frac{3 \, a e^{\left (-5\right )}}{32 \, d} - \frac{{\left (d x + c\right )}^{4} a e^{\left (-5\right )}}{64 \, d{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{4}} + \frac{{\left (d x + c\right )}^{3} b e^{\left (-5\right )}}{96 \, d{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{3}} - \frac{{\left (d x + c\right )}^{2} a e^{\left (-5\right )}}{16 \, d{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{2}} + \frac{3 \,{\left (d x + c\right )} b e^{\left (-5\right )}}{32 \, d{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )}} - \frac{3 \, b{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )} e^{\left (-5\right )}}{32 \,{\left (d x + c\right )} d} - \frac{a{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{2} e^{\left (-5\right )}}{16 \,{\left (d x + c\right )}^{2} d} - \frac{b{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{3} e^{\left (-5\right )}}{96 \,{\left (d x + c\right )}^{3} d} - \frac{a{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{4} e^{\left (-5\right )}}{64 \,{\left (d x + c\right )}^{4} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/32*b*arcsin(d*x + c)*e^(-5)/d - 1/64*(d*x + c)^4*b*arcsin(d*x + c)*e^(-5)/(d*(sqrt(-(d*x + c)^2 + 1) + 1)^4
) - 1/16*(d*x + c)^2*b*arcsin(d*x + c)*e^(-5)/(d*(sqrt(-(d*x + c)^2 + 1) + 1)^2) - 1/16*b*(sqrt(-(d*x + c)^2 +
1) + 1)^2*arcsin(d*x + c)*e^(-5)/((d*x + c)^2*d) - 1/64*b*(sqrt(-(d*x + c)^2 + 1) + 1)^4*arcsin(d*x + c)*e^(-
5)/((d*x + c)^4*d) - 3/32*a*e^(-5)/d - 1/64*(d*x + c)^4*a*e^(-5)/(d*(sqrt(-(d*x + c)^2 + 1) + 1)^4) + 1/96*(d*
x + c)^3*b*e^(-5)/(d*(sqrt(-(d*x + c)^2 + 1) + 1)^3) - 1/16*(d*x + c)^2*a*e^(-5)/(d*(sqrt(-(d*x + c)^2 + 1) +
1)^2) + 3/32*(d*x + c)*b*e^(-5)/(d*(sqrt(-(d*x + c)^2 + 1) + 1)) - 3/32*b*(sqrt(-(d*x + c)^2 + 1) + 1)*e^(-5)/
((d*x + c)*d) - 1/16*a*(sqrt(-(d*x + c)^2 + 1) + 1)^2*e^(-5)/((d*x + c)^2*d) - 1/96*b*(sqrt(-(d*x + c)^2 + 1)
+ 1)^3*e^(-5)/((d*x + c)^3*d) - 1/64*a*(sqrt(-(d*x + c)^2 + 1) + 1)^4*e^(-5)/((d*x + c)^4*d)