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

Optimal. Leaf size=88 \[ -\frac {a+b \sin ^{-1}(c+d x)}{3 d e^4 (c+d x)^3}-\frac {b \sqrt {1-(c+d x)^2}}{6 d e^4 (c+d x)^2}-\frac {b \tanh ^{-1}\left (\sqrt {1-(c+d x)^2}\right )}{6 d e^4} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4805, 12, 4627, 266, 51, 63, 206} \[ -\frac {a+b \sin ^{-1}(c+d x)}{3 d e^4 (c+d x)^3}-\frac {b \sqrt {1-(c+d x)^2}}{6 d e^4 (c+d x)^2}-\frac {b \tanh ^{-1}\left (\sqrt {1-(c+d x)^2}\right )}{6 d e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Sqrt[1 - (c + d*x)^2])/(6*d*e^4*(c + d*x)^2) - (a + b*ArcSin[c + d*x])/(3*d*e^4*(c + d*x)^3) - (b*ArcTanh[
Sqrt[1 - (c + d*x)^2]])/(6*d*e^4)

Rule 12

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

Rule 51

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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

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Mathematica [A]  time = 0.10, size = 86, normalized size = 0.98 \[ -\frac {2 a+b \sqrt {-c^2-2 c d x-d^2 x^2+1} (c+d x)+2 b \sin ^{-1}(c+d x)+b (c+d x)^3 \tanh ^{-1}\left (\sqrt {1-(c+d x)^2}\right )}{6 d e^4 (c+d x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(2*a + b*(c + d*x)*Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2] + 2*b*ArcSin[c + d*x] + b*(c + d*x)^3*ArcTanh[Sqrt[1
- (c + d*x)^2]])/(d*e^4*(c + d*x)^3)

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fricas [B]  time = 0.58, size = 209, normalized size = 2.38 \[ -\frac {4 \, b \arcsin \left (d x + c\right ) + {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (\sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} + 1\right ) - {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (\sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} - 1\right ) + 2 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} {\left (b d x + b c\right )} + 4 \, a}{12 \, {\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 2.34, size = 376, normalized size = 4.27 \[ -\frac {{\left (d x + c\right )}^{3} b \arcsin \left (d x + c\right ) e^{\left (-4\right )}}{24 \, d {\left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{3}} - \frac {{\left (d x + c\right )} b \arcsin \left (d x + c\right ) e^{\left (-4\right )}}{8 \, d {\left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )}} - \frac {b {\left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )} \arcsin \left (d x + c\right ) e^{\left (-4\right )}}{8 \, {\left (d x + c\right )} d} - \frac {b {\left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{3} \arcsin \left (d x + c\right ) e^{\left (-4\right )}}{24 \, {\left (d x + c\right )}^{3} d} - \frac {b e^{\left (-4\right )} \log \left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )}{6 \, d} + \frac {b e^{\left (-4\right )} \log \left ({\left | d x + c \right |}\right )}{6 \, d} - \frac {{\left (d x + c\right )}^{3} a e^{\left (-4\right )}}{24 \, d {\left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{3}} + \frac {{\left (d x + c\right )}^{2} b e^{\left (-4\right )}}{24 \, d {\left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{2}} - \frac {{\left (d x + c\right )} a e^{\left (-4\right )}}{8 \, d {\left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )}} - \frac {a {\left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )} e^{\left (-4\right )}}{8 \, {\left (d x + c\right )} d} - \frac {b {\left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{2} e^{\left (-4\right )}}{24 \, {\left (d x + c\right )}^{2} d} - \frac {a {\left (\sqrt {-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{3} e^{\left (-4\right )}}{24 \, {\left (d x + c\right )}^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(d*x + c)^3*b*arcsin(d*x + c)*e^(-4)/(d*(sqrt(-(d*x + c)^2 + 1) + 1)^3) - 1/8*(d*x + c)*b*arcsin(d*x + c
)*e^(-4)/(d*(sqrt(-(d*x + c)^2 + 1) + 1)) - 1/8*b*(sqrt(-(d*x + c)^2 + 1) + 1)*arcsin(d*x + c)*e^(-4)/((d*x +
c)*d) - 1/24*b*(sqrt(-(d*x + c)^2 + 1) + 1)^3*arcsin(d*x + c)*e^(-4)/((d*x + c)^3*d) - 1/6*b*e^(-4)*log(sqrt(-
(d*x + c)^2 + 1) + 1)/d + 1/6*b*e^(-4)*log(abs(d*x + c))/d - 1/24*(d*x + c)^3*a*e^(-4)/(d*(sqrt(-(d*x + c)^2 +
1) + 1)^3) + 1/24*(d*x + c)^2*b*e^(-4)/(d*(sqrt(-(d*x + c)^2 + 1) + 1)^2) - 1/8*(d*x + c)*a*e^(-4)/(d*(sqrt(-
(d*x + c)^2 + 1) + 1)) - 1/8*a*(sqrt(-(d*x + c)^2 + 1) + 1)*e^(-4)/((d*x + c)*d) - 1/24*b*(sqrt(-(d*x + c)^2 +
1) + 1)^2*e^(-4)/((d*x + c)^2*d) - 1/24*a*(sqrt(-(d*x + c)^2 + 1) + 1)^3*e^(-4)/((d*x + c)^3*d)

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maple [A]  time = 0.01, size = 78, normalized size = 0.89 \[ \frac {-\frac {a}{3 e^{4} \left (d x +c \right )^{3}}+\frac {b \left (-\frac {\arcsin \left (d x +c \right )}{3 \left (d x +c \right )^{3}}-\frac {\sqrt {1-\left (d x +c \right )^{2}}}{6 \left (d x +c \right )^{2}}-\frac {\arctanh \left (\frac {1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )}{6}\right )}{e^{4}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left ({\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )} \int \frac {e^{\left (\frac {1}{2} \, \log \left (d x + c + 1\right ) + \frac {1}{2} \, \log \left (-d x - c + 1\right )\right )}}{d^{7} e^{4} x^{7} + 7 \, c d^{6} e^{4} x^{6} + {\left (21 \, c^{2} - 1\right )} d^{5} e^{4} x^{5} + 5 \, {\left (7 \, c^{3} - c\right )} d^{4} e^{4} x^{4} + 5 \, {\left (7 \, c^{4} - 2 \, c^{2}\right )} d^{3} e^{4} x^{3} + {\left (21 \, c^{5} - 10 \, c^{3}\right )} d^{2} e^{4} x^{2} + {\left (7 \, c^{6} - 5 \, c^{4}\right )} d e^{4} x + {\left (c^{7} - c^{5}\right )} e^{4} - {\left (d^{5} e^{4} x^{5} + 5 \, c d^{4} e^{4} x^{4} + {\left (10 \, c^{2} - 1\right )} d^{3} e^{4} x^{3} + {\left (10 \, c^{3} - 3 \, c\right )} d^{2} e^{4} x^{2} + {\left (5 \, c^{4} - 3 \, c^{2}\right )} d e^{4} x + {\left (c^{5} - c^{3}\right )} e^{4}\right )} {\left (d x + c + 1\right )} {\left (d x + c - 1\right )}}\,{d x} + \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )\right )} b}{3 \, {\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )}} - \frac {a}{3 \, {\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4)*integrate(1/3*e^(1/2*log(d*x + c + 1) +
1/2*log(-d*x - c + 1))/(d^7*e^4*x^7 + 7*c*d^6*e^4*x^6 + (21*c^2 - 1)*d^5*e^4*x^5 + 5*(7*c^3 - c)*d^4*e^4*x^4 +
5*(7*c^4 - 2*c^2)*d^3*e^4*x^3 + (21*c^5 - 10*c^3)*d^2*e^4*x^2 + (7*c^6 - 5*c^4)*d*e^4*x + (c^7 - c^5)*e^4 + (
d^5*e^4*x^5 + 5*c*d^4*e^4*x^4 + (10*c^2 - 1)*d^3*e^4*x^3 + (10*c^3 - 3*c)*d^2*e^4*x^2 + (5*c^4 - 3*c^2)*d*e^4*
x + (c^5 - c^3)*e^4)*e^(log(d*x + c + 1) + log(-d*x - c + 1))), x) + arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-
d*x - c + 1)))*b/(d^4*e^4*x^3 + 3*c*d^3*e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4) - 1/3*a/(d^4*e^4*x^3 + 3*c*d^3*
e^4*x^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asin}\left (c+d\,x\right )}{{\left (c\,e+d\,e\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {b \operatorname {asin}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(b*asin(c + d*x)/
(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x))/e**4

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