3.183 \(\int \frac {a+b \sin ^{-1}(c+d x)}{(c e+d e x)^2} \, dx\)

Optimal. Leaf size=51 \[ -\frac {a+b \sin ^{-1}(c+d x)}{d e^2 (c+d x)}-\frac {b \tanh ^{-1}\left (\sqrt {1-(c+d x)^2}\right )}{d e^2} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4805, 12, 4627, 266, 63, 206} \[ -\frac {a+b \sin ^{-1}(c+d x)}{d e^2 (c+d x)}-\frac {b \tanh ^{-1}\left (\sqrt {1-(c+d x)^2}\right )}{d e^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])/(c*e + d*e*x)^2,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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)^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a+b \sin ^{-1}(x)}{e^2 x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {a+b \sin ^{-1}(x)}{x^2} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac {a+b \sin ^{-1}(c+d x)}{d e^2 (c+d x)}+\frac {b \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-x^2}} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac {a+b \sin ^{-1}(c+d x)}{d e^2 (c+d x)}+\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} x} \, dx,x,(c+d x)^2\right )}{2 d e^2}\\ &=-\frac {a+b \sin ^{-1}(c+d x)}{d e^2 (c+d x)}-\frac {b \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1-(c+d x)^2}\right )}{d e^2}\\ &=-\frac {a+b \sin ^{-1}(c+d x)}{d e^2 (c+d x)}-\frac {b \tanh ^{-1}\left (\sqrt {1-(c+d x)^2}\right )}{d e^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 45, normalized size = 0.88 \[ -\frac {\frac {a+b \sin ^{-1}(c+d x)}{c+d x}+b \tanh ^{-1}\left (\sqrt {1-(c+d x)^2}\right )}{d e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c + d*x])/(c*e + d*e*x)^2,x]

[Out]

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

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fricas [B]  time = 0.58, size = 101, normalized size = 1.98 \[ -\frac {2 \, b \arcsin \left (d x + c\right ) + {\left (b d x + b c\right )} \log \left (\sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} + 1\right ) - {\left (b d x + b c\right )} \log \left (\sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} - 1\right ) + 2 \, a}{2 \, {\left (d^{2} e^{2} x + c d e^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))/(d*e*x+c*e)^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.51, size = 110, normalized size = 2.16 \[ -\frac {1}{2} \, b d {\left (\frac {{\left (\log \left (\sqrt {-{\left (d x e + c e\right )}^{2} e^{\left (-2\right )} + 1} + 1\right ) - \log \left (-\sqrt {-{\left (d x e + c e\right )}^{2} e^{\left (-2\right )} + 1} + 1\right )\right )} e^{\left (-4\right )}}{d^{2}} + \frac {2 \, \arcsin \left (d x + c\right ) e^{\left (-3\right )}}{{\left (d x e + c e\right )} d^{2}}\right )} e^{2} - \frac {a e^{\left (-1\right )}}{{\left (d x e + c e\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))/(d*e*x+c*e)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 56, normalized size = 1.10 \[ \frac {-\frac {a}{e^{2} \left (d x +c \right )}+\frac {b \left (-\frac {\arcsin \left (d x +c \right )}{d x +c}-\arctanh \left (\frac {1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )\right )}{e^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(d*x+c))/(d*e*x+c*e)^2,x)

[Out]

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

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maxima [B]  time = 0.43, size = 120, normalized size = 2.35 \[ -b {\left (\frac {\arcsin \left (d x + c\right )}{d^{2} e^{2} x + c d e^{2}} + \frac {\log \left (\frac {2 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}{{\left | d^{2} e^{2} x + c d e^{2} \right |}} + \frac {2}{{\left | d^{2} e^{2} x + c d e^{2} \right |}}\right )}{d e^{2}}\right )} - \frac {a}{d^{2} e^{2} x + c d e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))/(d*e*x+c*e)^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {a+b\,\mathrm {asin}\left (c+d\,x\right )}{{\left (c\,e+d\,e\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c + d*x))/(c*e + d*e*x)^2,x)

[Out]

int((a + b*asin(c + d*x))/(c*e + d*e*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(d*x+c))/(d*e*x+c*e)**2,x)

[Out]

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

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