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

Optimal. Leaf size=85 \[ \frac {b c \tan ^{-1}\left (\frac {c^2 d x+e}{\sqrt {1-c^2 x^2} \sqrt {c^2 d^2-e^2}}\right )}{e \sqrt {c^2 d^2-e^2}}-\frac {a+b \sin ^{-1}(c x)}{e (d+e x)} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {4743, 725, 204} \[ \frac {b c \tan ^{-1}\left (\frac {c^2 d x+e}{\sqrt {1-c^2 x^2} \sqrt {c^2 d^2-e^2}}\right )}{e \sqrt {c^2 d^2-e^2}}-\frac {a+b \sin ^{-1}(c x)}{e (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 204

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

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 4743

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

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 83, normalized size = 0.98 \[ \frac {\frac {b c \tan ^{-1}\left (\frac {c^2 d x+e}{\sqrt {1-c^2 x^2} \sqrt {c^2 d^2-e^2}}\right )}{\sqrt {c^2 d^2-e^2}}-\frac {a+b \sin ^{-1}(c x)}{d+e x}}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.68, size = 371, normalized size = 4.36 \[ \left [-\frac {2 \, a c^{2} d^{2} - 2 \, a e^{2} + \sqrt {-c^{2} d^{2} + e^{2}} {\left (b c e x + b c d\right )} \log \left (\frac {2 \, c^{2} d e x - c^{2} d^{2} + {\left (2 \, c^{4} d^{2} - c^{2} e^{2}\right )} x^{2} - 2 \, \sqrt {-c^{2} d^{2} + e^{2}} {\left (c^{2} d x + e\right )} \sqrt {-c^{2} x^{2} + 1} + 2 \, e^{2}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, {\left (b c^{2} d^{2} - b e^{2}\right )} \arcsin \left (c x\right )}{2 \, {\left (c^{2} d^{3} e - d e^{3} + {\left (c^{2} d^{2} e^{2} - e^{4}\right )} x\right )}}, -\frac {a c^{2} d^{2} - a e^{2} - \sqrt {c^{2} d^{2} - e^{2}} {\left (b c e x + b c d\right )} \arctan \left (\frac {\sqrt {c^{2} d^{2} - e^{2}} {\left (c^{2} d x + e\right )} \sqrt {-c^{2} x^{2} + 1}}{c^{2} d^{2} - {\left (c^{4} d^{2} - c^{2} e^{2}\right )} x^{2} - e^{2}}\right ) + {\left (b c^{2} d^{2} - b e^{2}\right )} \arcsin \left (c x\right )}{c^{2} d^{3} e - d e^{3} + {\left (c^{2} d^{2} e^{2} - e^{4}\right )} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.39, size = 204, normalized size = 2.40 \[ -\frac {{\left (\frac {2 \, c^{2} \arctan \left (\frac {\frac {c d {\left (\sqrt {-{\left (x e + d\right )}^{2} {\left (c - \frac {c d}{x e + d}\right )}^{2} e^{\left (-2\right )} + 1} - 1\right )} e}{{\left (x e + d\right )} {\left (c - \frac {c d}{x e + d}\right )}} - e}{\sqrt {c^{2} d^{2} - e^{2}}}\right ) e^{\left (-3\right )}}{\sqrt {c^{2} d^{2} - e^{2}}} + \frac {c^{2} \arcsin \left ({\left ({\left (\frac {{\left (x e + d\right )} {\left (c - \frac {c d}{x e + d}\right )} e}{c} + d e\right )} e^{\left (-1\right )} - d\right )} c e^{\left (-1\right )}\right ) e^{\left (-3\right )}}{{\left (x e + d\right )} {\left (c - \frac {c d}{x e + d}\right )} + c d}\right )} b e^{2}}{c} - \frac {a e^{\left (-1\right )}}{x e + d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 191, normalized size = 2.25 \[ -\frac {c a}{\left (c e x +d c \right ) e}-\frac {c b \arcsin \left (c x \right )}{\left (c e x +d c \right ) e}-\frac {c b \ln \left (\frac {-\frac {2 \left (c^{2} d^{2}-e^{2}\right )}{e^{2}}+\frac {2 d c \left (c x +\frac {d c}{e}\right )}{e}+2 \sqrt {-\frac {c^{2} d^{2}-e^{2}}{e^{2}}}\, \sqrt {-\left (c x +\frac {d c}{e}\right )^{2}+\frac {2 d c \left (c x +\frac {d c}{e}\right )}{e}-\frac {c^{2} d^{2}-e^{2}}{e^{2}}}}{c x +\frac {d c}{e}}\right )}{e^{2} \sqrt {-\frac {c^{2} d^{2}-e^{2}}{e^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e-c*d>0)', see `assume?` for m
ore details)Is e-c*d positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{{\left (d+e\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))/(d + e*x)^2,x)

[Out]

int((a + b*asin(c*x))/(d + e*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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