3.7 \(\int \frac{a+b \sin ^{-1}(c x)}{(d+e x)^3} \, dx\)

Optimal. Leaf size=135 \[ -\frac{a+b \sin ^{-1}(c x)}{2 e (d+e x)^2}+\frac{b c \sqrt{1-c^2 x^2}}{2 \left (c^2 d^2-e^2\right ) (d+e x)}+\frac{b c^3 d \tan ^{-1}\left (\frac{c^2 d x+e}{\sqrt{1-c^2 x^2} \sqrt{c^2 d^2-e^2}}\right )}{2 e \left (c^2 d^2-e^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0852537, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4743, 731, 725, 204} \[ -\frac{a+b \sin ^{-1}(c x)}{2 e (d+e x)^2}+\frac{b c \sqrt{1-c^2 x^2}}{2 \left (c^2 d^2-e^2\right ) (d+e x)}+\frac{b c^3 d \tan ^{-1}\left (\frac{c^2 d x+e}{\sqrt{1-c^2 x^2} \sqrt{c^2 d^2-e^2}}\right )}{2 e \left (c^2 d^2-e^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])/(d + e*x)^3,x]

[Out]

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

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]

Rule 731

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)*(a + c*x^2)^(p
+ 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d)/(c*d^2 + a*e^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x]
/; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 3, 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 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])

Rubi steps

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

Mathematica [C]  time = 0.352993, size = 207, normalized size = 1.53 \[ \frac{1}{2} \left (-\frac{a}{e (d+e x)^2}+\frac{b c \sqrt{1-c^2 x^2}}{\left (c^2 d^2-e^2\right ) (d+e x)}-\frac{i b c^3 d \left (\log (4)+\log \left (\frac{e^2 \sqrt{c^2 d^2-e^2} \left (\sqrt{1-c^2 x^2} \sqrt{c^2 d^2-e^2}+i c^2 d x+i e\right )}{b c^3 d (d+e x)}\right )\right )}{e (c d-e) (c d+e) \sqrt{c^2 d^2-e^2}}-\frac{b \sin ^{-1}(c x)}{e (d+e x)^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c*x])/(d + e*x)^3,x]

[Out]

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

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Maple [B]  time = 0.033, size = 301, normalized size = 2.2 \begin{align*} -{\frac{{c}^{2}a}{2\, \left ( ecx+dc \right ) ^{2}e}}-{\frac{{c}^{2}b\arcsin \left ( cx \right ) }{2\, \left ( ecx+dc \right ) ^{2}e}}+{\frac{{c}^{2}b}{2\,e \left ({c}^{2}{d}^{2}-{e}^{2} \right ) }\sqrt{- \left ( cx+{\frac{dc}{e}} \right ) ^{2}+2\,{\frac{dc}{e} \left ( cx+{\frac{dc}{e}} \right ) }-{\frac{{c}^{2}{d}^{2}-{e}^{2}}{{e}^{2}}}} \left ( cx+{\frac{dc}{e}} \right ) ^{-1}}-{\frac{b{c}^{3}d}{2\,{e}^{2} \left ({c}^{2}{d}^{2}-{e}^{2} \right ) }\ln \left ({ \left ( -2\,{\frac{{c}^{2}{d}^{2}-{e}^{2}}{{e}^{2}}}+2\,{\frac{dc}{e} \left ( cx+{\frac{dc}{e}} \right ) }+2\,\sqrt{-{\frac{{c}^{2}{d}^{2}-{e}^{2}}{{e}^{2}}}}\sqrt{- \left ( cx+{\frac{dc}{e}} \right ) ^{2}+2\,{\frac{dc}{e} \left ( cx+{\frac{dc}{e}} \right ) }-{\frac{{c}^{2}{d}^{2}-{e}^{2}}{{e}^{2}}}} \right ) \left ( cx+{\frac{dc}{e}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{\frac{{c}^{2}{d}^{2}-{e}^{2}}{{e}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x))/(e*x+d)^3,x)

[Out]

-1/2*c^2*a/(c*e*x+c*d)^2/e-1/2*c^2*b/(c*e*x+c*d)^2/e*arcsin(c*x)+1/2*c^2*b/e/(c^2*d^2-e^2)/(c*x+d*c/e)*(-(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)-1/2*c^3*b/e^2*d/(c^2*d^2-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., 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((a+b*arcsin(c*x))/(e*x+d)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.97458, size = 1349, normalized size = 9.99 \begin{align*} \left [-\frac{2 \, a c^{4} d^{4} - 4 \, a c^{2} d^{2} e^{2} + 2 \, a e^{4} -{\left (b c^{3} d e^{2} x^{2} + 2 \, b c^{3} d^{2} e x + b c^{3} d^{3}\right )} \sqrt{-c^{2} d^{2} + e^{2}} \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^{4} d^{4} - 2 \, b c^{2} d^{2} e^{2} + b e^{4}\right )} \arcsin \left (c x\right ) - 2 \,{\left (b c^{3} d^{3} e - b c d e^{3} +{\left (b c^{3} d^{2} e^{2} - b c e^{4}\right )} x\right )} \sqrt{-c^{2} x^{2} + 1}}{4 \,{\left (c^{4} d^{6} e - 2 \, c^{2} d^{4} e^{3} + d^{2} e^{5} +{\left (c^{4} d^{4} e^{3} - 2 \, c^{2} d^{2} e^{5} + e^{7}\right )} x^{2} + 2 \,{\left (c^{4} d^{5} e^{2} - 2 \, c^{2} d^{3} e^{4} + d e^{6}\right )} x\right )}}, -\frac{a c^{4} d^{4} - 2 \, a c^{2} d^{2} e^{2} + a e^{4} -{\left (b c^{3} d e^{2} x^{2} + 2 \, b c^{3} d^{2} e x + b c^{3} d^{3}\right )} \sqrt{c^{2} d^{2} - e^{2}} \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^{4} d^{4} - 2 \, b c^{2} d^{2} e^{2} + b e^{4}\right )} \arcsin \left (c x\right ) -{\left (b c^{3} d^{3} e - b c d e^{3} +{\left (b c^{3} d^{2} e^{2} - b c e^{4}\right )} x\right )} \sqrt{-c^{2} x^{2} + 1}}{2 \,{\left (c^{4} d^{6} e - 2 \, c^{2} d^{4} e^{3} + d^{2} e^{5} +{\left (c^{4} d^{4} e^{3} - 2 \, c^{2} d^{2} e^{5} + e^{7}\right )} x^{2} + 2 \,{\left (c^{4} d^{5} e^{2} - 2 \, c^{2} d^{3} e^{4} + d e^{6}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/(e*x+d)^3,x, algorithm="fricas")

[Out]

[-1/4*(2*a*c^4*d^4 - 4*a*c^2*d^2*e^2 + 2*a*e^4 - (b*c^3*d*e^2*x^2 + 2*b*c^3*d^2*e*x + b*c^3*d^3)*sqrt(-c^2*d^2
+ e^2)*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^4*d^4 - 2*b*c^2*d^2*e^2 + b*e^4)*arcsin(c*x) - 2*(b*c^
3*d^3*e - b*c*d*e^3 + (b*c^3*d^2*e^2 - b*c*e^4)*x)*sqrt(-c^2*x^2 + 1))/(c^4*d^6*e - 2*c^2*d^4*e^3 + d^2*e^5 +
(c^4*d^4*e^3 - 2*c^2*d^2*e^5 + e^7)*x^2 + 2*(c^4*d^5*e^2 - 2*c^2*d^3*e^4 + d*e^6)*x), -1/2*(a*c^4*d^4 - 2*a*c^
2*d^2*e^2 + a*e^4 - (b*c^3*d*e^2*x^2 + 2*b*c^3*d^2*e*x + b*c^3*d^3)*sqrt(c^2*d^2 - e^2)*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^4*d^4 - 2*b*c^2*d^2*e^
2 + b*e^4)*arcsin(c*x) - (b*c^3*d^3*e - b*c*d*e^3 + (b*c^3*d^2*e^2 - b*c*e^4)*x)*sqrt(-c^2*x^2 + 1))/(c^4*d^6*
e - 2*c^2*d^4*e^3 + d^2*e^5 + (c^4*d^4*e^3 - 2*c^2*d^2*e^5 + e^7)*x^2 + 2*(c^4*d^5*e^2 - 2*c^2*d^3*e^4 + d*e^6
)*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x))/(e*x+d)**3,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \arcsin \left (c x\right ) + a}{{\left (e x + d\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/(e*x+d)^3,x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)/(e*x + d)^3, x)