3.195 \(\int \frac{(a+b \sin ^{-1}(c+d x))^2}{(c e+d e x)^3} \, dx\)

Optimal. Leaf size=87 \[ -\frac{b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{d e^3 (c+d x)}-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac{b^2 \log (c+d x)}{d e^3} \]

[Out]

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

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Rubi [A]  time = 0.135123, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4805, 12, 4627, 4681, 29} \[ -\frac{b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{d e^3 (c+d x)}-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^3 (c+d x)^2}+\frac{b^2 \log (c+d x)}{d e^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])^2/(c*e + d*e*x)^3,x]

[Out]

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

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 4681

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n)/(d*f*(m + 1)), x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x
^2)^FracPart[p])/(f*(m + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSi
n[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p
+ 3, 0] && NeQ[m, -1]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rubi steps

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

Mathematica [A]  time = 0.233506, size = 126, normalized size = 1.45 \[ -\frac{a \left (a+2 b (c+d x) \sqrt{-c^2-2 c d x-d^2 x^2+1}\right )+2 b \sin ^{-1}(c+d x) \left (a+b (c+d x) \sqrt{-c^2-2 c d x-d^2 x^2+1}\right )-2 b^2 (c+d x)^2 \log (c+d x)+b^2 \sin ^{-1}(c+d x)^2}{2 d e^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c + d*x])^2/(c*e + d*e*x)^3,x]

[Out]

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

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Maple [A]  time = 0.036, size = 152, normalized size = 1.8 \begin{align*} -{\frac{{a}^{2}}{2\,d{e}^{3} \left ( dx+c \right ) ^{2}}}-{\frac{{b}^{2} \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}}{2\,d{e}^{3} \left ( dx+c \right ) ^{2}}}-{\frac{{b}^{2}\arcsin \left ( dx+c \right ) }{d{e}^{3} \left ( dx+c \right ) }\sqrt{1- \left ( dx+c \right ) ^{2}}}+{\frac{{b}^{2}\ln \left ( dx+c \right ) }{d{e}^{3}}}-{\frac{ab\arcsin \left ( dx+c \right ) }{d{e}^{3} \left ( dx+c \right ) ^{2}}}-{\frac{ab}{d{e}^{3} \left ( dx+c \right ) }\sqrt{1- \left ( dx+c \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^3,x)

[Out]

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

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Maxima [B]  time = 1.53729, size = 319, normalized size = 3.67 \begin{align*} -{\left (\frac{\sqrt{-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} d \arcsin \left (d x + c\right )}{d^{3} e^{3} x + c d^{2} e^{3}} - \frac{\log \left (d x + c\right )}{d e^{3}}\right )} b^{2} - a b{\left (\frac{\sqrt{-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} d}{d^{3} e^{3} x + c d^{2} e^{3}} + \frac{\arcsin \left (d x + c\right )}{d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}}\right )} - \frac{b^{2} \arcsin \left (d x + c\right )^{2}}{2 \,{\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} - \frac{a^{2}}{2 \,{\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 3.01178, size = 338, normalized size = 3.89 \begin{align*} -\frac{b^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\right ) + a^{2} - 2 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (d x + c\right ) + 2 \,{\left (a b d x + a b c +{\left (b^{2} d x + b^{2} c\right )} \arcsin \left (d x + c\right )\right )} \sqrt{-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}{2 \,{\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^3,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(d*x+c))**2/(d*e*x+c*e)**3,x)

[Out]

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

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Giac [B]  time = 1.39003, size = 666, normalized size = 7.66 \begin{align*} -\frac{b^{2} \arcsin \left (d x + c\right )^{2} e^{\left (-3\right )}}{4 \, d} - \frac{{\left (d x + c\right )}^{2} b^{2} \arcsin \left (d x + c\right )^{2} e^{\left (-3\right )}}{8 \, d{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{2}} - \frac{b^{2}{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{2} \arcsin \left (d x + c\right )^{2} e^{\left (-3\right )}}{8 \,{\left (d x + c\right )}^{2} d} - \frac{a b \arcsin \left (d x + c\right ) e^{\left (-3\right )}}{2 \, d} - \frac{{\left (d x + c\right )}^{2} a b \arcsin \left (d x + c\right ) e^{\left (-3\right )}}{4 \, d{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{2}} + \frac{{\left (d x + c\right )} b^{2} \arcsin \left (d x + c\right ) e^{\left (-3\right )}}{2 \, d{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )}} - \frac{b^{2}{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )} \arcsin \left (d x + c\right ) e^{\left (-3\right )}}{2 \,{\left (d x + c\right )} d} - \frac{a b{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{2} \arcsin \left (d x + c\right ) e^{\left (-3\right )}}{4 \,{\left (d x + c\right )}^{2} d} + \frac{2 \, b^{2} e^{\left (-3\right )} \log \left (2\right )}{d} - \frac{b^{2} e^{\left (-3\right )} \log \left (2 \, \sqrt{-{\left (d x + c\right )}^{2} + 1} + 2\right )}{d} + \frac{b^{2} e^{\left (-3\right )} \log \left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )}{d} + \frac{b^{2} e^{\left (-3\right )} \log \left ({\left | d x + c \right |}\right )}{d} - \frac{a^{2} e^{\left (-3\right )}}{4 \, d} - \frac{{\left (d x + c\right )}^{2} a^{2} e^{\left (-3\right )}}{8 \, d{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{2}} + \frac{{\left (d x + c\right )} a b e^{\left (-3\right )}}{2 \, d{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )}} - \frac{a b{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )} e^{\left (-3\right )}}{2 \,{\left (d x + c\right )} d} - \frac{a^{2}{\left (\sqrt{-{\left (d x + c\right )}^{2} + 1} + 1\right )}^{2} e^{\left (-3\right )}}{8 \,{\left (d x + c\right )}^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))^2/(d*e*x+c*e)^3,x, algorithm="giac")

[Out]

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