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

Optimal. Leaf size=191 \[ -\frac{a+b \sin ^{-1}(c x)}{3 e (d+e x)^3}+\frac{b c^3 d \sqrt{1-c^2 x^2}}{2 \left (c^2 d^2-e^2\right )^2 (d+e x)}+\frac{b c \sqrt{1-c^2 x^2}}{6 \left (c^2 d^2-e^2\right ) (d+e x)^2}+\frac{b c^3 \left (2 c^2 d^2+e^2\right ) \tan ^{-1}\left (\frac{c^2 d x+e}{\sqrt{1-c^2 x^2} \sqrt{c^2 d^2-e^2}}\right )}{6 e \left (c^2 d^2-e^2\right )^{5/2}} \]

[Out]

(b*c*Sqrt[1 - c^2*x^2])/(6*(c^2*d^2 - e^2)*(d + e*x)^2) + (b*c^3*d*Sqrt[1 - c^2*x^2])/(2*(c^2*d^2 - e^2)^2*(d
+ e*x)) - (a + b*ArcSin[c*x])/(3*e*(d + e*x)^3) + (b*c^3*(2*c^2*d^2 + e^2)*ArcTan[(e + c^2*d*x)/(Sqrt[c^2*d^2
- e^2]*Sqrt[1 - c^2*x^2])])/(6*e*(c^2*d^2 - e^2)^(5/2))

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Rubi [A]  time = 0.140214, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {4743, 745, 807, 725, 204} \[ -\frac{a+b \sin ^{-1}(c x)}{3 e (d+e x)^3}+\frac{b c^3 d \sqrt{1-c^2 x^2}}{2 \left (c^2 d^2-e^2\right )^2 (d+e x)}+\frac{b c \sqrt{1-c^2 x^2}}{6 \left (c^2 d^2-e^2\right ) (d+e x)^2}+\frac{b c^3 \left (2 c^2 d^2+e^2\right ) \tan ^{-1}\left (\frac{c^2 d x+e}{\sqrt{1-c^2 x^2} \sqrt{c^2 d^2-e^2}}\right )}{6 e \left (c^2 d^2-e^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*Sqrt[1 - c^2*x^2])/(6*(c^2*d^2 - e^2)*(d + e*x)^2) + (b*c^3*d*Sqrt[1 - c^2*x^2])/(2*(c^2*d^2 - e^2)^2*(d
+ e*x)) - (a + b*ArcSin[c*x])/(3*e*(d + e*x)^3) + (b*c^3*(2*c^2*d^2 + e^2)*ArcTan[(e + c^2*d*x)/(Sqrt[c^2*d^2
- e^2]*Sqrt[1 - c^2*x^2])])/(6*e*(c^2*d^2 - e^2)^(5/2))

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 745

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/((m + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^(m + 1)*Simp[d*(m + 1)
- e*(m + 2*p + 3)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[
m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ[p]) || ILtQ
[Simplify[m + 2*p + 3], 0])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[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)^4} \, dx &=-\frac{a+b \sin ^{-1}(c x)}{3 e (d+e x)^3}+\frac{(b c) \int \frac{1}{(d+e x)^3 \sqrt{1-c^2 x^2}} \, dx}{3 e}\\ &=\frac{b c \sqrt{1-c^2 x^2}}{6 \left (c^2 d^2-e^2\right ) (d+e x)^2}-\frac{a+b \sin ^{-1}(c x)}{3 e (d+e x)^3}-\frac{\left (b c^3\right ) \int \frac{-2 d+e x}{(d+e x)^2 \sqrt{1-c^2 x^2}} \, dx}{6 e \left (c^2 d^2-e^2\right )}\\ &=\frac{b c \sqrt{1-c^2 x^2}}{6 \left (c^2 d^2-e^2\right ) (d+e x)^2}+\frac{b c^3 d \sqrt{1-c^2 x^2}}{2 \left (c^2 d^2-e^2\right )^2 (d+e x)}-\frac{a+b \sin ^{-1}(c x)}{3 e (d+e x)^3}+\frac{\left (b c^3 \left (2 c^2 d^2+e^2\right )\right ) \int \frac{1}{(d+e x) \sqrt{1-c^2 x^2}} \, dx}{6 e \left (c^2 d^2-e^2\right )^2}\\ &=\frac{b c \sqrt{1-c^2 x^2}}{6 \left (c^2 d^2-e^2\right ) (d+e x)^2}+\frac{b c^3 d \sqrt{1-c^2 x^2}}{2 \left (c^2 d^2-e^2\right )^2 (d+e x)}-\frac{a+b \sin ^{-1}(c x)}{3 e (d+e x)^3}-\frac{\left (b c^3 \left (2 c^2 d^2+e^2\right )\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 )}{6 e \left (c^2 d^2-e^2\right )^2}\\ &=\frac{b c \sqrt{1-c^2 x^2}}{6 \left (c^2 d^2-e^2\right ) (d+e x)^2}+\frac{b c^3 d \sqrt{1-c^2 x^2}}{2 \left (c^2 d^2-e^2\right )^2 (d+e x)}-\frac{a+b \sin ^{-1}(c x)}{3 e (d+e x)^3}+\frac{b c^3 \left (2 c^2 d^2+e^2\right ) \tan ^{-1}\left (\frac{e+c^2 d x}{\sqrt{c^2 d^2-e^2} \sqrt{1-c^2 x^2}}\right )}{6 e \left (c^2 d^2-e^2\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.521047, size = 241, normalized size = 1.26 \[ \frac{1}{6} \left (-\frac{2 a}{e (d+e x)^3}+\frac{b \sqrt{1-c^2 x^2} \left (c^3 d (4 d+3 e x)-c e^2\right )}{\left (e^2-c^2 d^2\right )^2 (d+e x)^2}-\frac{b c^3 \left (2 c^2 d^2+e^2\right ) \log \left (\sqrt{1-c^2 x^2} \sqrt{e^2-c^2 d^2}+c^2 d x+e\right )}{e (e-c d)^2 (c d+e)^2 \sqrt{e^2-c^2 d^2}}+\frac{b c^3 \left (2 c^2 d^2+e^2\right ) \log (d+e x)}{e (e-c d)^2 (c d+e)^2 \sqrt{e^2-c^2 d^2}}-\frac{2 b \sin ^{-1}(c x)}{e (d+e x)^3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.008, size = 560, normalized size = 2.9 \begin{align*} -{\frac{a{c}^{3}}{3\, \left ( ecx+dc \right ) ^{3}e}}-{\frac{{c}^{3}b\arcsin \left ( cx \right ) }{3\, \left ( ecx+dc \right ) ^{3}e}}+{\frac{{c}^{3}b}{6\,{e}^{2} \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 ) ^{-2}}+{\frac{b{c}^{4}d}{2\,e \left ({c}^{2}{d}^{2}-{e}^{2} \right ) ^{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}}}} \left ( cx+{\frac{dc}{e}} \right ) ^{-1}}-{\frac{{c}^{5}b{d}^{2}}{2\,{e}^{2} \left ({c}^{2}{d}^{2}-{e}^{2} \right ) ^{2}}\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}}}}}}}+{\frac{{c}^{3}b}{6\,{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)^4,x)

[Out]

-1/3*c^3*a/(c*e*x+c*d)^3/e-1/3*c^3*b/(c*e*x+c*d)^3/e*arcsin(c*x)+1/6*c^3*b/e^2/(c^2*d^2-e^2)/(c*x+d*c/e)^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)+1/2*c^4*b/e*d/(c^2*d^2-e^2)^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^5*b/e^2*d^2/(c^2*d^2-e^2)^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))+1/6*c^3*b/e^2/(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]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left ({\left (c e^{4} x^{3} + 3 \, c d e^{3} x^{2} + 3 \, c d^{2} e^{2} x + c d^{3} e\right )} \int \frac{e^{\left (\frac{1}{2} \, \log \left (c x + 1\right ) + \frac{1}{2} \, \log \left (-c x + 1\right )\right )}}{c^{4} e^{4} x^{7} + 3 \, c^{4} d e^{3} x^{6} - 3 \, c^{2} d^{2} e^{2} x^{3} - c^{2} d^{3} e x^{2} +{\left (3 \, c^{4} d^{2} e^{2} - c^{2} e^{4}\right )} x^{5} +{\left (c^{4} d^{3} e - 3 \, c^{2} d e^{3}\right )} x^{4} -{\left (c^{2} e^{4} x^{5} + 3 \, c^{2} d e^{3} x^{4} - 3 \, d^{2} e^{2} x - d^{3} e +{\left (3 \, c^{2} d^{2} e^{2} - e^{4}\right )} x^{3} +{\left (c^{2} d^{3} e - 3 \, d e^{3}\right )} x^{2}\right )}{\left (c x + 1\right )}{\left (c x - 1\right )}}\,{d x} + \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )\right )} b}{3 \,{\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}} - \frac{a}{3 \,{\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 12.2468, size = 2233, normalized size = 11.69 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(4*a*c^6*d^6 - 12*a*c^4*d^4*e^2 + 12*a*c^2*d^2*e^4 - 4*a*e^6 + (2*b*c^5*d^5 + b*c^3*d^3*e^2 + (2*b*c^5*
d^2*e^3 + b*c^3*e^5)*x^3 + 3*(2*b*c^5*d^3*e^2 + b*c^3*d*e^4)*x^2 + 3*(2*b*c^5*d^4*e + b*c^3*d^2*e^3)*x)*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)*s
qrt(-c^2*x^2 + 1) + 2*e^2)/(e^2*x^2 + 2*d*e*x + d^2)) + 4*(b*c^6*d^6 - 3*b*c^4*d^4*e^2 + 3*b*c^2*d^2*e^4 - b*e
^6)*arcsin(c*x) - 2*(4*b*c^5*d^5*e - 5*b*c^3*d^3*e^3 + b*c*d*e^5 + 3*(b*c^5*d^3*e^3 - b*c^3*d*e^5)*x^2 + (7*b*
c^5*d^4*e^2 - 8*b*c^3*d^2*e^4 + b*c*e^6)*x)*sqrt(-c^2*x^2 + 1))/(c^6*d^9*e - 3*c^4*d^7*e^3 + 3*c^2*d^5*e^5 - d
^3*e^7 + (c^6*d^6*e^4 - 3*c^4*d^4*e^6 + 3*c^2*d^2*e^8 - e^10)*x^3 + 3*(c^6*d^7*e^3 - 3*c^4*d^5*e^5 + 3*c^2*d^3
*e^7 - d*e^9)*x^2 + 3*(c^6*d^8*e^2 - 3*c^4*d^6*e^4 + 3*c^2*d^4*e^6 - d^2*e^8)*x), -1/6*(2*a*c^6*d^6 - 6*a*c^4*
d^4*e^2 + 6*a*c^2*d^2*e^4 - 2*a*e^6 - (2*b*c^5*d^5 + b*c^3*d^3*e^2 + (2*b*c^5*d^2*e^3 + b*c^3*e^5)*x^3 + 3*(2*
b*c^5*d^3*e^2 + b*c^3*d*e^4)*x^2 + 3*(2*b*c^5*d^4*e + b*c^3*d^2*e^3)*x)*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)) + 2*(b*c^6*d^6 - 3*b*c^4*
d^4*e^2 + 3*b*c^2*d^2*e^4 - b*e^6)*arcsin(c*x) - (4*b*c^5*d^5*e - 5*b*c^3*d^3*e^3 + b*c*d*e^5 + 3*(b*c^5*d^3*e
^3 - b*c^3*d*e^5)*x^2 + (7*b*c^5*d^4*e^2 - 8*b*c^3*d^2*e^4 + b*c*e^6)*x)*sqrt(-c^2*x^2 + 1))/(c^6*d^9*e - 3*c^
4*d^7*e^3 + 3*c^2*d^5*e^5 - d^3*e^7 + (c^6*d^6*e^4 - 3*c^4*d^4*e^6 + 3*c^2*d^2*e^8 - e^10)*x^3 + 3*(c^6*d^7*e^
3 - 3*c^4*d^5*e^5 + 3*c^2*d^3*e^7 - d*e^9)*x^2 + 3*(c^6*d^8*e^2 - 3*c^4*d^6*e^4 + 3*c^2*d^4*e^6 - d^2*e^8)*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 )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asin(c*x))/(d + e*x)**4, 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 )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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