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

Optimal. Leaf size=257 \[ -\frac{(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{3 e^2 (d+e x)^3}-\frac{g \left (a+b \sin ^{-1}(c x)\right )}{2 e^2 (d+e x)^2}+\frac{b c \sqrt{1-c^2 x^2} \left (c^2 d f-e g\right )}{2 \left (c^2 d^2-e^2\right )^2 (d+e x)}+\frac{b c \sqrt{1-c^2 x^2} (e f-d g)}{6 e \left (c^2 d^2-e^2\right ) (d+e x)^2}+\frac{b c^3 \left (c^2 d^2 (d g+2 e f)+e^2 (e f-4 d g)\right ) \tan ^{-1}\left (\frac{c^2 d x+e}{\sqrt{1-c^2 x^2} \sqrt{c^2 d^2-e^2}}\right )}{6 e^2 \left (c^2 d^2-e^2\right )^{5/2}} \]

[Out]

(b*c*(e*f - d*g)*Sqrt[1 - c^2*x^2])/(6*e*(c^2*d^2 - e^2)*(d + e*x)^2) + (b*c*(c^2*d*f - e*g)*Sqrt[1 - c^2*x^2]
)/(2*(c^2*d^2 - e^2)^2*(d + e*x)) - ((e*f - d*g)*(a + b*ArcSin[c*x]))/(3*e^2*(d + e*x)^3) - (g*(a + b*ArcSin[c
*x]))/(2*e^2*(d + e*x)^2) + (b*c^3*(e^2*(e*f - 4*d*g) + c^2*d^2*(2*e*f + d*g))*ArcTan[(e + c^2*d*x)/(Sqrt[c^2*
d^2 - e^2]*Sqrt[1 - c^2*x^2])])/(6*e^2*(c^2*d^2 - e^2)^(5/2))

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Rubi [A]  time = 0.427468, antiderivative size = 257, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {43, 4753, 12, 835, 807, 725, 204} \[ -\frac{(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{3 e^2 (d+e x)^3}-\frac{g \left (a+b \sin ^{-1}(c x)\right )}{2 e^2 (d+e x)^2}+\frac{b c \sqrt{1-c^2 x^2} \left (c^2 d f-e g\right )}{2 \left (c^2 d^2-e^2\right )^2 (d+e x)}+\frac{b c \sqrt{1-c^2 x^2} (e f-d g)}{6 e \left (c^2 d^2-e^2\right ) (d+e x)^2}+\frac{b c^3 \left (c^2 d^2 (d g+2 e f)+e^2 (e f-4 d g)\right ) \tan ^{-1}\left (\frac{c^2 d x+e}{\sqrt{1-c^2 x^2} \sqrt{c^2 d^2-e^2}}\right )}{6 e^2 \left (c^2 d^2-e^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*(e*f - d*g)*Sqrt[1 - c^2*x^2])/(6*e*(c^2*d^2 - e^2)*(d + e*x)^2) + (b*c*(c^2*d*f - e*g)*Sqrt[1 - c^2*x^2]
)/(2*(c^2*d^2 - e^2)^2*(d + e*x)) - ((e*f - d*g)*(a + b*ArcSin[c*x]))/(3*e^2*(d + e*x)^3) - (g*(a + b*ArcSin[c
*x]))/(2*e^2*(d + e*x)^2) + (b*c^3*(e^2*(e*f - 4*d*g) + c^2*d^2*(2*e*f + d*g))*ArcTan[(e + c^2*d*x)/(Sqrt[c^2*
d^2 - e^2]*Sqrt[1 - c^2*x^2])])/(6*e^2*(c^2*d^2 - e^2)^(5/2))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 4753

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(Px_)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> With[{u = IntHide[Px*(d
+ e*x)^m, x]}, Dist[a + b*ArcSin[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 - c^2*x^2], x], x], x]
] /; FreeQ[{a, b, c, d, e, m}, x] && PolynomialQ[Px, 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 835

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))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

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{(f+g x) \left (a+b \sin ^{-1}(c x)\right )}{(d+e x)^4} \, dx &=-\frac{(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{3 e^2 (d+e x)^3}-\frac{g \left (a+b \sin ^{-1}(c x)\right )}{2 e^2 (d+e x)^2}-(b c) \int \frac{-2 e f-d g-3 e g x}{6 e^2 (d+e x)^3 \sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{3 e^2 (d+e x)^3}-\frac{g \left (a+b \sin ^{-1}(c x)\right )}{2 e^2 (d+e x)^2}-\frac{(b c) \int \frac{-2 e f-d g-3 e g x}{(d+e x)^3 \sqrt{1-c^2 x^2}} \, dx}{6 e^2}\\ &=\frac{b c (e f-d g) \sqrt{1-c^2 x^2}}{6 e \left (c^2 d^2-e^2\right ) (d+e x)^2}-\frac{(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{3 e^2 (d+e x)^3}-\frac{g \left (a+b \sin ^{-1}(c x)\right )}{2 e^2 (d+e x)^2}-\frac{(b c) \int \frac{2 \left (3 e^2 g-c^2 d (2 e f+d g)\right )+2 c^2 e (e f-d g) x}{(d+e x)^2 \sqrt{1-c^2 x^2}} \, dx}{12 e^2 \left (c^2 d^2-e^2\right )}\\ &=\frac{b c (e f-d g) \sqrt{1-c^2 x^2}}{6 e \left (c^2 d^2-e^2\right ) (d+e x)^2}+\frac{b c \left (c^2 d f-e g\right ) \sqrt{1-c^2 x^2}}{2 \left (c^2 d^2-e^2\right )^2 (d+e x)}-\frac{(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{3 e^2 (d+e x)^3}-\frac{g \left (a+b \sin ^{-1}(c x)\right )}{2 e^2 (d+e x)^2}+\frac{\left (b c^3 \left (e^2 (e f-4 d g)+c^2 d^2 (2 e f+d g)\right )\right ) \int \frac{1}{(d+e x) \sqrt{1-c^2 x^2}} \, dx}{6 e^2 \left (c^2 d^2-e^2\right )^2}\\ &=\frac{b c (e f-d g) \sqrt{1-c^2 x^2}}{6 e \left (c^2 d^2-e^2\right ) (d+e x)^2}+\frac{b c \left (c^2 d f-e g\right ) \sqrt{1-c^2 x^2}}{2 \left (c^2 d^2-e^2\right )^2 (d+e x)}-\frac{(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{3 e^2 (d+e x)^3}-\frac{g \left (a+b \sin ^{-1}(c x)\right )}{2 e^2 (d+e x)^2}-\frac{\left (b c^3 \left (e^2 (e f-4 d g)+c^2 d^2 (2 e f+d g)\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^2 \left (c^2 d^2-e^2\right )^2}\\ &=\frac{b c (e f-d g) \sqrt{1-c^2 x^2}}{6 e \left (c^2 d^2-e^2\right ) (d+e x)^2}+\frac{b c \left (c^2 d f-e g\right ) \sqrt{1-c^2 x^2}}{2 \left (c^2 d^2-e^2\right )^2 (d+e x)}-\frac{(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{3 e^2 (d+e x)^3}-\frac{g \left (a+b \sin ^{-1}(c x)\right )}{2 e^2 (d+e x)^2}+\frac{b c^3 \left (e^2 (e f-4 d g)+c^2 d^2 (2 e f+d g)\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^2 \left (c^2 d^2-e^2\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.773146, size = 321, normalized size = 1.25 \[ \frac{\frac{a (2 d g-2 e f)}{(d+e x)^3}-\frac{3 a g}{(d+e x)^2}+\frac{b c e \sqrt{1-c^2 x^2} \left (c^2 d \left (d^2 (-g)+4 d e f+3 e^2 f x\right )-e^2 (2 d g+e (f+3 g x))\right )}{\left (e^2-c^2 d^2\right )^2 (d+e x)^2}-\frac{b c^3 \left (c^2 d^2 (d g+2 e f)+e^2 (e f-4 d g)\right ) \log \left (\sqrt{1-c^2 x^2} \sqrt{e^2-c^2 d^2}+c^2 d x+e\right )}{(e-c d)^2 (c d+e)^2 \sqrt{e^2-c^2 d^2}}+\frac{b c^3 \log (d+e x) \left (c^2 d^2 (d g+2 e f)+e^2 (e f-4 d g)\right )}{(e-c d)^2 (c d+e)^2 \sqrt{e^2-c^2 d^2}}-\frac{b \sin ^{-1}(c x) (d g+2 e f+3 e g x)}{(d+e x)^3}}{6 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.013, size = 1269, normalized size = 4.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)*(a+b*arcsin(c*x))/(e*x+d)^4,x)

[Out]

-1/2*c^2*a*g/e^2/(c*e*x+c*d)^2+1/3*c^3*a/e^2/(c*e*x+c*d)^3*d*g-1/3*c^3*a/e/(c*e*x+c*d)^3*f-1/2*c^2*b*arcsin(c*
x)*g/e^2/(c*e*x+c*d)^2+1/3*c^3*b*arcsin(c*x)/e^2/(c*e*x+c*d)^3*d*g-1/3*c^3*b*arcsin(c*x)/e/(c*e*x+c*d)^3*f-1/6
*c^3*b/e^3/(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)*d*g+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)*f-1/2*c^4*b/e
^2*d^2/(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)*g+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)*f+1/2*c^5*b/e^3*d^3
/(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))*g-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))*f-2/3*c^3*b/e^3/(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))*d*g+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))*f+1/2*c^2*b/e^2*g/(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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{asin}{\left (c x \right )}\right ) \left (f + g x\right )}{\left (d + e x\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asin(c*x))*(f + g*x)/(d + e*x)**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (g x + f\right )}{\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (e x + d\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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