3.103 \(\int \frac{(f+g x+h x^2) (a+b \sin ^{-1}(c x))}{(d+e x)^4} \, dx\)

Optimal. Leaf size=349 \[ -\frac{\left (a+b \sin ^{-1}(c x)\right ) \left (d^2 h-d e g+e^2 f\right )}{3 e^3 (d+e x)^3}-\frac{(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right )}{2 e^3 (d+e x)^2}-\frac{h \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}+\frac{b c \sqrt{1-c^2 x^2} \left (d^2 h-d e g+e^2 f\right )}{6 e^2 \left (c^2 d^2-e^2\right ) (d+e x)^2}-\frac{b c \sqrt{1-c^2 x^2} \left (e^2 (e g-2 d h)-c^2 \left (d e^2 f-d^3 h\right )\right )}{2 e^2 \left (c^2 d^2-e^2\right )^2 (d+e x)}+\frac{b c \tan ^{-1}\left (\frac{c^2 d x+e}{\sqrt{1-c^2 x^2} \sqrt{c^2 d^2-e^2}}\right ) \left (c^4 d^2 \left (2 d^2 h+d e g+2 e^2 f\right )+c^2 e^2 \left (-5 d^2 h-4 d e g+e^2 f\right )+6 e^4 h\right )}{6 e^3 \left (c^2 d^2-e^2\right )^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.617012, antiderivative size = 349, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {698, 4753, 12, 1651, 807, 725, 204} \[ -\frac{\left (a+b \sin ^{-1}(c x)\right ) \left (d^2 h-d e g+e^2 f\right )}{3 e^3 (d+e x)^3}-\frac{(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right )}{2 e^3 (d+e x)^2}-\frac{h \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}+\frac{b c \sqrt{1-c^2 x^2} \left (d^2 h-d e g+e^2 f\right )}{6 e^2 \left (c^2 d^2-e^2\right ) (d+e x)^2}-\frac{b c \sqrt{1-c^2 x^2} \left (e^2 (e g-2 d h)-c^2 \left (d e^2 f-d^3 h\right )\right )}{2 e^2 \left (c^2 d^2-e^2\right )^2 (d+e x)}+\frac{b c \tan ^{-1}\left (\frac{c^2 d x+e}{\sqrt{1-c^2 x^2} \sqrt{c^2 d^2-e^2}}\right ) \left (c^4 d^2 \left (2 d^2 h+d e g+2 e^2 f\right )+c^2 e^2 \left (-5 d^2 h-4 d e g+e^2 f\right )+6 e^4 h\right )}{6 e^3 \left (c^2 d^2-e^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[((f + g*x + h*x^2)*(a + b*ArcSin[c*x]))/(d + e*x)^4,x]

[Out]

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

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

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 1651

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, d
+ e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[(e*R*(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*ExpandToSum[(m
+ 1)*(c*d^2 + a*e^2)*Q + c*d*R*(m + 1) - c*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, c, d, e, p}, x] && Po
lyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]

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

Mathematica [A]  time = 1.82451, size = 442, normalized size = 1.27 \[ -\frac{\frac{2 a \left (d^2 h-d e g+e^2 f\right )}{(d+e x)^3}+\frac{3 a (e g-2 d h)}{(d+e x)^2}+\frac{6 a h}{d+e x}+\frac{b c e \sqrt{1-c^2 x^2} \left (c^2 d \left (d^2 e (g+3 h x)+2 d^3 h-4 d e^2 f-3 e^3 f x\right )+e^2 \left (-5 d^2 h+2 d e (g-3 h x)+e^2 (f+3 g x)\right )\right )}{\left (e^2-c^2 d^2\right )^2 (d+e x)^2}+\frac{b c \log \left (\sqrt{1-c^2 x^2} \sqrt{e^2-c^2 d^2}+c^2 d x+e\right ) \left (c^4 d^2 \left (2 d^2 h+d e g+2 e^2 f\right )+c^2 e^2 \left (-5 d^2 h-4 d e g+e^2 f\right )+6 e^4 h\right )}{(e-c d)^2 (c d+e)^2 \sqrt{e^2-c^2 d^2}}-\frac{b c \log (d+e x) \left (c^4 d^2 \left (2 d^2 h+d e g+2 e^2 f\right )+c^2 e^2 \left (-5 d^2 h-4 d e g+e^2 f\right )+6 e^4 h\right )}{(e-c d)^2 (c d+e)^2 \sqrt{e^2-c^2 d^2}}+\frac{b \sin ^{-1}(c x) \left (2 d^2 h+d e (g+6 h x)+e^2 (2 f+3 x (g+2 h x))\right )}{(d+e x)^3}}{6 e^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x + h*x^2)*(a + b*ArcSin[c*x]))/(d + e*x)^4,x]

[Out]

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

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Maple [B]  time = 0.015, size = 2173, normalized size = 6.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((h*x^2+g*x+f)*(a+b*arcsin(c*x))/(e*x+d)^4,x)

[Out]

-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/
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-c*b*arcs
in(c*x)*h/e^3/(c*e*x+c*d)-c*b/e^4*h/(-(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))+c^2*a/e^
3/(c*e*x+c*d)^2*d*h-1/3*c^3*a/e^3/(c*e*x+c*d)^3*d^2*h+1/3*c^3*a/e^2/(c*e*x+c*d)^3*d*g-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/(c*e*x+c*d)^3*f-c*a*h/e^3/(c*e*x+c*d)-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+7/6*c^3*b/e^4/(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^2*h-1/2*c^5*b/e^4*d^4/(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))*h-c^2*b/e^3/(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)*d*h+1/3*c^3*b*arcsin(c*x)/e^2/(c*e*x+c*d)^3*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+1/6*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)+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+1/6*c^3*b/e^4/(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^2*h+1/2*c^4*b/e^3*d^3/(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)*h-1/2*c^2*a*g/e^2/(c*e*x+c*d)^2-1/3*c^3*a/e/(c*e*x+c*d)^3*f-1/3*c^3*b*a
rcsin(c*x)/e^3/(c*e*x+c*d)^3*d^2*h+c^2*b*arcsin(c*x)/e^3/(c*e*x+c*d)^2*d*h

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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{{\left (3 \, e^{2} x^{2} + 3 \, d e x + d^{2}\right )} a h}{3 \,{\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\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 (6 \, b e^{2} h x^{2} + 2 \, b e^{2} f + b d e g + 2 \, b d^{2} h + 3 \,{\left (b e^{2} g + 2 \, b d e h\right )} x\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) +{\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )} \int \frac{{\left (6 \, b c e^{2} h x^{2} + 2 \, b c e^{2} f + b c d e g + 2 \, b c d^{2} h + 3 \,{\left (b c e^{2} g + 2 \, b c d e h\right )} x\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^{6} x^{7} + 3 \, c^{4} d e^{5} x^{6} - 3 \, c^{2} d^{2} e^{4} x^{3} - c^{2} d^{3} e^{3} x^{2} +{\left (3 \, c^{4} d^{2} e^{4} - c^{2} e^{6}\right )} x^{5} +{\left (c^{4} d^{3} e^{3} - 3 \, c^{2} d e^{5}\right )} x^{4} -{\left (c^{2} e^{6} x^{5} + 3 \, c^{2} d e^{5} x^{4} - 3 \, d^{2} e^{4} x - d^{3} e^{3} +{\left (3 \, c^{2} d^{2} e^{4} - e^{6}\right )} x^{3} +{\left (c^{2} d^{3} e^{3} - 3 \, d e^{5}\right )} x^{2}\right )}{\left (c x + 1\right )}{\left (c x - 1\right )}}\,{d x}}{6 \,{\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + d^{3} e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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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((h*x^2+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 + h x^{2}\right )}{\left (d + e x\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x**2+g*x+f)*(a+b*asin(c*x))/(e*x+d)**4,x)

[Out]

Integral((a + b*asin(c*x))*(f + g*x + h*x**2)/(d + e*x)**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (h x^{2} + 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((h*x^2+g*x+f)*(a+b*arcsin(c*x))/(e*x+d)^4,x, algorithm="giac")

[Out]

integrate((h*x^2 + g*x + f)*(b*arcsin(c*x) + a)/(e*x + d)^4, x)