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]

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

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Rubi [A]  time = 0.43, antiderivative size = 257, normalized size of antiderivative = 1.00, 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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])

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 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 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 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]

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*}

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Mathematica [A]  time = 0.77, 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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fricas [B]  time = 47.73, size = 1920, normalized size = 7.47 \[ \text {result too large to display} \]

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]

[-1/12*(6*(a*c^6*d^6*e - 3*a*c^4*d^4*e^3 + 3*a*c^2*d^2*e^5 - a*e^7)*g*x - sqrt(-c^2*d^2 + e^2)*(((2*b*c^5*d^2*
e^4 + b*c^3*e^6)*f + (b*c^5*d^3*e^3 - 4*b*c^3*d*e^5)*g)*x^3 + 3*((2*b*c^5*d^3*e^3 + b*c^3*d*e^5)*f + (b*c^5*d^
4*e^2 - 4*b*c^3*d^2*e^4)*g)*x^2 + (2*b*c^5*d^5*e + b*c^3*d^3*e^3)*f + (b*c^5*d^6 - 4*b*c^3*d^4*e^2)*g + 3*((2*
b*c^5*d^4*e^2 + b*c^3*d^2*e^4)*f + (b*c^5*d^5*e - 4*b*c^3*d^3*e^3)*g)*x)*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)
) + 4*(a*c^6*d^6*e - 3*a*c^4*d^4*e^3 + 3*a*c^2*d^2*e^5 - a*e^7)*f + 2*(a*c^6*d^7 - 3*a*c^4*d^5*e^2 + 3*a*c^2*d
^3*e^4 - a*d*e^6)*g + 2*(3*(b*c^6*d^6*e - 3*b*c^4*d^4*e^3 + 3*b*c^2*d^2*e^5 - b*e^7)*g*x + 2*(b*c^6*d^6*e - 3*
b*c^4*d^4*e^3 + 3*b*c^2*d^2*e^5 - b*e^7)*f + (b*c^6*d^7 - 3*b*c^4*d^5*e^2 + 3*b*c^2*d^3*e^4 - b*d*e^6)*g)*arcs
in(c*x) - 2*sqrt(-c^2*x^2 + 1)*(3*((b*c^5*d^3*e^4 - b*c^3*d*e^6)*f - (b*c^3*d^2*e^5 - b*c*e^7)*g)*x^2 + (4*b*c
^5*d^5*e^2 - 5*b*c^3*d^3*e^4 + b*c*d*e^6)*f - (b*c^5*d^6*e + b*c^3*d^4*e^3 - 2*b*c*d^2*e^5)*g + ((7*b*c^5*d^4*
e^3 - 8*b*c^3*d^2*e^5 + b*c*e^7)*f - (b*c^5*d^5*e^2 + 4*b*c^3*d^3*e^4 - 5*b*c*d*e^6)*g)*x))/(c^6*d^9*e^2 - 3*c
^4*d^7*e^4 + 3*c^2*d^5*e^6 - d^3*e^8 + (c^6*d^6*e^5 - 3*c^4*d^4*e^7 + 3*c^2*d^2*e^9 - e^11)*x^3 + 3*(c^6*d^7*e
^4 - 3*c^4*d^5*e^6 + 3*c^2*d^3*e^8 - d*e^10)*x^2 + 3*(c^6*d^8*e^3 - 3*c^4*d^6*e^5 + 3*c^2*d^4*e^7 - d^2*e^9)*x
), -1/6*(3*(a*c^6*d^6*e - 3*a*c^4*d^4*e^3 + 3*a*c^2*d^2*e^5 - a*e^7)*g*x - sqrt(c^2*d^2 - e^2)*(((2*b*c^5*d^2*
e^4 + b*c^3*e^6)*f + (b*c^5*d^3*e^3 - 4*b*c^3*d*e^5)*g)*x^3 + 3*((2*b*c^5*d^3*e^3 + b*c^3*d*e^5)*f + (b*c^5*d^
4*e^2 - 4*b*c^3*d^2*e^4)*g)*x^2 + (2*b*c^5*d^5*e + b*c^3*d^3*e^3)*f + (b*c^5*d^6 - 4*b*c^3*d^4*e^2)*g + 3*((2*
b*c^5*d^4*e^2 + b*c^3*d^2*e^4)*f + (b*c^5*d^5*e - 4*b*c^3*d^3*e^3)*g)*x)*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*(a*c^6*d^6*e - 3*a*c^4*d^4*e^3 + 3*a*c^2
*d^2*e^5 - a*e^7)*f + (a*c^6*d^7 - 3*a*c^4*d^5*e^2 + 3*a*c^2*d^3*e^4 - a*d*e^6)*g + (3*(b*c^6*d^6*e - 3*b*c^4*
d^4*e^3 + 3*b*c^2*d^2*e^5 - b*e^7)*g*x + 2*(b*c^6*d^6*e - 3*b*c^4*d^4*e^3 + 3*b*c^2*d^2*e^5 - b*e^7)*f + (b*c^
6*d^7 - 3*b*c^4*d^5*e^2 + 3*b*c^2*d^3*e^4 - b*d*e^6)*g)*arcsin(c*x) - sqrt(-c^2*x^2 + 1)*(3*((b*c^5*d^3*e^4 -
b*c^3*d*e^6)*f - (b*c^3*d^2*e^5 - b*c*e^7)*g)*x^2 + (4*b*c^5*d^5*e^2 - 5*b*c^3*d^3*e^4 + b*c*d*e^6)*f - (b*c^5
*d^6*e + b*c^3*d^4*e^3 - 2*b*c*d^2*e^5)*g + ((7*b*c^5*d^4*e^3 - 8*b*c^3*d^2*e^5 + b*c*e^7)*f - (b*c^5*d^5*e^2
+ 4*b*c^3*d^3*e^4 - 5*b*c*d*e^6)*g)*x))/(c^6*d^9*e^2 - 3*c^4*d^7*e^4 + 3*c^2*d^5*e^6 - d^3*e^8 + (c^6*d^6*e^5
- 3*c^4*d^4*e^7 + 3*c^2*d^2*e^9 - e^11)*x^3 + 3*(c^6*d^7*e^4 - 3*c^4*d^5*e^6 + 3*c^2*d^3*e^8 - d*e^10)*x^2 + 3
*(c^6*d^8*e^3 - 3*c^4*d^6*e^5 + 3*c^2*d^4*e^7 - d^2*e^9)*x)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (g x + f\right )} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (e x + d\right )}^{4}}\,{d x} \]

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)

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maple [B]  time = 0.02, size = 1269, normalized size = 4.94 \[ \text {result too large to display} \]

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+1/2
*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)-2/3*c^3*b/
e^3*g*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))-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+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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (3 \, e x + d\right )} a g}{6 \, {\left (e^{5} x^{3} + 3 \, d e^{4} x^{2} + 3 \, d^{2} e^{3} x + d^{3} e^{2}\right )}} - \frac {a f}{3 \, {\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}} - \frac {{\left (3 \, b e g x + 2 \, b e f + b d g\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + {\left (e^{5} x^{3} + 3 \, d e^{4} x^{2} + 3 \, d^{2} e^{3} x + d^{3} e^{2}\right )} \int \frac {{\left (3 \, b c e g x + 2 \, b c e f + b c d g\right )} e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (-c x + 1\right )\right )}}{c^{4} e^{5} x^{7} + 3 \, c^{4} d e^{4} x^{6} - 3 \, c^{2} d^{2} e^{3} x^{3} - c^{2} d^{3} e^{2} x^{2} + {\left (3 \, c^{4} d^{2} e^{3} - c^{2} e^{5}\right )} x^{5} + {\left (c^{4} d^{3} e^{2} - 3 \, c^{2} d e^{4}\right )} x^{4} - {\left (c^{2} e^{5} x^{5} + 3 \, c^{2} d e^{4} x^{4} - 3 \, d^{2} e^{3} x - d^{3} e^{2} + {\left (3 \, c^{2} d^{2} e^{3} - e^{5}\right )} x^{3} + {\left (c^{2} d^{3} e^{2} - 3 \, d e^{4}\right )} x^{2}\right )} {\left (c x + 1\right )} {\left (c x - 1\right )}}\,{d x}}{6 \, {\left (e^{5} x^{3} + 3 \, d e^{4} x^{2} + 3 \, d^{2} e^{3} x + d^{3} e^{2}\right )}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (f+g\,x\right )\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d+e\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right ) \left (f + g x\right )}{\left (d + e x\right )^{4}}\, dx \]

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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