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

Optimal. Leaf size=202 \[ -\frac {(f+g x)^2 \left (a+b \sin ^{-1}(c x)\right )}{2 (d+e x)^2 (e f-d g)}+\frac {b c \sqrt {1-c^2 x^2} (e f-d g)}{2 e \left (c^2 d^2-e^2\right ) (d+e x)}-\frac {b c \left (2 e^2 g-c^2 d (d g+e f)\right ) \tan ^{-1}\left (\frac {c^2 d x+e}{\sqrt {1-c^2 x^2} \sqrt {c^2 d^2-e^2}}\right )}{2 e^2 \left (c^2 d^2-e^2\right )^{3/2}}+\frac {b g^2 \sin ^{-1}(c x)}{2 e^2 (e f-d g)} \]

[Out]

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

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Rubi [A]  time = 0.36, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {37, 4753, 12, 1651, 844, 216, 725, 204} \[ -\frac {(f+g x)^2 \left (a+b \sin ^{-1}(c x)\right )}{2 (d+e x)^2 (e f-d g)}+\frac {b c \sqrt {1-c^2 x^2} (e f-d g)}{2 e \left (c^2 d^2-e^2\right ) (d+e x)}-\frac {b c \left (2 e^2 g-c^2 d (d g+e f)\right ) \tan ^{-1}\left (\frac {c^2 d x+e}{\sqrt {1-c^2 x^2} \sqrt {c^2 d^2-e^2}}\right )}{2 e^2 \left (c^2 d^2-e^2\right )^{3/2}}+\frac {b g^2 \sin ^{-1}(c x)}{2 e^2 (e f-d g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

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 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 0]

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

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Mathematica [A]  time = 0.54, size = 263, normalized size = 1.30 \[ \frac {\frac {a (d g-e f)}{(d+e x)^2}-\frac {2 a g}{d+e x}-\frac {b c e \sqrt {1-c^2 x^2} (e f-d g)}{\left (e^2-c^2 d^2\right ) (d+e x)}+\frac {b c \left (c^2 d (d g+e f)-2 e^2 g\right ) \log \left (\sqrt {1-c^2 x^2} \sqrt {e^2-c^2 d^2}+c^2 d x+e\right )}{(e-c d) (c d+e) \sqrt {e^2-c^2 d^2}}+\frac {b c \log (d+e x) \left (c^2 d (d g+e f)-2 e^2 g\right )}{(c d-e) (c d+e) \sqrt {e^2-c^2 d^2}}-\frac {b \sin ^{-1}(c x) (d g+e (f+2 g x))}{(d+e x)^2}}{2 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 13.68, size = 1184, normalized size = 5.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(4*(a*c^4*d^4*e - 2*a*c^2*d^2*e^3 + a*e^5)*g*x + (b*c^3*d^3*e*f + (b*c^3*d*e^3*f + (b*c^3*d^2*e^2 - 2*b*
c*e^4)*g)*x^2 + (b*c^3*d^4 - 2*b*c*d^2*e^2)*g + 2*(b*c^3*d^2*e^2*f + (b*c^3*d^3*e - 2*b*c*d*e^3)*g)*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)*sq
rt(-c^2*x^2 + 1) + 2*e^2)/(e^2*x^2 + 2*d*e*x + d^2)) + 2*(a*c^4*d^4*e - 2*a*c^2*d^2*e^3 + a*e^5)*f + 2*(a*c^4*
d^5 - 2*a*c^2*d^3*e^2 + a*d*e^4)*g + 2*(2*(b*c^4*d^4*e - 2*b*c^2*d^2*e^3 + b*e^5)*g*x + (b*c^4*d^4*e - 2*b*c^2
*d^2*e^3 + b*e^5)*f + (b*c^4*d^5 - 2*b*c^2*d^3*e^2 + b*d*e^4)*g)*arcsin(c*x) - 2*sqrt(-c^2*x^2 + 1)*((b*c^3*d^
3*e^2 - b*c*d*e^4)*f - (b*c^3*d^4*e - b*c*d^2*e^3)*g + ((b*c^3*d^2*e^3 - b*c*e^5)*f - (b*c^3*d^3*e^2 - b*c*d*e
^4)*g)*x))/(c^4*d^6*e^2 - 2*c^2*d^4*e^4 + d^2*e^6 + (c^4*d^4*e^4 - 2*c^2*d^2*e^6 + e^8)*x^2 + 2*(c^4*d^5*e^3 -
2*c^2*d^3*e^5 + d*e^7)*x), -1/2*(2*(a*c^4*d^4*e - 2*a*c^2*d^2*e^3 + a*e^5)*g*x - (b*c^3*d^3*e*f + (b*c^3*d*e^
3*f + (b*c^3*d^2*e^2 - 2*b*c*e^4)*g)*x^2 + (b*c^3*d^4 - 2*b*c*d^2*e^2)*g + 2*(b*c^3*d^2*e^2*f + (b*c^3*d^3*e -
2*b*c*d*e^3)*g)*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)) + (a*c^4*d^4*e - 2*a*c^2*d^2*e^3 + a*e^5)*f + (a*c^4*d^5 - 2*a*c^2*d^3*e^2 +
a*d*e^4)*g + (2*(b*c^4*d^4*e - 2*b*c^2*d^2*e^3 + b*e^5)*g*x + (b*c^4*d^4*e - 2*b*c^2*d^2*e^3 + b*e^5)*f + (b*c
^4*d^5 - 2*b*c^2*d^3*e^2 + b*d*e^4)*g)*arcsin(c*x) - sqrt(-c^2*x^2 + 1)*((b*c^3*d^3*e^2 - b*c*d*e^4)*f - (b*c^
3*d^4*e - b*c*d^2*e^3)*g + ((b*c^3*d^2*e^3 - b*c*e^5)*f - (b*c^3*d^3*e^2 - b*c*d*e^4)*g)*x))/(c^4*d^6*e^2 - 2*
c^2*d^4*e^4 + d^2*e^6 + (c^4*d^4*e^4 - 2*c^2*d^2*e^6 + e^8)*x^2 + 2*(c^4*d^5*e^3 - 2*c^2*d^3*e^5 + d*e^7)*x)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (g x + f\right )} {\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (e x + d\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((g*x + f)*(b*arcsin(c*x) + a)/(e*x + d)^3, x)

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maple [B]  time = 0.02, size = 805, normalized size = 3.99 \[ -\frac {c a g}{e^{2} \left (c e x +d c \right )}+\frac {c^{2} a d g}{2 e^{2} \left (c e x +d c \right )^{2}}-\frac {c^{2} a f}{2 e \left (c e x +d c \right )^{2}}-\frac {c b \arcsin \left (c x \right ) g}{e^{2} \left (c e x +d c \right )}+\frac {c^{2} b \arcsin \left (c x \right ) d g}{2 e^{2} \left (c e x +d c \right )^{2}}-\frac {c^{2} b \arcsin \left (c x \right ) f}{2 e \left (c e x +d c \right )^{2}}-\frac {c^{2} b \sqrt {-\left (c x +\frac {d c}{e}\right )^{2}+\frac {2 d c \left (c x +\frac {d c}{e}\right )}{e}-\frac {c^{2} d^{2}-e^{2}}{e^{2}}}\, d g}{2 e^{2} \left (c^{2} d^{2}-e^{2}\right ) \left (c x +\frac {d c}{e}\right )}+\frac {c^{2} b \sqrt {-\left (c x +\frac {d c}{e}\right )^{2}+\frac {2 d c \left (c x +\frac {d c}{e}\right )}{e}-\frac {c^{2} d^{2}-e^{2}}{e^{2}}}\, f}{2 e \left (c^{2} d^{2}-e^{2}\right ) \left (c x +\frac {d c}{e}\right )}+\frac {c^{3} b \,d^{2} \ln \left (\frac {-\frac {2 \left (c^{2} d^{2}-e^{2}\right )}{e^{2}}+\frac {2 d c \left (c x +\frac {d c}{e}\right )}{e}+2 \sqrt {-\frac {c^{2} d^{2}-e^{2}}{e^{2}}}\, \sqrt {-\left (c x +\frac {d c}{e}\right )^{2}+\frac {2 d c \left (c x +\frac {d c}{e}\right )}{e}-\frac {c^{2} d^{2}-e^{2}}{e^{2}}}}{c x +\frac {d c}{e}}\right ) g}{2 e^{3} \left (c^{2} d^{2}-e^{2}\right ) \sqrt {-\frac {c^{2} d^{2}-e^{2}}{e^{2}}}}-\frac {c^{3} b d \ln \left (\frac {-\frac {2 \left (c^{2} d^{2}-e^{2}\right )}{e^{2}}+\frac {2 d c \left (c x +\frac {d c}{e}\right )}{e}+2 \sqrt {-\frac {c^{2} d^{2}-e^{2}}{e^{2}}}\, \sqrt {-\left (c x +\frac {d c}{e}\right )^{2}+\frac {2 d c \left (c x +\frac {d c}{e}\right )}{e}-\frac {c^{2} d^{2}-e^{2}}{e^{2}}}}{c x +\frac {d c}{e}}\right ) f}{2 e^{2} \left (c^{2} d^{2}-e^{2}\right ) \sqrt {-\frac {c^{2} d^{2}-e^{2}}{e^{2}}}}-\frac {c b g \ln \left (\frac {-\frac {2 \left (c^{2} d^{2}-e^{2}\right )}{e^{2}}+\frac {2 d c \left (c x +\frac {d c}{e}\right )}{e}+2 \sqrt {-\frac {c^{2} d^{2}-e^{2}}{e^{2}}}\, \sqrt {-\left (c x +\frac {d c}{e}\right )^{2}+\frac {2 d c \left (c x +\frac {d c}{e}\right )}{e}-\frac {c^{2} d^{2}-e^{2}}{e^{2}}}}{c x +\frac {d c}{e}}\right )}{e^{3} \sqrt {-\frac {c^{2} d^{2}-e^{2}}{e^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*a*g/e^2/(c*e*x+c*d)+1/2*c^2*a/e^2/(c*e*x+c*d)^2*d*g-1/2*c^2*a/e/(c*e*x+c*d)^2*f-c*b*arcsin(c*x)*g/e^2/(c*e*
x+c*d)+1/2*c^2*b*arcsin(c*x)/e^2/(c*e*x+c*d)^2*d*g-1/2*c^2*b*arcsin(c*x)/e/(c*e*x+c*d)^2*f-1/2*c^2*b/e^2/(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*g+1/2*c^2*b/e/(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)*f+1/2*c^3*b/e^3*d^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))*g-1/2*c^3*b/e^2*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))*f-c*b/e^3*g/(-(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(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e-c*d>0)', see `assume?` for m
ore details)Is e-c*d positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (f+g\,x\right )\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d+e\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)*(a + b*asin(c*x)))/(d + e*x)^3,x)

[Out]

int(((f + g*x)*(a + b*asin(c*x)))/(d + e*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right ) \left (f + g x\right )}{\left (d + e x\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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