3.9.25 \(\int \frac {a+b x+c x^2}{(d+e x)^3 \sqrt {f+g x}} \, dx\) [825]

Optimal. Leaf size=206 \[ -\frac {\left (a+\frac {d (c d-b e)}{e^2}\right ) \sqrt {f+g x}}{2 (e f-d g) (d+e x)^2}+\frac {(c d (8 e f-5 d g)-e (4 b e f-b d g-3 a e g)) \sqrt {f+g x}}{4 e^2 (e f-d g)^2 (d+e x)}+\frac {\left (e g (4 b e f-b d g-3 a e g)-c \left (8 e^2 f^2-8 d e f g+3 d^2 g^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{4 e^{5/2} (e f-d g)^{5/2}} \]

[Out]

1/4*(e*g*(-3*a*e*g-b*d*g+4*b*e*f)-c*(3*d^2*g^2-8*d*e*f*g+8*e^2*f^2))*arctanh(e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^
(1/2))/e^(5/2)/(-d*g+e*f)^(5/2)-1/2*(a+d*(-b*e+c*d)/e^2)*(g*x+f)^(1/2)/(-d*g+e*f)/(e*x+d)^2+1/4*(c*d*(-5*d*g+8
*e*f)-e*(-3*a*e*g-b*d*g+4*b*e*f))*(g*x+f)^(1/2)/e^2/(-d*g+e*f)^2/(e*x+d)

________________________________________________________________________________________

Rubi [A]
time = 0.24, antiderivative size = 210, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {911, 1171, 393, 214} \begin {gather*} -\frac {\sqrt {f+g x} \left (a e^2-b d e+c d^2\right )}{2 e^2 (d+e x)^2 (e f-d g)}+\frac {\tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (e g (-3 a e g-b d g+4 b e f)-c \left (3 d^2 g^2-8 d e f g+8 e^2 f^2\right )\right )}{4 e^{5/2} (e f-d g)^{5/2}}+\frac {\sqrt {f+g x} (c d (8 e f-5 d g)-e (-3 a e g-b d g+4 b e f))}{4 e^2 (d+e x) (e f-d g)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)/((d + e*x)^3*Sqrt[f + g*x]),x]

[Out]

-1/2*((c*d^2 - b*d*e + a*e^2)*Sqrt[f + g*x])/(e^2*(e*f - d*g)*(d + e*x)^2) + ((c*d*(8*e*f - 5*d*g) - e*(4*b*e*
f - b*d*g - 3*a*e*g))*Sqrt[f + g*x])/(4*e^2*(e*f - d*g)^2*(d + e*x)) + ((e*g*(4*b*e*f - b*d*g - 3*a*e*g) - c*(
8*e^2*f^2 - 8*d*e*f*g + 3*d^2*g^2))*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/(4*e^(5/2)*(e*f - d*g)^(
5/2))

Rule 214

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

Rule 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 911

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + g*(x^q/e))^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - (2*c*d - b*e)*(x^q/e^2) + c*(x^(2*q)/e^2))^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1171

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1
)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rubi steps

\begin {align*} \int \frac {a+b x+c x^2}{(d+e x)^3 \sqrt {f+g x}} \, dx &=\frac {2 \text {Subst}\left (\int \frac {\frac {c f^2-b f g+a g^2}{g^2}-\frac {(2 c f-b g) x^2}{g^2}+\frac {c x^4}{g^2}}{\left (\frac {-e f+d g}{g}+\frac {e x^2}{g}\right )^3} \, dx,x,\sqrt {f+g x}\right )}{g}\\ &=-\frac {\left (a+\frac {d (c d-b e)}{e^2}\right ) \sqrt {f+g x}}{2 (e f-d g) (d+e x)^2}+\frac {\text {Subst}\left (\int \frac {-3 a+\frac {c d^2}{e^2}-\frac {b d}{e}-\frac {4 c f^2}{g^2}+\frac {4 b f}{g}+\frac {4 c (e f-d g) x^2}{e g^2}}{\left (\frac {-e f+d g}{g}+\frac {e x^2}{g}\right )^2} \, dx,x,\sqrt {f+g x}\right )}{2 (e f-d g)}\\ &=-\frac {\left (a+\frac {d (c d-b e)}{e^2}\right ) \sqrt {f+g x}}{2 (e f-d g) (d+e x)^2}+\frac {(c d (8 e f-5 d g)-e (4 b e f-b d g-3 a e g)) \sqrt {f+g x}}{4 e^2 (e f-d g)^2 (d+e x)}-\frac {\left (e g (4 b e f-b d g-3 a e g)-c \left (8 e^2 f^2-8 d e f g+3 d^2 g^2\right )\right ) \text {Subst}\left (\int \frac {1}{\frac {-e f+d g}{g}+\frac {e x^2}{g}} \, dx,x,\sqrt {f+g x}\right )}{4 e^2 g (e f-d g)^2}\\ &=-\frac {\left (a+\frac {d (c d-b e)}{e^2}\right ) \sqrt {f+g x}}{2 (e f-d g) (d+e x)^2}+\frac {(c d (8 e f-5 d g)-e (4 b e f-b d g-3 a e g)) \sqrt {f+g x}}{4 e^2 (e f-d g)^2 (d+e x)}+\frac {\left (e g (4 b e f-b d g-3 a e g)-c \left (8 e^2 f^2-8 d e f g+3 d^2 g^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{4 e^{5/2} (e f-d g)^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.83, size = 203, normalized size = 0.99 \begin {gather*} \frac {\frac {\sqrt {e} \sqrt {f+g x} \left (c d \left (-3 d^2 g+8 e^2 f x+d e (6 f-5 g x)\right )+e \left (a e (-2 e f+5 d g+3 e g x)-b \left (2 d e f+d^2 g+4 e^2 f x-d e g x\right )\right )\right )}{(e f-d g)^2 (d+e x)^2}+\frac {\left (e g (-4 b e f+b d g+3 a e g)+c \left (8 e^2 f^2-8 d e f g+3 d^2 g^2\right )\right ) \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {-e f+d g}}\right )}{(-e f+d g)^{5/2}}}{4 e^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)/((d + e*x)^3*Sqrt[f + g*x]),x]

[Out]

((Sqrt[e]*Sqrt[f + g*x]*(c*d*(-3*d^2*g + 8*e^2*f*x + d*e*(6*f - 5*g*x)) + e*(a*e*(-2*e*f + 5*d*g + 3*e*g*x) -
b*(2*d*e*f + d^2*g + 4*e^2*f*x - d*e*g*x))))/((e*f - d*g)^2*(d + e*x)^2) + ((e*g*(-4*b*e*f + b*d*g + 3*a*e*g)
+ c*(8*e^2*f^2 - 8*d*e*f*g + 3*d^2*g^2))*ArcTan[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[-(e*f) + d*g]])/(-(e*f) + d*g)^(5
/2))/(4*e^(5/2))

________________________________________________________________________________________

Maple [A]
time = 0.11, size = 260, normalized size = 1.26

method	result	size
derivativedivides	\(\frac {\frac {g \left (3 a \,e^{2} g +b d e g -4 b \,e^{2} f -5 c \,d^{2} g +8 c d e f \right ) \left (g x +f \right )^{\frac {3}{2}}}{4 e \left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right )}+\frac {\left (5 a \,e^{2} g -b d e g -4 b \,e^{2} f -3 c \,d^{2} g +8 c d e f \right ) g \sqrt {g x +f}}{4 e^{2} \left (d g -e f \right )}}{\left (e \left (g x +f \right )+d g -e f \right )^{2}}+\frac {\left (3 a \,e^{2} g^{2}+b d e \,g^{2}-4 b \,e^{2} f g +3 c \,d^{2} g^{2}-8 c d e f g +8 c \,e^{2} f^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{4 \left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) e^{2} \sqrt {\left (d g -e f \right ) e}}\)	\(260\)
default	\(\frac {\frac {g \left (3 a \,e^{2} g +b d e g -4 b \,e^{2} f -5 c \,d^{2} g +8 c d e f \right ) \left (g x +f \right )^{\frac {3}{2}}}{4 e \left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right )}+\frac {\left (5 a \,e^{2} g -b d e g -4 b \,e^{2} f -3 c \,d^{2} g +8 c d e f \right ) g \sqrt {g x +f}}{4 e^{2} \left (d g -e f \right )}}{\left (e \left (g x +f \right )+d g -e f \right )^{2}}+\frac {\left (3 a \,e^{2} g^{2}+b d e \,g^{2}-4 b \,e^{2} f g +3 c \,d^{2} g^{2}-8 c d e f g +8 c \,e^{2} f^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{4 \left (d^{2} g^{2}-2 d e f g +e^{2} f^{2}\right ) e^{2} \sqrt {\left (d g -e f \right ) e}}\)	\(260\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)/(e*x+d)^3/(g*x+f)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2*(1/8*g*(3*a*e^2*g+b*d*e*g-4*b*e^2*f-5*c*d^2*g+8*c*d*e*f)/e/(d^2*g^2-2*d*e*f*g+e^2*f^2)*(g*x+f)^(3/2)+1/8*(5*
a*e^2*g-b*d*e*g-4*b*e^2*f-3*c*d^2*g+8*c*d*e*f)/e^2*g/(d*g-e*f)*(g*x+f)^(1/2))/(e*(g*x+f)+d*g-e*f)^2+1/4*(3*a*e
^2*g^2+b*d*e*g^2-4*b*e^2*f*g+3*c*d^2*g^2-8*c*d*e*f*g+8*c*e^2*f^2)/(d^2*g^2-2*d*e*f*g+e^2*f^2)/e^2/((d*g-e*f)*e
)^(1/2)*arctan(e*(g*x+f)^(1/2)/((d*g-e*f)*e)^(1/2))

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/(e*x+d)^3/(g*x+f)^(1/2),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(4*%e^2*f-4*%e*d*g>0)', see `as
sume?` for m

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 523 vs. \(2 (193) = 386\).
time = 1.76, size = 1062, normalized size = 5.16 \begin {gather*} \left [-\frac {{\left (3 \, c d^{4} g^{2} + {\left (8 \, c f^{2} - 4 \, b f g + 3 \, a g^{2}\right )} x^{2} e^{4} - {\left ({\left (8 \, c d f g - b d g^{2}\right )} x^{2} - 2 \, {\left (8 \, c d f^{2} - 4 \, b d f g + 3 \, a d g^{2}\right )} x\right )} e^{3} + {\left (3 \, c d^{2} g^{2} x^{2} + 8 \, c d^{2} f^{2} - 4 \, b d^{2} f g + 3 \, a d^{2} g^{2} - 2 \, {\left (8 \, c d^{2} f g - b d^{2} g^{2}\right )} x\right )} e^{2} + {\left (6 \, c d^{3} g^{2} x - 8 \, c d^{3} f g + b d^{3} g^{2}\right )} e\right )} \sqrt {-d g e + f e^{2}} \log \left (-\frac {d g - {\left (g x + 2 \, f\right )} e + 2 \, \sqrt {-d g e + f e^{2}} \sqrt {g x + f}}{x e + d}\right ) + 2 \, {\left (3 \, c d^{4} g^{2} e - {\left (2 \, a f^{2} + {\left (4 \, b f^{2} - 3 \, a f g\right )} x\right )} e^{5} - {\left (2 \, b d f^{2} - 7 \, a d f g - {\left (8 \, c d f^{2} + 5 \, b d f g - 3 \, a d g^{2}\right )} x\right )} e^{4} + {\left (6 \, c d^{2} f^{2} + b d^{2} f g - 5 \, a d^{2} g^{2} - {\left (13 \, c d^{2} f g + b d^{2} g^{2}\right )} x\right )} e^{3} + {\left (5 \, c d^{3} g^{2} x - 9 \, c d^{3} f g + b d^{3} g^{2}\right )} e^{2}\right )} \sqrt {g x + f}}{8 \, {\left (d^{5} g^{3} e^{3} - f^{3} x^{2} e^{8} + {\left (3 \, d f^{2} g x^{2} - 2 \, d f^{3} x\right )} e^{7} - {\left (3 \, d^{2} f g^{2} x^{2} - 6 \, d^{2} f^{2} g x + d^{2} f^{3}\right )} e^{6} + {\left (d^{3} g^{3} x^{2} - 6 \, d^{3} f g^{2} x + 3 \, d^{3} f^{2} g\right )} e^{5} + {\left (2 \, d^{4} g^{3} x - 3 \, d^{4} f g^{2}\right )} e^{4}\right )}}, -\frac {{\left (3 \, c d^{4} g^{2} + {\left (8 \, c f^{2} - 4 \, b f g + 3 \, a g^{2}\right )} x^{2} e^{4} - {\left ({\left (8 \, c d f g - b d g^{2}\right )} x^{2} - 2 \, {\left (8 \, c d f^{2} - 4 \, b d f g + 3 \, a d g^{2}\right )} x\right )} e^{3} + {\left (3 \, c d^{2} g^{2} x^{2} + 8 \, c d^{2} f^{2} - 4 \, b d^{2} f g + 3 \, a d^{2} g^{2} - 2 \, {\left (8 \, c d^{2} f g - b d^{2} g^{2}\right )} x\right )} e^{2} + {\left (6 \, c d^{3} g^{2} x - 8 \, c d^{3} f g + b d^{3} g^{2}\right )} e\right )} \sqrt {d g e - f e^{2}} \arctan \left (-\frac {\sqrt {d g e - f e^{2}} \sqrt {g x + f}}{d g - f e}\right ) + {\left (3 \, c d^{4} g^{2} e - {\left (2 \, a f^{2} + {\left (4 \, b f^{2} - 3 \, a f g\right )} x\right )} e^{5} - {\left (2 \, b d f^{2} - 7 \, a d f g - {\left (8 \, c d f^{2} + 5 \, b d f g - 3 \, a d g^{2}\right )} x\right )} e^{4} + {\left (6 \, c d^{2} f^{2} + b d^{2} f g - 5 \, a d^{2} g^{2} - {\left (13 \, c d^{2} f g + b d^{2} g^{2}\right )} x\right )} e^{3} + {\left (5 \, c d^{3} g^{2} x - 9 \, c d^{3} f g + b d^{3} g^{2}\right )} e^{2}\right )} \sqrt {g x + f}}{4 \, {\left (d^{5} g^{3} e^{3} - f^{3} x^{2} e^{8} + {\left (3 \, d f^{2} g x^{2} - 2 \, d f^{3} x\right )} e^{7} - {\left (3 \, d^{2} f g^{2} x^{2} - 6 \, d^{2} f^{2} g x + d^{2} f^{3}\right )} e^{6} + {\left (d^{3} g^{3} x^{2} - 6 \, d^{3} f g^{2} x + 3 \, d^{3} f^{2} g\right )} e^{5} + {\left (2 \, d^{4} g^{3} x - 3 \, d^{4} f g^{2}\right )} e^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/(e*x+d)^3/(g*x+f)^(1/2),x, algorithm="fricas")

[Out]

[-1/8*((3*c*d^4*g^2 + (8*c*f^2 - 4*b*f*g + 3*a*g^2)*x^2*e^4 - ((8*c*d*f*g - b*d*g^2)*x^2 - 2*(8*c*d*f^2 - 4*b*
d*f*g + 3*a*d*g^2)*x)*e^3 + (3*c*d^2*g^2*x^2 + 8*c*d^2*f^2 - 4*b*d^2*f*g + 3*a*d^2*g^2 - 2*(8*c*d^2*f*g - b*d^
2*g^2)*x)*e^2 + (6*c*d^3*g^2*x - 8*c*d^3*f*g + b*d^3*g^2)*e)*sqrt(-d*g*e + f*e^2)*log(-(d*g - (g*x + 2*f)*e +
2*sqrt(-d*g*e + f*e^2)*sqrt(g*x + f))/(x*e + d)) + 2*(3*c*d^4*g^2*e - (2*a*f^2 + (4*b*f^2 - 3*a*f*g)*x)*e^5 -
(2*b*d*f^2 - 7*a*d*f*g - (8*c*d*f^2 + 5*b*d*f*g - 3*a*d*g^2)*x)*e^4 + (6*c*d^2*f^2 + b*d^2*f*g - 5*a*d^2*g^2 -
(13*c*d^2*f*g + b*d^2*g^2)*x)*e^3 + (5*c*d^3*g^2*x - 9*c*d^3*f*g + b*d^3*g^2)*e^2)*sqrt(g*x + f))/(d^5*g^3*e^
3 - f^3*x^2*e^8 + (3*d*f^2*g*x^2 - 2*d*f^3*x)*e^7 - (3*d^2*f*g^2*x^2 - 6*d^2*f^2*g*x + d^2*f^3)*e^6 + (d^3*g^3
*x^2 - 6*d^3*f*g^2*x + 3*d^3*f^2*g)*e^5 + (2*d^4*g^3*x - 3*d^4*f*g^2)*e^4), -1/4*((3*c*d^4*g^2 + (8*c*f^2 - 4*
b*f*g + 3*a*g^2)*x^2*e^4 - ((8*c*d*f*g - b*d*g^2)*x^2 - 2*(8*c*d*f^2 - 4*b*d*f*g + 3*a*d*g^2)*x)*e^3 + (3*c*d^
2*g^2*x^2 + 8*c*d^2*f^2 - 4*b*d^2*f*g + 3*a*d^2*g^2 - 2*(8*c*d^2*f*g - b*d^2*g^2)*x)*e^2 + (6*c*d^3*g^2*x - 8*
c*d^3*f*g + b*d^3*g^2)*e)*sqrt(d*g*e - f*e^2)*arctan(-sqrt(d*g*e - f*e^2)*sqrt(g*x + f)/(d*g - f*e)) + (3*c*d^
4*g^2*e - (2*a*f^2 + (4*b*f^2 - 3*a*f*g)*x)*e^5 - (2*b*d*f^2 - 7*a*d*f*g - (8*c*d*f^2 + 5*b*d*f*g - 3*a*d*g^2)
*x)*e^4 + (6*c*d^2*f^2 + b*d^2*f*g - 5*a*d^2*g^2 - (13*c*d^2*f*g + b*d^2*g^2)*x)*e^3 + (5*c*d^3*g^2*x - 9*c*d^
3*f*g + b*d^3*g^2)*e^2)*sqrt(g*x + f))/(d^5*g^3*e^3 - f^3*x^2*e^8 + (3*d*f^2*g*x^2 - 2*d*f^3*x)*e^7 - (3*d^2*f
*g^2*x^2 - 6*d^2*f^2*g*x + d^2*f^3)*e^6 + (d^3*g^3*x^2 - 6*d^3*f*g^2*x + 3*d^3*f^2*g)*e^5 + (2*d^4*g^3*x - 3*d
^4*f*g^2)*e^4)]

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)/(e*x+d)**3/(g*x+f)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]
time = 4.01, size = 373, normalized size = 1.81 \begin {gather*} \frac {{\left (3 \, c d^{2} g^{2} - 8 \, c d f g e + b d g^{2} e + 8 \, c f^{2} e^{2} - 4 \, b f g e^{2} + 3 \, a g^{2} e^{2}\right )} \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {d g e - f e^{2}}}\right )}{4 \, {\left (d^{2} g^{2} e^{2} - 2 \, d f g e^{3} + f^{2} e^{4}\right )} \sqrt {d g e - f e^{2}}} - \frac {3 \, \sqrt {g x + f} c d^{3} g^{3} + 5 \, {\left (g x + f\right )}^{\frac {3}{2}} c d^{2} g^{2} e - 11 \, \sqrt {g x + f} c d^{2} f g^{2} e + \sqrt {g x + f} b d^{2} g^{3} e - 8 \, {\left (g x + f\right )}^{\frac {3}{2}} c d f g e^{2} + 8 \, \sqrt {g x + f} c d f^{2} g e^{2} - {\left (g x + f\right )}^{\frac {3}{2}} b d g^{2} e^{2} + 3 \, \sqrt {g x + f} b d f g^{2} e^{2} - 5 \, \sqrt {g x + f} a d g^{3} e^{2} + 4 \, {\left (g x + f\right )}^{\frac {3}{2}} b f g e^{3} - 4 \, \sqrt {g x + f} b f^{2} g e^{3} - 3 \, {\left (g x + f\right )}^{\frac {3}{2}} a g^{2} e^{3} + 5 \, \sqrt {g x + f} a f g^{2} e^{3}}{4 \, {\left (d^{2} g^{2} e^{2} - 2 \, d f g e^{3} + f^{2} e^{4}\right )} {\left (d g + {\left (g x + f\right )} e - f e\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/(e*x+d)^3/(g*x+f)^(1/2),x, algorithm="giac")

[Out]

1/4*(3*c*d^2*g^2 - 8*c*d*f*g*e + b*d*g^2*e + 8*c*f^2*e^2 - 4*b*f*g*e^2 + 3*a*g^2*e^2)*arctan(sqrt(g*x + f)*e/s
qrt(d*g*e - f*e^2))/((d^2*g^2*e^2 - 2*d*f*g*e^3 + f^2*e^4)*sqrt(d*g*e - f*e^2)) - 1/4*(3*sqrt(g*x + f)*c*d^3*g
^3 + 5*(g*x + f)^(3/2)*c*d^2*g^2*e - 11*sqrt(g*x + f)*c*d^2*f*g^2*e + sqrt(g*x + f)*b*d^2*g^3*e - 8*(g*x + f)^
(3/2)*c*d*f*g*e^2 + 8*sqrt(g*x + f)*c*d*f^2*g*e^2 - (g*x + f)^(3/2)*b*d*g^2*e^2 + 3*sqrt(g*x + f)*b*d*f*g^2*e^
2 - 5*sqrt(g*x + f)*a*d*g^3*e^2 + 4*(g*x + f)^(3/2)*b*f*g*e^3 - 4*sqrt(g*x + f)*b*f^2*g*e^3 - 3*(g*x + f)^(3/2
)*a*g^2*e^3 + 5*sqrt(g*x + f)*a*f*g^2*e^3)/((d^2*g^2*e^2 - 2*d*f*g*e^3 + f^2*e^4)*(d*g + (g*x + f)*e - f*e)^2)

________________________________________________________________________________________

Mupad [B]
time = 0.28, size = 270, normalized size = 1.31 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {e}\,\sqrt {f+g\,x}}{\sqrt {d\,g-e\,f}}\right )\,\left (3\,c\,d^2\,g^2-8\,c\,d\,e\,f\,g+b\,d\,e\,g^2+8\,c\,e^2\,f^2-4\,b\,e^2\,f\,g+3\,a\,e^2\,g^2\right )}{4\,e^{5/2}\,{\left (d\,g-e\,f\right )}^{5/2}}-\frac {\frac {\sqrt {f+g\,x}\,\left (3\,c\,d^2\,g^2+b\,d\,e\,g^2-8\,c\,f\,d\,e\,g-5\,a\,e^2\,g^2+4\,b\,f\,e^2\,g\right )}{4\,e^2\,\left (d\,g-e\,f\right )}-\frac {{\left (f+g\,x\right )}^{3/2}\,\left (-5\,c\,d^2\,g^2+b\,d\,e\,g^2+8\,c\,f\,d\,e\,g+3\,a\,e^2\,g^2-4\,b\,f\,e^2\,g\right )}{4\,e\,{\left (d\,g-e\,f\right )}^2}}{e^2\,{\left (f+g\,x\right )}^2-\left (f+g\,x\right )\,\left (2\,e^2\,f-2\,d\,e\,g\right )+d^2\,g^2+e^2\,f^2-2\,d\,e\,f\,g} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((e^(1/2)*(f + g*x)^(1/2))/(d*g - e*f)^(1/2))*(3*a*e^2*g^2 + 3*c*d^2*g^2 + 8*c*e^2*f^2 + b*d*e*g^2 - 4*b*
e^2*f*g - 8*c*d*e*f*g))/(4*e^(5/2)*(d*g - e*f)^(5/2)) - (((f + g*x)^(1/2)*(3*c*d^2*g^2 - 5*a*e^2*g^2 + b*d*e*g
^2 + 4*b*e^2*f*g - 8*c*d*e*f*g))/(4*e^2*(d*g - e*f)) - ((f + g*x)^(3/2)*(3*a*e^2*g^2 - 5*c*d^2*g^2 + b*d*e*g^2
- 4*b*e^2*f*g + 8*c*d*e*f*g))/(4*e*(d*g - e*f)^2))/(e^2*(f + g*x)^2 - (f + g*x)*(2*e^2*f - 2*d*e*g) + d^2*g^2
+ e^2*f^2 - 2*d*e*f*g)

________________________________________________________________________________________