3.9.32 \(\int \frac {a+b x+c x^2}{(d+e x)^3 (f+g x)^{3/2}} \, dx\) [832]

Optimal. Leaf size=248 \[ \frac {2 \left (c f^2-b f g+a g^2\right )}{(e f-d g)^3 \sqrt {f+g x}}-\frac {\left (c d^2-b d e+a e^2\right ) \sqrt {f+g x}}{2 e (e f-d g)^2 (d+e x)^2}+\frac {(c d (8 e f-d g)-e (4 b e f+3 b d g-7 a e g)) \sqrt {f+g x}}{4 e (e f-d g)^3 (d+e x)}-\frac {\left (c \left (8 e^2 f^2+8 d e f g-d^2 g^2\right )+3 e g (5 a e g-b (4 e f+d g))\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{4 e^{3/2} (e f-d g)^{7/2}} \]

[Out]

-1/4*(c*(-d^2*g^2+8*d*e*f*g+8*e^2*f^2)+3*e*g*(5*a*e*g-b*(d*g+4*e*f)))*arctanh(e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)
^(1/2))/e^(3/2)/(-d*g+e*f)^(7/2)+2*(a*g^2-b*f*g+c*f^2)/(-d*g+e*f)^3/(g*x+f)^(1/2)-1/2*(a*e^2-b*d*e+c*d^2)*(g*x
+f)^(1/2)/e/(-d*g+e*f)^2/(e*x+d)^2+1/4*(c*d*(-d*g+8*e*f)-e*(-7*a*e*g+3*b*d*g+4*b*e*f))*(g*x+f)^(1/2)/e/(-d*g+e
*f)^3/(e*x+d)

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Rubi [A]
time = 0.40, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {911, 1273, 467, 464, 214} \begin {gather*} -\frac {\sqrt {f+g x} \left (a e^2-b d e+c d^2\right )}{2 e (d+e x)^2 (e f-d g)^2}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (3 e g (5 a e g-b (d g+4 e f))+c \left (-d^2 g^2+8 d e f g+8 e^2 f^2\right )\right )}{4 e^{3/2} (e f-d g)^{7/2}}+\frac {2 \left (a g^2-b f g+c f^2\right )}{\sqrt {f+g x} (e f-d g)^3}+\frac {\sqrt {f+g x} (c d (8 e f-d g)-e (-7 a e g+3 b d g+4 b e f))}{4 e (d+e x) (e f-d g)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(c*f^2 - b*f*g + a*g^2))/((e*f - d*g)^3*Sqrt[f + g*x]) - ((c*d^2 - b*d*e + a*e^2)*Sqrt[f + g*x])/(2*e*(e*f
- d*g)^2*(d + e*x)^2) + ((c*d*(8*e*f - d*g) - e*(4*b*e*f + 3*b*d*g - 7*a*e*g))*Sqrt[f + g*x])/(4*e*(e*f - d*g)
^3*(d + e*x)) - ((c*(8*e^2*f^2 + 8*d*e*f*g - d^2*g^2) + 3*e*g*(5*a*e*g - b*(4*e*f + d*g)))*ArcTanh[(Sqrt[e]*Sq
rt[f + g*x])/Sqrt[e*f - d*g]])/(4*e^(3/2)*(e*f - d*g)^(7/2))

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 464

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

Rule 467

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x
*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 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 1273

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[(-d)^(m
/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2*(-d)^(-m/2 + 1)*e^(2*
p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x],
x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rubi steps

\begin {align*} \int \frac {a+b x+c x^2}{(d+e x)^3 (f+g x)^{3/2}} \, 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}}{x^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 (c d^2-b d e+a e^2\right ) \sqrt {f+g x}}{2 e (e f-d g)^2 (d+e x)^2}-\frac {g^3 \text {Subst}\left (\int \frac {\frac {4 e^2 (e f-d g) \left (c f^2-b f g+a g^2\right )}{g^5}-\frac {e \left (3 e (b d-a e) g^2+c \left (4 e^2 f^2-8 d e f g+d^2 g^2\right )\right ) x^2}{g^5}}{x^2 \left (\frac {-e f+d g}{g}+\frac {e x^2}{g}\right )^2} \, dx,x,\sqrt {f+g x}\right )}{2 e^2 (e f-d g)^2}\\ &=-\frac {\left (c d^2-b d e+a e^2\right ) \sqrt {f+g x}}{2 e (e f-d g)^2 (d+e x)^2}+\frac {(c d (8 e f-d g)-e (4 b e f+3 b d g-7 a e g)) \sqrt {f+g x}}{4 e (e f-d g)^3 (d+e x)}+\frac {g^3 \text {Subst}\left (\int \frac {\frac {8 e^2 \left (c f^2-b f g+a g^2\right )}{g^4}+\frac {e (c d (8 e f-d g)-e (4 b e f+3 b d g-7 a e g)) x^2}{g^3 (e f-d g)}}{x^2 \left (\frac {-e f+d g}{g}+\frac {e x^2}{g}\right )} \, dx,x,\sqrt {f+g x}\right )}{4 e^2 (e f-d g)^2}\\ &=\frac {2 \left (c f^2-b f g+a g^2\right )}{(e f-d g)^3 \sqrt {f+g x}}-\frac {\left (c d^2-b d e+a e^2\right ) \sqrt {f+g x}}{2 e (e f-d g)^2 (d+e x)^2}+\frac {(c d (8 e f-d g)-e (4 b e f+3 b d g-7 a e g)) \sqrt {f+g x}}{4 e (e f-d g)^3 (d+e x)}+\frac {\left (c \left (8 e^2 f^2+8 d e f g-d^2 g^2\right )+3 e g (5 a e g-b (4 e f+d g))\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 g (e f-d g)^3}\\ &=\frac {2 \left (c f^2-b f g+a g^2\right )}{(e f-d g)^3 \sqrt {f+g x}}-\frac {\left (c d^2-b d e+a e^2\right ) \sqrt {f+g x}}{2 e (e f-d g)^2 (d+e x)^2}+\frac {(c d (8 e f-d g)-e (4 b e f+3 b d g-7 a e g)) \sqrt {f+g x}}{4 e (e f-d g)^3 (d+e x)}-\frac {\left (c \left (8 e^2 f^2+8 d e f g-d^2 g^2\right )+3 e g (5 a e g-b (4 e f+d g))\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{4 e^{3/2} (e f-d g)^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 1.31, size = 297, normalized size = 1.20 \begin {gather*} \frac {\frac {\sqrt {e} \left (c \left (8 e^3 f^2 x^2+d^3 g (f+g x)+8 d e^2 f x (3 f+g x)+d^2 e \left (14 f^2+5 f g x-g^2 x^2\right )\right )-e \left (a \left (-8 d^2 g^2-d e g (9 f+25 g x)+e^2 \left (2 f^2-5 f g x-15 g^2 x^2\right )\right )+b \left (4 e^2 f x (f+3 g x)+d^2 g (13 f+5 g x)+d e \left (2 f^2+21 f g x+3 g^2 x^2\right )\right )\right )\right )}{(e f-d g)^3 (d+e x)^2 \sqrt {f+g x}}-\frac {\left (c \left (8 e^2 f^2+8 d e f g-d^2 g^2\right )+3 e g (5 a e g-b (4 e f+d g))\right ) \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {-e f+d g}}\right )}{(-e f+d g)^{7/2}}}{4 e^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[e]*(c*(8*e^3*f^2*x^2 + d^3*g*(f + g*x) + 8*d*e^2*f*x*(3*f + g*x) + d^2*e*(14*f^2 + 5*f*g*x - g^2*x^2))
- e*(a*(-8*d^2*g^2 - d*e*g*(9*f + 25*g*x) + e^2*(2*f^2 - 5*f*g*x - 15*g^2*x^2)) + b*(4*e^2*f*x*(f + 3*g*x) + d
^2*g*(13*f + 5*g*x) + d*e*(2*f^2 + 21*f*g*x + 3*g^2*x^2)))))/((e*f - d*g)^3*(d + e*x)^2*Sqrt[f + g*x]) - ((c*(
8*e^2*f^2 + 8*d*e*f*g - d^2*g^2) + 3*e*g*(5*a*e*g - b*(4*e*f + d*g)))*ArcTan[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[-(e*
f) + d*g]])/(-(e*f) + d*g)^(7/2))/(4*e^(3/2))

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Maple [A]
time = 0.10, size = 294, normalized size = 1.19

method	result	size
derivativedivides	\(-\frac {2 \left (\frac {\left (\frac {7}{8} a \,e^{2} g^{2}-\frac {3}{8} b d e \,g^{2}-\frac {1}{2} b \,e^{2} f g -\frac {1}{8} c \,d^{2} g^{2}+c d e f g \right ) \left (g x +f \right )^{\frac {3}{2}}+\frac {g \left (9 a d \,e^{2} g^{2}-9 a \,e^{3} f g -5 b \,d^{2} e \,g^{2}+b d \,e^{2} f g +4 b \,e^{3} f^{2}+c \,d^{3} g^{2}+7 c \,d^{2} e f g -8 c d \,e^{2} f^{2}\right ) \sqrt {g x +f}}{8 e}}{\left (e \left (g x +f \right )+d g -e f \right )^{2}}+\frac {\left (15 a \,e^{2} g^{2}-3 b d e \,g^{2}-12 b \,e^{2} f g -c \,d^{2} g^{2}+8 c d e f g +8 c \,e^{2} f^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{8 e \sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right )^{3}}-\frac {2 \left (a \,g^{2}-b f g +c \,f^{2}\right )}{\left (d g -e f \right )^{3} \sqrt {g x +f}}\)	\(294\)
default	\(-\frac {2 \left (\frac {\left (\frac {7}{8} a \,e^{2} g^{2}-\frac {3}{8} b d e \,g^{2}-\frac {1}{2} b \,e^{2} f g -\frac {1}{8} c \,d^{2} g^{2}+c d e f g \right ) \left (g x +f \right )^{\frac {3}{2}}+\frac {g \left (9 a d \,e^{2} g^{2}-9 a \,e^{3} f g -5 b \,d^{2} e \,g^{2}+b d \,e^{2} f g +4 b \,e^{3} f^{2}+c \,d^{3} g^{2}+7 c \,d^{2} e f g -8 c d \,e^{2} f^{2}\right ) \sqrt {g x +f}}{8 e}}{\left (e \left (g x +f \right )+d g -e f \right )^{2}}+\frac {\left (15 a \,e^{2} g^{2}-3 b d e \,g^{2}-12 b \,e^{2} f g -c \,d^{2} g^{2}+8 c d e f g +8 c \,e^{2} f^{2}\right ) \arctan \left (\frac {e \sqrt {g x +f}}{\sqrt {\left (d g -e f \right ) e}}\right )}{8 e \sqrt {\left (d g -e f \right ) e}}\right )}{\left (d g -e f \right )^{3}}-\frac {2 \left (a \,g^{2}-b f g +c \,f^{2}\right )}{\left (d g -e f \right )^{3} \sqrt {g x +f}}\)	\(294\)

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)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/(d*g-e*f)^3*(((7/8*a*e^2*g^2-3/8*b*d*e*g^2-1/2*b*e^2*f*g-1/8*c*d^2*g^2+c*d*e*f*g)*(g*x+f)^(3/2)+1/8*g*(9*a*
d*e^2*g^2-9*a*e^3*f*g-5*b*d^2*e*g^2+b*d*e^2*f*g+4*b*e^3*f^2+c*d^3*g^2+7*c*d^2*e*f*g-8*c*d*e^2*f^2)/e*(g*x+f)^(
1/2))/(e*(g*x+f)+d*g-e*f)^2+1/8*(15*a*e^2*g^2-3*b*d*e*g^2-12*b*e^2*f*g-c*d^2*g^2+8*c*d*e*f*g+8*c*e^2*f^2)/e/((
d*g-e*f)*e)^(1/2)*arctan(e*(g*x+f)^(1/2)/((d*g-e*f)*e)^(1/2)))-2*(a*g^2-b*f*g+c*f^2)/(d*g-e*f)^3/(g*x+f)^(1/2)

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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)^(3/2),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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 936 vs. \(2 (234) = 468\).
time = 6.47, size = 1888, normalized size = 7.61 \begin {gather*} \text {Too large to display} \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)^(3/2),x, algorithm="fricas")

[Out]

[-1/8*((c*d^4*g^3*x + c*d^4*f*g^2 - ((8*c*f^2*g - 12*b*f*g^2 + 15*a*g^3)*x^3 + (8*c*f^3 - 12*b*f^2*g + 15*a*f*
g^2)*x^2)*e^4 - ((8*c*d*f*g^2 - 3*b*d*g^3)*x^3 + 3*(8*c*d*f^2*g - 9*b*d*f*g^2 + 10*a*d*g^3)*x^2 + 2*(8*c*d*f^3
- 12*b*d*f^2*g + 15*a*d*f*g^2)*x)*e^3 + (c*d^2*g^3*x^3 - 8*c*d^2*f^3 + 12*b*d^2*f^2*g - 15*a*d^2*f*g^2 - 3*(5
*c*d^2*f*g^2 - 2*b*d^2*g^3)*x^2 - 3*(8*c*d^2*f^2*g - 6*b*d^2*f*g^2 + 5*a*d^2*g^3)*x)*e^2 + (2*c*d^3*g^3*x^2 -
8*c*d^3*f^2*g + 3*b*d^3*f*g^2 - 3*(2*c*d^3*f*g^2 - b*d^3*g^3)*x)*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*sqrt(g*x + f)*((2*a*f^3 - (8*c*f^3 - 12*b*f^2*g +
15*a*f*g^2)*x^2 + (4*b*f^3 - 5*a*f^2*g)*x)*e^5 + (2*b*d*f^3 - 11*a*d*f^2*g - 3*(3*b*d*f*g^2 - 5*a*d*g^3)*x^2 -
(24*c*d*f^3 - 17*b*d*f^2*g + 20*a*d*f*g^2)*x)*e^4 - (14*c*d^2*f^3 - 11*b*d^2*f^2*g - a*d^2*f*g^2 - 3*(3*c*d^2
*f*g^2 - b*d^2*g^3)*x^2 - (19*c*d^2*f^2*g - 16*b*d^2*f*g^2 + 25*a*d^2*g^3)*x)*e^3 - (c*d^3*g^3*x^2 - 13*c*d^3*
f^2*g + 13*b*d^3*f*g^2 - 8*a*d^3*g^3 - (4*c*d^3*f*g^2 - 5*b*d^3*g^3)*x)*e^2 + (c*d^4*g^3*x + c*d^4*f*g^2)*e))/
((f^4*g*x^3 + f^5*x^2)*e^8 - 2*(2*d*f^3*g^2*x^3 + d*f^4*g*x^2 - d*f^5*x)*e^7 + (6*d^2*f^2*g^3*x^3 - 2*d^2*f^3*
g^2*x^2 - 7*d^2*f^4*g*x + d^2*f^5)*e^6 - 4*(d^3*f*g^4*x^3 - 2*d^3*f^2*g^3*x^2 - 2*d^3*f^3*g^2*x + d^3*f^4*g)*e
^5 + (d^4*g^5*x^3 - 7*d^4*f*g^4*x^2 - 2*d^4*f^2*g^3*x + 6*d^4*f^3*g^2)*e^4 + 2*(d^5*g^5*x^2 - d^5*f*g^4*x - 2*
d^5*f^2*g^3)*e^3 + (d^6*g^5*x + d^6*f*g^4)*e^2), -1/4*((c*d^4*g^3*x + c*d^4*f*g^2 - ((8*c*f^2*g - 12*b*f*g^2 +
15*a*g^3)*x^3 + (8*c*f^3 - 12*b*f^2*g + 15*a*f*g^2)*x^2)*e^4 - ((8*c*d*f*g^2 - 3*b*d*g^3)*x^3 + 3*(8*c*d*f^2*
g - 9*b*d*f*g^2 + 10*a*d*g^3)*x^2 + 2*(8*c*d*f^3 - 12*b*d*f^2*g + 15*a*d*f*g^2)*x)*e^3 + (c*d^2*g^3*x^3 - 8*c*
d^2*f^3 + 12*b*d^2*f^2*g - 15*a*d^2*f*g^2 - 3*(5*c*d^2*f*g^2 - 2*b*d^2*g^3)*x^2 - 3*(8*c*d^2*f^2*g - 6*b*d^2*f
*g^2 + 5*a*d^2*g^3)*x)*e^2 + (2*c*d^3*g^3*x^2 - 8*c*d^3*f^2*g + 3*b*d^3*f*g^2 - 3*(2*c*d^3*f*g^2 - b*d^3*g^3)*
x)*e)*sqrt(d*g*e - f*e^2)*arctan(-sqrt(d*g*e - f*e^2)*sqrt(g*x + f)/(d*g - f*e)) + sqrt(g*x + f)*((2*a*f^3 - (
8*c*f^3 - 12*b*f^2*g + 15*a*f*g^2)*x^2 + (4*b*f^3 - 5*a*f^2*g)*x)*e^5 + (2*b*d*f^3 - 11*a*d*f^2*g - 3*(3*b*d*f
*g^2 - 5*a*d*g^3)*x^2 - (24*c*d*f^3 - 17*b*d*f^2*g + 20*a*d*f*g^2)*x)*e^4 - (14*c*d^2*f^3 - 11*b*d^2*f^2*g - a
*d^2*f*g^2 - 3*(3*c*d^2*f*g^2 - b*d^2*g^3)*x^2 - (19*c*d^2*f^2*g - 16*b*d^2*f*g^2 + 25*a*d^2*g^3)*x)*e^3 - (c*
d^3*g^3*x^2 - 13*c*d^3*f^2*g + 13*b*d^3*f*g^2 - 8*a*d^3*g^3 - (4*c*d^3*f*g^2 - 5*b*d^3*g^3)*x)*e^2 + (c*d^4*g^
3*x + c*d^4*f*g^2)*e))/((f^4*g*x^3 + f^5*x^2)*e^8 - 2*(2*d*f^3*g^2*x^3 + d*f^4*g*x^2 - d*f^5*x)*e^7 + (6*d^2*f
^2*g^3*x^3 - 2*d^2*f^3*g^2*x^2 - 7*d^2*f^4*g*x + d^2*f^5)*e^6 - 4*(d^3*f*g^4*x^3 - 2*d^3*f^2*g^3*x^2 - 2*d^3*f
^3*g^2*x + d^3*f^4*g)*e^5 + (d^4*g^5*x^3 - 7*d^4*f*g^4*x^2 - 2*d^4*f^2*g^3*x + 6*d^4*f^3*g^2)*e^4 + 2*(d^5*g^5
*x^2 - d^5*f*g^4*x - 2*d^5*f^2*g^3)*e^3 + (d^6*g^5*x + d^6*f*g^4)*e^2)]

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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)**(3/2),x)

[Out]

Timed out

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Giac [A]
time = 5.43, size = 462, normalized size = 1.86 \begin {gather*} \frac {{\left (c d^{2} g^{2} - 8 \, c d f g e + 3 \, b d g^{2} e - 8 \, c f^{2} e^{2} + 12 \, b f g e^{2} - 15 \, 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^{3} g^{3} e - 3 \, d^{2} f g^{2} e^{2} + 3 \, d f^{2} g e^{3} - f^{3} e^{4}\right )} \sqrt {d g e - f e^{2}}} - \frac {2 \, {\left (c f^{2} - b f g + a g^{2}\right )}}{{\left (d^{3} g^{3} - 3 \, d^{2} f g^{2} e + 3 \, d f^{2} g e^{2} - f^{3} e^{3}\right )} \sqrt {g x + f}} - \frac {\sqrt {g x + f} c d^{3} g^{3} - {\left (g x + f\right )}^{\frac {3}{2}} c d^{2} g^{2} e + 7 \, \sqrt {g x + f} c d^{2} f g^{2} e - 5 \, \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} - 3 \, {\left (g x + f\right )}^{\frac {3}{2}} b d g^{2} e^{2} + \sqrt {g x + f} b d f g^{2} e^{2} + 9 \, \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} + 7 \, {\left (g x + f\right )}^{\frac {3}{2}} a g^{2} e^{3} - 9 \, \sqrt {g x + f} a f g^{2} e^{3}}{4 \, {\left (d^{3} g^{3} e - 3 \, d^{2} f g^{2} e^{2} + 3 \, d f^{2} g e^{3} - f^{3} 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)^(3/2),x, algorithm="giac")

[Out]

1/4*(c*d^2*g^2 - 8*c*d*f*g*e + 3*b*d*g^2*e - 8*c*f^2*e^2 + 12*b*f*g*e^2 - 15*a*g^2*e^2)*arctan(sqrt(g*x + f)*e
/sqrt(d*g*e - f*e^2))/((d^3*g^3*e - 3*d^2*f*g^2*e^2 + 3*d*f^2*g*e^3 - f^3*e^4)*sqrt(d*g*e - f*e^2)) - 2*(c*f^2
- b*f*g + a*g^2)/((d^3*g^3 - 3*d^2*f*g^2*e + 3*d*f^2*g*e^2 - f^3*e^3)*sqrt(g*x + f)) - 1/4*(sqrt(g*x + f)*c*d
^3*g^3 - (g*x + f)^(3/2)*c*d^2*g^2*e + 7*sqrt(g*x + f)*c*d^2*f*g^2*e - 5*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 - 3*(g*x + f)^(3/2)*b*d*g^2*e^2 + sqrt(g*x + f)*b*d*f*g^2
*e^2 + 9*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 + 7*(g*x + f)^(
3/2)*a*g^2*e^3 - 9*sqrt(g*x + f)*a*f*g^2*e^3)/((d^3*g^3*e - 3*d^2*f*g^2*e^2 + 3*d*f^2*g*e^3 - f^3*e^4)*(d*g +
(g*x + f)*e - f*e)^2)

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Mupad [B]
time = 3.41, size = 363, normalized size = 1.46 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {f+g\,x}\,\left (-d^3\,e\,g^3+3\,d^2\,e^2\,f\,g^2-3\,d\,e^3\,f^2\,g+e^4\,f^3\right )}{\sqrt {e}\,{\left (d\,g-e\,f\right )}^{7/2}}\right )\,\left (-c\,d^2\,g^2+8\,c\,d\,e\,f\,g-3\,b\,d\,e\,g^2+8\,c\,e^2\,f^2-12\,b\,e^2\,f\,g+15\,a\,e^2\,g^2\right )}{4\,e^{3/2}\,{\left (d\,g-e\,f\right )}^{7/2}}-\frac {\frac {2\,\left (c\,f^2-b\,f\,g+a\,g^2\right )}{d\,g-e\,f}+\frac {{\left (f+g\,x\right )}^2\,\left (-c\,d^2\,g^2+8\,c\,d\,e\,f\,g-3\,b\,d\,e\,g^2+8\,c\,e^2\,f^2-12\,b\,e^2\,f\,g+15\,a\,e^2\,g^2\right )}{4\,{\left (d\,g-e\,f\right )}^3}+\frac {\left (f+g\,x\right )\,\left (c\,d^2\,g^2+8\,c\,d\,e\,f\,g-5\,b\,d\,e\,g^2+16\,c\,e^2\,f^2-20\,b\,e^2\,f\,g+25\,a\,e^2\,g^2\right )}{4\,e\,{\left (d\,g-e\,f\right )}^2}}{e^2\,{\left (f+g\,x\right )}^{5/2}-{\left (f+g\,x\right )}^{3/2}\,\left (2\,e^2\,f-2\,d\,e\,g\right )+\sqrt {f+g\,x}\,\left (d^2\,g^2-2\,d\,e\,f\,g+e^2\,f^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan(((f + g*x)^(1/2)*(e^4*f^3 - d^3*e*g^3 + 3*d^2*e^2*f*g^2 - 3*d*e^3*f^2*g))/(e^(1/2)*(d*g - e*f)^(7/2)))*(
15*a*e^2*g^2 - c*d^2*g^2 + 8*c*e^2*f^2 - 3*b*d*e*g^2 - 12*b*e^2*f*g + 8*c*d*e*f*g))/(4*e^(3/2)*(d*g - e*f)^(7/
2)) - ((2*(a*g^2 + c*f^2 - b*f*g))/(d*g - e*f) + ((f + g*x)^2*(15*a*e^2*g^2 - c*d^2*g^2 + 8*c*e^2*f^2 - 3*b*d*
e*g^2 - 12*b*e^2*f*g + 8*c*d*e*f*g))/(4*(d*g - e*f)^3) + ((f + g*x)*(25*a*e^2*g^2 + c*d^2*g^2 + 16*c*e^2*f^2 -
5*b*d*e*g^2 - 20*b*e^2*f*g + 8*c*d*e*f*g))/(4*e*(d*g - e*f)^2))/(e^2*(f + g*x)^(5/2) - (f + g*x)^(3/2)*(2*e^2
*f - 2*d*e*g) + (f + g*x)^(1/2)*(d^2*g^2 + e^2*f^2 - 2*d*e*f*g))

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