3.287 \(\int \frac {d+e x+f x^2+g x^3}{x^5 \sqrt {a+b x+c x^2}} \, dx\)

Optimal. Leaf size=270 \[ \frac {\sqrt {a+b x+c x^2} (7 b d-8 a e)}{24 a^2 x^3}-\frac {\tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right ) \left (32 a^2 b (3 c e-2 a g)+16 a^2 c (3 c d-4 a f)-40 a b^3 e-24 a b^2 (5 c d-2 a f)+35 b^4 d\right )}{128 a^{9/2}}+\frac {\sqrt {a+b x+c x^2} \left (64 a^2 (2 c e-3 a g)-120 a b^2 e-4 a b (55 c d-36 a f)+105 b^3 d\right )}{192 a^4 x}-\frac {\sqrt {a+b x+c x^2} \left (48 a^2 f-40 a b e-36 a c d+35 b^2 d\right )}{96 a^3 x^2}-\frac {d \sqrt {a+b x+c x^2}}{4 a x^4} \]

[Out]

-1/128*(35*b^4*d-40*a*b^3*e+16*a^2*c*(-4*a*f+3*c*d)-24*a*b^2*(-2*a*f+5*c*d)+32*a^2*b*(-2*a*g+3*c*e))*arctanh(1
/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))/a^(9/2)-1/4*d*(c*x^2+b*x+a)^(1/2)/a/x^4+1/24*(-8*a*e+7*b*d)*(c*x^2+b
*x+a)^(1/2)/a^2/x^3-1/96*(48*a^2*f-40*a*b*e-36*a*c*d+35*b^2*d)*(c*x^2+b*x+a)^(1/2)/a^3/x^2+1/192*(105*b^3*d-12
0*a*b^2*e-4*a*b*(-36*a*f+55*c*d)+64*a^2*(-3*a*g+2*c*e))*(c*x^2+b*x+a)^(1/2)/a^4/x

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Rubi [A]  time = 0.49, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {1650, 834, 806, 724, 206} \[ \frac {\sqrt {a+b x+c x^2} \left (64 a^2 (2 c e-3 a g)-120 a b^2 e-4 a b (55 c d-36 a f)+105 b^3 d\right )}{192 a^4 x}-\frac {\tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right ) \left (32 a^2 b (3 c e-2 a g)+16 a^2 c (3 c d-4 a f)-24 a b^2 (5 c d-2 a f)-40 a b^3 e+35 b^4 d\right )}{128 a^{9/2}}-\frac {\sqrt {a+b x+c x^2} \left (48 a^2 f-40 a b e-36 a c d+35 b^2 d\right )}{96 a^3 x^2}+\frac {\sqrt {a+b x+c x^2} (7 b d-8 a e)}{24 a^2 x^3}-\frac {d \sqrt {a+b x+c x^2}}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x + f*x^2 + g*x^3)/(x^5*Sqrt[a + b*x + c*x^2]),x]

[Out]

-(d*Sqrt[a + b*x + c*x^2])/(4*a*x^4) + ((7*b*d - 8*a*e)*Sqrt[a + b*x + c*x^2])/(24*a^2*x^3) - ((35*b^2*d - 36*
a*c*d - 40*a*b*e + 48*a^2*f)*Sqrt[a + b*x + c*x^2])/(96*a^3*x^2) + ((105*b^3*d - 120*a*b^2*e - 4*a*b*(55*c*d -
36*a*f) + 64*a^2*(2*c*e - 3*a*g))*Sqrt[a + b*x + c*x^2])/(192*a^4*x) - ((35*b^4*d - 40*a*b^3*e + 16*a^2*c*(3*
c*d - 4*a*f) - 24*a*b^2*(5*c*d - 2*a*f) + 32*a^2*b*(3*c*e - 2*a*g))*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*
x + c*x^2])])/(128*a^(9/2))

Rule 206

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

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x],
x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 834

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m
+ 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[(c*d*f - f*b*e + a*e*g)*(m + 1)
+ b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&
NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p])

Rule 1650

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomia
lQuotient[Pq, d + e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[(e*R*(d + e*x)^(m + 1)*(a + b*x + c*
x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^
(m + 1)*(a + b*x + c*x^2)^p*ExpandToSum[(m + 1)*(c*d^2 - b*d*e + a*e^2)*Q + c*d*R*(m + 1) - b*e*R*(m + p + 2)
- c*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, e, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {d+e x+f x^2+g x^3}{x^5 \sqrt {a+b x+c x^2}} \, dx &=-\frac {d \sqrt {a+b x+c x^2}}{4 a x^4}-\frac {\int \frac {\frac {1}{2} (7 b d-8 a e)+(3 c d-4 a f) x-4 a g x^2}{x^4 \sqrt {a+b x+c x^2}} \, dx}{4 a}\\ &=-\frac {d \sqrt {a+b x+c x^2}}{4 a x^4}+\frac {(7 b d-8 a e) \sqrt {a+b x+c x^2}}{24 a^2 x^3}+\frac {\int \frac {\frac {1}{4} \left (35 b^2 d-40 a b e-12 a (3 c d-4 a f)\right )+\left (7 b c d-8 a c e+12 a^2 g\right ) x}{x^3 \sqrt {a+b x+c x^2}} \, dx}{12 a^2}\\ &=-\frac {d \sqrt {a+b x+c x^2}}{4 a x^4}+\frac {(7 b d-8 a e) \sqrt {a+b x+c x^2}}{24 a^2 x^3}-\frac {\left (35 b^2 d-36 a c d-40 a b e+48 a^2 f\right ) \sqrt {a+b x+c x^2}}{96 a^3 x^2}-\frac {\int \frac {\frac {1}{8} \left (105 b^3 d-220 a b c d-120 a b^2 e+128 a^2 c e+144 a^2 b f-192 a^3 g\right )+\frac {1}{4} c \left (35 b^2 d-40 a b e-12 a (3 c d-4 a f)\right ) x}{x^2 \sqrt {a+b x+c x^2}} \, dx}{24 a^3}\\ &=-\frac {d \sqrt {a+b x+c x^2}}{4 a x^4}+\frac {(7 b d-8 a e) \sqrt {a+b x+c x^2}}{24 a^2 x^3}-\frac {\left (35 b^2 d-36 a c d-40 a b e+48 a^2 f\right ) \sqrt {a+b x+c x^2}}{96 a^3 x^2}+\frac {\left (105 b^3 d-120 a b^2 e-4 a b (55 c d-36 a f)+64 a^2 (2 c e-3 a g)\right ) \sqrt {a+b x+c x^2}}{192 a^4 x}+\frac {\left (35 b^4 d-40 a b^3 e+16 a^2 c (3 c d-4 a f)-24 a b^2 (5 c d-2 a f)+32 a^2 b (3 c e-2 a g)\right ) \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx}{128 a^4}\\ &=-\frac {d \sqrt {a+b x+c x^2}}{4 a x^4}+\frac {(7 b d-8 a e) \sqrt {a+b x+c x^2}}{24 a^2 x^3}-\frac {\left (35 b^2 d-36 a c d-40 a b e+48 a^2 f\right ) \sqrt {a+b x+c x^2}}{96 a^3 x^2}+\frac {\left (105 b^3 d-120 a b^2 e-4 a b (55 c d-36 a f)+64 a^2 (2 c e-3 a g)\right ) \sqrt {a+b x+c x^2}}{192 a^4 x}-\frac {\left (35 b^4 d-40 a b^3 e+16 a^2 c (3 c d-4 a f)-24 a b^2 (5 c d-2 a f)+32 a^2 b (3 c e-2 a g)\right ) \operatorname {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b x}{\sqrt {a+b x+c x^2}}\right )}{64 a^4}\\ &=-\frac {d \sqrt {a+b x+c x^2}}{4 a x^4}+\frac {(7 b d-8 a e) \sqrt {a+b x+c x^2}}{24 a^2 x^3}-\frac {\left (35 b^2 d-36 a c d-40 a b e+48 a^2 f\right ) \sqrt {a+b x+c x^2}}{96 a^3 x^2}+\frac {\left (105 b^3 d-120 a b^2 e-4 a b (55 c d-36 a f)+64 a^2 (2 c e-3 a g)\right ) \sqrt {a+b x+c x^2}}{192 a^4 x}-\frac {\left (35 b^4 d-40 a b^3 e+16 a^2 c (3 c d-4 a f)-24 a b^2 (5 c d-2 a f)+32 a^2 b (3 c e-2 a g)\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{128 a^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.52, size = 212, normalized size = 0.79 \[ \frac {\sqrt {a+x (b+c x)} \left (-16 a^3 \left (3 d+4 e x+6 x^2 (f+2 g x)\right )+8 a^2 x (7 b d+2 b x (5 e+9 f x)+c x (9 d+16 e x))-10 a b x^2 (7 b d+12 b e x+22 c d x)+105 b^3 d x^3\right )}{192 a^4 x^4}-\frac {\tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right ) \left (32 a^2 b (3 c e-2 a g)+16 a^2 c (3 c d-4 a f)-40 a b^3 e+24 a b^2 (2 a f-5 c d)+35 b^4 d\right )}{128 a^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x + f*x^2 + g*x^3)/(x^5*Sqrt[a + b*x + c*x^2]),x]

[Out]

(Sqrt[a + x*(b + c*x)]*(105*b^3*d*x^3 - 10*a*b*x^2*(7*b*d + 22*c*d*x + 12*b*e*x) + 8*a^2*x*(7*b*d + c*x*(9*d +
16*e*x) + 2*b*x*(5*e + 9*f*x)) - 16*a^3*(3*d + 4*e*x + 6*x^2*(f + 2*g*x))))/(192*a^4*x^4) - ((35*b^4*d - 40*a
*b^3*e + 16*a^2*c*(3*c*d - 4*a*f) + 24*a*b^2*(-5*c*d + 2*a*f) + 32*a^2*b*(3*c*e - 2*a*g))*ArcTanh[(2*a + b*x)/
(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])])/(128*a^(9/2))

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fricas [A]  time = 10.83, size = 525, normalized size = 1.94 \[ \left [\frac {3 \, {\left (64 \, a^{3} b g - {\left (35 \, b^{4} - 120 \, a b^{2} c + 48 \, a^{2} c^{2}\right )} d + 8 \, {\left (5 \, a b^{3} - 12 \, a^{2} b c\right )} e - 16 \, {\left (3 \, a^{2} b^{2} - 4 \, a^{3} c\right )} f\right )} \sqrt {a} x^{4} \log \left (-\frac {8 \, a b x + {\left (b^{2} + 4 \, a c\right )} x^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (b x + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{2}}\right ) - 4 \, {\left (48 \, a^{4} d - {\left (144 \, a^{3} b f - 192 \, a^{4} g + 5 \, {\left (21 \, a b^{3} - 44 \, a^{2} b c\right )} d - 8 \, {\left (15 \, a^{2} b^{2} - 16 \, a^{3} c\right )} e\right )} x^{3} - 2 \, {\left (40 \, a^{3} b e - 48 \, a^{4} f - {\left (35 \, a^{2} b^{2} - 36 \, a^{3} c\right )} d\right )} x^{2} - 8 \, {\left (7 \, a^{3} b d - 8 \, a^{4} e\right )} x\right )} \sqrt {c x^{2} + b x + a}}{768 \, a^{5} x^{4}}, -\frac {3 \, {\left (64 \, a^{3} b g - {\left (35 \, b^{4} - 120 \, a b^{2} c + 48 \, a^{2} c^{2}\right )} d + 8 \, {\left (5 \, a b^{3} - 12 \, a^{2} b c\right )} e - 16 \, {\left (3 \, a^{2} b^{2} - 4 \, a^{3} c\right )} f\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{2} + a b x + a^{2}\right )}}\right ) + 2 \, {\left (48 \, a^{4} d - {\left (144 \, a^{3} b f - 192 \, a^{4} g + 5 \, {\left (21 \, a b^{3} - 44 \, a^{2} b c\right )} d - 8 \, {\left (15 \, a^{2} b^{2} - 16 \, a^{3} c\right )} e\right )} x^{3} - 2 \, {\left (40 \, a^{3} b e - 48 \, a^{4} f - {\left (35 \, a^{2} b^{2} - 36 \, a^{3} c\right )} d\right )} x^{2} - 8 \, {\left (7 \, a^{3} b d - 8 \, a^{4} e\right )} x\right )} \sqrt {c x^{2} + b x + a}}{384 \, a^{5} x^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^3+f*x^2+e*x+d)/x^5/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/768*(3*(64*a^3*b*g - (35*b^4 - 120*a*b^2*c + 48*a^2*c^2)*d + 8*(5*a*b^3 - 12*a^2*b*c)*e - 16*(3*a^2*b^2 - 4
*a^3*c)*f)*sqrt(a)*x^4*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 + 4*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(a) + 8*a^2
)/x^2) - 4*(48*a^4*d - (144*a^3*b*f - 192*a^4*g + 5*(21*a*b^3 - 44*a^2*b*c)*d - 8*(15*a^2*b^2 - 16*a^3*c)*e)*x
^3 - 2*(40*a^3*b*e - 48*a^4*f - (35*a^2*b^2 - 36*a^3*c)*d)*x^2 - 8*(7*a^3*b*d - 8*a^4*e)*x)*sqrt(c*x^2 + b*x +
a))/(a^5*x^4), -1/384*(3*(64*a^3*b*g - (35*b^4 - 120*a*b^2*c + 48*a^2*c^2)*d + 8*(5*a*b^3 - 12*a^2*b*c)*e - 1
6*(3*a^2*b^2 - 4*a^3*c)*f)*sqrt(-a)*x^4*arctan(1/2*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x
+ a^2)) + 2*(48*a^4*d - (144*a^3*b*f - 192*a^4*g + 5*(21*a*b^3 - 44*a^2*b*c)*d - 8*(15*a^2*b^2 - 16*a^3*c)*e)
*x^3 - 2*(40*a^3*b*e - 48*a^4*f - (35*a^2*b^2 - 36*a^3*c)*d)*x^2 - 8*(7*a^3*b*d - 8*a^4*e)*x)*sqrt(c*x^2 + b*x
+ a))/(a^5*x^4)]

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giac [B]  time = 0.30, size = 1448, normalized size = 5.36 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^3+f*x^2+e*x+d)/x^5/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

1/64*(35*b^4*d - 120*a*b^2*c*d + 48*a^2*c^2*d + 48*a^2*b^2*f - 64*a^3*c*f - 64*a^3*b*g - 40*a*b^3*e + 96*a^2*b
*c*e)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt(-a))/(sqrt(-a)*a^4) - 1/192*(105*(sqrt(c)*x - sqrt(c*x^
2 + b*x + a))^7*b^4*d - 360*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*a*b^2*c*d + 144*(sqrt(c)*x - sqrt(c*x^2 + b*
x + a))^7*a^2*c^2*d + 144*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*a^2*b^2*f - 192*(sqrt(c)*x - sqrt(c*x^2 + b*x
+ a))^7*a^3*c*f - 192*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*a^3*b*g - 120*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^
7*a*b^3*e + 288*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*a^2*b*c*e - 384*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*a^
4*sqrt(c)*g - 385*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*a*b^4*d + 1320*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*a
^2*b^2*c*d - 528*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*a^3*c^2*d - 528*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*a
^3*b^2*f + 192*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*a^4*c*f + 576*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*a^4*b
*g + 440*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*a^2*b^3*e - 1056*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*a^3*b*c*
e - 384*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*a^4*b*sqrt(c)*f + 1152*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*a^5
*sqrt(c)*g - 768*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*a^4*c^(3/2)*e + 511*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))
^3*a^2*b^4*d - 1752*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a^3*b^2*c*d - 528*(sqrt(c)*x - sqrt(c*x^2 + b*x + a)
)^3*a^4*c^2*d + 624*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a^4*b^2*f + 192*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^
3*a^5*c*f - 576*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a^5*b*g - 584*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a^3*
b^3*e + 480*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a^4*b*c*e - 2048*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a^4*b
*c^(3/2)*d + 768*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a^5*b*sqrt(c)*f - 1152*(sqrt(c)*x - sqrt(c*x^2 + b*x +
a))^2*a^6*sqrt(c)*g - 384*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a^4*b^2*sqrt(c)*e + 1024*(sqrt(c)*x - sqrt(c*x
^2 + b*x + a))^2*a^5*c^(3/2)*e - 279*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^3*b^4*d - 360*(sqrt(c)*x - sqrt(c*x
^2 + b*x + a))*a^4*b^2*c*d + 144*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^5*c^2*d - 240*(sqrt(c)*x - sqrt(c*x^2 +
b*x + a))*a^5*b^2*f - 192*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^6*c*f + 192*(sqrt(c)*x - sqrt(c*x^2 + b*x + a
))*a^6*b*g + 264*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^4*b^3*e + 288*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^5*b
*c*e - 384*a^4*b^3*sqrt(c)*d + 512*a^5*b*c^(3/2)*d - 384*a^6*b*sqrt(c)*f + 384*a^7*sqrt(c)*g + 384*a^5*b^2*sqr
t(c)*e - 256*a^6*c^(3/2)*e)/(((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2 - a)^4*a^4)

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maple [B]  time = 0.01, size = 591, normalized size = 2.19 \[ \frac {b g \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{2 a^{\frac {3}{2}}}+\frac {c f \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{2 a^{\frac {3}{2}}}-\frac {3 b^{2} f \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{8 a^{\frac {5}{2}}}-\frac {3 b c e \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{4 a^{\frac {5}{2}}}-\frac {3 c^{2} d \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{8 a^{\frac {5}{2}}}+\frac {5 b^{3} e \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{16 a^{\frac {7}{2}}}+\frac {15 b^{2} c d \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{16 a^{\frac {7}{2}}}-\frac {35 b^{4} d \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{128 a^{\frac {9}{2}}}-\frac {\sqrt {c \,x^{2}+b x +a}\, g}{a x}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, b f}{4 a^{2} x}+\frac {2 \sqrt {c \,x^{2}+b x +a}\, c e}{3 a^{2} x}-\frac {5 \sqrt {c \,x^{2}+b x +a}\, b^{2} e}{8 a^{3} x}-\frac {55 \sqrt {c \,x^{2}+b x +a}\, b c d}{48 a^{3} x}+\frac {35 \sqrt {c \,x^{2}+b x +a}\, b^{3} d}{64 a^{4} x}-\frac {\sqrt {c \,x^{2}+b x +a}\, f}{2 a \,x^{2}}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, b e}{12 a^{2} x^{2}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, c d}{8 a^{2} x^{2}}-\frac {35 \sqrt {c \,x^{2}+b x +a}\, b^{2} d}{96 a^{3} x^{2}}-\frac {\sqrt {c \,x^{2}+b x +a}\, e}{3 a \,x^{3}}+\frac {7 \sqrt {c \,x^{2}+b x +a}\, b d}{24 a^{2} x^{3}}-\frac {\sqrt {c \,x^{2}+b x +a}\, d}{4 a \,x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x^3+f*x^2+e*x+d)/x^5/(c*x^2+b*x+a)^(1/2),x)

[Out]

-1/2*f/a/x^2*(c*x^2+b*x+a)^(1/2)+3/4*f*b/a^2/x*(c*x^2+b*x+a)^(1/2)-3/8*f*b^2/a^(5/2)*ln((b*x+2*a+2*(c*x^2+b*x+
a)^(1/2)*a^(1/2))/x)+1/2*f*c/a^(3/2)*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)-1/3*e/a/x^3*(c*x^2+b*x+a)^(
1/2)+5/12*e*b/a^2/x^2*(c*x^2+b*x+a)^(1/2)-5/8*e*b^2/a^3/x*(c*x^2+b*x+a)^(1/2)+5/16*e*b^3/a^(7/2)*ln((b*x+2*a+2
*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)-3/4*e*b/a^(5/2)*c*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)+2/3*e*c/a^2/x
*(c*x^2+b*x+a)^(1/2)-1/4*d*(c*x^2+b*x+a)^(1/2)/a/x^4+7/24*d*b/a^2/x^3*(c*x^2+b*x+a)^(1/2)-35/96*d*b^2/a^3/x^2*
(c*x^2+b*x+a)^(1/2)+35/64*d*b^3/a^4/x*(c*x^2+b*x+a)^(1/2)-35/128*d*b^4/a^(9/2)*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/
2)*a^(1/2))/x)+15/16*d*b^2/a^(7/2)*c*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)-55/48*d*b/a^3*c/x*(c*x^2+b*
x+a)^(1/2)+3/8*d*c/a^2/x^2*(c*x^2+b*x+a)^(1/2)-3/8*d*c^2/a^(5/2)*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)
-g/a/x*(c*x^2+b*x+a)^(1/2)+1/2*g*b/a^(3/2)*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)

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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^3+f*x^2+e*x+d)/x^5/(c*x^2+b*x+a)^(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*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {g\,x^3+f\,x^2+e\,x+d}{x^5\,\sqrt {c\,x^2+b\,x+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x + f*x^2 + g*x^3)/(x^5*(a + b*x + c*x^2)^(1/2)),x)

[Out]

int((d + e*x + f*x^2 + g*x^3)/(x^5*(a + b*x + c*x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d + e x + f x^{2} + g x^{3}}{x^{5} \sqrt {a + b x + c x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x**3+f*x**2+e*x+d)/x**5/(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral((d + e*x + f*x**2 + g*x**3)/(x**5*sqrt(a + b*x + c*x**2)), x)

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