3.14.95 \(\int \frac {b+2 c x}{(d+e x)^2 (a+b x+c x^2)^{3/2}} \, dx\)
Optimal. Leaf size=248 \[ -\frac {e \left (-4 c e (a e+2 b d)+3 b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{2 \left (a e^2-b d e+c d^2\right )^{5/2}}-\frac {2 \left (\left (b^2-4 a c\right ) (c d-b e)-c e x \left (b^2-4 a c\right )\right )}{\left (b^2-4 a c\right ) (d+e x) \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}+\frac {3 e^2 \sqrt {a+b x+c x^2} (2 c d-b e)}{(d+e x) \left (a e^2-b d e+c d^2\right )^2} \]
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Rubi [A] time = 0.27, antiderivative size = 248, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used =
{822, 806, 724, 206} \begin {gather*} -\frac {e \left (-4 c e (a e+2 b d)+3 b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{2 \left (a e^2-b d e+c d^2\right )^{5/2}}-\frac {2 \left (\left (b^2-4 a c\right ) (c d-b e)-c e x \left (b^2-4 a c\right )\right )}{\left (b^2-4 a c\right ) (d+e x) \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}+\frac {3 e^2 \sqrt {a+b x+c x^2} (2 c d-b e)}{(d+e x) \left (a e^2-b d e+c d^2\right )^2} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(b + 2*c*x)/((d + e*x)^2*(a + b*x + c*x^2)^(3/2)),x]
[Out]
(-2*((b^2 - 4*a*c)*(c*d - b*e) - c*(b^2 - 4*a*c)*e*x))/((b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)*(d + e*x)*Sqrt[a
+ b*x + c*x^2]) + (3*e^2*(2*c*d - b*e)*Sqrt[a + b*x + c*x^2])/((c*d^2 - b*d*e + a*e^2)^2*(d + e*x)) - (e*(8*c
^2*d^2 + 3*b^2*e^2 - 4*c*e*(2*b*d + a*e))*ArcTanh[(b*d - 2*a*e + (2*c*d - b*e)*x)/(2*Sqrt[c*d^2 - b*d*e + a*e^
2]*Sqrt[a + b*x + c*x^2])])/(2*(c*d^2 - b*d*e + a*e^2)^(5/2))
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 822
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x
)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b,
c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])
Rubi steps
\begin {align*} \int \frac {b+2 c x}{(d+e x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac {2 \left (\left (b^2-4 a c\right ) (c d-b e)-c \left (b^2-4 a c\right ) e x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (d+e x) \sqrt {a+b x+c x^2}}-\frac {2 \int \frac {\frac {1}{2} \left (b^2-4 a c\right ) e (4 c d-3 b e)-c \left (b^2-4 a c\right ) e^2 x}{(d+e x)^2 \sqrt {a+b x+c x^2}} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {2 \left (\left (b^2-4 a c\right ) (c d-b e)-c \left (b^2-4 a c\right ) e x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (d+e x) \sqrt {a+b x+c x^2}}+\frac {3 e^2 (2 c d-b e) \sqrt {a+b x+c x^2}}{\left (c d^2-b d e+a e^2\right )^2 (d+e x)}-\frac {\left (e \left (8 c^2 d^2+3 b^2 e^2-4 c e (2 b d+a e)\right )\right ) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{2 \left (c d^2-b d e+a e^2\right )^2}\\ &=-\frac {2 \left (\left (b^2-4 a c\right ) (c d-b e)-c \left (b^2-4 a c\right ) e x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (d+e x) \sqrt {a+b x+c x^2}}+\frac {3 e^2 (2 c d-b e) \sqrt {a+b x+c x^2}}{\left (c d^2-b d e+a e^2\right )^2 (d+e x)}+\frac {\left (e \left (8 c^2 d^2+3 b^2 e^2-4 c e (2 b d+a e)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 c d^2-4 b d e+4 a e^2-x^2} \, dx,x,\frac {-b d+2 a e-(2 c d-b e) x}{\sqrt {a+b x+c x^2}}\right )}{\left (c d^2-b d e+a e^2\right )^2}\\ &=-\frac {2 \left (\left (b^2-4 a c\right ) (c d-b e)-c \left (b^2-4 a c\right ) e x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (d+e x) \sqrt {a+b x+c x^2}}+\frac {3 e^2 (2 c d-b e) \sqrt {a+b x+c x^2}}{\left (c d^2-b d e+a e^2\right )^2 (d+e x)}-\frac {e \left (8 c^2 d^2+3 b^2 e^2-4 c e (2 b d+a e)\right ) \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{2 \left (c d^2-b d e+a e^2\right )^{5/2}}\\ \end {align*}
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Mathematica [A] time = 0.54, size = 215, normalized size = 0.87 \begin {gather*} \frac {e \left (-4 c e (a e+2 b d)+3 b^2 e^2+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {2 a e-b d+b e x-2 c d x}{2 \sqrt {a+x (b+c x)} \sqrt {e (a e-b d)+c d^2}}\right )}{2 \left (e (a e-b d)+c d^2\right )^{5/2}}+\frac {3 e^2 \sqrt {a+x (b+c x)} (2 c d-b e)}{(d+e x) \left (e (a e-b d)+c d^2\right )^2}+\frac {2 (b e-c d+c e x)}{(d+e x) \sqrt {a+x (b+c x)} \left (e (a e-b d)+c d^2\right )} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(b + 2*c*x)/((d + e*x)^2*(a + b*x + c*x^2)^(3/2)),x]
[Out]
(2*(-(c*d) + b*e + c*e*x))/((c*d^2 + e*(-(b*d) + a*e))*(d + e*x)*Sqrt[a + x*(b + c*x)]) + (3*e^2*(2*c*d - b*e)
*Sqrt[a + x*(b + c*x)])/((c*d^2 + e*(-(b*d) + a*e))^2*(d + e*x)) + (e*(8*c^2*d^2 + 3*b^2*e^2 - 4*c*e*(2*b*d +
a*e))*ArcTanh[(-(b*d) + 2*a*e - 2*c*d*x + b*e*x)/(2*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*Sqrt[a + x*(b + c*x)])])/(2
*(c*d^2 + e*(-(b*d) + a*e))^(5/2))
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IntegrateAlgebraic [B] time = 22.85, size = 5483, normalized size = 22.11 \begin {gather*} \text {Result too large to show} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[(b + 2*c*x)/((d + e*x)^2*(a + b*x + c*x^2)^(3/2)),x]
[Out]
Result too large to show
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fricas [B] time = 2.55, size = 2102, normalized size = 8.48
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((2*c*x+b)/(e*x+d)^2/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")
[Out]
[-1/4*((8*a*c^2*d^3*e - 8*a*b*c*d^2*e^2 + (3*a*b^2 - 4*a^2*c)*d*e^3 + (8*c^3*d^2*e^2 - 8*b*c^2*d*e^3 + (3*b^2*
c - 4*a*c^2)*e^4)*x^3 + (8*c^3*d^3*e - (5*b^2*c + 4*a*c^2)*d*e^3 + (3*b^3 - 4*a*b*c)*e^4)*x^2 + (8*b*c^2*d^3*e
- 8*(b^2*c - a*c^2)*d^2*e^2 + 3*(b^3 - 4*a*b*c)*d*e^3 + (3*a*b^2 - 4*a^2*c)*e^4)*x)*sqrt(c*d^2 - b*d*e + a*e^
2)*log((8*a*b*d*e - 8*a^2*e^2 - (b^2 + 4*a*c)*d^2 - (8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e^2)*x^2 - 4*sqrt(c
*d^2 - b*d*e + a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x) - 2*(4*b*c*d^2 + 4*a*b*e^2 - (3*b^
2 + 4*a*c)*d*e)*x)/(e^2*x^2 + 2*d*e*x + d^2)) + 4*(2*c^3*d^5 - 6*b*c^2*d^4*e + a^2*b*e^5 + 2*(3*b^2*c - a*c^2)
*d^3*e^2 - (2*b^3 - a*b*c)*d^2*e^3 + (a*b^2 - 4*a^2*c)*d*e^4 - 3*(2*c^3*d^3*e^2 - 3*b*c^2*d^2*e^3 - a*b*c*e^5
+ (b^2*c + 2*a*c^2)*d*e^4)*x^2 - (2*c^3*d^4*e + 2*b*c^2*d^3*e^2 - (7*b^2*c - 4*a*c^2)*d^2*e^3 + (3*b^3 + 2*a*b
*c)*d*e^4 - (3*a*b^2 - 2*a^2*c)*e^5)*x)*sqrt(c*x^2 + b*x + a))/(a*c^3*d^7 - 3*a*b*c^2*d^6*e - 3*a^3*b*d^2*e^5
+ a^4*d*e^6 + 3*(a*b^2*c + a^2*c^2)*d^5*e^2 - (a*b^3 + 6*a^2*b*c)*d^4*e^3 + 3*(a^2*b^2 + a^3*c)*d^3*e^4 + (c^4
*d^6*e - 3*b*c^3*d^5*e^2 - 3*a^2*b*c*d*e^6 + a^3*c*e^7 + 3*(b^2*c^2 + a*c^3)*d^4*e^3 - (b^3*c + 6*a*b*c^2)*d^3
*e^4 + 3*(a*b^2*c + a^2*c^2)*d^2*e^5)*x^3 + (c^4*d^7 - 2*b*c^3*d^6*e + 3*a*c^3*d^5*e^2 + 3*a*b^3*d^2*e^5 + a^3
*b*e^7 + (2*b^3*c - 3*a*b*c^2)*d^4*e^3 - (b^4 + 3*a*b^2*c - 3*a^2*c^2)*d^3*e^4 - (3*a^2*b^2 - a^3*c)*d*e^6)*x^
2 + (b*c^3*d^7 + 3*b^3*c*d^5*e^2 + 3*a^3*c*d^2*e^5 - 2*a^3*b*d*e^6 + a^4*e^7 - (3*b^2*c^2 - a*c^3)*d^6*e - (b^
4 + 3*a*b^2*c - 3*a^2*c^2)*d^4*e^3 + (2*a*b^3 - 3*a^2*b*c)*d^3*e^4)*x), -1/2*((8*a*c^2*d^3*e - 8*a*b*c*d^2*e^2
+ (3*a*b^2 - 4*a^2*c)*d*e^3 + (8*c^3*d^2*e^2 - 8*b*c^2*d*e^3 + (3*b^2*c - 4*a*c^2)*e^4)*x^3 + (8*c^3*d^3*e -
(5*b^2*c + 4*a*c^2)*d*e^3 + (3*b^3 - 4*a*b*c)*e^4)*x^2 + (8*b*c^2*d^3*e - 8*(b^2*c - a*c^2)*d^2*e^2 + 3*(b^3 -
4*a*b*c)*d*e^3 + (3*a*b^2 - 4*a^2*c)*e^4)*x)*sqrt(-c*d^2 + b*d*e - a*e^2)*arctan(-1/2*sqrt(-c*d^2 + b*d*e - a
*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x)/(a*c*d^2 - a*b*d*e + a^2*e^2 + (c^2*d^2 - b*c*d*e
+ a*c*e^2)*x^2 + (b*c*d^2 - b^2*d*e + a*b*e^2)*x)) + 2*(2*c^3*d^5 - 6*b*c^2*d^4*e + a^2*b*e^5 + 2*(3*b^2*c - a
*c^2)*d^3*e^2 - (2*b^3 - a*b*c)*d^2*e^3 + (a*b^2 - 4*a^2*c)*d*e^4 - 3*(2*c^3*d^3*e^2 - 3*b*c^2*d^2*e^3 - a*b*c
*e^5 + (b^2*c + 2*a*c^2)*d*e^4)*x^2 - (2*c^3*d^4*e + 2*b*c^2*d^3*e^2 - (7*b^2*c - 4*a*c^2)*d^2*e^3 + (3*b^3 +
2*a*b*c)*d*e^4 - (3*a*b^2 - 2*a^2*c)*e^5)*x)*sqrt(c*x^2 + b*x + a))/(a*c^3*d^7 - 3*a*b*c^2*d^6*e - 3*a^3*b*d^2
*e^5 + a^4*d*e^6 + 3*(a*b^2*c + a^2*c^2)*d^5*e^2 - (a*b^3 + 6*a^2*b*c)*d^4*e^3 + 3*(a^2*b^2 + a^3*c)*d^3*e^4 +
(c^4*d^6*e - 3*b*c^3*d^5*e^2 - 3*a^2*b*c*d*e^6 + a^3*c*e^7 + 3*(b^2*c^2 + a*c^3)*d^4*e^3 - (b^3*c + 6*a*b*c^2
)*d^3*e^4 + 3*(a*b^2*c + a^2*c^2)*d^2*e^5)*x^3 + (c^4*d^7 - 2*b*c^3*d^6*e + 3*a*c^3*d^5*e^2 + 3*a*b^3*d^2*e^5
+ a^3*b*e^7 + (2*b^3*c - 3*a*b*c^2)*d^4*e^3 - (b^4 + 3*a*b^2*c - 3*a^2*c^2)*d^3*e^4 - (3*a^2*b^2 - a^3*c)*d*e^
6)*x^2 + (b*c^3*d^7 + 3*b^3*c*d^5*e^2 + 3*a^3*c*d^2*e^5 - 2*a^3*b*d*e^6 + a^4*e^7 - (3*b^2*c^2 - a*c^3)*d^6*e
- (b^4 + 3*a*b^2*c - 3*a^2*c^2)*d^4*e^3 + (2*a*b^3 - 3*a^2*b*c)*d^3*e^4)*x)]
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giac [F(-1)] 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((2*c*x+b)/(e*x+d)^2/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")
[Out]
Timed out
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maple [B] time = 0.06, size = 2293, normalized size = 9.25
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((2*c*x+b)/(e*x+d)^2/(c*x^2+b*x+a)^(3/2),x)
[Out]
18/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)/((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*b^2*c^2
*d^2+2/e/(a*e^2-b*d*e+c*d^2)/(x+d/e)/((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*c*d+6*e
/(a*e^2-b*d*e+c*d^2)^2/((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*b*c*d-12*c^2/(a*e^2-b
*d*e+c*d^2)/(4*a*c-b^2)/((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*x*b-18*e/(a*e^2-b*d*
e+c*d^2)^2/(4*a*c-b^2)/((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*x*b^2*c^2*d-3/2*e^2/(
a*e^2-b*d*e+c*d^2)^2/((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*b^2-1/(a*e^2-b*d*e+c*d^
2)/(x+d/e)/((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*b-6/(a*e^2-b*d*e+c*d^2)^2/((x+d/e
)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*c^2*d^2-2*c/(a*e^2-b*d*e+c*d^2)/((a*e^2-b*d*e+c*d^2
)/e^2)^(1/2)*ln(((b*e-2*c*d)*(x+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*
c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))+2*c/(a*e^2-b*d*e+c*d^2)/((x+d/e)^2*c+(b*e-2*c
*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)+6/(a*e^2-b*d*e+c*d^2)^2/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-
2*c*d)*(x+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/
e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*c^2*d^2+3/2*e^2/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)/((x+d/e)^2*c+(b*e
-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*b^4+3/2*e^2/(a*e^2-b*d*e+c*d^2)^2/((a*e^2-b*d*e+c*d^2)/e^2)^(
1/2)*ln(((b*e-2*c*d)*(x+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2
*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*b^2-24/e/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)/((x+d/e)^2
*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*x*c^4*d^3-9*e/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)/((x+d/
e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*b^3*c*d+24/e*c^3/(a*e^2-b*d*e+c*d^2)/(4*a*c-b^2)/(
(x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*x*d+12/e*c^2/(a*e^2-b*d*e+c*d^2)/(4*a*c-b^2)/
((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*b*d-6*c/(a*e^2-b*d*e+c*d^2)/(4*a*c-b^2)/((x+
d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*b^2+36/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)/((x+d/e
)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*x*b*c^3*d^2-12/e/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)/
((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*b*c^3*d^3-6*e/(a*e^2-b*d*e+c*d^2)^2/((a*e^2-
b*d*e+c*d^2)/e^2)^(1/2)*ln(((b*e-2*c*d)*(x+d/e)/e+2*(a*e^2-b*d*e+c*d^2)/e^2+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*
((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*b*c*d+3*e^2/(a*e^2-b*d*e+c*d^2)^2/
(4*a*c-b^2)/((x+d/e)^2*c+(b*e-2*c*d)*(x+d/e)/e+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*x*b^3*c
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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((2*c*x+b)/(e*x+d)^2/(c*x^2+b*x+a)^(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(a*e^2-b*d*e>0)', see `assume?`
for more details)Is a*e^2-b*d*e +c*d^2 positive, negative or zero?
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {b+2\,c\,x}{{\left (d+e\,x\right )}^2\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b + 2*c*x)/((d + e*x)^2*(a + b*x + c*x^2)^(3/2)),x)
[Out]
int((b + 2*c*x)/((d + e*x)^2*(a + b*x + c*x^2)^(3/2)), x)
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sympy [F(-1)] 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((2*c*x+b)/(e*x+d)**2/(c*x**2+b*x+a)**(3/2),x)
[Out]
Timed out
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