3.14.70 \(\int \frac {(b+2 c x) (a+b x+c x^2)^{3/2}}{(d+e x)^2} \, dx\)
Optimal. Leaf size=303 \[ -\frac {(2 c d-b e) \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{4 \sqrt {c} e^5}+\frac {\sqrt {a e^2-b d e+c d^2} \left (-4 c e (4 b d-a e)+3 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{2 e^5}+\frac {\sqrt {a+b x+c x^2} \left (-4 c e (5 b d-a e)+5 b^2 e^2-4 c e x (2 c d-b e)+16 c^2 d^2\right )}{2 e^4}+\frac {\left (a+b x+c x^2\right )^{3/2} (-3 b e+8 c d+2 c e x)}{3 e^2 (d+e x)} \]
________________________________________________________________________________________
Rubi [A] time = 0.48, antiderivative size = 303, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used =
{812, 814, 843, 621, 206, 724} \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (-4 c e (5 b d-a e)+5 b^2 e^2-4 c e x (2 c d-b e)+16 c^2 d^2\right )}{2 e^4}-\frac {(2 c d-b e) \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{4 \sqrt {c} e^5}+\frac {\sqrt {a e^2-b d e+c d^2} \left (-4 c e (4 b d-a e)+3 b^2 e^2+16 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 e^5}+\frac {\left (a+b x+c x^2\right )^{3/2} (-3 b e+8 c d+2 c e x)}{3 e^2 (d+e x)} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[((b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/(d + e*x)^2,x]
[Out]
((16*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(5*b*d - a*e) - 4*c*e*(2*c*d - b*e)*x)*Sqrt[a + b*x + c*x^2])/(2*e^4) + ((8*c
*d - 3*b*e + 2*c*e*x)*(a + b*x + c*x^2)^(3/2))/(3*e^2*(d + e*x)) - ((2*c*d - b*e)*(16*c^2*d^2 + b^2*e^2 - 4*c*
e*(4*b*d - 3*a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(4*Sqrt[c]*e^5) + (Sqrt[c*d^2 - b*d
*e + a*e^2]*(16*c^2*d^2 + 3*b^2*e^2 - 4*c*e*(4*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*e^5)
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 621
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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 812
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + b*x + c*x^2)^p)/(e^2*(m + 1)*(m
+ 2*p + 2)), x] + Dist[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1)*Simp[g*(
b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m + 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p
+ 2))*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] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] && !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Rule 814
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*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] && GtQ[p, 0] && (IntegerQ[p] || !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) && !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Rule 843
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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]
&& !IGtQ[m, 0]
Rubi steps
\begin {align*} \int \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{(d+e x)^2} \, dx &=\frac {(8 c d-3 b e+2 c e x) \left (a+b x+c x^2\right )^{3/2}}{3 e^2 (d+e x)}-\frac {\int \frac {\left (8 b c d-3 b^2 e-4 a c e+8 c (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{d+e x} \, dx}{2 e^2}\\ &=\frac {\left (16 c^2 d^2+5 b^2 e^2-4 c e (5 b d-a e)-4 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{2 e^4}+\frac {(8 c d-3 b e+2 c e x) \left (a+b x+c x^2\right )^{3/2}}{3 e^2 (d+e x)}+\frac {\int \frac {2 c \left (e (b d-2 a e) \left (8 b c d-3 b^2 e-4 a c e\right )-2 d (2 c d-b e) \left (4 b c d-b^2 e-4 a c e\right )\right )-2 c (2 c d-b e) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{8 c e^4}\\ &=\frac {\left (16 c^2 d^2+5 b^2 e^2-4 c e (5 b d-a e)-4 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{2 e^4}+\frac {(8 c d-3 b e+2 c e x) \left (a+b x+c x^2\right )^{3/2}}{3 e^2 (d+e x)}-\frac {\left ((2 c d-b e) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{4 e^5}+\frac {\left (2 c d (2 c d-b e) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right )+2 c e \left (e (b d-2 a e) \left (8 b c d-3 b^2 e-4 a c e\right )-2 d (2 c d-b e) \left (4 b c d-b^2 e-4 a c e\right )\right )\right ) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{8 c e^5}\\ &=\frac {\left (16 c^2 d^2+5 b^2 e^2-4 c e (5 b d-a e)-4 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{2 e^4}+\frac {(8 c d-3 b e+2 c e x) \left (a+b x+c x^2\right )^{3/2}}{3 e^2 (d+e x)}-\frac {\left ((2 c d-b e) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{2 e^5}-\frac {\left (2 c d (2 c d-b e) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right )+2 c e \left (e (b d-2 a e) \left (8 b c d-3 b^2 e-4 a c e\right )-2 d (2 c d-b e) \left (4 b c d-b^2 e-4 a c e\right )\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 )}{4 c e^5}\\ &=\frac {\left (16 c^2 d^2+5 b^2 e^2-4 c e (5 b d-a e)-4 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{2 e^4}+\frac {(8 c d-3 b e+2 c e x) \left (a+b x+c x^2\right )^{3/2}}{3 e^2 (d+e x)}-\frac {(2 c d-b e) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{4 \sqrt {c} e^5}+\frac {\sqrt {c d^2-b d e+a e^2} \left (16 c^2 d^2-16 b c d e+3 b^2 e^2+4 a c e^2\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 e^5}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 1.06, size = 395, normalized size = 1.30 \begin {gather*} \frac {\frac {(a+x (b+c x))^{3/2} \left (c e (-2 a e+11 b d-3 b e x)-3 b^2 e^2+c^2 \left (6 d e x-8 d^2\right )\right )}{3 e^2}+\frac {-2 c^2 e \sqrt {a+x (b+c x)} \left (e (a e-b d)+c d^2\right ) \left (4 c e (a e-5 b d+b e x)+5 b^2 e^2+8 c^2 d (2 d-e x)\right )+2 c^2 \left (4 c e (a e-4 b d)+3 b^2 e^2+16 c^2 d^2\right ) \left (e (a e-b d)+c d^2\right )^{3/2} \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 )+c^{3/2} (2 c d-b e) \left (e (a e-b d)+c d^2\right ) \left (4 c e (3 a e-4 b d)+b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{4 c^2 e^5}+\frac {(a+x (b+c x))^{5/2} (b e-2 c d)}{d+e x}}{e (b d-a e)-c d^2} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[((b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/(d + e*x)^2,x]
[Out]
(((-2*c*d + b*e)*(a + x*(b + c*x))^(5/2))/(d + e*x) + ((a + x*(b + c*x))^(3/2)*(-3*b^2*e^2 + c*e*(11*b*d - 2*a
*e - 3*b*e*x) + c^2*(-8*d^2 + 6*d*e*x)))/(3*e^2) + (-2*c^2*e*(c*d^2 + e*(-(b*d) + a*e))*Sqrt[a + x*(b + c*x)]*
(5*b^2*e^2 + 8*c^2*d*(2*d - e*x) + 4*c*e*(-5*b*d + a*e + b*e*x)) + c^(3/2)*(2*c*d - b*e)*(c*d^2 + e*(-(b*d) +
a*e))*(16*c^2*d^2 + b^2*e^2 + 4*c*e*(-4*b*d + 3*a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] +
2*c^2*(16*c^2*d^2 + 3*b^2*e^2 + 4*c*e*(-4*b*d + a*e))*(c*d^2 + e*(-(b*d) + a*e))^(3/2)*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)])])/(4*c^2*e^5))/(-(c*d^2) + e*(b*
d - a*e))
________________________________________________________________________________________
IntegrateAlgebraic [B] time = 11.55, size = 6952, normalized size = 22.94 \begin {gather*} \text {Result too large to show} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[((b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/(d + e*x)^2,x]
[Out]
Result too large to show
________________________________________________________________________________________
fricas [A] time = 127.99, size = 2231, normalized size = 7.36
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((2*c*x+b)*(c*x^2+b*x+a)^(3/2)/(e*x+d)^2,x, algorithm="fricas")
[Out]
[-1/24*(3*(32*c^3*d^4 - 48*b*c^2*d^3*e + 6*(3*b^2*c + 4*a*c^2)*d^2*e^2 - (b^3 + 12*a*b*c)*d*e^3 + (32*c^3*d^3*
e - 48*b*c^2*d^2*e^2 + 6*(3*b^2*c + 4*a*c^2)*d*e^3 - (b^3 + 12*a*b*c)*e^4)*x)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x
- b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 6*(16*c^3*d^3 - 16*b*c^2*d^2*e + (3*b^2*c + 4*
a*c^2)*d*e^2 + (16*c^3*d^2*e - 16*b*c^2*d*e^2 + (3*b^2*c + 4*a*c^2)*e^3)*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*(4*c^3*e^4*x^3 + 48*c^3*d^3*e - 60*b*c^2*d^2*e^2 - 6*a*b*c*e^4 + (15*
b^2*c + 28*a*c^2)*d*e^3 - 2*(4*c^3*d*e^3 - 5*b*c^2*e^4)*x^2 + (24*c^3*d^2*e^2 - 32*b*c^2*d*e^3 + (9*b^2*c + 16
*a*c^2)*e^4)*x)*sqrt(c*x^2 + b*x + a))/(c*e^6*x + c*d*e^5), 1/12*(3*(32*c^3*d^4 - 48*b*c^2*d^3*e + 6*(3*b^2*c
+ 4*a*c^2)*d^2*e^2 - (b^3 + 12*a*b*c)*d*e^3 + (32*c^3*d^3*e - 48*b*c^2*d^2*e^2 + 6*(3*b^2*c + 4*a*c^2)*d*e^3 -
(b^3 + 12*a*b*c)*e^4)*x)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*
c)) + 3*(16*c^3*d^3 - 16*b*c^2*d^2*e + (3*b^2*c + 4*a*c^2)*d*e^2 + (16*c^3*d^2*e - 16*b*c^2*d*e^2 + (3*b^2*c +
4*a*c^2)*e^3)*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)) + 2*(4*c^3*e^4*x^3 +
48*c^3*d^3*e - 60*b*c^2*d^2*e^2 - 6*a*b*c*e^4 + (15*b^2*c + 28*a*c^2)*d*e^3 - 2*(4*c^3*d*e^3 - 5*b*c^2*e^4)*x^
2 + (24*c^3*d^2*e^2 - 32*b*c^2*d*e^3 + (9*b^2*c + 16*a*c^2)*e^4)*x)*sqrt(c*x^2 + b*x + a))/(c*e^6*x + c*d*e^5)
, 1/24*(12*(16*c^3*d^3 - 16*b*c^2*d^2*e + (3*b^2*c + 4*a*c^2)*d*e^2 + (16*c^3*d^2*e - 16*b*c^2*d*e^2 + (3*b^2*
c + 4*a*c^2)*e^3)*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)) - 3*(32*c^3*d^4 - 48*b*c^2*d^3*e + 6*(3*b^2*c + 4*a*c^2)*d^2*e^2 - (b^3 + 12*a*b*c)*
d*e^3 + (32*c^3*d^3*e - 48*b*c^2*d^2*e^2 + 6*(3*b^2*c + 4*a*c^2)*d*e^3 - (b^3 + 12*a*b*c)*e^4)*x)*sqrt(c)*log(
-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(4*c^3*e^4*x^3 + 48*c^3*
d^3*e - 60*b*c^2*d^2*e^2 - 6*a*b*c*e^4 + (15*b^2*c + 28*a*c^2)*d*e^3 - 2*(4*c^3*d*e^3 - 5*b*c^2*e^4)*x^2 + (24
*c^3*d^2*e^2 - 32*b*c^2*d*e^3 + (9*b^2*c + 16*a*c^2)*e^4)*x)*sqrt(c*x^2 + b*x + a))/(c*e^6*x + c*d*e^5), 1/12*
(6*(16*c^3*d^3 - 16*b*c^2*d^2*e + (3*b^2*c + 4*a*c^2)*d*e^2 + (16*c^3*d^2*e - 16*b*c^2*d*e^2 + (3*b^2*c + 4*a*
c^2)*e^3)*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)) + 3*(32*c^3*d^4 - 48*b*c^2*d^3*e + 6*(3*b^2*c + 4*a*c^2)*d^2*e^2 - (b^3 + 12*a*b*c)*d*e^3 +
(32*c^3*d^3*e - 48*b*c^2*d^2*e^2 + 6*(3*b^2*c + 4*a*c^2)*d*e^3 - (b^3 + 12*a*b*c)*e^4)*x)*sqrt(-c)*arctan(1/2*
sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(4*c^3*e^4*x^3 + 48*c^3*d^3*e - 60*b*c
^2*d^2*e^2 - 6*a*b*c*e^4 + (15*b^2*c + 28*a*c^2)*d*e^3 - 2*(4*c^3*d*e^3 - 5*b*c^2*e^4)*x^2 + (24*c^3*d^2*e^2 -
32*b*c^2*d*e^3 + (9*b^2*c + 16*a*c^2)*e^4)*x)*sqrt(c*x^2 + b*x + a))/(c*e^6*x + c*d*e^5)]
________________________________________________________________________________________
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)*(c*x^2+b*x+a)^(3/2)/(e*x+d)^2,x, algorithm="giac")
[Out]
Timed out
________________________________________________________________________________________
maple [B] time = 0.07, size = 6898, normalized size = 22.77 \begin {gather*} \text {output too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((2*c*x+b)*(c*x^2+b*x+a)^(3/2)/(e*x+d)^2,x)
[Out]
result too large to display
________________________________________________________________________________________
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)*(c*x^2+b*x+a)^(3/2)/(e*x+d)^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 zero or nonzero?
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (b+2\,c\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{{\left (d+e\,x\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/(d + e*x)^2,x)
[Out]
int(((b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/(d + e*x)^2, x)
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b + 2 c x\right ) \left (a + b x + c x^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((2*c*x+b)*(c*x**2+b*x+a)**(3/2)/(e*x+d)**2,x)
[Out]
Integral((b + 2*c*x)*(a + b*x + c*x**2)**(3/2)/(d + e*x)**2, x)
________________________________________________________________________________________