3.1561 \(\int \frac{(b+2 c x) (a+b x+c x^2)^{3/2}}{d+e x} \, dx\)

Optimal. Leaf size=360 \[ \frac{\left (48 c^2 e^2 \left (a^2 e^2-4 a b d e+3 b^2 d^2\right )-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)-b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{64 c^{3/2} e^5}-\frac{\sqrt{a+b x+c x^2} \left (-2 c e x \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right )-16 c^2 d e (7 b d-4 a e)+4 b c e^2 (12 b d-11 a e)-b^3 e^3+64 c^3 d^3\right )}{32 c e^4}-\frac{(2 c d-b e) \left (a e^2-b d e+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{e^5}-\frac{\left (a+b x+c x^2\right )^{3/2} (-7 b e+8 c d-6 c e x)}{12 e^2} \]

[Out]

-((64*c^3*d^3 - b^3*e^3 + 4*b*c*e^2*(12*b*d - 11*a*e) - 16*c^2*d*e*(7*b*d - 4*a*e) - 2*c*e*(16*c^2*d^2 + b^2*e
^2 - 4*c*e*(4*b*d - 3*a*e))*x)*Sqrt[a + b*x + c*x^2])/(32*c*e^4) - ((8*c*d - 7*b*e - 6*c*e*x)*(a + b*x + c*x^2
)^(3/2))/(12*e^2) + ((128*c^4*d^4 - b^4*e^4 - 8*b^2*c*e^3*(2*b*d - 3*a*e) - 64*c^3*d^2*e*(4*b*d - 3*a*e) + 48*
c^2*e^2*(3*b^2*d^2 - 4*a*b*d*e + a^2*e^2))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(64*c^(3/2)
*e^5) - ((2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2)^(3/2)*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])])/e^5

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Rubi [A]  time = 0.601157, antiderivative size = 360, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {814, 843, 621, 206, 724} \[ \frac{\left (48 c^2 e^2 \left (a^2 e^2-4 a b d e+3 b^2 d^2\right )-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)-b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{64 c^{3/2} e^5}-\frac{\sqrt{a+b x+c x^2} \left (-2 c e x \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right )-16 c^2 d e (7 b d-4 a e)+4 b c e^2 (12 b d-11 a e)-b^3 e^3+64 c^3 d^3\right )}{32 c e^4}-\frac{(2 c d-b e) \left (a e^2-b d e+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{e^5}-\frac{\left (a+b x+c x^2\right )^{3/2} (-7 b e+8 c d-6 c e x)}{12 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((64*c^3*d^3 - b^3*e^3 + 4*b*c*e^2*(12*b*d - 11*a*e) - 16*c^2*d*e*(7*b*d - 4*a*e) - 2*c*e*(16*c^2*d^2 + b^2*e
^2 - 4*c*e*(4*b*d - 3*a*e))*x)*Sqrt[a + b*x + c*x^2])/(32*c*e^4) - ((8*c*d - 7*b*e - 6*c*e*x)*(a + b*x + c*x^2
)^(3/2))/(12*e^2) + ((128*c^4*d^4 - b^4*e^4 - 8*b^2*c*e^3*(2*b*d - 3*a*e) - 64*c^3*d^2*e*(4*b*d - 3*a*e) + 48*
c^2*e^2*(3*b^2*d^2 - 4*a*b*d*e + a^2*e^2))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(64*c^(3/2)
*e^5) - ((2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2)^(3/2)*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])])/e^5

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]

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 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]

Rubi steps

\begin{align*} \int \frac{(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx &=-\frac{(8 c d-7 b e-6 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^2}-\frac{\int \frac{\left (c \left (7 b^2 d e+4 a c d e-8 b \left (c d^2+a e^2\right )\right )-c \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{d+e x} \, dx}{8 c e^2}\\ &=-\frac{\left (64 c^3 d^3-b^3 e^3+4 b c e^2 (12 b d-11 a e)-16 c^2 d e (7 b d-4 a e)-2 c e \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{32 c e^4}-\frac{(8 c d-7 b e-6 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^2}+\frac{\int \frac{\frac{1}{2} c \left (d \left (4 b c d-b^2 e-4 a c e\right ) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right )+4 c e (b d-2 a e) \left (7 b^2 d e+4 a c d e-8 b \left (c d^2+a e^2\right )\right )\right )+\frac{1}{2} c \left (128 c^4 d^4-b^4 e^4-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)+48 c^2 e^2 \left (3 b^2 d^2-4 a b d e+a^2 e^2\right )\right ) x}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{32 c^2 e^4}\\ &=-\frac{\left (64 c^3 d^3-b^3 e^3+4 b c e^2 (12 b d-11 a e)-16 c^2 d e (7 b d-4 a e)-2 c e \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{32 c e^4}-\frac{(8 c d-7 b e-6 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^2}-\frac{\left ((2 c d-b e) \left (c d^2-b d e+a e^2\right )^2\right ) \int \frac{1}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{e^5}+\frac{\left (128 c^4 d^4-b^4 e^4-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)+48 c^2 e^2 \left (3 b^2 d^2-4 a b d e+a^2 e^2\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{64 c e^5}\\ &=-\frac{\left (64 c^3 d^3-b^3 e^3+4 b c e^2 (12 b d-11 a e)-16 c^2 d e (7 b d-4 a e)-2 c e \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{32 c e^4}-\frac{(8 c d-7 b e-6 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^2}+\frac{\left (2 (2 c d-b e) \left (c d^2-b d e+a e^2\right )^2\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 )}{e^5}+\frac{\left (128 c^4 d^4-b^4 e^4-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)+48 c^2 e^2 \left (3 b^2 d^2-4 a b d e+a^2 e^2\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 )}{32 c e^5}\\ &=-\frac{\left (64 c^3 d^3-b^3 e^3+4 b c e^2 (12 b d-11 a e)-16 c^2 d e (7 b d-4 a e)-2 c e \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{32 c e^4}-\frac{(8 c d-7 b e-6 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^2}+\frac{\left (128 c^4 d^4-b^4 e^4-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)+48 c^2 e^2 \left (3 b^2 d^2-4 a b d e+a^2 e^2\right )\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{64 c^{3/2} e^5}-\frac{(2 c d-b e) \left (c d^2-b d e+a e^2\right )^{3/2} \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 )}{e^5}\\ \end{align*}

Mathematica [A]  time = 0.780454, size = 342, normalized size = 0.95 \[ \frac{3 \left (48 c^2 e^2 \left (a^2 e^2-4 a b d e+3 b^2 d^2\right )-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)-b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )+2 \sqrt{c} \left (e \sqrt{a+x (b+c x)} \left (8 c^2 e \left (a e (15 e x-32 d)+b \left (42 d^2-20 d e x+13 e^2 x^2\right )\right )+2 b c e^2 (94 a e-72 b d+31 b e x)+3 b^3 e^3-16 c^3 \left (-6 d^2 e x+12 d^3+4 d e^2 x^2-3 e^3 x^3\right )\right )+96 c (2 c d-b e) \left (e (a e-b d)+c d^2\right )^{3/2} \tanh ^{-1}\left (\frac{2 a e-b d+b e x-2 c d x}{2 \sqrt{a+x (b+c x)} \sqrt{e (a e-b d)+c d^2}}\right )\right )}{192 c^{3/2} e^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(128*c^4*d^4 - b^4*e^4 - 8*b^2*c*e^3*(2*b*d - 3*a*e) - 64*c^3*d^2*e*(4*b*d - 3*a*e) + 48*c^2*e^2*(3*b^2*d^2
- 4*a*b*d*e + a^2*e^2))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] + 2*Sqrt[c]*(e*Sqrt[a + x*(b +
c*x)]*(3*b^3*e^3 + 2*b*c*e^2*(-72*b*d + 94*a*e + 31*b*e*x) - 16*c^3*(12*d^3 - 6*d^2*e*x + 4*d*e^2*x^2 - 3*e^3
*x^3) + 8*c^2*e*(a*e*(-32*d + 15*e*x) + b*(42*d^2 - 20*d*e*x + 13*e^2*x^2))) + 96*c*(2*c*d - b*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)])]))/(192*c^(3/2)*e^5)

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Maple [B]  time = 0.011, size = 3119, normalized size = 8.7 \begin{align*} \text{output too large to display} \end{align*}

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),x)

[Out]

3/4*c/e*(c*x^2+b*x+a)^(1/2)*x*a-3/8/c^(1/2)/e*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^2*a-6/e^3/((a*e^2-
b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*
((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*a*b*d^2*c+4/e^4/((a*e^2-b*d*e+c*d^
2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2
*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*b^2*d^3*c-1/e^2*((x+d/e)^2*c+(b*e-2*c*d)/e*(
x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*x*b*c*d+2/e^2/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/
e^2+(b*e-2*c*d)/e*(x+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*
d^2)/e^2)^(1/2))/(x+d/e))*a^2*c*d+4/e^4/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c
*d)/e*(x+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1
/2))/(x+d/e))*a*c^2*d^3+2/e^2/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x+d
/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/
e))*a*b^2*d-5/e^5/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*((a*e^
2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*b*d^4*c^
2-3/e^2*c^(1/2)*ln((1/2*(b*e-2*c*d)/e+(x+d/e)*c)/c^(1/2)+((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2
)/e^2)^(1/2))*a*b*d+3/8/e*(c*x^2+b*x+a)^(1/2)*b*a-3/32/c/e*(c*x^2+b*x+a)^(1/2)*b^3+1/2*c/e*(c*x^2+b*x+a)^(3/2)
*x-3/16/e*(c*x^2+b*x+a)^(1/2)*x*b^2+2/e^5*ln((1/2*(b*e-2*c*d)/e+(x+d/e)*c)/c^(1/2)+((x+d/e)^2*c+(b*e-2*c*d)/e*
(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*c^(5/2)*d^4+3/64/c^(3/2)/e*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))
*b^4+3/4*c^(1/2)/e*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^2-2/3/e^2*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+
(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)*c*d-2/e^4*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*c^2
*d^3+1/4/e*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*x*b^2+1/8/e/c*((x+d/e)^2*c+(b*e-2
*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*b^3-1/16/e/c^(3/2)*ln((1/2*(b*e-2*c*d)/e+(x+d/e)*c)/c^(1/2)+((x
+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*b^4+1/e*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*
e^2-b*d*e+c*d^2)/e^2)^(1/2)*a*b-3/2/e^2*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*b^2*
d+1/3/e*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)*b+1/4/e*(c*x^2+b*x+a)^(3/2)*b+1/e^3*
((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*x*c^2*d^2+7/2/e^3*((x+d/e)^2*c+(b*e-2*c*d)/e
*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*b*d^2*c+3/4/e/c^(1/2)*ln((1/2*(b*e-2*c*d)/e+(x+d/e)*c)/c^(1/2)+((x+d/e
)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*a*b^2+3/e^3*ln((1/2*(b*e-2*c*d)/e+(x+d/e)*c)/c^(1/
2)+((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*c^(3/2)*d^2*a-4/e^4*ln((1/2*(b*e-2*c*d)/
e+(x+d/e)*c)/c^(1/2)+((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*c^(3/2)*d^3*b+9/4/e^3*
ln((1/2*(b*e-2*c*d)/e+(x+d/e)*c)/c^(1/2)+((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*c^
(1/2)*b^2*d^2-1/4/e^2/c^(1/2)*ln((1/2*(b*e-2*c*d)/e+(x+d/e)*c)/c^(1/2)+((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e
^2-b*d*e+c*d^2)/e^2)^(1/2))*b^3*d-2/e^2*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*a*c*
d-1/e^3/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*((a*e^2-b*d*e+c*
d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*b^3*d^2+2/e^6/((a*
e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1
/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*c^3*d^5-1/e/((a*e^2-b*d*e+c*d^
2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2
*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))*a^2*b

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

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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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),x, algorithm="fricas")

[Out]

Timed out

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

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),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

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),x, algorithm="giac")

[Out]

Exception raised: TypeError