3.1564 \(\int \frac{(b+2 c x) (a+b x+c x^2)^{3/2}}{(d+e x)^4} \, dx\)

Optimal. Leaf size=462 \[ \frac{\left (48 c^2 e^2 \left (a^2 e^2-4 a b d e+3 b^2 d^2\right )-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)-b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{16 e^5 \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac{\left (a+b x+c x^2\right )^{3/2} \left (3 e x \left (-4 c e (2 b d-a e)+b^2 e^2+8 c^2 d^2\right )-4 c d e (3 b d-a e)-b e^2 (b d-4 a e)+16 c^2 d^3\right )}{12 e^2 (d+e x)^3 \left (a e^2-b d e+c d^2\right )}+\frac{\sqrt{a+b x+c x^2} \left (2 c e x \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right )-16 c^2 d e (5 b d-4 a e)+4 b c e^2 (4 b d-5 a e)+b^3 e^3+64 c^3 d^3\right )}{8 e^4 (d+e x) \left (a e^2-b d e+c d^2\right )}-\frac{4 c^{3/2} (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{e^5} \]

[Out]

((64*c^3*d^3 + b^3*e^3 + 4*b*c*e^2*(4*b*d - 5*a*e) - 16*c^2*d*e*(5*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])/(8*e^4*(c*d^2 - b*d*e + a*e^2)*(d + e*x)) - ((16*c^2*d^3 -
b*e^2*(b*d - 4*a*e) - 4*c*d*e*(3*b*d - a*e) + 3*e*(8*c^2*d^2 + b^2*e^2 - 4*c*e*(2*b*d - a*e))*x)*(a + b*x + c*
x^2)^(3/2))/(12*e^2*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^3) - (4*c^(3/2)*(2*c*d - b*e)*ArcTanh[(b + 2*c*x)/(2*Sqr
t[c]*Sqrt[a + b*x + c*x^2])])/e^5 + ((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*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])])/(16*e^5*(c*d^2 - b*d*e + a*e^2)^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.625628, antiderivative size = 462, normalized size of antiderivative = 1., 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 = {810, 812, 843, 621, 206, 724} \[ \frac{\left (48 c^2 e^2 \left (a^2 e^2-4 a b d e+3 b^2 d^2\right )-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)-b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{16 e^5 \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac{\left (a+b x+c x^2\right )^{3/2} \left (3 e x \left (-4 c e (2 b d-a e)+b^2 e^2+8 c^2 d^2\right )-4 c d e (3 b d-a e)-b e^2 (b d-4 a e)+16 c^2 d^3\right )}{12 e^2 (d+e x)^3 \left (a e^2-b d e+c d^2\right )}+\frac{\sqrt{a+b x+c x^2} \left (2 c e x \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right )-16 c^2 d e (5 b d-4 a e)+4 b c e^2 (4 b d-5 a e)+b^3 e^3+64 c^3 d^3\right )}{8 e^4 (d+e x) \left (a e^2-b d e+c d^2\right )}-\frac{4 c^{3/2} (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{e^5} \]

Antiderivative was successfully verified.

[In]

Int[((b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/(d + e*x)^4,x]

[Out]

((64*c^3*d^3 + b^3*e^3 + 4*b*c*e^2*(4*b*d - 5*a*e) - 16*c^2*d*e*(5*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])/(8*e^4*(c*d^2 - b*d*e + a*e^2)*(d + e*x)) - ((16*c^2*d^3 -
b*e^2*(b*d - 4*a*e) - 4*c*d*e*(3*b*d - a*e) + 3*e*(8*c^2*d^2 + b^2*e^2 - 4*c*e*(2*b*d - a*e))*x)*(a + b*x + c*
x^2)^(3/2))/(12*e^2*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^3) - (4*c^(3/2)*(2*c*d - b*e)*ArcTanh[(b + 2*c*x)/(2*Sqr
t[c]*Sqrt[a + b*x + c*x^2])])/e^5 + ((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*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])])/(16*e^5*(c*d^2 - b*d*e + a*e^2)^(3/2))

Rule 810

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 - b*d*e + a*e^2) - d*p*(2*c*d - b*e)*(e*
f - d*g) - e*(g*(m + 1)*(c*d^2 - b*d*e + a*e^2) + p*(2*c*d - b*e)*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2
- b*d*e + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x
+ c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) + b^2*e*(d*g*(p + 1) - e*f*(m + p + 2)) + b*(a*e^2*g*(m + 1)
- c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2))) - c*(2*c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2)) - e*(2*a*e*g*(m + 1
) - b*(d*g*(m - 2*p) + e*f*(m + 2*p + 2))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*
c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 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 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)^4} \, dx &=-\frac{\left (16 c^2 d^3-b e^2 (b d-4 a e)-4 c d e (3 b d-a e)+3 e \left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{12 e^2 \left (c d^2-b d e+a e^2\right ) (d+e x)^3}-\frac{\int \frac{\left (\frac{1}{2} \left (12 b^2 c d e+16 a c^2 d e+b^3 e^2-4 b c \left (4 c d^2+5 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)^2} \, dx}{4 e^2 \left (c d^2-b d e+a e^2\right )}\\ &=\frac{\left (64 c^3 d^3+b^3 e^3+4 b c e^2 (4 b d-5 a e)-16 c^2 d e (5 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}}{8 e^4 \left (c d^2-b d e+a e^2\right ) (d+e x)}-\frac{\left (16 c^2 d^3-b e^2 (b d-4 a e)-4 c d e (3 b d-a e)+3 e \left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{12 e^2 \left (c d^2-b d e+a e^2\right ) (d+e x)^3}+\frac{\int \frac{\frac{1}{2} \left (-16 b^3 c d e^2-b^4 e^3-64 b c^2 d \left (c d^2+2 a e^2\right )+16 a c^2 e \left (4 c d^2+3 a e^2\right )+8 b^2 c e \left (10 c d^2+3 a e^2\right )\right )-32 c^2 (2 c d-b e) \left (c d^2-b d e+a e^2\right ) x}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{8 e^4 \left (c d^2-b d e+a e^2\right )}\\ &=\frac{\left (64 c^3 d^3+b^3 e^3+4 b c e^2 (4 b d-5 a e)-16 c^2 d e (5 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}}{8 e^4 \left (c d^2-b d e+a e^2\right ) (d+e x)}-\frac{\left (16 c^2 d^3-b e^2 (b d-4 a e)-4 c d e (3 b d-a e)+3 e \left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{12 e^2 \left (c d^2-b d e+a e^2\right ) (d+e x)^3}-\frac{\left (4 c^2 (2 c d-b e)\right ) \int \frac{1}{\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}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{16 e^5 \left (c d^2-b d e+a e^2\right )}\\ &=\frac{\left (64 c^3 d^3+b^3 e^3+4 b c e^2 (4 b d-5 a e)-16 c^2 d e (5 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}}{8 e^4 \left (c d^2-b d e+a e^2\right ) (d+e x)}-\frac{\left (16 c^2 d^3-b e^2 (b d-4 a e)-4 c d e (3 b d-a e)+3 e \left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{12 e^2 \left (c d^2-b d e+a e^2\right ) (d+e x)^3}-\frac{\left (8 c^2 (2 c d-b e)\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 )}{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 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 )}{8 e^5 \left (c d^2-b d e+a e^2\right )}\\ &=\frac{\left (64 c^3 d^3+b^3 e^3+4 b c e^2 (4 b d-5 a e)-16 c^2 d e (5 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}}{8 e^4 \left (c d^2-b d e+a e^2\right ) (d+e x)}-\frac{\left (16 c^2 d^3-b e^2 (b d-4 a e)-4 c d e (3 b d-a e)+3 e \left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{12 e^2 \left (c d^2-b d e+a e^2\right ) (d+e x)^3}-\frac{4 c^{3/2} (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \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 ) \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 )}{16 e^5 \left (c d^2-b d e+a e^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 3.0375, size = 485, normalized size = 1.05 \[ \frac{\frac{2 e \sqrt{a+x (b+c x)} \left (2 c e^2 \left (-4 a^2 e^2 (d+3 e x)-2 a b e \left (9 d^2+20 d e x+23 e^2 x^2\right )+b^2 d \left (24 d^2+63 d e x+55 e^2 x^2\right )\right )+b e^3 \left (-8 a^2 e^2+2 a b e (d-7 e x)+b^2 \left (3 d^2+8 d e x-3 e^2 x^2\right )\right )-8 c^2 e \left (b d \left (76 d^2 e x+30 d^3+57 d e^2 x^2+6 e^3 x^3\right )-a e \left (51 d^2 e x+20 d^3+41 d e^2 x^2+6 e^3 x^3\right )\right )+16 c^3 d^2 \left (30 d^2 e x+12 d^3+22 d e^2 x^2+3 e^3 x^3\right )\right )}{(d+e x)^3 \left (e (a e-b d)+c d^2\right )}-\frac{3 \left (48 c^2 e^2 \left (a^2 e^2-4 a b d e+3 b^2 d^2\right )-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)-b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac{2 a e-b d+b e x-2 c d x}{2 \sqrt{a+x (b+c x)} \sqrt{e (a e-b d)+c d^2}}\right )}{\left (e (a e-b d)+c d^2\right )^{3/2}}-192 c^{3/2} (2 c d-b e) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{48 e^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/(d + e*x)^4,x]

[Out]

((2*e*Sqrt[a + x*(b + c*x)]*(16*c^3*d^2*(12*d^3 + 30*d^2*e*x + 22*d*e^2*x^2 + 3*e^3*x^3) + b*e^3*(-8*a^2*e^2 +
2*a*b*e*(d - 7*e*x) + b^2*(3*d^2 + 8*d*e*x - 3*e^2*x^2)) + 2*c*e^2*(-4*a^2*e^2*(d + 3*e*x) - 2*a*b*e*(9*d^2 +
20*d*e*x + 23*e^2*x^2) + b^2*d*(24*d^2 + 63*d*e*x + 55*e^2*x^2)) - 8*c^2*e*(-(a*e*(20*d^3 + 51*d^2*e*x + 41*d
*e^2*x^2 + 6*e^3*x^3)) + b*d*(30*d^3 + 76*d^2*e*x + 57*d*e^2*x^2 + 6*e^3*x^3))))/((c*d^2 + e*(-(b*d) + a*e))*(
d + e*x)^3) - 192*c^(3/2)*(2*c*d - b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] - (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*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)]
)])/(c*d^2 + e*(-(b*d) + a*e))^(3/2))/(48*e^5)

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Maple [B]  time = 0.02, size = 15982, normalized size = 34.6 \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)^4,x)

[Out]

result too large to display

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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)^4,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)^4,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)**4,x)

[Out]

Timed out

________________________________________________________________________________________

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)^4,x, algorithm="giac")

[Out]

Exception raised: TypeError