3.16.53 \(\int \frac {(b+2 c x) \sqrt {a+b x+c x^2}}{(d+e x)^3} \, dx\) [1553]

Optimal. Leaf size=280 \[ -\frac {\left (8 c^2 d^3-b e^2 (b d-2 a e)-2 c d e (3 b d-2 a e)+e \left (12 c^2 d^2+b^2 e^2-4 c e (3 b d-2 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{4 e^2 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}+\frac {2 c^{3/2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{e^3}-\frac {(2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 a e)\right ) \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{8 e^3 \left (c d^2-b d e+a e^2\right )^{3/2}} \]

[Out]

2*c^(3/2)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/e^3-1/8*(-b*e+2*c*d)*(8*c^2*d^2-b^2*e^2-4*c*e*(-3
*a*e+2*b*d))*arctanh(1/2*(b*d-2*a*e+(-b*e+2*c*d)*x)/(a*e^2-b*d*e+c*d^2)^(1/2)/(c*x^2+b*x+a)^(1/2))/e^3/(a*e^2-
b*d*e+c*d^2)^(3/2)-1/4*(8*c^2*d^3-b*e^2*(-2*a*e+b*d)-2*c*d*e*(-2*a*e+3*b*d)+e*(12*c^2*d^2+b^2*e^2-4*c*e*(-2*a*
e+3*b*d))*x)*(c*x^2+b*x+a)^(1/2)/e^2/(a*e^2-b*d*e+c*d^2)/(e*x+d)^2

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Rubi [A]
time = 0.20, antiderivative size = 280, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {824, 857, 635, 212, 738} \begin {gather*} -\frac {(2 c d-b e) \left (-4 c e (2 b d-3 a e)-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 )}{8 e^3 \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac {\sqrt {a+b x+c x^2} \left (e x \left (-4 c e (3 b d-2 a e)+b^2 e^2+12 c^2 d^2\right )-2 c d e (3 b d-2 a e)-b e^2 (b d-2 a e)+8 c^2 d^3\right )}{4 e^2 (d+e x)^2 \left (a e^2-b d e+c d^2\right )}+\frac {2 c^{3/2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*((8*c^2*d^3 - b*e^2*(b*d - 2*a*e) - 2*c*d*e*(3*b*d - 2*a*e) + e*(12*c^2*d^2 + b^2*e^2 - 4*c*e*(3*b*d - 2*
a*e))*x)*Sqrt[a + b*x + c*x^2])/(e^2*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^2) + (2*c^(3/2)*ArcTanh[(b + 2*c*x)/(2*
Sqrt[c]*Sqrt[a + b*x + c*x^2])])/e^3 - ((2*c*d - b*e)*(8*c^2*d^2 - b^2*e^2 - 4*c*e*(2*b*d - 3*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])])/(8*e^3*(c*d^2 - b*d*e +
a*e^2)^(3/2))

Rule 212

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

Rule 635

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 738

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 824

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(-(d + e*x)^(m + 1))*((a + b*x + c*x^2)^p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)))*((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), 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 857

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) \sqrt {a+b x+c x^2}}{(d+e x)^3} \, dx &=-\frac {\left (8 c^2 d^3-b e^2 (b d-2 a e)-2 c d e (3 b d-2 a e)+e \left (12 c^2 d^2+b^2 e^2-4 c e (3 b d-2 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{4 e^2 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}-\frac {\int \frac {\frac {1}{2} \left (6 b^2 c d e+8 a c^2 d e+b^3 e^2-4 b c \left (2 c d^2+3 a e^2\right )\right )-8 c^2 \left (c d^2-b d e+a e^2\right ) x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{4 e^2 \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {\left (8 c^2 d^3-b e^2 (b d-2 a e)-2 c d e (3 b d-2 a e)+e \left (12 c^2 d^2+b^2 e^2-4 c e (3 b d-2 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{4 e^2 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}+\frac {\left (2 c^2\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{e^3}-\frac {\left (8 c^2 d \left (c d^2-b d e+a e^2\right )+\frac {1}{2} e \left (6 b^2 c d e+8 a c^2 d e+b^3 e^2-4 b c \left (2 c d^2+3 a e^2\right )\right )\right ) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{4 e^3 \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {\left (8 c^2 d^3-b e^2 (b d-2 a e)-2 c d e (3 b d-2 a e)+e \left (12 c^2 d^2+b^2 e^2-4 c e (3 b d-2 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{4 e^2 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}+\frac {\left (4 c^2\right ) \text {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^3}+\frac {\left (8 c^2 d \left (c d^2-b d e+a e^2\right )+\frac {1}{2} e \left (6 b^2 c d e+8 a c^2 d e+b^3 e^2-4 b c \left (2 c d^2+3 a e^2\right )\right )\right ) \text {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 )}{2 e^3 \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {\left (8 c^2 d^3-b e^2 (b d-2 a e)-2 c d e (3 b d-2 a e)+e \left (12 c^2 d^2+b^2 e^2-4 c e (3 b d-2 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{4 e^2 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}+\frac {2 c^{3/2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{e^3}-\frac {(2 c d-b e) \left (8 c^2 d^2-b^2 e^2-4 c e (2 b d-3 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 )}{8 e^3 \left (c d^2-b d e+a e^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 10.85, size = 389, normalized size = 1.39 \begin {gather*} \frac {\frac {4 (2 c d-b e) \left (c d^2+e (-b d+a e)\right ) (a+x (b+c x))^{3/2}}{(d+e x)^2}+\frac {2 \left (-4 c^2 d^2+b^2 e^2+4 c e (b d-2 a e)\right ) (a+x (b+c x))^{3/2}}{d+e x}+\frac {-2 c e \sqrt {a+x (b+c x)} \left (b^3 e^3+4 c^3 d^2 (2 d-e x)+b c e^2 (5 b d-10 a e+b e x)-2 c^2 e (b d (7 d-2 e x)+2 a e (-3 d+2 e x))\right )+16 c^{5/2} \left (c d^2+e (-b d+a e)\right )^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )+c (2 c d-b e) \sqrt {c d^2+e (-b d+a e)} \left (8 c^2 d^2-b^2 e^2+4 c e (-2 b d+3 a e)\right ) \tanh ^{-1}\left (\frac {-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)}}\right )}{c e^3}}{8 \left (c d^2+e (-b d+a e)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*(2*c*d - b*e)*(c*d^2 + e*(-(b*d) + a*e))*(a + x*(b + c*x))^(3/2))/(d + e*x)^2 + (2*(-4*c^2*d^2 + b^2*e^2 +
4*c*e*(b*d - 2*a*e))*(a + x*(b + c*x))^(3/2))/(d + e*x) + (-2*c*e*Sqrt[a + x*(b + c*x)]*(b^3*e^3 + 4*c^3*d^2*
(2*d - e*x) + b*c*e^2*(5*b*d - 10*a*e + b*e*x) - 2*c^2*e*(b*d*(7*d - 2*e*x) + 2*a*e*(-3*d + 2*e*x))) + 16*c^(5
/2)*(c*d^2 + e*(-(b*d) + a*e))^2*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] + c*(2*c*d - b*e)*Sqrt
[c*d^2 + e*(-(b*d) + a*e)]*(8*c^2*d^2 - b^2*e^2 + 4*c*e*(-2*b*d + 3*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)])])/(c*e^3))/(8*(c*d^2 + e*(-(b*d) + a*e))^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1822\) vs. \(2(258)=516\).
time = 1.09, size = 1823, normalized size = 6.51

method	result	size
default	\(\text {Expression too large to display}\)	\(1823\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*x+b)*(c*x^2+b*x+a)^(1/2)/(e*x+d)^3,x,method=_RETURNVERBOSE)

[Out]

1/e^4*(b*e-2*c*d)*(-1/2/(a*e^2-b*d*e+c*d^2)*e^2/(x+d/e)^2*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*
d^2)/e^2)^(3/2)-1/4*e*(b*e-2*c*d)/(a*e^2-b*d*e+c*d^2)*(-1/(a*e^2-b*d*e+c*d^2)*e^2/(x+d/e)*(c*(x+d/e)^2+1/e*(b*
e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)+1/2*e*(b*e-2*c*d)/(a*e^2-b*d*e+c*d^2)*((c*(x+d/e)^2+1/e*(b*e-2
*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)+1/2/e*(b*e-2*c*d)*ln((1/2/e*(b*e-2*c*d)+c*(x+d/e))/c^(1/2)+(c*(x+
d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/c^(1/2)-(a*e^2-b*d*e+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+1/e*(b*e-2*c*d)*(x+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(x
+d/e)^2+1/e*(b*e-2*c*d)*(x+d/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)*e^2*(1/4*(2*
c*(x+d/e)+1/e*(b*e-2*c*d))/c*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)+1/8*(4*c*(a*e
^2-b*d*e+c*d^2)/e^2-1/e^2*(b*e-2*c*d)^2)/c^(3/2)*ln((1/2/e*(b*e-2*c*d)+c*(x+d/e))/c^(1/2)+(c*(x+d/e)^2+1/e*(b*
e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))))+1/2*c/(a*e^2-b*d*e+c*d^2)*e^2*((c*(x+d/e)^2+1/e*(b*e-2*c*d)
*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)+1/2/e*(b*e-2*c*d)*ln((1/2/e*(b*e-2*c*d)+c*(x+d/e))/c^(1/2)+(c*(x+d/e)^
2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/c^(1/2)-(a*e^2-b*d*e+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+1/e*(b*e-2*c*d)*(x+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(x+d/e)
^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e))))+2*c/e^3*(-1/(a*e^2-b*d*e+c*d^2)*e^2/(x+d
/e)*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)+1/2*e*(b*e-2*c*d)/(a*e^2-b*d*e+c*d^2)*
((c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)+1/2/e*(b*e-2*c*d)*ln((1/2/e*(b*e-2*c*d)+c
*(x+d/e))/c^(1/2)+(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/c^(1/2)-(a*e^2-b*d*e+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+1/e*(b*e-2*c*d)*(x+d/e)+2*((a*e^2-b*d*e
+c*d^2)/e^2)^(1/2)*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/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)*e^2*(1/4*(2*c*(x+d/e)+1/e*(b*e-2*c*d))/c*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/
e^2)^(1/2)+1/8*(4*c*(a*e^2-b*d*e+c*d^2)/e^2-1/e^2*(b*e-2*c*d)^2)/c^(3/2)*ln((1/2/e*(b*e-2*c*d)+c*(x+d/e))/c^(1
/2)+(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))))

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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)*(c*x^2+b*x+a)^(1/2)/(e*x+d)^3,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(c*d^2-%e*b*d+%e^2*a>0)', see `
assume?` for

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Fricas [F(-1)] Timed out
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)^(1/2)/(e*x+d)^3,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b + 2 c x\right ) \sqrt {a + b x + c x^{2}}}{\left (d + e x\right )^{3}}\, 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)**(1/2)/(e*x+d)**3,x)

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{1,[6,0,0,6,0]%%%}+%%%{%%{[-6,0]:[1,0,%%%{-1,[1]%%%}]%%},
[5,0,0,5,1]

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (b+2\,c\,x\right )\,\sqrt {c\,x^2+b\,x+a}}{{\left (d+e\,x\right )}^3} \,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)^(1/2))/(d + e*x)^3,x)

[Out]

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

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