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

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

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Rubi [A]  time = 0.32, 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 = {810, 843, 621, 206, 724} \begin {gather*} -\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 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 {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]

-((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])/(4*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*Sq
rt[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 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 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 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) \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 ) \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^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 ) \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 )}{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 = 1.21, size = 389, normalized size = 1.39 \begin {gather*} \frac {\frac {2 (a+x (b+c x))^{3/2} \left (4 c e (b d-2 a e)+b^2 e^2-4 c^2 d^2\right )}{d+e x}+\frac {-2 c e \sqrt {a+x (b+c x)} \left (-2 c^2 e (2 a e (2 e x-3 d)+b d (7 d-2 e x))+b c e^2 (-10 a e+5 b d+b e x)+b^3 e^3+4 c^3 d^2 (2 d-e x)\right )+c (2 c d-b e) \sqrt {e (a e-b d)+c d^2} \left (4 c e (3 a e-2 b d)-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 )+16 c^{5/2} \left (e (a e-b d)+c d^2\right )^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{c e^3}+\frac {4 (a+x (b+c x))^{3/2} (2 c d-b e) \left (e (a e-b d)+c d^2\right )}{(d+e x)^2}}{8 \left (e (a e-b d)+c d^2\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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IntegrateAlgebraic [F]  time = 180.32, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [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)^(1/2)/(e*x+d)^3,x, algorithm="fricas")

[Out]

Timed out

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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,x):;OUTPUT:Eval
uation time: 0.51Unable to divide, perhaps due to rounding error%%%{1,[6,0,0,6,0]%%%}+%%%{%%{[-6,0]:[1,0,%%%{-
1,[1]%%%}]%%},[5,0,0,5,1]%%%}+%%%{3,[4,1,0,5,1]%%%}+%%%{-3,[4,0,1,6,0]%%%}+%%%{%%%{12,[1]%%%},[4,0,0,4,2]%%%}+
%%%{%%{[-12,0]:[1,0,%%%{-1,[1]%%%}]%%},[3,1,0,4,2]%%%}+%%%{%%{[12,0]:[1,0,%%%{-1,[1]%%%}]%%},[3,0,1,5,1]%%%}+%
%%{%%{[%%%{-8,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[3,0,0,3,3]%%%}+%%%{3,[2,2,0,4,2]%%%}+%%%{-6,[2,1,1,5,1]%%%}+
%%%{%%%{12,[1]%%%},[2,1,0,3,3]%%%}+%%%{3,[2,0,2,6,0]%%%}+%%%{%%%{-12,[1]%%%},[2,0,1,4,2]%%%}+%%%{%%{[-6,0]:[1,
0,%%%{-1,[1]%%%}]%%},[1,2,0,3,3]%%%}+%%%{%%{[12,0]:[1,0,%%%{-1,[1]%%%}]%%},[1,1,1,4,2]%%%}+%%%{%%{[-6,0]:[1,0,
%%%{-1,[1]%%%}]%%},[1,0,2,5,1]%%%}+%%%{1,[0,3,0,3,3]%%%}+%%%{-3,[0,2,1,4,2]%%%}+%%%{3,[0,1,2,5,1]%%%}+%%%{-1,[
0,0,3,6,0]%%%} / %%%{%%{poly1[%%%{1,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[6,0,0,3,0]%%%}+%%%{%%%{-6,[2]%%%},[5,0
,0,2,1]%%%}+%%%{%%{[%%%{3,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[4,1,0,2,1]%%%}+%%%{%%{poly1[%%%{-3,[1]%%%},0]:[1
,0,%%%{-1,[1]%%%}]%%},[4,0,1,3,0]%%%}+%%%{%%{poly1[%%%{12,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[4,0,0,1,2]%%%}+%
%%{%%%{-12,[2]%%%},[3,1,0,1,2]%%%}+%%%{%%%{12,[2]%%%},[3,0,1,2,1]%%%}+%%%{%%%{-8,[3]%%%},[3,0,0,0,3]%%%}+%%%{%
%{[%%%{3,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[2,2,0,1,2]%%%}+%%%{%%{[%%%{-6,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},
[2,1,1,2,1]%%%}+%%%{%%{[%%%{12,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[2,1,0,0,3]%%%}+%%%{%%{poly1[%%%{3,[1]%%%},0
]:[1,0,%%%{-1,[1]%%%}]%%},[2,0,2,3,0]%%%}+%%%{%%{poly1[%%%{-12,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[2,0,1,1,2]%
%%}+%%%{%%%{-6,[2]%%%},[1,2,0,0,3]%%%}+%%%{%%%{12,[2]%%%},[1,1,1,1,2]%%%}+%%%{%%%{-6,[2]%%%},[1,0,2,2,1]%%%}+%
%%{%%{[%%%{1,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[0,3,0,0,3]%%%}+%%%{%%{[%%%{-3,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]
%%},[0,2,1,1,2]%%%}+%%%{%%{[%%%{3,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[0,1,2,2,1]%%%}+%%%{%%{poly1[%%%{-1,[1]%%
%},0]:[1,0,%%%{-1,[1]%%%}]%%},[0,0,3,3,0]%%%} Error: Bad Argument Value

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maple [B]  time = 0.07, size = 5046, normalized size = 18.02 \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)^(1/2)/(e*x+d)^3,x)

[Out]

result too large to display

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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(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?

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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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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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