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

Optimal. Leaf size=309 \[ \frac {3 \sqrt {c} \left (-4 c e (4 b d-a e)+3 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{4 e^5}-\frac {3 (2 c d-b e) \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{8 e^5 \sqrt {a e^2-b d e+c d^2}}-\frac {3 \sqrt {a+b x+c x^2} \left (-4 c e (3 b d-a e)+b^2 e^2+4 c e x (2 c d-b e)+16 c^2 d^2\right )}{4 e^4 (d+e x)}+\frac {\left (a+b x+c x^2\right )^{3/2} (-b e+4 c d+2 c e x)}{2 e^2 (d+e x)^2} \]

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Rubi [A]  time = 0.37, antiderivative size = 309, normalized size of antiderivative = 1.00, 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 = {812, 843, 621, 206, 724} \begin {gather*} -\frac {3 \sqrt {a+b x+c x^2} \left (-4 c e (3 b d-a e)+b^2 e^2+4 c e x (2 c d-b e)+16 c^2 d^2\right )}{4 e^4 (d+e x)}+\frac {3 \sqrt {c} \left (-4 c e (4 b d-a e)+3 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 e^5}-\frac {3 (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 {-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^5 \sqrt {a e^2-b d e+c d^2}}+\frac {\left (a+b x+c x^2\right )^{3/2} (-b e+4 c d+2 c e x)}{2 e^2 (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 2.31, size = 544, normalized size = 1.76 \begin {gather*} \frac {-\frac {(a+x (b+c x))^{5/2} \left (4 c e (2 a e-3 b d)+b^2 e^2+12 c^2 d^2\right )}{2 (d+e x) \left (e (a e-b d)+c d^2\right )}-\frac {2 e^3 (a+x (b+c x))^{3/2} \left (2 c^2 e (2 a e (2 e x-3 d)+3 b d (5 d-2 e x))+b c e^2 (10 a e-15 b d+b e x)+b^3 e^3-4 c^3 d^2 (4 d-3 e x)\right )+6 e \sqrt {a+x (b+c x)} \left (e (a e-b d)+c d^2\right ) \left (4 c^2 e (a e (e x-3 d)+b d (7 d-2 e x))+b c e^2 (8 a e-13 b d+b e x)+b^3 e^3+8 c^3 d^2 (e x-2 d)\right )+6 \sqrt {c} \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 )^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )+3 (2 c d-b e) \left (e (a e-b d)+c d^2\right )^{3/2} \left (4 c e (3 a e-4 b d)+b^2 e^2+16 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 )}{4 e^5 \left (e (b d-a e)-c d^2\right )}+\frac {(a+x (b+c x))^{5/2} (2 c d-b e)}{(d+e x)^2}}{2 \left (e (a e-b d)+c d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*c*d - b*e)*(a + x*(b + c*x))^(5/2))/(d + e*x)^2 - ((12*c^2*d^2 + b^2*e^2 + 4*c*e*(-3*b*d + 2*a*e))*(a + x
*(b + c*x))^(5/2))/(2*(c*d^2 + e*(-(b*d) + a*e))*(d + e*x)) - (6*e*(c*d^2 + e*(-(b*d) + a*e))*Sqrt[a + x*(b +
c*x)]*(b^3*e^3 + 8*c^3*d^2*(-2*d + e*x) + b*c*e^2*(-13*b*d + 8*a*e + b*e*x) + 4*c^2*e*(b*d*(7*d - 2*e*x) + a*e
*(-3*d + e*x))) + 2*e^3*(a + x*(b + c*x))^(3/2)*(b^3*e^3 - 4*c^3*d^2*(4*d - 3*e*x) + b*c*e^2*(-15*b*d + 10*a*e
+ b*e*x) + 2*c^2*e*(3*b*d*(5*d - 2*e*x) + 2*a*e*(-3*d + 2*e*x))) + 6*Sqrt[c]*(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))^2*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] + 3*(2*c*d
- b*e)*(c*d^2 + e*(-(b*d) + a*e))^(3/2)*(16*c^2*d^2 + b^2*e^2 + 4*c*e*(-4*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)])])/(4*e^5*(-(c*d^2) + e*(b*d - a
*e))))/(2*(c*d^2 + e*(-(b*d) + a*e)))

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IntegrateAlgebraic [F]  time = 180.04, 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)*(a + b*x + c*x^2)^(3/2))/(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)^(3/2)/(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)^(3/2)/(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: 0.82Unable to divide, perhaps due to rounding error%%%{1,[6,0,8,0,0]%%%}+%%%{%%{[-6,0]:[1,0,%%%{-
1,[1]%%%}]%%},[5,0,7,0,1]%%%}+%%%{-3,[4,1,8,0,0]%%%}+%%%{3,[4,0,7,1,1]%%%}+%%%{%%%{12,[1]%%%},[4,0,6,0,2]%%%}+
%%%{%%{[12,0]:[1,0,%%%{-1,[1]%%%}]%%},[3,1,7,0,1]%%%}+%%%{%%{[-12,0]:[1,0,%%%{-1,[1]%%%}]%%},[3,0,6,1,2]%%%}+%
%%{%%{[%%%{-8,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[3,0,5,0,3]%%%}+%%%{3,[2,2,8,0,0]%%%}+%%%{-6,[2,1,7,1,1]%%%}+
%%%{%%%{-12,[1]%%%},[2,1,6,0,2]%%%}+%%%{3,[2,0,6,2,2]%%%}+%%%{%%%{12,[1]%%%},[2,0,5,1,3]%%%}+%%%{%%{[-6,0]:[1,
0,%%%{-1,[1]%%%}]%%},[1,2,7,0,1]%%%}+%%%{%%{[12,0]:[1,0,%%%{-1,[1]%%%}]%%},[1,1,6,1,2]%%%}+%%%{%%{[-6,0]:[1,0,
%%%{-1,[1]%%%}]%%},[1,0,5,2,3]%%%}+%%%{-1,[0,3,8,0,0]%%%}+%%%{3,[0,2,7,1,1]%%%}+%%%{-3,[0,1,6,2,2]%%%}+%%%{1,[
0,0,5,3,3]%%%} / %%%{%%{poly1[%%%{1,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[6,0,3,0,0]%%%}+%%%{%%%{-6,[2]%%%},[5,0
,2,0,1]%%%}+%%%{%%{[%%%{-3,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[4,1,3,0,0]%%%}+%%%{%%{poly1[%%%{3,[1]%%%},0]:[1
,0,%%%{-1,[1]%%%}]%%},[4,0,2,1,1]%%%}+%%%{%%{poly1[%%%{12,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[4,0,1,0,2]%%%}+%
%%{%%%{12,[2]%%%},[3,1,2,0,1]%%%}+%%%{%%%{-12,[2]%%%},[3,0,1,1,2]%%%}+%%%{%%%{-8,[3]%%%},[3,0,0,0,3]%%%}+%%%{%
%{[%%%{3,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[2,2,3,0,0]%%%}+%%%{%%{[%%%{-6,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},
[2,1,2,1,1]%%%}+%%%{%%{[%%%{-12,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[2,1,1,0,2]%%%}+%%%{%%{poly1[%%%{3,[1]%%%},
0]:[1,0,%%%{-1,[1]%%%}]%%},[2,0,1,2,2]%%%}+%%%{%%{poly1[%%%{12,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[2,0,0,1,3]%
%%}+%%%{%%%{-6,[2]%%%},[1,2,2,0,1]%%%}+%%%{%%%{12,[2]%%%},[1,1,1,1,2]%%%}+%%%{%%%{-6,[2]%%%},[1,0,0,2,3]%%%}+%
%%{%%{[%%%{-1,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[0,3,3,0,0]%%%}+%%%{%%{[%%%{3,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]
%%},[0,2,2,1,1]%%%}+%%%{%%{[%%%{-3,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[0,1,1,2,2]%%%}+%%%{%%{poly1[%%%{1,[1]%%
%},0]:[1,0,%%%{-1,[1]%%%}]%%},[0,0,0,3,3]%%%} Error: Bad Argument Value

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maple [B]  time = 0.08, size = 10362, normalized size = 33.53 \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)^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)^(3/2)/(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 )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{{\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)^(3/2))/(d + e*x)^3,x)

[Out]

int(((b + 2*c*x)*(a + b*x + c*x^2)^(3/2))/(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 ) \left (a + b x + c x^{2}\right )^{\frac {3}{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)**(3/2)/(e*x+d)**3,x)

[Out]

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

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