3.1248 \(\int \frac {1}{(b d+2 c d x)^3 (a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=132 \[ -\frac {12 c \sqrt {a+b x+c x^2}}{d^3 \left (b^2-4 a c\right )^2 (b+2 c x)^2}-\frac {2}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2 \sqrt {a+b x+c x^2}}-\frac {6 \sqrt {c} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{d^3 \left (b^2-4 a c\right )^{5/2}} \]

[Out]

-6*arctan(2*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^(1/2))*c^(1/2)/(-4*a*c+b^2)^(5/2)/d^3-2/(-4*a*c+b^2)/d^3/
(2*c*x+b)^2/(c*x^2+b*x+a)^(1/2)-12*c*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^2/d^3/(2*c*x+b)^2

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Rubi [A]  time = 0.08, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {687, 693, 688, 205} \[ -\frac {12 c \sqrt {a+b x+c x^2}}{d^3 \left (b^2-4 a c\right )^2 (b+2 c x)^2}-\frac {2}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2 \sqrt {a+b x+c x^2}}-\frac {6 \sqrt {c} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{d^3 \left (b^2-4 a c\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)^(3/2)),x]

[Out]

-2/((b^2 - 4*a*c)*d^3*(b + 2*c*x)^2*Sqrt[a + b*x + c*x^2]) - (12*c*Sqrt[a + b*x + c*x^2])/((b^2 - 4*a*c)^2*d^3
*(b + 2*c*x)^2) - (6*Sqrt[c]*ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]])/((b^2 - 4*a*c)^(5/2)
*d^3)

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 687

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*c*(d + e*x)^(m +
1)*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(2*c*e*(m + 2*p + 3))/(e*(p + 1)*(b^2 - 4*a
*c)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[p, -1] &&  !GtQ[m, 1] && RationalQ[m] && IntegerQ[2*p]

Rule 688

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[4*c, Subst[Int[1/(b^2*e
- 4*a*c*e + 4*c*e*x^2), x], x, Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rule 693

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-2*b*d*(d + e*x)^(m
+ 1)*(a + b*x + c*x^2)^(p + 1))/(d^2*(m + 1)*(b^2 - 4*a*c)), x] + Dist[(b^2*(m + 2*p + 3))/(d^2*(m + 1)*(b^2
- 4*a*c)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*
c, 0] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

Rubi steps

\begin {align*} \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac {2}{\left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \sqrt {a+b x+c x^2}}-\frac {(12 c) \int \frac {1}{(b d+2 c d x)^3 \sqrt {a+b x+c x^2}} \, dx}{b^2-4 a c}\\ &=-\frac {2}{\left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \sqrt {a+b x+c x^2}}-\frac {12 c \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right )^2 d^3 (b+2 c x)^2}-\frac {(6 c) \int \frac {1}{(b d+2 c d x) \sqrt {a+b x+c x^2}} \, dx}{\left (b^2-4 a c\right )^2 d^2}\\ &=-\frac {2}{\left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \sqrt {a+b x+c x^2}}-\frac {12 c \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right )^2 d^3 (b+2 c x)^2}-\frac {\left (24 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{2 b^2 c d-8 a c^2 d+8 c^2 d x^2} \, dx,x,\sqrt {a+b x+c x^2}\right )}{\left (b^2-4 a c\right )^2 d^2}\\ &=-\frac {2}{\left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \sqrt {a+b x+c x^2}}-\frac {12 c \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right )^2 d^3 (b+2 c x)^2}-\frac {6 \sqrt {c} \tan ^{-1}\left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2} d^3}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 60, normalized size = 0.45 \[ -\frac {2 \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};\frac {4 c (a+x (b+c x))}{4 a c-b^2}\right )}{d^3 \left (b^2-4 a c\right )^2 \sqrt {a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)^(3/2)),x]

[Out]

(-2*Hypergeometric2F1[-1/2, 2, 1/2, (4*c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])/((b^2 - 4*a*c)^2*d^3*Sqrt[a + x*(
b + c*x)])

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fricas [B]  time = 2.73, size = 702, normalized size = 5.32 \[ \left [\frac {3 \, {\left (4 \, c^{3} x^{4} + 8 \, b c^{2} x^{3} + a b^{2} + {\left (5 \, b^{2} c + 4 \, a c^{2}\right )} x^{2} + {\left (b^{3} + 4 \, a b c\right )} x\right )} \sqrt {-\frac {c}{b^{2} - 4 \, a c}} \log \left (-\frac {4 \, c^{2} x^{2} + 4 \, b c x - b^{2} + 8 \, a c - 4 \, \sqrt {c x^{2} + b x + a} {\left (b^{2} - 4 \, a c\right )} \sqrt {-\frac {c}{b^{2} - 4 \, a c}}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}\right ) - 2 \, {\left (6 \, c^{2} x^{2} + 6 \, b c x + b^{2} + 2 \, a c\right )} \sqrt {c x^{2} + b x + a}}{4 \, {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} d^{3} x^{4} + 8 \, {\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} d^{3} x^{3} + {\left (5 \, b^{6} c - 36 \, a b^{4} c^{2} + 48 \, a^{2} b^{2} c^{3} + 64 \, a^{3} c^{4}\right )} d^{3} x^{2} + {\left (b^{7} - 4 \, a b^{5} c - 16 \, a^{2} b^{3} c^{2} + 64 \, a^{3} b c^{3}\right )} d^{3} x + {\left (a b^{6} - 8 \, a^{2} b^{4} c + 16 \, a^{3} b^{2} c^{2}\right )} d^{3}}, -\frac {2 \, {\left (3 \, {\left (4 \, c^{3} x^{4} + 8 \, b c^{2} x^{3} + a b^{2} + {\left (5 \, b^{2} c + 4 \, a c^{2}\right )} x^{2} + {\left (b^{3} + 4 \, a b c\right )} x\right )} \sqrt {\frac {c}{b^{2} - 4 \, a c}} \arctan \left (-\frac {\sqrt {c x^{2} + b x + a} {\left (b^{2} - 4 \, a c\right )} \sqrt {\frac {c}{b^{2} - 4 \, a c}}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + {\left (6 \, c^{2} x^{2} + 6 \, b c x + b^{2} + 2 \, a c\right )} \sqrt {c x^{2} + b x + a}\right )}}{4 \, {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} d^{3} x^{4} + 8 \, {\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} d^{3} x^{3} + {\left (5 \, b^{6} c - 36 \, a b^{4} c^{2} + 48 \, a^{2} b^{2} c^{3} + 64 \, a^{3} c^{4}\right )} d^{3} x^{2} + {\left (b^{7} - 4 \, a b^{5} c - 16 \, a^{2} b^{3} c^{2} + 64 \, a^{3} b c^{3}\right )} d^{3} x + {\left (a b^{6} - 8 \, a^{2} b^{4} c + 16 \, a^{3} b^{2} c^{2}\right )} d^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(3*(4*c^3*x^4 + 8*b*c^2*x^3 + a*b^2 + (5*b^2*c + 4*a*c^2)*x^2 + (b^3 + 4*a*b*c)*x)*sqrt(-c/(b^2 - 4*a*c))*log
(-(4*c^2*x^2 + 4*b*c*x - b^2 + 8*a*c - 4*sqrt(c*x^2 + b*x + a)*(b^2 - 4*a*c)*sqrt(-c/(b^2 - 4*a*c)))/(4*c^2*x^
2 + 4*b*c*x + b^2)) - 2*(6*c^2*x^2 + 6*b*c*x + b^2 + 2*a*c)*sqrt(c*x^2 + b*x + a))/(4*(b^4*c^3 - 8*a*b^2*c^4 +
16*a^2*c^5)*d^3*x^4 + 8*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^3*x^3 + (5*b^6*c - 36*a*b^4*c^2 + 48*a^2*b^2
*c^3 + 64*a^3*c^4)*d^3*x^2 + (b^7 - 4*a*b^5*c - 16*a^2*b^3*c^2 + 64*a^3*b*c^3)*d^3*x + (a*b^6 - 8*a^2*b^4*c +
16*a^3*b^2*c^2)*d^3), -2*(3*(4*c^3*x^4 + 8*b*c^2*x^3 + a*b^2 + (5*b^2*c + 4*a*c^2)*x^2 + (b^3 + 4*a*b*c)*x)*sq
rt(c/(b^2 - 4*a*c))*arctan(-1/2*sqrt(c*x^2 + b*x + a)*(b^2 - 4*a*c)*sqrt(c/(b^2 - 4*a*c))/(c^2*x^2 + b*c*x + a
*c)) + (6*c^2*x^2 + 6*b*c*x + b^2 + 2*a*c)*sqrt(c*x^2 + b*x + a))/(4*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^3*
x^4 + 8*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^3*x^3 + (5*b^6*c - 36*a*b^4*c^2 + 48*a^2*b^2*c^3 + 64*a^3*c^4
)*d^3*x^2 + (b^7 - 4*a*b^5*c - 16*a^2*b^3*c^2 + 64*a^3*b*c^3)*d^3*x + (a*b^6 - 8*a^2*b^4*c + 16*a^3*b^2*c^2)*d
^3)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to divide, perhaps due to rounding error%%%{%%%{128,[5]%%%},[6,2,3,0]%%%}+%%%{%%%{-64,[4]%%%},[6,1,3,2]%%%}
+%%%{%%%{8,[3]%%%},[6,0,3,4]%%%}+%%%{%%{[%%%{-384,[4]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[5,2,3,1]%%%}+%%%{%%{[%%%
{192,[3]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[5,1,3,3]%%%}+%%%{%%{[%%%{-24,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[5,0,
3,5]%%%}+%%%{%%%{-384,[5]%%%},[4,3,3,0]%%%}+%%%{%%%{768,[4]%%%},[4,2,3,2]%%%}+%%%{%%%{-312,[3]%%%},[4,1,3,4]%%
%}+%%%{%%%{36,[2]%%%},[4,0,3,6]%%%}+%%%{%%{[%%%{768,[4]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[3,3,3,1]%%%}+%%%{%%{[%
%%{-896,[3]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[3,2,3,3]%%%}+%%%{%%{[%%%{304,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[3
,1,3,5]%%%}+%%%{%%{[%%%{-32,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[3,0,3,7]%%%}+%%%{%%%{384,[5]%%%},[2,4,3,0]%%%}
+%%%{%%%{-960,[4]%%%},[2,3,3,2]%%%}+%%%{%%%{696,[3]%%%},[2,2,3,4]%%%}+%%%{%%%{-192,[2]%%%},[2,1,3,6]%%%}+%%%{%
%%{18,[1]%%%},[2,0,3,8]%%%}+%%%{%%{[%%%{-384,[4]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[1,4,3,1]%%%}+%%%{%%{[%%%{576,
[3]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[1,3,3,3]%%%}+%%%{%%{[%%%{-312,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[1,2,3,5]
%%%}+%%%{%%{[%%%{72,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[1,1,3,7]%%%}+%%%{%%{[-6,0]:[1,0,%%%{-1,[1]%%%}]%%},[1,
0,3,9]%%%}+%%%{%%%{-128,[5]%%%},[0,5,3,0]%%%}+%%%{%%%{256,[4]%%%},[0,4,3,2]%%%}+%%%{%%%{-200,[3]%%%},[0,3,3,4]
%%%}+%%%{%%%{76,[2]%%%},[0,2,3,6]%%%}+%%%{%%%{-14,[1]%%%},[0,1,3,8]%%%}+%%%{1,[0,0,3,10]%%%} / %%%{%%{poly1[%%
%{-8,[4]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[6,0,0,0]%%%}+%%%{%%%{24,[4]%%%},[5,0,0,1]%%%}+%%%{%%{[%%%{24,[4]%%%},
0]:[1,0,%%%{-1,[1]%%%}]%%},[4,1,0,0]%%%}+%%%{%%{poly1[%%%{-36,[3]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[4,0,0,2]%%%}
+%%%{%%%{-48,[4]%%%},[3,1,0,1]%%%}+%%%{%%%{32,[3]%%%},[3,0,0,3]%%%}+%%%{%%{[%%%{-24,[4]%%%},0]:[1,0,%%%{-1,[1]
%%%}]%%},[2,2,0,0]%%%}+%%%{%%{[%%%{48,[3]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[2,1,0,2]%%%}+%%%{%%{poly1[%%%{-18,[2
]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[2,0,0,4]%%%}+%%%{%%%{24,[4]%%%},[1,2,0,1]%%%}+%%%{%%%{-24,[3]%%%},[1,1,0,3]%
%%}+%%%{%%%{6,[2]%%%},[1,0,0,5]%%%}+%%%{%%{[%%%{8,[4]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[0,3,0,0]%%%}+%%%{%%{[%%%
{-12,[3]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[0,2,0,2]%%%}+%%%{%%{[%%%{6,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[0,1,0,
4]%%%}+%%%{%%{poly1[%%%{-1,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[0,0,0,6]%%%} Error: Bad Argument Value

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maple [A]  time = 0.05, size = 218, normalized size = 1.65 \[ \frac {6 \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 \left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{\left (4 a c -b^{2}\right )^{2} \sqrt {\frac {4 a c -b^{2}}{c}}\, d^{3}}-\frac {3}{\left (4 a c -b^{2}\right )^{2} \sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}\, d^{3}}-\frac {1}{4 \left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{2} \sqrt {\left (x +\frac {b}{2 c}\right )^{2} c +\frac {4 a c -b^{2}}{4 c}}\, c^{2} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d^3/c^2/(4*a*c-b^2)/(x+1/2*b/c)^2/((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)-3/d^3/(4*a*c-b^2)^2/((x+1/2*b
/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)+6/d^3/(4*a*c-b^2)^2/((4*a*c-b^2)/c)^(1/2)*ln((1/2*(4*a*c-b^2)/c+1/2*((4*a*c-b
^2)/c)^(1/2)*(4*(x+1/2*b/c)^2*c+(4*a*c-b^2)/c)^(1/2))/(x+1/2*b/c))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(3/2),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(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (b\,d+2\,c\,d\,x\right )}^3\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{a b^{3} \sqrt {a + b x + c x^{2}} + 6 a b^{2} c x \sqrt {a + b x + c x^{2}} + 12 a b c^{2} x^{2} \sqrt {a + b x + c x^{2}} + 8 a c^{3} x^{3} \sqrt {a + b x + c x^{2}} + b^{4} x \sqrt {a + b x + c x^{2}} + 7 b^{3} c x^{2} \sqrt {a + b x + c x^{2}} + 18 b^{2} c^{2} x^{3} \sqrt {a + b x + c x^{2}} + 20 b c^{3} x^{4} \sqrt {a + b x + c x^{2}} + 8 c^{4} x^{5} \sqrt {a + b x + c x^{2}}}\, dx}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)**3/(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral(1/(a*b**3*sqrt(a + b*x + c*x**2) + 6*a*b**2*c*x*sqrt(a + b*x + c*x**2) + 12*a*b*c**2*x**2*sqrt(a + b*
x + c*x**2) + 8*a*c**3*x**3*sqrt(a + b*x + c*x**2) + b**4*x*sqrt(a + b*x + c*x**2) + 7*b**3*c*x**2*sqrt(a + b*
x + c*x**2) + 18*b**2*c**2*x**3*sqrt(a + b*x + c*x**2) + 20*b*c**3*x**4*sqrt(a + b*x + c*x**2) + 8*c**4*x**5*s
qrt(a + b*x + c*x**2)), x)/d**3

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