\(\int \frac {(b d+2 c d x)^4}{a+b x+c x^2} \, dx\) [1158]

Optimal result

Integrand size = 24, antiderivative size = 72 \[ \int \frac {(b d+2 c d x)^4}{a+b x+c x^2} \, dx=2 \left (b^2-4 a c\right ) d^4 (b+2 c x)+\frac {2}{3} d^4 (b+2 c x)^3-2 \left (b^2-4 a c\right )^{3/2} d^4 \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {706, 632, 212} \[ \int \frac {(b d+2 c d x)^4}{a+b x+c x^2} \, dx=-2 d^4 \left (b^2-4 a c\right )^{3/2} \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )+2 d^4 \left (b^2-4 a c\right ) (b+2 c x)+\frac {2}{3} d^4 (b+2 c x)^3 \]

[In]

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

[Out]

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

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 632

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

Rule 706

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2*d*(d + e*x)^(m - 1
)*((a + b*x + c*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] + Dist[d^2*(m - 1)*((b^2 - 4*a*c)/(b^2*(m + 2*p + 1))), In
t[(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] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

Rubi steps \begin{align*} \text {integral}& = \frac {2}{3} d^4 (b+2 c x)^3+\left (\left (b^2-4 a c\right ) d^2\right ) \int \frac {(b d+2 c d x)^2}{a+b x+c x^2} \, dx \\ & = 2 \left (b^2-4 a c\right ) d^4 (b+2 c x)+\frac {2}{3} d^4 (b+2 c x)^3+\left (\left (b^2-4 a c\right )^2 d^4\right ) \int \frac {1}{a+b x+c x^2} \, dx \\ & = 2 \left (b^2-4 a c\right ) d^4 (b+2 c x)+\frac {2}{3} d^4 (b+2 c x)^3-\left (2 \left (b^2-4 a c\right )^2 d^4\right ) \text {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right ) \\ & = 2 \left (b^2-4 a c\right ) d^4 (b+2 c x)+\frac {2}{3} d^4 (b+2 c x)^3-2 \left (b^2-4 a c\right )^{3/2} d^4 \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00 \[ \int \frac {(b d+2 c d x)^4}{a+b x+c x^2} \, dx=d^4 \left (\frac {8}{3} c x \left (3 b^2+3 b c x+2 c \left (-3 a+c x^2\right )\right )+2 \left (-b^2+4 a c\right )^{3/2} \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )\right ) \]

[In]

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

[Out]

d^4*((8*c*x*(3*b^2 + 3*b*c*x + 2*c*(-3*a + c*x^2)))/3 + 2*(-b^2 + 4*a*c)^(3/2)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 +
4*a*c]])

Maple [A] (verified)

Time = 2.44 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.25

method	result	size
default	\(d^{4} \left (\frac {16 c^{3} x^{3}}{3}+8 b \,c^{2} x^{2}-16 a \,c^{2} x +8 b^{2} c x +\frac {2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}\right )\)	\(90\)
risch	\(\frac {16 d^{4} c^{3} x^{3}}{3}+8 c^{2} d^{4} b \,x^{2}-16 a \,c^{2} d^{4} x +8 c \,d^{4} b^{2} x -d^{4} \left (-4 a c +b^{2}\right )^{\frac {3}{2}} \ln \left (2 \left (-4 a c +b^{2}\right )^{\frac {3}{2}} c x +\left (-4 a c +b^{2}\right )^{\frac {3}{2}} b +16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )+d^{4} \left (-4 a c +b^{2}\right )^{\frac {3}{2}} \ln \left (-2 \left (-4 a c +b^{2}\right )^{\frac {3}{2}} c x -\left (-4 a c +b^{2}\right )^{\frac {3}{2}} b +16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )\)	\(167\)

[In]

int((2*c*d*x+b*d)^4/(c*x^2+b*x+a),x,method=_RETURNVERBOSE)

[Out]

d^4*(16/3*c^3*x^3+8*b*c^2*x^2-16*a*c^2*x+8*b^2*c*x+2*(16*a^2*c^2-8*a*b^2*c+b^4)/(4*a*c-b^2)^(1/2)*arctan((2*c*
x+b)/(4*a*c-b^2)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.89 \[ \int \frac {(b d+2 c d x)^4}{a+b x+c x^2} \, dx=\left [\frac {16}{3} \, c^{3} d^{4} x^{3} + 8 \, b c^{2} d^{4} x^{2} - {\left (b^{2} - 4 \, a c\right )}^{\frac {3}{2}} d^{4} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + 8 \, {\left (b^{2} c - 2 \, a c^{2}\right )} d^{4} x, \frac {16}{3} \, c^{3} d^{4} x^{3} + 8 \, b c^{2} d^{4} x^{2} - 2 \, {\left (b^{2} - 4 \, a c\right )} \sqrt {-b^{2} + 4 \, a c} d^{4} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 8 \, {\left (b^{2} c - 2 \, a c^{2}\right )} d^{4} x\right ] \]

[In]

integrate((2*c*d*x+b*d)^4/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

[16/3*c^3*d^4*x^3 + 8*b*c^2*d^4*x^2 - (b^2 - 4*a*c)^(3/2)*d^4*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c + sqrt(b^
2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) + 8*(b^2*c - 2*a*c^2)*d^4*x, 16/3*c^3*d^4*x^3 + 8*b*c^2*d^4*x^2 - 2
*(b^2 - 4*a*c)*sqrt(-b^2 + 4*a*c)*d^4*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + 8*(b^2*c - 2*a*c
^2)*d^4*x]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (71) = 142\).

Time = 0.35 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.83 \[ \int \frac {(b d+2 c d x)^4}{a+b x+c x^2} \, dx=8 b c^{2} d^{4} x^{2} + \frac {16 c^{3} d^{4} x^{3}}{3} - d^{4} \sqrt {- \left (4 a c - b^{2}\right )^{3}} \log {\left (x + \frac {4 a b c d^{4} - b^{3} d^{4} - d^{4} \sqrt {- \left (4 a c - b^{2}\right )^{3}}}{8 a c^{2} d^{4} - 2 b^{2} c d^{4}} \right )} + d^{4} \sqrt {- \left (4 a c - b^{2}\right )^{3}} \log {\left (x + \frac {4 a b c d^{4} - b^{3} d^{4} + d^{4} \sqrt {- \left (4 a c - b^{2}\right )^{3}}}{8 a c^{2} d^{4} - 2 b^{2} c d^{4}} \right )} + x \left (- 16 a c^{2} d^{4} + 8 b^{2} c d^{4}\right ) \]

[In]

integrate((2*c*d*x+b*d)**4/(c*x**2+b*x+a),x)

[Out]

8*b*c**2*d**4*x**2 + 16*c**3*d**4*x**3/3 - d**4*sqrt(-(4*a*c - b**2)**3)*log(x + (4*a*b*c*d**4 - b**3*d**4 - d
**4*sqrt(-(4*a*c - b**2)**3))/(8*a*c**2*d**4 - 2*b**2*c*d**4)) + d**4*sqrt(-(4*a*c - b**2)**3)*log(x + (4*a*b*
c*d**4 - b**3*d**4 + d**4*sqrt(-(4*a*c - b**2)**3))/(8*a*c**2*d**4 - 2*b**2*c*d**4)) + x*(-16*a*c**2*d**4 + 8*
b**2*c*d**4)

Maxima [F(-2)]

Exception generated. \[ \int \frac {(b d+2 c d x)^4}{a+b x+c x^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((2*c*d*x+b*d)^4/(c*x^2+b*x+a),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 deta

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.60 \[ \int \frac {(b d+2 c d x)^4}{a+b x+c x^2} \, dx=\frac {2 \, {\left (b^{4} d^{4} - 8 \, a b^{2} c d^{4} + 16 \, a^{2} c^{2} d^{4}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c}} + \frac {8 \, {\left (2 \, c^{6} d^{4} x^{3} + 3 \, b c^{5} d^{4} x^{2} + 3 \, b^{2} c^{4} d^{4} x - 6 \, a c^{5} d^{4} x\right )}}{3 \, c^{3}} \]

[In]

integrate((2*c*d*x+b*d)^4/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

2*(b^4*d^4 - 8*a*b^2*c*d^4 + 16*a^2*c^2*d^4)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/sqrt(-b^2 + 4*a*c) + 8/3*(
2*c^6*d^4*x^3 + 3*b*c^5*d^4*x^2 + 3*b^2*c^4*d^4*x - 6*a*c^5*d^4*x)/c^3

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.85 \[ \int \frac {(b d+2 c d x)^4}{a+b x+c x^2} \, dx=\frac {16\,c^3\,d^4\,x^3}{3}-x\,\left (16\,a\,c^2\,d^4-8\,b^2\,c\,d^4\right )+2\,d^4\,\mathrm {atan}\left (\frac {b\,d^4\,{\left (4\,a\,c-b^2\right )}^{3/2}+2\,c\,d^4\,x\,{\left (4\,a\,c-b^2\right )}^{3/2}}{16\,a^2\,c^2\,d^4-8\,a\,b^2\,c\,d^4+b^4\,d^4}\right )\,{\left (4\,a\,c-b^2\right )}^{3/2}+8\,b\,c^2\,d^4\,x^2 \]

[In]

int((b*d + 2*c*d*x)^4/(a + b*x + c*x^2),x)

[Out]

(16*c^3*d^4*x^3)/3 - x*(16*a*c^2*d^4 - 8*b^2*c*d^4) + 2*d^4*atan((b*d^4*(4*a*c - b^2)^(3/2) + 2*c*d^4*x*(4*a*c
- b^2)^(3/2))/(b^4*d^4 + 16*a^2*c^2*d^4 - 8*a*b^2*c*d^4))*(4*a*c - b^2)^(3/2) + 8*b*c^2*d^4*x^2