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

Optimal. Leaf size=72 \[ 2 d^4 \left (b^2-4 a c\right ) (b+2 c x)-2 d^4 \left (b^2-4 a c\right )^{3/2} \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )+\frac {2}{3} d^4 (b+2 c x)^3 \]

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {692, 618, 206} \begin {gather*} 2 d^4 \left (b^2-4 a c\right ) (b+2 c x)-2 d^4 \left (b^2-4 a c\right )^{3/2} \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )+\frac {2}{3} d^4 (b+2 c x)^3 \end {gather*}

Antiderivative was successfully verified.

[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 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 618

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 692

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*} \int \frac {(b d+2 c d x)^4}{a+b x+c x^2} \, dx &=\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 ) \operatorname {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]  time = 0.04, size = 72, normalized size = 1.00 \begin {gather*} d^4 \left (\frac {8}{3} c x \left (2 c \left (c x^2-3 a\right )+3 b^2+3 b c x\right )+2 \left (4 a c-b^2\right )^{3/2} \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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]])

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(b d+2 c d x)^4}{a+b x+c x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 208, normalized size = 2.89 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

________________________________________________________________________________________

giac [A]  time = 0.17, size = 115, normalized size = 1.60 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

maple [B]  time = 0.05, size = 170, normalized size = 2.36 \begin {gather*} \frac {16 c^{3} d^{4} x^{3}}{3}+\frac {32 a^{2} c^{2} d^{4} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}-\frac {16 a \,b^{2} c \,d^{4} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}+\frac {2 b^{4} d^{4} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}+8 b \,c^{2} d^{4} x^{2}-16 a \,c^{2} d^{4} x +8 b^{2} c \,d^{4} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/3*d^4*c^3*x^3+8*d^4*b*c^2*x^2-16*d^4*a*c^2*x+8*d^4*b^2*c*x+32*d^4/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c
-b^2)^(1/2))*a^2*c^2-16*d^4/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*b^2*c+2*d^4/(4*a*c-b^2)^(1
/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b^4

________________________________________________________________________________________

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*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 details)Is 4*a*c-b^2 positive or negative?

________________________________________________________________________________________

mupad [B]  time = 0.07, size = 133, normalized size = 1.85 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

sympy [B]  time = 0.52, size = 204, normalized size = 2.83 \begin {gather*} 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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________