3.10.77 \(\int \frac {1}{(b d+2 c d x)^4 (a+b x+c x^2)} \, dx\)

Optimal. Leaf size=86 \[ \frac {2}{d^4 \left (b^2-4 a c\right )^2 (b+2 c x)}+\frac {2}{3 d^4 \left (b^2-4 a c\right ) (b+2 c x)^3}-\frac {2 \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{d^4 \left (b^2-4 a c\right )^{5/2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 86, 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 = {693, 618, 206} \begin {gather*} \frac {2}{d^4 \left (b^2-4 a c\right )^2 (b+2 c x)}+\frac {2}{3 d^4 \left (b^2-4 a c\right ) (b+2 c x)^3}-\frac {2 \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{d^4 \left (b^2-4 a c\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((b*d + 2*c*d*x)^4*(a + b*x + c*x^2)),x]

[Out]

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

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

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Mathematica [A]  time = 0.05, size = 83, normalized size = 0.97 \begin {gather*} \frac {2 \left (\frac {b^2-4 a c}{(b+2 c x)^3}+\frac {3 \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+\frac {3}{b+2 c x}\right )}{3 d^4 \left (b^2-4 a c\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((b*d + 2*c*d*x)^4*(a + b*x + c*x^2)),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(b d+2 c d x)^4 \left (a+b x+c x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/((b*d + 2*c*d*x)^4*(a + b*x + c*x^2)),x]

[Out]

IntegrateAlgebraic[1/((b*d + 2*c*d*x)^4*(a + b*x + c*x^2)), x]

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fricas [B]  time = 0.43, size = 627, normalized size = 7.29 \begin {gather*} \left [\frac {8 \, b^{4} - 40 \, a b^{2} c + 32 \, a^{2} c^{2} + 24 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} + 3 \, {\left (8 \, c^{3} x^{3} + 12 \, b c^{2} x^{2} + 6 \, b^{2} c x + b^{3}\right )} \sqrt {b^{2} - 4 \, a c} \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 ) + 24 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x}{3 \, {\left (8 \, {\left (b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}\right )} d^{4} x^{3} + 12 \, {\left (b^{7} c^{2} - 12 \, a b^{5} c^{3} + 48 \, a^{2} b^{3} c^{4} - 64 \, a^{3} b c^{5}\right )} d^{4} x^{2} + 6 \, {\left (b^{8} c - 12 \, a b^{6} c^{2} + 48 \, a^{2} b^{4} c^{3} - 64 \, a^{3} b^{2} c^{4}\right )} d^{4} x + {\left (b^{9} - 12 \, a b^{7} c + 48 \, a^{2} b^{5} c^{2} - 64 \, a^{3} b^{3} c^{3}\right )} d^{4}\right )}}, \frac {2 \, {\left (4 \, b^{4} - 20 \, a b^{2} c + 16 \, a^{2} c^{2} + 12 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{2} - 3 \, {\left (8 \, c^{3} x^{3} + 12 \, b c^{2} x^{2} + 6 \, b^{2} c x + b^{3}\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 12 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x\right )}}{3 \, {\left (8 \, {\left (b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}\right )} d^{4} x^{3} + 12 \, {\left (b^{7} c^{2} - 12 \, a b^{5} c^{3} + 48 \, a^{2} b^{3} c^{4} - 64 \, a^{3} b c^{5}\right )} d^{4} x^{2} + 6 \, {\left (b^{8} c - 12 \, a b^{6} c^{2} + 48 \, a^{2} b^{4} c^{3} - 64 \, a^{3} b^{2} c^{4}\right )} d^{4} x + {\left (b^{9} - 12 \, a b^{7} c + 48 \, a^{2} b^{5} c^{2} - 64 \, a^{3} b^{3} c^{3}\right )} d^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^4/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

[1/3*(8*b^4 - 40*a*b^2*c + 32*a^2*c^2 + 24*(b^2*c^2 - 4*a*c^3)*x^2 + 3*(8*c^3*x^3 + 12*b*c^2*x^2 + 6*b^2*c*x +
b^3)*sqrt(b^2 - 4*a*c)*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)) + 24*(b^3*c - 4*a*b*c^2)*x)/(8*(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*d^4*x^3 + 12*(b^7*c
^2 - 12*a*b^5*c^3 + 48*a^2*b^3*c^4 - 64*a^3*b*c^5)*d^4*x^2 + 6*(b^8*c - 12*a*b^6*c^2 + 48*a^2*b^4*c^3 - 64*a^3
*b^2*c^4)*d^4*x + (b^9 - 12*a*b^7*c + 48*a^2*b^5*c^2 - 64*a^3*b^3*c^3)*d^4), 2/3*(4*b^4 - 20*a*b^2*c + 16*a^2*
c^2 + 12*(b^2*c^2 - 4*a*c^3)*x^2 - 3*(8*c^3*x^3 + 12*b*c^2*x^2 + 6*b^2*c*x + b^3)*sqrt(-b^2 + 4*a*c)*arctan(-s
qrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + 12*(b^3*c - 4*a*b*c^2)*x)/(8*(b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b
^2*c^5 - 64*a^3*c^6)*d^4*x^3 + 12*(b^7*c^2 - 12*a*b^5*c^3 + 48*a^2*b^3*c^4 - 64*a^3*b*c^5)*d^4*x^2 + 6*(b^8*c
- 12*a*b^6*c^2 + 48*a^2*b^4*c^3 - 64*a^3*b^2*c^4)*d^4*x + (b^9 - 12*a*b^7*c + 48*a^2*b^5*c^2 - 64*a^3*b^3*c^3)
*d^4)]

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giac [A]  time = 0.16, size = 128, normalized size = 1.49 \begin {gather*} \frac {2 \, \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{4} d^{4} - 8 \, a b^{2} c d^{4} + 16 \, a^{2} c^{2} d^{4}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {8 \, {\left (3 \, c^{2} x^{2} + 3 \, b c x + b^{2} - a c\right )}}{3 \, {\left (b^{4} d^{4} - 8 \, a b^{2} c d^{4} + 16 \, a^{2} c^{2} d^{4}\right )} {\left (2 \, c x + b\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^4/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 89, normalized size = 1.03 \begin {gather*} \frac {2 \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {5}{2}} d^{4}}+\frac {2}{\left (4 a c -b^{2}\right )^{2} \left (2 c x +b \right ) d^{4}}-\frac {2}{3 \left (4 a c -b^{2}\right ) \left (2 c x +b \right )^{3} d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*c*d*x+b*d)^4/(c*x^2+b*x+a),x)

[Out]

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

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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(1/(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?

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mupad [B]  time = 0.59, size = 219, normalized size = 2.55 \begin {gather*} \frac {\frac {8\,c^2\,x^2}{{\left (4\,a\,c-b^2\right )}^2}-\frac {8\,\left (a\,c-b^2\right )}{3\,{\left (4\,a\,c-b^2\right )}^2}+\frac {8\,b\,c\,x}{{\left (4\,a\,c-b^2\right )}^2}}{b^3\,d^4+6\,b^2\,c\,d^4\,x+12\,b\,c^2\,d^4\,x^2+8\,c^3\,d^4\,x^3}+\frac {2\,\mathrm {atan}\left (\frac {16\,a^2\,b\,c^2\,d^4-8\,a\,b^3\,c\,d^4+b^5\,d^4}{d^4\,{\left (4\,a\,c-b^2\right )}^{5/2}}+\frac {2\,c\,x\,\left (16\,a^2\,c^2\,d^4-8\,a\,b^2\,c\,d^4+b^4\,d^4\right )}{d^4\,{\left (4\,a\,c-b^2\right )}^{5/2}}\right )}{d^4\,{\left (4\,a\,c-b^2\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b*d + 2*c*d*x)^4*(a + b*x + c*x^2)),x)

[Out]

((8*c^2*x^2)/(4*a*c - b^2)^2 - (8*(a*c - b^2))/(3*(4*a*c - b^2)^2) + (8*b*c*x)/(4*a*c - b^2)^2)/(b^3*d^4 + 8*c
^3*d^4*x^3 + 12*b*c^2*d^4*x^2 + 6*b^2*c*d^4*x) + (2*atan((b^5*d^4 + 16*a^2*b*c^2*d^4 - 8*a*b^3*c*d^4)/(d^4*(4*
a*c - b^2)^(5/2)) + (2*c*x*(b^4*d^4 + 16*a^2*c^2*d^4 - 8*a*b^2*c*d^4))/(d^4*(4*a*c - b^2)^(5/2))))/(d^4*(4*a*c
- b^2)^(5/2))

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sympy [B]  time = 1.74, size = 442, normalized size = 5.14 \begin {gather*} \frac {- 8 a c + 8 b^{2} + 24 b c x + 24 c^{2} x^{2}}{48 a^{2} b^{3} c^{2} d^{4} - 24 a b^{5} c d^{4} + 3 b^{7} d^{4} + x^{3} \left (384 a^{2} c^{5} d^{4} - 192 a b^{2} c^{4} d^{4} + 24 b^{4} c^{3} d^{4}\right ) + x^{2} \left (576 a^{2} b c^{4} d^{4} - 288 a b^{3} c^{3} d^{4} + 36 b^{5} c^{2} d^{4}\right ) + x \left (288 a^{2} b^{2} c^{3} d^{4} - 144 a b^{4} c^{2} d^{4} + 18 b^{6} c d^{4}\right )} - \frac {\sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \log {\left (x + \frac {- 64 a^{3} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 48 a^{2} b^{2} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 12 a b^{4} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + b^{6} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + b}{2 c} \right )}}{d^{4}} + \frac {\sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \log {\left (x + \frac {64 a^{3} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 48 a^{2} b^{2} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 12 a b^{4} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - b^{6} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + b}{2 c} \right )}}{d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)**4/(c*x**2+b*x+a),x)

[Out]

(-8*a*c + 8*b**2 + 24*b*c*x + 24*c**2*x**2)/(48*a**2*b**3*c**2*d**4 - 24*a*b**5*c*d**4 + 3*b**7*d**4 + x**3*(3
84*a**2*c**5*d**4 - 192*a*b**2*c**4*d**4 + 24*b**4*c**3*d**4) + x**2*(576*a**2*b*c**4*d**4 - 288*a*b**3*c**3*d
**4 + 36*b**5*c**2*d**4) + x*(288*a**2*b**2*c**3*d**4 - 144*a*b**4*c**2*d**4 + 18*b**6*c*d**4)) - sqrt(-1/(4*a
*c - b**2)**5)*log(x + (-64*a**3*c**3*sqrt(-1/(4*a*c - b**2)**5) + 48*a**2*b**2*c**2*sqrt(-1/(4*a*c - b**2)**5
) - 12*a*b**4*c*sqrt(-1/(4*a*c - b**2)**5) + b**6*sqrt(-1/(4*a*c - b**2)**5) + b)/(2*c))/d**4 + sqrt(-1/(4*a*c
- b**2)**5)*log(x + (64*a**3*c**3*sqrt(-1/(4*a*c - b**2)**5) - 48*a**2*b**2*c**2*sqrt(-1/(4*a*c - b**2)**5) +
12*a*b**4*c*sqrt(-1/(4*a*c - b**2)**5) - b**6*sqrt(-1/(4*a*c - b**2)**5) + b)/(2*c))/d**4

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