3.1156 \(\int \frac{(b d+2 c d x)^6}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=97 \[ \frac{2}{3} d^6 \left (b^2-4 a c\right ) (b+2 c x)^3+2 d^6 \left (b^2-4 a c\right )^2 (b+2 c x)-2 d^6 \left (b^2-4 a c\right )^{5/2} \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )+\frac{2}{5} d^6 (b+2 c x)^5 \]

[Out]

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

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Rubi [A]  time = 0.0804153, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {692, 618, 206} \[ \frac{2}{3} d^6 \left (b^2-4 a c\right ) (b+2 c x)^3+2 d^6 \left (b^2-4 a c\right )^2 (b+2 c x)-2 d^6 \left (b^2-4 a c\right )^{5/2} \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )+\frac{2}{5} d^6 (b+2 c x)^5 \]

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^6/(a + b*x + c*x^2),x]

[Out]

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

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])

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 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])

Rubi steps

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

Mathematica [A]  time = 0.0757721, size = 120, normalized size = 1.24 \[ d^6 \left (\frac{4}{15} c x \left (16 c^2 \left (15 a^2-5 a c x^2+3 c^2 x^4\right )+20 b^2 c \left (7 c x^2-9 a\right )+120 b c^2 x \left (c x^2-a\right )+90 b^3 c x+45 b^4\right )-2 \left (4 a c-b^2\right )^{5/2} \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^6/(a + b*x + c*x^2),x]

[Out]

d^6*((4*c*x*(45*b^4 + 90*b^3*c*x + 120*b*c^2*x*(-a + c*x^2) + 20*b^2*c*(-9*a + 7*c*x^2) + 16*c^2*(15*a^2 - 5*a
*c*x^2 + 3*c^2*x^4)))/15 - 2*(-b^2 + 4*a*c)^(5/2)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])

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Maple [B]  time = 0.149, size = 284, normalized size = 2.9 \begin{align*}{\frac{64\,{d}^{6}{c}^{5}{x}^{5}}{5}}+32\,{d}^{6}b{c}^{4}{x}^{4}-{\frac{64\,{d}^{6}{x}^{3}a{c}^{4}}{3}}+{\frac{112\,{d}^{6}{x}^{3}{b}^{2}{c}^{3}}{3}}-32\,{d}^{6}{x}^{2}ab{c}^{3}+24\,{d}^{6}{x}^{2}{b}^{3}{c}^{2}+64\,{d}^{6}{a}^{2}{c}^{3}x-48\,{d}^{6}a{b}^{2}{c}^{2}x+12\,{d}^{6}c{b}^{4}x-128\,{\frac{{d}^{6}{a}^{3}{c}^{3}}{\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+96\,{\frac{{d}^{6}{a}^{2}{b}^{2}{c}^{2}}{\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-24\,{\frac{{d}^{6}a{b}^{4}c}{\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+2\,{\frac{{d}^{6}{b}^{6}}{\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^6/(c*x^2+b*x+a),x)

[Out]

64/5*d^6*c^5*x^5+32*d^6*b*c^4*x^4-64/3*d^6*x^3*a*c^4+112/3*d^6*x^3*b^2*c^3-32*d^6*x^2*a*b*c^3+24*d^6*x^2*b^3*c
^2+64*d^6*a^2*c^3*x-48*d^6*a*b^2*c^2*x+12*d^6*c*b^4*x-128*d^6/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(
1/2))*a^3*c^3+96*d^6/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a^2*b^2*c^2-24*d^6/(4*a*c-b^2)^(1/2
)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*b^4*c+2*d^6/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b^6

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^6/(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.75762, size = 780, normalized size = 8.04 \begin{align*} \left [\frac{64}{5} \, c^{5} d^{6} x^{5} + 32 \, b c^{4} d^{6} x^{4} + \frac{16}{3} \,{\left (7 \, b^{2} c^{3} - 4 \, a c^{4}\right )} d^{6} x^{3} + 8 \,{\left (3 \, b^{3} c^{2} - 4 \, a b c^{3}\right )} d^{6} x^{2} +{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt{b^{2} - 4 \, a c} d^{6} \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 ) + 4 \,{\left (3 \, b^{4} c - 12 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d^{6} x, \frac{64}{5} \, c^{5} d^{6} x^{5} + 32 \, b c^{4} d^{6} x^{4} + \frac{16}{3} \,{\left (7 \, b^{2} c^{3} - 4 \, a c^{4}\right )} d^{6} x^{3} + 8 \,{\left (3 \, b^{3} c^{2} - 4 \, a b c^{3}\right )} d^{6} x^{2} - 2 \,{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt{-b^{2} + 4 \, a c} d^{6} \arctan \left (-\frac{\sqrt{-b^{2} + 4 \, a c}{\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + 4 \,{\left (3 \, b^{4} c - 12 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d^{6} x\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^6/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

[64/5*c^5*d^6*x^5 + 32*b*c^4*d^6*x^4 + 16/3*(7*b^2*c^3 - 4*a*c^4)*d^6*x^3 + 8*(3*b^3*c^2 - 4*a*b*c^3)*d^6*x^2
+ (b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(b^2 - 4*a*c)*d^6*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)) + 4*(3*b^4*c - 12*a*b^2*c^2 + 16*a^2*c^3)*d^6*x, 64/5*c^5*d^6*x^5 + 32*b*c
^4*d^6*x^4 + 16/3*(7*b^2*c^3 - 4*a*c^4)*d^6*x^3 + 8*(3*b^3*c^2 - 4*a*b*c^3)*d^6*x^2 - 2*(b^4 - 8*a*b^2*c + 16*
a^2*c^2)*sqrt(-b^2 + 4*a*c)*d^6*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + 4*(3*b^4*c - 12*a*b^2*
c^2 + 16*a^2*c^3)*d^6*x]

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Sympy [B]  time = 0.948615, size = 337, normalized size = 3.47 \begin{align*} 32 b c^{4} d^{6} x^{4} + \frac{64 c^{5} d^{6} x^{5}}{5} + d^{6} \sqrt{- \left (4 a c - b^{2}\right )^{5}} \log{\left (x + \frac{16 a^{2} b c^{2} d^{6} - 8 a b^{3} c d^{6} + b^{5} d^{6} - d^{6} \sqrt{- \left (4 a c - b^{2}\right )^{5}}}{32 a^{2} c^{3} d^{6} - 16 a b^{2} c^{2} d^{6} + 2 b^{4} c d^{6}} \right )} - d^{6} \sqrt{- \left (4 a c - b^{2}\right )^{5}} \log{\left (x + \frac{16 a^{2} b c^{2} d^{6} - 8 a b^{3} c d^{6} + b^{5} d^{6} + d^{6} \sqrt{- \left (4 a c - b^{2}\right )^{5}}}{32 a^{2} c^{3} d^{6} - 16 a b^{2} c^{2} d^{6} + 2 b^{4} c d^{6}} \right )} + x^{3} \left (- \frac{64 a c^{4} d^{6}}{3} + \frac{112 b^{2} c^{3} d^{6}}{3}\right ) + x^{2} \left (- 32 a b c^{3} d^{6} + 24 b^{3} c^{2} d^{6}\right ) + x \left (64 a^{2} c^{3} d^{6} - 48 a b^{2} c^{2} d^{6} + 12 b^{4} c d^{6}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)**6/(c*x**2+b*x+a),x)

[Out]

32*b*c**4*d**6*x**4 + 64*c**5*d**6*x**5/5 + d**6*sqrt(-(4*a*c - b**2)**5)*log(x + (16*a**2*b*c**2*d**6 - 8*a*b
**3*c*d**6 + b**5*d**6 - d**6*sqrt(-(4*a*c - b**2)**5))/(32*a**2*c**3*d**6 - 16*a*b**2*c**2*d**6 + 2*b**4*c*d*
*6)) - d**6*sqrt(-(4*a*c - b**2)**5)*log(x + (16*a**2*b*c**2*d**6 - 8*a*b**3*c*d**6 + b**5*d**6 + d**6*sqrt(-(
4*a*c - b**2)**5))/(32*a**2*c**3*d**6 - 16*a*b**2*c**2*d**6 + 2*b**4*c*d**6)) + x**3*(-64*a*c**4*d**6/3 + 112*
b**2*c**3*d**6/3) + x**2*(-32*a*b*c**3*d**6 + 24*b**3*c**2*d**6) + x*(64*a**2*c**3*d**6 - 48*a*b**2*c**2*d**6
+ 12*b**4*c*d**6)

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Giac [B]  time = 1.2476, size = 266, normalized size = 2.74 \begin{align*} \frac{2 \,{\left (b^{6} d^{6} - 12 \, a b^{4} c d^{6} + 48 \, a^{2} b^{2} c^{2} d^{6} - 64 \, a^{3} c^{3} d^{6}\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{\sqrt{-b^{2} + 4 \, a c}} + \frac{4 \,{\left (48 \, c^{10} d^{6} x^{5} + 120 \, b c^{9} d^{6} x^{4} + 140 \, b^{2} c^{8} d^{6} x^{3} - 80 \, a c^{9} d^{6} x^{3} + 90 \, b^{3} c^{7} d^{6} x^{2} - 120 \, a b c^{8} d^{6} x^{2} + 45 \, b^{4} c^{6} d^{6} x - 180 \, a b^{2} c^{7} d^{6} x + 240 \, a^{2} c^{8} d^{6} x\right )}}{15 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^6/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

2*(b^6*d^6 - 12*a*b^4*c*d^6 + 48*a^2*b^2*c^2*d^6 - 64*a^3*c^3*d^6)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/sqrt
(-b^2 + 4*a*c) + 4/15*(48*c^10*d^6*x^5 + 120*b*c^9*d^6*x^4 + 140*b^2*c^8*d^6*x^3 - 80*a*c^9*d^6*x^3 + 90*b^3*c
^7*d^6*x^2 - 120*a*b*c^8*d^6*x^2 + 45*b^4*c^6*d^6*x - 180*a*b^2*c^7*d^6*x + 240*a^2*c^8*d^6*x)/c^5