### 3.1166 $$\int \frac{(b d+2 c d x)^8}{(a+b x+c x^2)^2} \, dx$$

Optimal. Leaf size=126 $\frac{28}{3} c d^8 \left (b^2-4 a c\right ) (b+2 c x)^3+28 c d^8 \left (b^2-4 a c\right )^2 (b+2 c x)-28 c d^8 \left (b^2-4 a c\right )^{5/2} \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )-\frac{d^8 (b+2 c x)^7}{a+b x+c x^2}+\frac{28}{5} c d^8 (b+2 c x)^5$

[Out]

28*c*(b^2 - 4*a*c)^2*d^8*(b + 2*c*x) + (28*c*(b^2 - 4*a*c)*d^8*(b + 2*c*x)^3)/3 + (28*c*d^8*(b + 2*c*x)^5)/5 -
(d^8*(b + 2*c*x)^7)/(a + b*x + c*x^2) - 28*c*(b^2 - 4*a*c)^(5/2)*d^8*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]]

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Rubi [A]  time = 0.104732, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {686, 692, 618, 206} $\frac{28}{3} c d^8 \left (b^2-4 a c\right ) (b+2 c x)^3+28 c d^8 \left (b^2-4 a c\right )^2 (b+2 c x)-28 c d^8 \left (b^2-4 a c\right )^{5/2} \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )-\frac{d^8 (b+2 c x)^7}{a+b x+c x^2}+\frac{28}{5} c d^8 (b+2 c x)^5$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*d + 2*c*d*x)^8/(a + b*x + c*x^2)^2,x]

[Out]

28*c*(b^2 - 4*a*c)^2*d^8*(b + 2*c*x) + (28*c*(b^2 - 4*a*c)*d^8*(b + 2*c*x)^3)/3 + (28*c*d^8*(b + 2*c*x)^5)/5 -
(d^8*(b + 2*c*x)^7)/(a + b*x + c*x^2) - 28*c*(b^2 - 4*a*c)^(5/2)*d^8*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]]

Rule 686

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

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)^8}{\left (a+b x+c x^2\right )^2} \, dx &=-\frac{d^8 (b+2 c x)^7}{a+b x+c x^2}+\left (14 c d^2\right ) \int \frac{(b d+2 c d x)^6}{a+b x+c x^2} \, dx\\ &=\frac{28}{5} c d^8 (b+2 c x)^5-\frac{d^8 (b+2 c x)^7}{a+b x+c x^2}+\left (14 c \left (b^2-4 a c\right ) d^4\right ) \int \frac{(b d+2 c d x)^4}{a+b x+c x^2} \, dx\\ &=\frac{28}{3} c \left (b^2-4 a c\right ) d^8 (b+2 c x)^3+\frac{28}{5} c d^8 (b+2 c x)^5-\frac{d^8 (b+2 c x)^7}{a+b x+c x^2}+\left (14 c \left (b^2-4 a c\right )^2 d^6\right ) \int \frac{(b d+2 c d x)^2}{a+b x+c x^2} \, dx\\ &=28 c \left (b^2-4 a c\right )^2 d^8 (b+2 c x)+\frac{28}{3} c \left (b^2-4 a c\right ) d^8 (b+2 c x)^3+\frac{28}{5} c d^8 (b+2 c x)^5-\frac{d^8 (b+2 c x)^7}{a+b x+c x^2}+\left (14 c \left (b^2-4 a c\right )^3 d^8\right ) \int \frac{1}{a+b x+c x^2} \, dx\\ &=28 c \left (b^2-4 a c\right )^2 d^8 (b+2 c x)+\frac{28}{3} c \left (b^2-4 a c\right ) d^8 (b+2 c x)^3+\frac{28}{5} c d^8 (b+2 c x)^5-\frac{d^8 (b+2 c x)^7}{a+b x+c x^2}-\left (28 c \left (b^2-4 a c\right )^3 d^8\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )\\ &=28 c \left (b^2-4 a c\right )^2 d^8 (b+2 c x)+\frac{28}{3} c \left (b^2-4 a c\right ) d^8 (b+2 c x)^3+\frac{28}{5} c d^8 (b+2 c x)^5-\frac{d^8 (b+2 c x)^7}{a+b x+c x^2}-28 c \left (b^2-4 a c\right )^{5/2} d^8 \tanh ^{-1}\left (\frac{b+2 c x}{\sqrt{b^2-4 a c}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0819658, size = 155, normalized size = 1.23 $d^8 \left (32 c^2 x \left (24 a^2 c^2-16 a b^2 c+3 b^4\right )-\frac{512}{3} c^4 x^3 \left (a c-b^2\right )+128 b c^3 x^2 \left (b^2-2 a c\right )-\frac{\left (b^2-4 a c\right )^3 (b+2 c x)}{a+x (b+c x)}-28 c \left (4 a c-b^2\right )^{5/2} \tan ^{-1}\left (\frac{b+2 c x}{\sqrt{4 a c-b^2}}\right )+128 b c^5 x^4+\frac{256 c^6 x^5}{5}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*d + 2*c*d*x)^8/(a + b*x + c*x^2)^2,x]

[Out]

d^8*(32*c^2*(3*b^4 - 16*a*b^2*c + 24*a^2*c^2)*x + 128*b*c^3*(b^2 - 2*a*c)*x^2 - (512*c^4*(-b^2 + a*c)*x^3)/3 +
128*b*c^5*x^4 + (256*c^6*x^5)/5 - ((b^2 - 4*a*c)^3*(b + 2*c*x))/(a + x*(b + c*x)) - 28*c*(-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 = 0.155, size = 479, normalized size = 3.8 \begin{align*}{\frac{256\,{d}^{8}{c}^{6}{x}^{5}}{5}}+128\,{d}^{8}b{c}^{5}{x}^{4}-{\frac{512\,{d}^{8}{x}^{3}a{c}^{5}}{3}}+{\frac{512\,{d}^{8}{x}^{3}{b}^{2}{c}^{4}}{3}}-256\,{d}^{8}{x}^{2}ab{c}^{4}+128\,{d}^{8}{x}^{2}{b}^{3}{c}^{3}+768\,{d}^{8}{a}^{2}{c}^{4}x-512\,{d}^{8}{b}^{2}a{c}^{3}x+96\,{d}^{8}{b}^{4}{c}^{2}x+128\,{\frac{{d}^{8}{a}^{3}{c}^{4}x}{c{x}^{2}+bx+a}}-96\,{\frac{{d}^{8}{b}^{2}{a}^{2}{c}^{3}x}{c{x}^{2}+bx+a}}+24\,{\frac{{d}^{8}a{b}^{4}{c}^{2}x}{c{x}^{2}+bx+a}}-2\,{\frac{{d}^{8}{b}^{6}cx}{c{x}^{2}+bx+a}}+64\,{\frac{{d}^{8}{a}^{3}b{c}^{3}}{c{x}^{2}+bx+a}}-48\,{\frac{{d}^{8}{a}^{2}{b}^{3}{c}^{2}}{c{x}^{2}+bx+a}}+12\,{\frac{{d}^{8}a{b}^{5}c}{c{x}^{2}+bx+a}}-{\frac{{d}^{8}{b}^{7}}{c{x}^{2}+bx+a}}-1792\,{\frac{{d}^{8}{a}^{3}{c}^{4}}{\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+1344\,{\frac{{d}^{8}{b}^{2}{a}^{2}{c}^{3}}{\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-336\,{\frac{{d}^{8}a{b}^{4}{c}^{2}}{\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,cx+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+28\,{\frac{{d}^{8}{b}^{6}c}{\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)^8/(c*x^2+b*x+a)^2,x)

[Out]

256/5*d^8*c^6*x^5+128*d^8*b*c^5*x^4-512/3*d^8*x^3*a*c^5+512/3*d^8*x^3*b^2*c^4-256*d^8*x^2*a*b*c^4+128*d^8*x^2*
b^3*c^3+768*d^8*a^2*c^4*x-512*d^8*b^2*a*c^3*x+96*d^8*b^4*c^2*x+128*d^8/(c*x^2+b*x+a)*a^3*c^4*x-96*d^8/(c*x^2+b
*x+a)*b^2*a^2*c^3*x+24*d^8/(c*x^2+b*x+a)*a*b^4*c^2*x-2*d^8/(c*x^2+b*x+a)*b^6*c*x+64*d^8/(c*x^2+b*x+a)*a^3*b*c^
3-48*d^8/(c*x^2+b*x+a)*a^2*b^3*c^2+12*d^8/(c*x^2+b*x+a)*a*b^5*c-d^8/(c*x^2+b*x+a)*b^7-1792*d^8*c^4/(4*a*c-b^2)
^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a^3+1344*d^8*c^3/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/
2))*a^2*b^2-336*d^8*c^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*b^4+28*d^8*c/(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)^8/(c*x^2+b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.08082, size = 1634, normalized size = 12.97 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^8/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

[1/15*(768*c^7*d^8*x^7 + 2688*b*c^6*d^8*x^6 + 896*(5*b^2*c^5 - 2*a*c^6)*d^8*x^5 + 4480*(b^3*c^4 - a*b*c^5)*d^8
*x^4 + 1120*(3*b^4*c^3 - 8*a*b^2*c^4 + 8*a^2*c^5)*d^8*x^3 + 480*(3*b^5*c^2 - 12*a*b^3*c^3 + 16*a^2*b*c^4)*d^8*
x^2 - 30*(b^6*c - 60*a*b^4*c^2 + 304*a^2*b^2*c^3 - 448*a^3*c^4)*d^8*x - 15*(b^7 - 12*a*b^5*c + 48*a^2*b^3*c^2
- 64*a^3*b*c^3)*d^8 + 210*((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d^8*x^2 + (b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)
*d^8*x + (a*b^4*c - 8*a^2*b^2*c^2 + 16*a^3*c^3)*d^8)*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)))/(c*x^2 + b*x + a), 1/15*(768*c^7*d^8*x^7 + 2688*b*c^6*d^8
*x^6 + 896*(5*b^2*c^5 - 2*a*c^6)*d^8*x^5 + 4480*(b^3*c^4 - a*b*c^5)*d^8*x^4 + 1120*(3*b^4*c^3 - 8*a*b^2*c^4 +
8*a^2*c^5)*d^8*x^3 + 480*(3*b^5*c^2 - 12*a*b^3*c^3 + 16*a^2*b*c^4)*d^8*x^2 - 30*(b^6*c - 60*a*b^4*c^2 + 304*a^
2*b^2*c^3 - 448*a^3*c^4)*d^8*x - 15*(b^7 - 12*a*b^5*c + 48*a^2*b^3*c^2 - 64*a^3*b*c^3)*d^8 - 420*((b^4*c^2 - 8
*a*b^2*c^3 + 16*a^2*c^4)*d^8*x^2 + (b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d^8*x + (a*b^4*c - 8*a^2*b^2*c^2 + 16*
a^3*c^3)*d^8)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)))/(c*x^2 + b*x + a)]

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Sympy [B]  time = 3.36178, size = 476, normalized size = 3.78 \begin{align*} 128 b c^{5} d^{8} x^{4} + \frac{256 c^{6} d^{8} x^{5}}{5} + 14 c d^{8} \sqrt{- \left (4 a c - b^{2}\right )^{5}} \log{\left (x + \frac{224 a^{2} b c^{3} d^{8} - 112 a b^{3} c^{2} d^{8} + 14 b^{5} c d^{8} - 14 c d^{8} \sqrt{- \left (4 a c - b^{2}\right )^{5}}}{448 a^{2} c^{4} d^{8} - 224 a b^{2} c^{3} d^{8} + 28 b^{4} c^{2} d^{8}} \right )} - 14 c d^{8} \sqrt{- \left (4 a c - b^{2}\right )^{5}} \log{\left (x + \frac{224 a^{2} b c^{3} d^{8} - 112 a b^{3} c^{2} d^{8} + 14 b^{5} c d^{8} + 14 c d^{8} \sqrt{- \left (4 a c - b^{2}\right )^{5}}}{448 a^{2} c^{4} d^{8} - 224 a b^{2} c^{3} d^{8} + 28 b^{4} c^{2} d^{8}} \right )} + x^{3} \left (- \frac{512 a c^{5} d^{8}}{3} + \frac{512 b^{2} c^{4} d^{8}}{3}\right ) + x^{2} \left (- 256 a b c^{4} d^{8} + 128 b^{3} c^{3} d^{8}\right ) + x \left (768 a^{2} c^{4} d^{8} - 512 a b^{2} c^{3} d^{8} + 96 b^{4} c^{2} d^{8}\right ) + \frac{64 a^{3} b c^{3} d^{8} - 48 a^{2} b^{3} c^{2} d^{8} + 12 a b^{5} c d^{8} - b^{7} d^{8} + x \left (128 a^{3} c^{4} d^{8} - 96 a^{2} b^{2} c^{3} d^{8} + 24 a b^{4} c^{2} d^{8} - 2 b^{6} c d^{8}\right )}{a + b x + c x^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)**8/(c*x**2+b*x+a)**2,x)

[Out]

128*b*c**5*d**8*x**4 + 256*c**6*d**8*x**5/5 + 14*c*d**8*sqrt(-(4*a*c - b**2)**5)*log(x + (224*a**2*b*c**3*d**8
- 112*a*b**3*c**2*d**8 + 14*b**5*c*d**8 - 14*c*d**8*sqrt(-(4*a*c - b**2)**5))/(448*a**2*c**4*d**8 - 224*a*b**
2*c**3*d**8 + 28*b**4*c**2*d**8)) - 14*c*d**8*sqrt(-(4*a*c - b**2)**5)*log(x + (224*a**2*b*c**3*d**8 - 112*a*b
**3*c**2*d**8 + 14*b**5*c*d**8 + 14*c*d**8*sqrt(-(4*a*c - b**2)**5))/(448*a**2*c**4*d**8 - 224*a*b**2*c**3*d**
8 + 28*b**4*c**2*d**8)) + x**3*(-512*a*c**5*d**8/3 + 512*b**2*c**4*d**8/3) + x**2*(-256*a*b*c**4*d**8 + 128*b*
*3*c**3*d**8) + x*(768*a**2*c**4*d**8 - 512*a*b**2*c**3*d**8 + 96*b**4*c**2*d**8) + (64*a**3*b*c**3*d**8 - 48*
a**2*b**3*c**2*d**8 + 12*a*b**5*c*d**8 - b**7*d**8 + x*(128*a**3*c**4*d**8 - 96*a**2*b**2*c**3*d**8 + 24*a*b**
4*c**2*d**8 - 2*b**6*c*d**8))/(a + b*x + c*x**2)

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Giac [B]  time = 1.1537, size = 416, normalized size = 3.3 \begin{align*} \frac{28 \,{\left (b^{6} c d^{8} - 12 \, a b^{4} c^{2} d^{8} + 48 \, a^{2} b^{2} c^{3} d^{8} - 64 \, a^{3} c^{4} d^{8}\right )} \arctan \left (\frac{2 \, c x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{\sqrt{-b^{2} + 4 \, a c}} - \frac{2 \, b^{6} c d^{8} x - 24 \, a b^{4} c^{2} d^{8} x + 96 \, a^{2} b^{2} c^{3} d^{8} x - 128 \, a^{3} c^{4} d^{8} x + b^{7} d^{8} - 12 \, a b^{5} c d^{8} + 48 \, a^{2} b^{3} c^{2} d^{8} - 64 \, a^{3} b c^{3} d^{8}}{c x^{2} + b x + a} + \frac{32 \,{\left (24 \, c^{16} d^{8} x^{5} + 60 \, b c^{15} d^{8} x^{4} + 80 \, b^{2} c^{14} d^{8} x^{3} - 80 \, a c^{15} d^{8} x^{3} + 60 \, b^{3} c^{13} d^{8} x^{2} - 120 \, a b c^{14} d^{8} x^{2} + 45 \, b^{4} c^{12} d^{8} x - 240 \, a b^{2} c^{13} d^{8} x + 360 \, a^{2} c^{14} d^{8} x\right )}}{15 \, c^{10}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^8/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

28*(b^6*c*d^8 - 12*a*b^4*c^2*d^8 + 48*a^2*b^2*c^3*d^8 - 64*a^3*c^4*d^8)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))
/sqrt(-b^2 + 4*a*c) - (2*b^6*c*d^8*x - 24*a*b^4*c^2*d^8*x + 96*a^2*b^2*c^3*d^8*x - 128*a^3*c^4*d^8*x + b^7*d^8
- 12*a*b^5*c*d^8 + 48*a^2*b^3*c^2*d^8 - 64*a^3*b*c^3*d^8)/(c*x^2 + b*x + a) + 32/15*(24*c^16*d^8*x^5 + 60*b*c
^15*d^8*x^4 + 80*b^2*c^14*d^8*x^3 - 80*a*c^15*d^8*x^3 + 60*b^3*c^13*d^8*x^2 - 120*a*b*c^14*d^8*x^2 + 45*b^4*c^
12*d^8*x - 240*a*b^2*c^13*d^8*x + 360*a^2*c^14*d^8*x)/c^10