3.197 \(\int \frac{\text{b1}+\text{c1} x}{(a+2 b x+c x^2)^4} \, dx\)

Optimal. Leaf size=173 \[ \frac{5 c^2 (\text{b1} c-b \text{c1}) \tanh ^{-1}\left (\frac{b+c x}{\sqrt{b^2-a c}}\right )}{16 \left (b^2-a c\right )^{7/2}}-\frac{5 c (b+c x) (\text{b1} c-b \text{c1})}{16 \left (b^2-a c\right )^3 \left (a+2 b x+c x^2\right )}+\frac{5 (b+c x) (\text{b1} c-b \text{c1})}{24 \left (b^2-a c\right )^2 \left (a+2 b x+c x^2\right )^2}-\frac{-a \text{c1}+x (\text{b1} c-b \text{c1})+b \text{b1}}{6 \left (b^2-a c\right ) \left (a+2 b x+c x^2\right )^3} \]

[Out]

-(b*b1 - a*c1 + (b1*c - b*c1)*x)/(6*(b^2 - a*c)*(a + 2*b*x + c*x^2)^3) + (5*(b1*c - b*c1)*(b + c*x))/(24*(b^2
- a*c)^2*(a + 2*b*x + c*x^2)^2) - (5*c*(b1*c - b*c1)*(b + c*x))/(16*(b^2 - a*c)^3*(a + 2*b*x + c*x^2)) + (5*c^
2*(b1*c - b*c1)*ArcTanh[(b + c*x)/Sqrt[b^2 - a*c]])/(16*(b^2 - a*c)^(7/2))

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Rubi [A]  time = 0.109845, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {638, 614, 618, 206} \[ \frac{5 c^2 (\text{b1} c-b \text{c1}) \tanh ^{-1}\left (\frac{b+c x}{\sqrt{b^2-a c}}\right )}{16 \left (b^2-a c\right )^{7/2}}-\frac{5 c (b+c x) (\text{b1} c-b \text{c1})}{16 \left (b^2-a c\right )^3 \left (a+2 b x+c x^2\right )}+\frac{5 (b+c x) (\text{b1} c-b \text{c1})}{24 \left (b^2-a c\right )^2 \left (a+2 b x+c x^2\right )^2}-\frac{-a \text{c1}+x (\text{b1} c-b \text{c1})+b \text{b1}}{6 \left (b^2-a c\right ) \left (a+2 b x+c x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(b1 + c1*x)/(a + 2*b*x + c*x^2)^4,x]

[Out]

-(b*b1 - a*c1 + (b1*c - b*c1)*x)/(6*(b^2 - a*c)*(a + 2*b*x + c*x^2)^3) + (5*(b1*c - b*c1)*(b + c*x))/(24*(b^2
- a*c)^2*(a + 2*b*x + c*x^2)^2) - (5*c*(b1*c - b*c1)*(b + c*x))/(16*(b^2 - a*c)^3*(a + 2*b*x + c*x^2)) + (5*c^
2*(b1*c - b*c1)*ArcTanh[(b + c*x)/Sqrt[b^2 - a*c]])/(16*(b^2 - a*c)^(7/2))

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

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{\text{b1}+\text{c1} x}{\left (a+2 b x+c x^2\right )^4} \, dx &=-\frac{b \text{b1}-a \text{c1}+(\text{b1} c-b \text{c1}) x}{6 \left (b^2-a c\right ) \left (a+2 b x+c x^2\right )^3}-\frac{(5 (\text{b1} c-b \text{c1})) \int \frac{1}{\left (a+2 b x+c x^2\right )^3} \, dx}{6 \left (b^2-a c\right )}\\ &=-\frac{b \text{b1}-a \text{c1}+(\text{b1} c-b \text{c1}) x}{6 \left (b^2-a c\right ) \left (a+2 b x+c x^2\right )^3}+\frac{5 (\text{b1} c-b \text{c1}) (b+c x)}{24 \left (b^2-a c\right )^2 \left (a+2 b x+c x^2\right )^2}+\frac{(5 c (\text{b1} c-b \text{c1})) \int \frac{1}{\left (a+2 b x+c x^2\right )^2} \, dx}{8 \left (b^2-a c\right )^2}\\ &=-\frac{b \text{b1}-a \text{c1}+(\text{b1} c-b \text{c1}) x}{6 \left (b^2-a c\right ) \left (a+2 b x+c x^2\right )^3}+\frac{5 (\text{b1} c-b \text{c1}) (b+c x)}{24 \left (b^2-a c\right )^2 \left (a+2 b x+c x^2\right )^2}-\frac{5 c (\text{b1} c-b \text{c1}) (b+c x)}{16 \left (b^2-a c\right )^3 \left (a+2 b x+c x^2\right )}-\frac{\left (5 c^2 (\text{b1} c-b \text{c1})\right ) \int \frac{1}{a+2 b x+c x^2} \, dx}{16 \left (b^2-a c\right )^3}\\ &=-\frac{b \text{b1}-a \text{c1}+(\text{b1} c-b \text{c1}) x}{6 \left (b^2-a c\right ) \left (a+2 b x+c x^2\right )^3}+\frac{5 (\text{b1} c-b \text{c1}) (b+c x)}{24 \left (b^2-a c\right )^2 \left (a+2 b x+c x^2\right )^2}-\frac{5 c (\text{b1} c-b \text{c1}) (b+c x)}{16 \left (b^2-a c\right )^3 \left (a+2 b x+c x^2\right )}+\frac{\left (5 c^2 (\text{b1} c-b \text{c1})\right ) \operatorname{Subst}\left (\int \frac{1}{4 \left (b^2-a c\right )-x^2} \, dx,x,2 b+2 c x\right )}{8 \left (b^2-a c\right )^3}\\ &=-\frac{b \text{b1}-a \text{c1}+(\text{b1} c-b \text{c1}) x}{6 \left (b^2-a c\right ) \left (a+2 b x+c x^2\right )^3}+\frac{5 (\text{b1} c-b \text{c1}) (b+c x)}{24 \left (b^2-a c\right )^2 \left (a+2 b x+c x^2\right )^2}-\frac{5 c (\text{b1} c-b \text{c1}) (b+c x)}{16 \left (b^2-a c\right )^3 \left (a+2 b x+c x^2\right )}+\frac{5 c^2 (\text{b1} c-b \text{c1}) \tanh ^{-1}\left (\frac{b+c x}{\sqrt{b^2-a c}}\right )}{16 \left (b^2-a c\right )^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.21874, size = 168, normalized size = 0.97 \[ \frac{\frac{15 c^2 (b \text{c1}-\text{b1} c) \tan ^{-1}\left (\frac{b+c x}{\sqrt{a c-b^2}}\right )}{\sqrt{a c-b^2}}-\frac{10 \left (b^2-a c\right ) (b+c x) (b \text{c1}-\text{b1} c)}{(a+x (2 b+c x))^2}+\frac{8 \left (b^2-a c\right )^2 (a \text{c1}-b \text{b1}+b \text{c1} x-\text{b1} c x)}{(a+x (2 b+c x))^3}+\frac{15 c (b+c x) (b \text{c1}-\text{b1} c)}{a+x (2 b+c x)}}{48 \left (b^2-a c\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(b1 + c1*x)/(a + 2*b*x + c*x^2)^4,x]

[Out]

((8*(b^2 - a*c)^2*(-(b*b1) + a*c1 - b1*c*x + b*c1*x))/(a + x*(2*b + c*x))^3 - (10*(b^2 - a*c)*(-(b1*c) + b*c1)
*(b + c*x))/(a + x*(2*b + c*x))^2 + (15*c*(-(b1*c) + b*c1)*(b + c*x))/(a + x*(2*b + c*x)) + (15*c^2*(-(b1*c) +
b*c1)*ArcTan[(b + c*x)/Sqrt[-b^2 + a*c]])/Sqrt[-b^2 + a*c])/(48*(b^2 - a*c)^3)

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Maple [B]  time = 0.006, size = 405, normalized size = 2.3 \begin{align*}{\frac{ \left ( -2\,b{\it c1}+2\,{\it b1}\,c \right ) x+2\,b{\it b1}-2\,a{\it c1}}{ \left ( 12\,ac-12\,{b}^{2} \right ) \left ( c{x}^{2}+2\,bx+a \right ) ^{3}}}-{\frac{10\,cxb{\it c1}}{3\, \left ( 4\,ac-4\,{b}^{2} \right ) ^{2} \left ( c{x}^{2}+2\,bx+a \right ) ^{2}}}+{\frac{10\,x{c}^{2}{\it b1}}{3\, \left ( 4\,ac-4\,{b}^{2} \right ) ^{2} \left ( c{x}^{2}+2\,bx+a \right ) ^{2}}}-{\frac{10\,{b}^{2}{\it c1}}{3\, \left ( 4\,ac-4\,{b}^{2} \right ) ^{2} \left ( c{x}^{2}+2\,bx+a \right ) ^{2}}}+{\frac{10\,b{\it b1}\,c}{3\, \left ( 4\,ac-4\,{b}^{2} \right ) ^{2} \left ( c{x}^{2}+2\,bx+a \right ) ^{2}}}-20\,{\frac{x{c}^{2}b{\it c1}}{ \left ( 4\,ac-4\,{b}^{2} \right ) ^{3} \left ( c{x}^{2}+2\,bx+a \right ) }}+20\,{\frac{{c}^{3}x{\it b1}}{ \left ( 4\,ac-4\,{b}^{2} \right ) ^{3} \left ( c{x}^{2}+2\,bx+a \right ) }}-20\,{\frac{{b}^{2}c{\it c1}}{ \left ( 4\,ac-4\,{b}^{2} \right ) ^{3} \left ( c{x}^{2}+2\,bx+a \right ) }}+20\,{\frac{{\it b1}\,b{c}^{2}}{ \left ( 4\,ac-4\,{b}^{2} \right ) ^{3} \left ( c{x}^{2}+2\,bx+a \right ) }}-20\,{\frac{{\it c1}\,b{c}^{2}}{ \left ( 4\,ac-4\,{b}^{2} \right ) ^{3}\sqrt{ac-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,cx+2\,b}{\sqrt{ac-{b}^{2}}}} \right ) }+20\,{\frac{{\it b1}\,{c}^{3}}{ \left ( 4\,ac-4\,{b}^{2} \right ) ^{3}\sqrt{ac-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,cx+2\,b}{\sqrt{ac-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c1*x+b1)/(c*x^2+2*b*x+a)^4,x)

[Out]

1/3*((-2*b*c1+2*b1*c)*x+2*b*b1-2*a*c1)/(4*a*c-4*b^2)/(c*x^2+2*b*x+a)^3-10/3/(4*a*c-4*b^2)^2/(c*x^2+2*b*x+a)^2*
x*c*b*c1+10/3/(4*a*c-4*b^2)^2/(c*x^2+2*b*x+a)^2*x*c^2*b1-10/3/(4*a*c-4*b^2)^2/(c*x^2+2*b*x+a)^2*b^2*c1+10/3/(4
*a*c-4*b^2)^2/(c*x^2+2*b*x+a)^2*b*b1*c-20/(4*a*c-4*b^2)^3*c^2/(c*x^2+2*b*x+a)*x*b*c1+20/(4*a*c-4*b^2)^3*c^3/(c
*x^2+2*b*x+a)*x*b1-20/(4*a*c-4*b^2)^3*c/(c*x^2+2*b*x+a)*b^2*c1+20/(4*a*c-4*b^2)^3*c^2/(c*x^2+2*b*x+a)*b*b1-20/
(4*a*c-4*b^2)^3*c^2/(a*c-b^2)^(1/2)*arctan(1/2*(2*c*x+2*b)/(a*c-b^2)^(1/2))*b*c1+20/(4*a*c-4*b^2)^3*c^3/(a*c-b
^2)^(1/2)*arctan(1/2*(2*c*x+2*b)/(a*c-b^2)^(1/2))*b1

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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((c1*x+b1)/(c*x^2+2*b*x+a)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.10304, size = 4097, normalized size = 23.68 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c1*x+b1)/(c*x^2+2*b*x+a)^4,x, algorithm="fricas")

[Out]

[-1/96*(16*b^7*b1 - 68*a*b^5*b1*c + 118*a^2*b^3*b1*c^2 - 66*a^3*b*b1*c^3 + 30*(b^2*b1*c^5 - a*b1*c^6 - (b^3*c^
4 - a*b*c^5)*c1)*x^5 + 150*(b^3*b1*c^4 - a*b*b1*c^5 - (b^4*c^3 - a*b^2*c^4)*c1)*x^4 + 20*(11*b^4*b1*c^3 - 7*a*
b^2*b1*c^4 - 4*a^2*b1*c^5 - (11*b^5*c^2 - 7*a*b^3*c^3 - 4*a^2*b*c^4)*c1)*x^3 + 60*(b^5*b1*c^2 + 3*a*b^3*b1*c^3
- 4*a^2*b*b1*c^4 - (b^6*c + 3*a*b^4*c^2 - 4*a^2*b^2*c^3)*c1)*x^2 - 15*(a^3*b1*c^3 - a^3*b*c^2*c1 + (b1*c^6 -
b*c^5*c1)*x^6 + 6*(b*b1*c^5 - b^2*c^4*c1)*x^5 + 3*(4*b^2*b1*c^4 + a*b1*c^5 - (4*b^3*c^3 + a*b*c^4)*c1)*x^4 + 4
*(2*b^3*b1*c^3 + 3*a*b*b1*c^4 - (2*b^4*c^2 + 3*a*b^2*c^3)*c1)*x^3 + 3*(4*a*b^2*b1*c^3 + a^2*b1*c^4 - (4*a*b^3*
c^2 + a^2*b*c^3)*c1)*x^2 + 6*(a^2*b*b1*c^3 - a^2*b^2*c^2*c1)*x)*sqrt(b^2 - a*c)*log((c^2*x^2 + 2*b*c*x + 2*b^2
- a*c + 2*sqrt(b^2 - a*c)*(c*x + b))/(c*x^2 + 2*b*x + a)) + 2*(2*a*b^6 - 11*a^2*b^4*c + a^3*b^2*c^2 + 8*a^4*c
^3)*c1 - 6*(4*b^6*b1*c - 22*a*b^4*b1*c^2 + 7*a^2*b^2*b1*c^3 + 11*a^3*b1*c^4 - (4*b^7 - 22*a*b^5*c + 7*a^2*b^3*
c^2 + 11*a^3*b*c^3)*c1)*x)/(a^3*b^8 - 4*a^4*b^6*c + 6*a^5*b^4*c^2 - 4*a^6*b^2*c^3 + a^7*c^4 + (b^8*c^3 - 4*a*b
^6*c^4 + 6*a^2*b^4*c^5 - 4*a^3*b^2*c^6 + a^4*c^7)*x^6 + 6*(b^9*c^2 - 4*a*b^7*c^3 + 6*a^2*b^5*c^4 - 4*a^3*b^3*c
^5 + a^4*b*c^6)*x^5 + 3*(4*b^10*c - 15*a*b^8*c^2 + 20*a^2*b^6*c^3 - 10*a^3*b^4*c^4 + a^5*c^6)*x^4 + 4*(2*b^11
- 5*a*b^9*c + 10*a^3*b^5*c^3 - 10*a^4*b^3*c^4 + 3*a^5*b*c^5)*x^3 + 3*(4*a*b^10 - 15*a^2*b^8*c + 20*a^3*b^6*c^2
- 10*a^4*b^4*c^3 + a^6*c^5)*x^2 + 6*(a^2*b^9 - 4*a^3*b^7*c + 6*a^4*b^5*c^2 - 4*a^5*b^3*c^3 + a^6*b*c^4)*x), -
1/48*(8*b^7*b1 - 34*a*b^5*b1*c + 59*a^2*b^3*b1*c^2 - 33*a^3*b*b1*c^3 + 15*(b^2*b1*c^5 - a*b1*c^6 - (b^3*c^4 -
a*b*c^5)*c1)*x^5 + 75*(b^3*b1*c^4 - a*b*b1*c^5 - (b^4*c^3 - a*b^2*c^4)*c1)*x^4 + 10*(11*b^4*b1*c^3 - 7*a*b^2*b
1*c^4 - 4*a^2*b1*c^5 - (11*b^5*c^2 - 7*a*b^3*c^3 - 4*a^2*b*c^4)*c1)*x^3 + 30*(b^5*b1*c^2 + 3*a*b^3*b1*c^3 - 4*
a^2*b*b1*c^4 - (b^6*c + 3*a*b^4*c^2 - 4*a^2*b^2*c^3)*c1)*x^2 - 15*(a^3*b1*c^3 - a^3*b*c^2*c1 + (b1*c^6 - b*c^5
*c1)*x^6 + 6*(b*b1*c^5 - b^2*c^4*c1)*x^5 + 3*(4*b^2*b1*c^4 + a*b1*c^5 - (4*b^3*c^3 + a*b*c^4)*c1)*x^4 + 4*(2*b
^3*b1*c^3 + 3*a*b*b1*c^4 - (2*b^4*c^2 + 3*a*b^2*c^3)*c1)*x^3 + 3*(4*a*b^2*b1*c^3 + a^2*b1*c^4 - (4*a*b^3*c^2 +
a^2*b*c^3)*c1)*x^2 + 6*(a^2*b*b1*c^3 - a^2*b^2*c^2*c1)*x)*sqrt(-b^2 + a*c)*arctan(-sqrt(-b^2 + a*c)*(c*x + b)
/(b^2 - a*c)) + (2*a*b^6 - 11*a^2*b^4*c + a^3*b^2*c^2 + 8*a^4*c^3)*c1 - 3*(4*b^6*b1*c - 22*a*b^4*b1*c^2 + 7*a^
2*b^2*b1*c^3 + 11*a^3*b1*c^4 - (4*b^7 - 22*a*b^5*c + 7*a^2*b^3*c^2 + 11*a^3*b*c^3)*c1)*x)/(a^3*b^8 - 4*a^4*b^6
*c + 6*a^5*b^4*c^2 - 4*a^6*b^2*c^3 + a^7*c^4 + (b^8*c^3 - 4*a*b^6*c^4 + 6*a^2*b^4*c^5 - 4*a^3*b^2*c^6 + a^4*c^
7)*x^6 + 6*(b^9*c^2 - 4*a*b^7*c^3 + 6*a^2*b^5*c^4 - 4*a^3*b^3*c^5 + a^4*b*c^6)*x^5 + 3*(4*b^10*c - 15*a*b^8*c^
2 + 20*a^2*b^6*c^3 - 10*a^3*b^4*c^4 + a^5*c^6)*x^4 + 4*(2*b^11 - 5*a*b^9*c + 10*a^3*b^5*c^3 - 10*a^4*b^3*c^4 +
3*a^5*b*c^5)*x^3 + 3*(4*a*b^10 - 15*a^2*b^8*c + 20*a^3*b^6*c^2 - 10*a^4*b^4*c^3 + a^6*c^5)*x^2 + 6*(a^2*b^9 -
4*a^3*b^7*c + 6*a^4*b^5*c^2 - 4*a^5*b^3*c^3 + a^6*b*c^4)*x)]

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Sympy [B]  time = 3.87369, size = 1027, normalized size = 5.94 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c1*x+b1)/(c*x**2+2*b*x+a)**4,x)

[Out]

5*c**2*sqrt(-1/(a*c - b**2)**7)*(b*c1 - b1*c)*log(x + (-5*a**4*c**6*sqrt(-1/(a*c - b**2)**7)*(b*c1 - b1*c) + 2
0*a**3*b**2*c**5*sqrt(-1/(a*c - b**2)**7)*(b*c1 - b1*c) - 30*a**2*b**4*c**4*sqrt(-1/(a*c - b**2)**7)*(b*c1 - b
1*c) + 20*a*b**6*c**3*sqrt(-1/(a*c - b**2)**7)*(b*c1 - b1*c) - 5*b**8*c**2*sqrt(-1/(a*c - b**2)**7)*(b*c1 - b1
*c) + 5*b**2*c**2*c1 - 5*b*b1*c**3)/(5*b*c**3*c1 - 5*b1*c**4))/32 - 5*c**2*sqrt(-1/(a*c - b**2)**7)*(b*c1 - b1
*c)*log(x + (5*a**4*c**6*sqrt(-1/(a*c - b**2)**7)*(b*c1 - b1*c) - 20*a**3*b**2*c**5*sqrt(-1/(a*c - b**2)**7)*(
b*c1 - b1*c) + 30*a**2*b**4*c**4*sqrt(-1/(a*c - b**2)**7)*(b*c1 - b1*c) - 20*a*b**6*c**3*sqrt(-1/(a*c - b**2)*
*7)*(b*c1 - b1*c) + 5*b**8*c**2*sqrt(-1/(a*c - b**2)**7)*(b*c1 - b1*c) + 5*b**2*c**2*c1 - 5*b*b1*c**3)/(5*b*c*
*3*c1 - 5*b1*c**4))/32 - (8*a**3*c**2*c1 + 9*a**2*b**2*c*c1 - 33*a**2*b*b1*c**2 - 2*a*b**4*c1 + 26*a*b**3*b1*c
- 8*b**5*b1 + x**5*(15*b*c**4*c1 - 15*b1*c**5) + x**4*(75*b**2*c**3*c1 - 75*b*b1*c**4) + x**3*(40*a*b*c**3*c1
- 40*a*b1*c**4 + 110*b**3*c**2*c1 - 110*b**2*b1*c**3) + x**2*(120*a*b**2*c**2*c1 - 120*a*b*b1*c**3 + 30*b**4*
c*c1 - 30*b**3*b1*c**2) + x*(33*a**2*b*c**2*c1 - 33*a**2*b1*c**3 + 54*a*b**3*c*c1 - 54*a*b**2*b1*c**2 - 12*b**
5*c1 + 12*b**4*b1*c))/(48*a**6*c**3 - 144*a**5*b**2*c**2 + 144*a**4*b**4*c - 48*a**3*b**6 + x**6*(48*a**3*c**6
- 144*a**2*b**2*c**5 + 144*a*b**4*c**4 - 48*b**6*c**3) + x**5*(288*a**3*b*c**5 - 864*a**2*b**3*c**4 + 864*a*b
**5*c**3 - 288*b**7*c**2) + x**4*(144*a**4*c**5 + 144*a**3*b**2*c**4 - 1296*a**2*b**4*c**3 + 1584*a*b**6*c**2
- 576*b**8*c) + x**3*(576*a**4*b*c**4 - 1344*a**3*b**3*c**3 + 576*a**2*b**5*c**2 + 576*a*b**7*c - 384*b**9) +
x**2*(144*a**5*c**4 + 144*a**4*b**2*c**3 - 1296*a**3*b**4*c**2 + 1584*a**2*b**6*c - 576*a*b**8) + x*(288*a**5*
b*c**3 - 864*a**4*b**3*c**2 + 864*a**3*b**5*c - 288*a**2*b**7))

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Giac [B]  time = 1.07513, size = 490, normalized size = 2.83 \begin{align*} -\frac{5 \,{\left (b_{1} c^{3} - b c^{2} c_{1}\right )} \arctan \left (\frac{c x + b}{\sqrt{-b^{2} + a c}}\right )}{16 \,{\left (b^{6} - 3 \, a b^{4} c + 3 \, a^{2} b^{2} c^{2} - a^{3} c^{3}\right )} \sqrt{-b^{2} + a c}} - \frac{15 \, b_{1} c^{5} x^{5} - 15 \, b c^{4} c_{1} x^{5} + 75 \, b b_{1} c^{4} x^{4} - 75 \, b^{2} c^{3} c_{1} x^{4} + 110 \, b^{2} b_{1} c^{3} x^{3} + 40 \, a b_{1} c^{4} x^{3} - 110 \, b^{3} c^{2} c_{1} x^{3} - 40 \, a b c^{3} c_{1} x^{3} + 30 \, b^{3} b_{1} c^{2} x^{2} + 120 \, a b b_{1} c^{3} x^{2} - 30 \, b^{4} c c_{1} x^{2} - 120 \, a b^{2} c^{2} c_{1} x^{2} - 12 \, b^{4} b_{1} c x + 54 \, a b^{2} b_{1} c^{2} x + 33 \, a^{2} b_{1} c^{3} x + 12 \, b^{5} c_{1} x - 54 \, a b^{3} c c_{1} x - 33 \, a^{2} b c^{2} c_{1} x + 8 \, b^{5} b_{1} - 26 \, a b^{3} b_{1} c + 33 \, a^{2} b b_{1} c^{2} + 2 \, a b^{4} c_{1} - 9 \, a^{2} b^{2} c c_{1} - 8 \, a^{3} c^{2} c_{1}}{48 \,{\left (b^{6} - 3 \, a b^{4} c + 3 \, a^{2} b^{2} c^{2} - a^{3} c^{3}\right )}{\left (c x^{2} + 2 \, b x + a\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c1*x+b1)/(c*x^2+2*b*x+a)^4,x, algorithm="giac")

[Out]

-5/16*(b1*c^3 - b*c^2*c1)*arctan((c*x + b)/sqrt(-b^2 + a*c))/((b^6 - 3*a*b^4*c + 3*a^2*b^2*c^2 - a^3*c^3)*sqrt
(-b^2 + a*c)) - 1/48*(15*b1*c^5*x^5 - 15*b*c^4*c1*x^5 + 75*b*b1*c^4*x^4 - 75*b^2*c^3*c1*x^4 + 110*b^2*b1*c^3*x
^3 + 40*a*b1*c^4*x^3 - 110*b^3*c^2*c1*x^3 - 40*a*b*c^3*c1*x^3 + 30*b^3*b1*c^2*x^2 + 120*a*b*b1*c^3*x^2 - 30*b^
4*c*c1*x^2 - 120*a*b^2*c^2*c1*x^2 - 12*b^4*b1*c*x + 54*a*b^2*b1*c^2*x + 33*a^2*b1*c^3*x + 12*b^5*c1*x - 54*a*b
^3*c*c1*x - 33*a^2*b*c^2*c1*x + 8*b^5*b1 - 26*a*b^3*b1*c + 33*a^2*b*b1*c^2 + 2*a*b^4*c1 - 9*a^2*b^2*c*c1 - 8*a
^3*c^2*c1)/((b^6 - 3*a*b^4*c + 3*a^2*b^2*c^2 - a^3*c^3)*(c*x^2 + 2*b*x + a)^3)