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

Optimal. Leaf size=172 \[ \frac {2 e^2 \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{c^3}-\frac {4 e (2 c d-b e) \left (-c e (3 a e+b d)+b^2 e^2+c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^3 \sqrt {b^2-4 a c}}-\frac {(d+e x)^4}{a+b x+c x^2}+\frac {4 e^3 x (3 c d-b e)}{c^2}+\frac {2 e^4 x^2}{c} \]

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Rubi [A]  time = 0.22, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {768, 701, 634, 618, 206, 628} \begin {gather*} \frac {2 e^2 \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{c^3}-\frac {4 e (2 c d-b e) \left (-c e (3 a e+b d)+b^2 e^2+c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^3 \sqrt {b^2-4 a c}}-\frac {(d+e x)^4}{a+b x+c x^2}+\frac {4 e^3 x (3 c d-b e)}{c^2}+\frac {2 e^4 x^2}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

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

Rule 701

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)
^m, a + b*x + c*x^2, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2,
0] && NeQ[2*c*d - b*e, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 768

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 0]

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 241, normalized size = 1.40 \begin {gather*} \frac {\frac {-c e^3 \left (a^2 e+2 a b (2 d+e x)+4 b^2 d x\right )+b^2 e^4 (a+b x)+2 c^2 d e^2 (3 a d+2 a e x+3 b d x)-c^3 d^3 (d+4 e x)}{a+x (b+c x)}+2 e^2 \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right ) \log (a+x (b+c x))+\frac {4 e (b e-2 c d) \left (c e (3 a e+b d)-b^2 e^2-c^2 d^2\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+c e^3 x (8 c d-3 b e)+c^2 e^4 x^2}{c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*e^3*(8*c*d - 3*b*e)*x + c^2*e^4*x^2 + (b^2*e^4*(a + b*x) - c^3*d^3*(d + 4*e*x) + 2*c^2*d*e^2*(3*a*d + 3*b*d
*x + 2*a*e*x) - c*e^3*(a^2*e + 4*b^2*d*x + 2*a*b*(2*d + e*x)))/(a + x*(b + c*x)) + (4*e*(-2*c*d + b*e)*(-(c^2*
d^2) - b^2*e^2 + c*e*(b*d + 3*a*e))*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] + 2*e^2*(3*c^2*
d^2 + b^2*e^2 - c*e*(3*b*d + a*e))*Log[a + x*(b + c*x)])/c^3

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.47, size = 1701, normalized size = 9.89

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[((b^2*c^3 - 4*a*c^4)*e^4*x^4 - (b^2*c^3 - 4*a*c^4)*d^4 + 6*(a*b^2*c^2 - 4*a^2*c^3)*d^2*e^2 - 4*(a*b^3*c - 4*a
^2*b*c^2)*d*e^3 + (a*b^4 - 5*a^2*b^2*c + 4*a^3*c^2)*e^4 + 2*(4*(b^2*c^3 - 4*a*c^4)*d*e^3 - (b^3*c^2 - 4*a*b*c^
3)*e^4)*x^3 + (8*(b^3*c^2 - 4*a*b*c^3)*d*e^3 - (3*b^4*c - 13*a*b^2*c^2 + 4*a^2*c^3)*e^4)*x^2 + 2*(2*a*c^3*d^3*
e - 3*a*b*c^2*d^2*e^2 + 3*(a*b^2*c - 2*a^2*c^2)*d*e^3 - (a*b^3 - 3*a^2*b*c)*e^4 + (2*c^4*d^3*e - 3*b*c^3*d^2*e
^2 + 3*(b^2*c^2 - 2*a*c^3)*d*e^3 - (b^3*c - 3*a*b*c^2)*e^4)*x^2 + (2*b*c^3*d^3*e - 3*b^2*c^2*d^2*e^2 + 3*(b^3*
c - 2*a*b*c^2)*d*e^3 - (b^4 - 3*a*b^2*c)*e^4)*x)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c - sq
rt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) - (4*(b^2*c^3 - 4*a*c^4)*d^3*e - 6*(b^3*c^2 - 4*a*b*c^3)*d^2*e
^2 + 4*(b^4*c - 7*a*b^2*c^2 + 12*a^2*c^3)*d*e^3 - (b^5 - 9*a*b^3*c + 20*a^2*b*c^2)*e^4)*x + 2*(3*(a*b^2*c^2 -
4*a^2*c^3)*d^2*e^2 - 3*(a*b^3*c - 4*a^2*b*c^2)*d*e^3 + (a*b^4 - 5*a^2*b^2*c + 4*a^3*c^2)*e^4 + (3*(b^2*c^3 - 4
*a*c^4)*d^2*e^2 - 3*(b^3*c^2 - 4*a*b*c^3)*d*e^3 + (b^4*c - 5*a*b^2*c^2 + 4*a^2*c^3)*e^4)*x^2 + (3*(b^3*c^2 - 4
*a*b*c^3)*d^2*e^2 - 3*(b^4*c - 4*a*b^2*c^2)*d*e^3 + (b^5 - 5*a*b^3*c + 4*a^2*b*c^2)*e^4)*x)*log(c*x^2 + b*x +
a))/(a*b^2*c^3 - 4*a^2*c^4 + (b^2*c^4 - 4*a*c^5)*x^2 + (b^3*c^3 - 4*a*b*c^4)*x), ((b^2*c^3 - 4*a*c^4)*e^4*x^4
- (b^2*c^3 - 4*a*c^4)*d^4 + 6*(a*b^2*c^2 - 4*a^2*c^3)*d^2*e^2 - 4*(a*b^3*c - 4*a^2*b*c^2)*d*e^3 + (a*b^4 - 5*a
^2*b^2*c + 4*a^3*c^2)*e^4 + 2*(4*(b^2*c^3 - 4*a*c^4)*d*e^3 - (b^3*c^2 - 4*a*b*c^3)*e^4)*x^3 + (8*(b^3*c^2 - 4*
a*b*c^3)*d*e^3 - (3*b^4*c - 13*a*b^2*c^2 + 4*a^2*c^3)*e^4)*x^2 - 4*(2*a*c^3*d^3*e - 3*a*b*c^2*d^2*e^2 + 3*(a*b
^2*c - 2*a^2*c^2)*d*e^3 - (a*b^3 - 3*a^2*b*c)*e^4 + (2*c^4*d^3*e - 3*b*c^3*d^2*e^2 + 3*(b^2*c^2 - 2*a*c^3)*d*e
^3 - (b^3*c - 3*a*b*c^2)*e^4)*x^2 + (2*b*c^3*d^3*e - 3*b^2*c^2*d^2*e^2 + 3*(b^3*c - 2*a*b*c^2)*d*e^3 - (b^4 -
3*a*b^2*c)*e^4)*x)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) - (4*(b^2*c^3 - 4*
a*c^4)*d^3*e - 6*(b^3*c^2 - 4*a*b*c^3)*d^2*e^2 + 4*(b^4*c - 7*a*b^2*c^2 + 12*a^2*c^3)*d*e^3 - (b^5 - 9*a*b^3*c
+ 20*a^2*b*c^2)*e^4)*x + 2*(3*(a*b^2*c^2 - 4*a^2*c^3)*d^2*e^2 - 3*(a*b^3*c - 4*a^2*b*c^2)*d*e^3 + (a*b^4 - 5*
a^2*b^2*c + 4*a^3*c^2)*e^4 + (3*(b^2*c^3 - 4*a*c^4)*d^2*e^2 - 3*(b^3*c^2 - 4*a*b*c^3)*d*e^3 + (b^4*c - 5*a*b^2
*c^2 + 4*a^2*c^3)*e^4)*x^2 + (3*(b^3*c^2 - 4*a*b*c^3)*d^2*e^2 - 3*(b^4*c - 4*a*b^2*c^2)*d*e^3 + (b^5 - 5*a*b^3
*c + 4*a^2*b*c^2)*e^4)*x)*log(c*x^2 + b*x + a))/(a*b^2*c^3 - 4*a^2*c^4 + (b^2*c^4 - 4*a*c^5)*x^2 + (b^3*c^3 -
4*a*b*c^4)*x)]

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giac [A]  time = 0.16, size = 285, normalized size = 1.66 \begin {gather*} \frac {2 \, {\left (3 \, c^{2} d^{2} e^{2} - 3 \, b c d e^{3} + b^{2} e^{4} - a c e^{4}\right )} \log \left (c x^{2} + b x + a\right )}{c^{3}} + \frac {4 \, {\left (2 \, c^{3} d^{3} e - 3 \, b c^{2} d^{2} e^{2} + 3 \, b^{2} c d e^{3} - 6 \, a c^{2} d e^{3} - b^{3} e^{4} + 3 \, a b c e^{4}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} c^{3}} + \frac {c^{3} x^{2} e^{4} + 8 \, c^{3} d x e^{3} - 3 \, b c^{2} x e^{4}}{c^{4}} - \frac {c^{3} d^{4} - 6 \, a c^{2} d^{2} e^{2} + 4 \, a b c d e^{3} - a b^{2} e^{4} + a^{2} c e^{4} + {\left (4 \, c^{3} d^{3} e - 6 \, b c^{2} d^{2} e^{2} + 4 \, b^{2} c d e^{3} - 4 \, a c^{2} d e^{3} - b^{3} e^{4} + 2 \, a b c e^{4}\right )} x}{{\left (c x^{2} + b x + a\right )} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(3*c^2*d^2*e^2 - 3*b*c*d*e^3 + b^2*e^4 - a*c*e^4)*log(c*x^2 + b*x + a)/c^3 + 4*(2*c^3*d^3*e - 3*b*c^2*d^2*e^
2 + 3*b^2*c*d*e^3 - 6*a*c^2*d*e^3 - b^3*e^4 + 3*a*b*c*e^4)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 +
4*a*c)*c^3) + (c^3*x^2*e^4 + 8*c^3*d*x*e^3 - 3*b*c^2*x*e^4)/c^4 - (c^3*d^4 - 6*a*c^2*d^2*e^2 + 4*a*b*c*d*e^3
- a*b^2*e^4 + a^2*c*e^4 + (4*c^3*d^3*e - 6*b*c^2*d^2*e^2 + 4*b^2*c*d*e^3 - 4*a*c^2*d*e^3 - b^3*e^4 + 2*a*b*c*e
^4)*x)/((c*x^2 + b*x + a)*c^3)

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maple [B]  time = 0.06, size = 618, normalized size = 3.59 \begin {gather*} -\frac {2 a b \,e^{4} x}{\left (c \,x^{2}+b x +a \right ) c^{2}}+\frac {12 a b \,e^{4} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{2}}+\frac {4 a d \,e^{3} x}{\left (c \,x^{2}+b x +a \right ) c}-\frac {24 a d \,e^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}+\frac {b^{3} e^{4} x}{\left (c \,x^{2}+b x +a \right ) c^{3}}-\frac {4 b^{3} e^{4} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{3}}-\frac {4 b^{2} d \,e^{3} x}{\left (c \,x^{2}+b x +a \right ) c^{2}}+\frac {12 b^{2} d \,e^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{2}}+\frac {6 b \,d^{2} e^{2} x}{\left (c \,x^{2}+b x +a \right ) c}-\frac {12 b \,d^{2} e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}+\frac {e^{4} x^{2}}{c}-\frac {4 d^{3} e x}{c \,x^{2}+b x +a}+\frac {8 d^{3} e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}-\frac {a^{2} e^{4}}{\left (c \,x^{2}+b x +a \right ) c^{2}}+\frac {a \,b^{2} e^{4}}{\left (c \,x^{2}+b x +a \right ) c^{3}}-\frac {4 a b d \,e^{3}}{\left (c \,x^{2}+b x +a \right ) c^{2}}+\frac {6 a \,d^{2} e^{2}}{\left (c \,x^{2}+b x +a \right ) c}-\frac {2 a \,e^{4} \ln \left (c \,x^{2}+b x +a \right )}{c^{2}}+\frac {2 b^{2} e^{4} \ln \left (c \,x^{2}+b x +a \right )}{c^{3}}-\frac {6 b d \,e^{3} \ln \left (c \,x^{2}+b x +a \right )}{c^{2}}-\frac {3 b \,e^{4} x}{c^{2}}+\frac {6 d^{2} e^{2} \ln \left (c \,x^{2}+b x +a \right )}{c}+\frac {8 d \,e^{3} x}{c}-\frac {d^{4}}{c \,x^{2}+b x +a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^4*x^2/c-3*e^4/c^2*b*x+8*e^3/c*x*d-2/c^2/(c*x^2+b*x+a)*e^4*x*a*b+4/c/(c*x^2+b*x+a)*e^3*x*a*d+1/c^3/(c*x^2+b*x
+a)*e^4*x*b^3-4/c^2/(c*x^2+b*x+a)*e^3*x*b^2*d+6/c/(c*x^2+b*x+a)*e^2*x*b*d^2-4/(c*x^2+b*x+a)*e*x*d^3-1/c^2/(c*x
^2+b*x+a)*a^2*e^4+1/c^3/(c*x^2+b*x+a)*a*b^2*e^4-4/c^2/(c*x^2+b*x+a)*a*b*d*e^3+6/c/(c*x^2+b*x+a)*a*d^2*e^2-1/(c
*x^2+b*x+a)*d^4-2/c^2*ln(c*x^2+b*x+a)*a*e^4+2/c^3*ln(c*x^2+b*x+a)*b^2*e^4-6/c^2*ln(c*x^2+b*x+a)*b*d*e^3+6/c*ln
(c*x^2+b*x+a)*d^2*e^2+12/c^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*b*e^4-24/c/(4*a*c-b^2)^(1
/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*d*e^3+8*e/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*d^3-
4/c^3/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b^3*e^4+12/c^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/
(4*a*c-b^2)^(1/2))*b^2*d*e^3-12/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b*d^2*e^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((2*c*x+b)*(e*x+d)^4/(c*x^2+b*x+a)^2,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 = 0.32, size = 358, normalized size = 2.08 \begin {gather*} x\,\left (\frac {b\,e^4+8\,c\,d\,e^3}{c^2}-\frac {4\,b\,e^4}{c^2}\right )-\frac {\frac {a^2\,c\,e^4-a\,b^2\,e^4+4\,a\,b\,c\,d\,e^3-6\,a\,c^2\,d^2\,e^2+c^3\,d^4}{c}-\frac {x\,\left (b^3\,e^4-4\,b^2\,c\,d\,e^3+6\,b\,c^2\,d^2\,e^2-2\,a\,b\,c\,e^4-4\,c^3\,d^3\,e+4\,a\,c^2\,d\,e^3\right )}{c}}{c^3\,x^2+b\,c^2\,x+a\,c^2}-\frac {\ln \left (c\,x^2+b\,x+a\right )\,\left (16\,a^2\,c^2\,e^4-20\,a\,b^2\,c\,e^4+48\,a\,b\,c^2\,d\,e^3-48\,a\,c^3\,d^2\,e^2+4\,b^4\,e^4-12\,b^3\,c\,d\,e^3+12\,b^2\,c^2\,d^2\,e^2\right )}{2\,\left (4\,a\,c^4-b^2\,c^3\right )}+\frac {e^4\,x^2}{c}-\frac {4\,e\,\mathrm {atan}\left (\frac {b+2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )\,\left (b\,e-2\,c\,d\right )\,\left (b^2\,e^2-b\,c\,d\,e+c^2\,d^2-3\,a\,c\,e^2\right )}{c^3\,\sqrt {4\,a\,c-b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*((b*e^4 + 8*c*d*e^3)/c^2 - (4*b*e^4)/c^2) - ((c^3*d^4 - a*b^2*e^4 + a^2*c*e^4 - 6*a*c^2*d^2*e^2 + 4*a*b*c*d*
e^3)/c - (x*(b^3*e^4 - 4*c^3*d^3*e + 6*b*c^2*d^2*e^2 - 2*a*b*c*e^4 + 4*a*c^2*d*e^3 - 4*b^2*c*d*e^3))/c)/(a*c^2
+ c^3*x^2 + b*c^2*x) - (log(a + b*x + c*x^2)*(4*b^4*e^4 + 16*a^2*c^2*e^4 - 48*a*c^3*d^2*e^2 + 12*b^2*c^2*d^2*
e^2 - 20*a*b^2*c*e^4 - 12*b^3*c*d*e^3 + 48*a*b*c^2*d*e^3))/(2*(4*a*c^4 - b^2*c^3)) + (e^4*x^2)/c - (4*e*atan((
b + 2*c*x)/(4*a*c - b^2)^(1/2))*(b*e - 2*c*d)*(b^2*e^2 + c^2*d^2 - 3*a*c*e^2 - b*c*d*e))/(c^3*(4*a*c - b^2)^(1
/2))

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sympy [B]  time = 25.12, size = 1071, normalized size = 6.23 \begin {gather*} x \left (- \frac {3 b e^{4}}{c^{2}} + \frac {8 d e^{3}}{c}\right ) + \left (- \frac {2 e^{2} \left (a c e^{2} - b^{2} e^{2} + 3 b c d e - 3 c^{2} d^{2}\right )}{c^{3}} - \frac {2 e \sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right ) \left (3 a c e^{2} - b^{2} e^{2} + b c d e - c^{2} d^{2}\right )}{c^{3} \left (4 a c - b^{2}\right )}\right ) \log {\left (x + \frac {8 a^{2} c e^{4} - 4 a b^{2} e^{4} + 12 a b c d e^{3} + 4 a c^{3} \left (- \frac {2 e^{2} \left (a c e^{2} - b^{2} e^{2} + 3 b c d e - 3 c^{2} d^{2}\right )}{c^{3}} - \frac {2 e \sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right ) \left (3 a c e^{2} - b^{2} e^{2} + b c d e - c^{2} d^{2}\right )}{c^{3} \left (4 a c - b^{2}\right )}\right ) - 24 a c^{2} d^{2} e^{2} - b^{2} c^{2} \left (- \frac {2 e^{2} \left (a c e^{2} - b^{2} e^{2} + 3 b c d e - 3 c^{2} d^{2}\right )}{c^{3}} - \frac {2 e \sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right ) \left (3 a c e^{2} - b^{2} e^{2} + b c d e - c^{2} d^{2}\right )}{c^{3} \left (4 a c - b^{2}\right )}\right ) + 4 b c^{2} d^{3} e}{12 a b c e^{4} - 24 a c^{2} d e^{3} - 4 b^{3} e^{4} + 12 b^{2} c d e^{3} - 12 b c^{2} d^{2} e^{2} + 8 c^{3} d^{3} e} \right )} + \left (- \frac {2 e^{2} \left (a c e^{2} - b^{2} e^{2} + 3 b c d e - 3 c^{2} d^{2}\right )}{c^{3}} + \frac {2 e \sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right ) \left (3 a c e^{2} - b^{2} e^{2} + b c d e - c^{2} d^{2}\right )}{c^{3} \left (4 a c - b^{2}\right )}\right ) \log {\left (x + \frac {8 a^{2} c e^{4} - 4 a b^{2} e^{4} + 12 a b c d e^{3} + 4 a c^{3} \left (- \frac {2 e^{2} \left (a c e^{2} - b^{2} e^{2} + 3 b c d e - 3 c^{2} d^{2}\right )}{c^{3}} + \frac {2 e \sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right ) \left (3 a c e^{2} - b^{2} e^{2} + b c d e - c^{2} d^{2}\right )}{c^{3} \left (4 a c - b^{2}\right )}\right ) - 24 a c^{2} d^{2} e^{2} - b^{2} c^{2} \left (- \frac {2 e^{2} \left (a c e^{2} - b^{2} e^{2} + 3 b c d e - 3 c^{2} d^{2}\right )}{c^{3}} + \frac {2 e \sqrt {- 4 a c + b^{2}} \left (b e - 2 c d\right ) \left (3 a c e^{2} - b^{2} e^{2} + b c d e - c^{2} d^{2}\right )}{c^{3} \left (4 a c - b^{2}\right )}\right ) + 4 b c^{2} d^{3} e}{12 a b c e^{4} - 24 a c^{2} d e^{3} - 4 b^{3} e^{4} + 12 b^{2} c d e^{3} - 12 b c^{2} d^{2} e^{2} + 8 c^{3} d^{3} e} \right )} + \frac {- a^{2} c e^{4} + a b^{2} e^{4} - 4 a b c d e^{3} + 6 a c^{2} d^{2} e^{2} - c^{3} d^{4} + x \left (- 2 a b c e^{4} + 4 a c^{2} d e^{3} + b^{3} e^{4} - 4 b^{2} c d e^{3} + 6 b c^{2} d^{2} e^{2} - 4 c^{3} d^{3} e\right )}{a c^{3} + b c^{3} x + c^{4} x^{2}} + \frac {e^{4} x^{2}}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(-3*b*e**4/c**2 + 8*d*e**3/c) + (-2*e**2*(a*c*e**2 - b**2*e**2 + 3*b*c*d*e - 3*c**2*d**2)/c**3 - 2*e*sqrt(-4
*a*c + b**2)*(b*e - 2*c*d)*(3*a*c*e**2 - b**2*e**2 + b*c*d*e - c**2*d**2)/(c**3*(4*a*c - b**2)))*log(x + (8*a*
*2*c*e**4 - 4*a*b**2*e**4 + 12*a*b*c*d*e**3 + 4*a*c**3*(-2*e**2*(a*c*e**2 - b**2*e**2 + 3*b*c*d*e - 3*c**2*d**
2)/c**3 - 2*e*sqrt(-4*a*c + b**2)*(b*e - 2*c*d)*(3*a*c*e**2 - b**2*e**2 + b*c*d*e - c**2*d**2)/(c**3*(4*a*c -
b**2))) - 24*a*c**2*d**2*e**2 - b**2*c**2*(-2*e**2*(a*c*e**2 - b**2*e**2 + 3*b*c*d*e - 3*c**2*d**2)/c**3 - 2*e
*sqrt(-4*a*c + b**2)*(b*e - 2*c*d)*(3*a*c*e**2 - b**2*e**2 + b*c*d*e - c**2*d**2)/(c**3*(4*a*c - b**2))) + 4*b
*c**2*d**3*e)/(12*a*b*c*e**4 - 24*a*c**2*d*e**3 - 4*b**3*e**4 + 12*b**2*c*d*e**3 - 12*b*c**2*d**2*e**2 + 8*c**
3*d**3*e)) + (-2*e**2*(a*c*e**2 - b**2*e**2 + 3*b*c*d*e - 3*c**2*d**2)/c**3 + 2*e*sqrt(-4*a*c + b**2)*(b*e - 2
*c*d)*(3*a*c*e**2 - b**2*e**2 + b*c*d*e - c**2*d**2)/(c**3*(4*a*c - b**2)))*log(x + (8*a**2*c*e**4 - 4*a*b**2*
e**4 + 12*a*b*c*d*e**3 + 4*a*c**3*(-2*e**2*(a*c*e**2 - b**2*e**2 + 3*b*c*d*e - 3*c**2*d**2)/c**3 + 2*e*sqrt(-4
*a*c + b**2)*(b*e - 2*c*d)*(3*a*c*e**2 - b**2*e**2 + b*c*d*e - c**2*d**2)/(c**3*(4*a*c - b**2))) - 24*a*c**2*d
**2*e**2 - b**2*c**2*(-2*e**2*(a*c*e**2 - b**2*e**2 + 3*b*c*d*e - 3*c**2*d**2)/c**3 + 2*e*sqrt(-4*a*c + b**2)*
(b*e - 2*c*d)*(3*a*c*e**2 - b**2*e**2 + b*c*d*e - c**2*d**2)/(c**3*(4*a*c - b**2))) + 4*b*c**2*d**3*e)/(12*a*b
*c*e**4 - 24*a*c**2*d*e**3 - 4*b**3*e**4 + 12*b**2*c*d*e**3 - 12*b*c**2*d**2*e**2 + 8*c**3*d**3*e)) + (-a**2*c
*e**4 + a*b**2*e**4 - 4*a*b*c*d*e**3 + 6*a*c**2*d**2*e**2 - c**3*d**4 + x*(-2*a*b*c*e**4 + 4*a*c**2*d*e**3 + b
**3*e**4 - 4*b**2*c*d*e**3 + 6*b*c**2*d**2*e**2 - 4*c**3*d**3*e))/(a*c**3 + b*c**3*x + c**4*x**2) + e**4*x**2/
c

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