3.14.33 \(\int \frac {(b+2 c x) (d+e x)^3}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=188 \[ \frac {e x \left (-c e (2 a e+3 b d)+b^2 e^2+6 c^2 d^2\right )}{c^2}+\frac {(2 c d-b e) \left (-c e (3 a e+b d)+b^2 e^2+c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 c^3}-\frac {e \sqrt {b^2-4 a c} \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^3}+\frac {e^2 x^2 (6 c d-b e)}{2 c}+\frac {2 e^3 x^3}{3} \]

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Rubi [A]  time = 0.24, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {800, 634, 618, 206, 628} \begin {gather*} \frac {(2 c d-b e) \left (-c e (3 a e+b d)+b^2 e^2+c^2 d^2\right ) \log \left (a+b x+c x^2\right )}{2 c^3}+\frac {e x \left (-c e (2 a e+3 b d)+b^2 e^2+6 c^2 d^2\right )}{c^2}-\frac {e \sqrt {b^2-4 a c} \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^3}+\frac {e^2 x^2 (6 c d-b e)}{2 c}+\frac {2 e^3 x^3}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((b + 2*c*x)*(d + e*x)^3)/(a + b*x + c*x^2),x]

[Out]

(e*(6*c^2*d^2 + b^2*e^2 - c*e*(3*b*d + 2*a*e))*x)/c^2 + (e^2*(6*c*d - b*e)*x^2)/(2*c) + (2*e^3*x^3)/3 - (Sqrt[
b^2 - 4*a*c]*e*(3*c^2*d^2 + b^2*e^2 - c*e*(3*b*d + a*e))*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/c^3 + ((2*c*d
- b*e)*(c^2*d^2 + b^2*e^2 - c*e*(b*d + 3*a*e))*Log[a + b*x + c*x^2])/(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 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 177, normalized size = 0.94 \begin {gather*} \frac {c e x \left (-3 c e (4 a e+6 b d+b e x)+6 b^2 e^2+2 c^2 \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )+3 (2 c d-b e) \left (-c e (3 a e+b d)+b^2 e^2+c^2 d^2\right ) \log (a+x (b+c x))-6 e \sqrt {4 a c-b^2} \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{6 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((b + 2*c*x)*(d + e*x)^3)/(a + b*x + c*x^2),x]

[Out]

(c*e*x*(6*b^2*e^2 - 3*c*e*(6*b*d + 4*a*e + b*e*x) + 2*c^2*(18*d^2 + 9*d*e*x + 2*e^2*x^2)) - 6*Sqrt[-b^2 + 4*a*
c]*e*(3*c^2*d^2 + b^2*e^2 - c*e*(3*b*d + a*e))*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]] + 3*(2*c*d - b*e)*(c^2*d
^2 + b^2*e^2 - c*e*(b*d + 3*a*e))*Log[a + x*(b + c*x)])/(6*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)^3}{a+b x+c x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((b + 2*c*x)*(d + e*x)^3)/(a + b*x + c*x^2),x]

[Out]

IntegrateAlgebraic[((b + 2*c*x)*(d + e*x)^3)/(a + b*x + c*x^2), x]

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fricas [A]  time = 0.44, size = 458, normalized size = 2.44 \begin {gather*} \left [\frac {4 \, c^{3} e^{3} x^{3} + 3 \, {\left (6 \, c^{3} d e^{2} - b c^{2} e^{3}\right )} x^{2} - 3 \, {\left (3 \, c^{2} d^{2} e - 3 \, b c d e^{2} + {\left (b^{2} - a c\right )} e^{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 ) + 6 \, {\left (6 \, c^{3} d^{2} e - 3 \, b c^{2} d e^{2} + {\left (b^{2} c - 2 \, a c^{2}\right )} e^{3}\right )} x + 3 \, {\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e + 3 \, {\left (b^{2} c - 2 \, a c^{2}\right )} d e^{2} - {\left (b^{3} - 3 \, a b c\right )} e^{3}\right )} \log \left (c x^{2} + b x + a\right )}{6 \, c^{3}}, \frac {4 \, c^{3} e^{3} x^{3} + 3 \, {\left (6 \, c^{3} d e^{2} - b c^{2} e^{3}\right )} x^{2} - 6 \, {\left (3 \, c^{2} d^{2} e - 3 \, b c d e^{2} + {\left (b^{2} - a c\right )} e^{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 ) + 6 \, {\left (6 \, c^{3} d^{2} e - 3 \, b c^{2} d e^{2} + {\left (b^{2} c - 2 \, a c^{2}\right )} e^{3}\right )} x + 3 \, {\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e + 3 \, {\left (b^{2} c - 2 \, a c^{2}\right )} d e^{2} - {\left (b^{3} - 3 \, a b c\right )} e^{3}\right )} \log \left (c x^{2} + b x + a\right )}{6 \, c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^3/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

[1/6*(4*c^3*e^3*x^3 + 3*(6*c^3*d*e^2 - b*c^2*e^3)*x^2 - 3*(3*c^2*d^2*e - 3*b*c*d*e^2 + (b^2 - a*c)*e^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)) + 6*(6*
c^3*d^2*e - 3*b*c^2*d*e^2 + (b^2*c - 2*a*c^2)*e^3)*x + 3*(2*c^3*d^3 - 3*b*c^2*d^2*e + 3*(b^2*c - 2*a*c^2)*d*e^
2 - (b^3 - 3*a*b*c)*e^3)*log(c*x^2 + b*x + a))/c^3, 1/6*(4*c^3*e^3*x^3 + 3*(6*c^3*d*e^2 - b*c^2*e^3)*x^2 - 6*(
3*c^2*d^2*e - 3*b*c*d*e^2 + (b^2 - a*c)*e^3)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 -
4*a*c)) + 6*(6*c^3*d^2*e - 3*b*c^2*d*e^2 + (b^2*c - 2*a*c^2)*e^3)*x + 3*(2*c^3*d^3 - 3*b*c^2*d^2*e + 3*(b^2*c
- 2*a*c^2)*d*e^2 - (b^3 - 3*a*b*c)*e^3)*log(c*x^2 + b*x + a))/c^3]

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giac [A]  time = 0.16, size = 251, normalized size = 1.34 \begin {gather*} \frac {{\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e + 3 \, b^{2} c d e^{2} - 6 \, a c^{2} d e^{2} - b^{3} e^{3} + 3 \, a b c e^{3}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{3}} + \frac {{\left (3 \, b^{2} c^{2} d^{2} e - 12 \, a c^{3} d^{2} e - 3 \, b^{3} c d e^{2} + 12 \, a b c^{2} d e^{2} + b^{4} e^{3} - 5 \, a b^{2} c e^{3} + 4 \, a^{2} c^{2} e^{3}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} c^{3}} + \frac {4 \, c^{3} x^{3} e^{3} + 18 \, c^{3} d x^{2} e^{2} + 36 \, c^{3} d^{2} x e - 3 \, b c^{2} x^{2} e^{3} - 18 \, b c^{2} d x e^{2} + 6 \, b^{2} c x e^{3} - 12 \, a c^{2} x e^{3}}{6 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)^3/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

1/2*(2*c^3*d^3 - 3*b*c^2*d^2*e + 3*b^2*c*d*e^2 - 6*a*c^2*d*e^2 - b^3*e^3 + 3*a*b*c*e^3)*log(c*x^2 + b*x + a)/c
^3 + (3*b^2*c^2*d^2*e - 12*a*c^3*d^2*e - 3*b^3*c*d*e^2 + 12*a*b*c^2*d*e^2 + b^4*e^3 - 5*a*b^2*c*e^3 + 4*a^2*c^
2*e^3)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*c^3) + 1/6*(4*c^3*x^3*e^3 + 18*c^3*d*x^2*e^2
+ 36*c^3*d^2*x*e - 3*b*c^2*x^2*e^3 - 18*b*c^2*d*x*e^2 + 6*b^2*c*x*e^3 - 12*a*c^2*x*e^3)/c^3

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maple [B]  time = 0.05, size = 492, normalized size = 2.62 \begin {gather*} \frac {2 e^{3} x^{3}}{3}+\frac {4 a^{2} e^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}-\frac {5 a \,b^{2} 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 {12 a b d \,e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}-\frac {12 a \,d^{2} e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}+\frac {b^{4} e^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{3}}-\frac {3 b^{3} d \,e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{2}}+\frac {3 b^{2} d^{2} e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}-\frac {b \,e^{3} x^{2}}{2 c}+3 d \,e^{2} x^{2}+\frac {3 a b \,e^{3} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{2}}-\frac {3 a d \,e^{2} \ln \left (c \,x^{2}+b x +a \right )}{c}-\frac {2 a \,e^{3} x}{c}-\frac {b^{3} e^{3} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{3}}+\frac {3 b^{2} d \,e^{2} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{2}}+\frac {b^{2} e^{3} x}{c^{2}}-\frac {3 b \,d^{2} e \ln \left (c \,x^{2}+b x +a \right )}{2 c}-\frac {3 b d \,e^{2} x}{c}+d^{3} \ln \left (c \,x^{2}+b x +a \right )+6 d^{2} e x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*x+b)*(e*x+d)^3/(c*x^2+b*x+a),x)

[Out]

2/3*e^3*x^3-1/2/c*e^3*x^2*b+3*e^2*x^2*d-2/c*e^3*x*a+1/c^2*e^3*x*b^2-3/c*e^2*x*b*d+6*e*x*d^2+3/2/c^2*ln(c*x^2+b
*x+a)*a*b*e^3-3/c*ln(c*x^2+b*x+a)*a*d*e^2-1/2/c^3*ln(c*x^2+b*x+a)*b^3*e^3+3/2/c^2*ln(c*x^2+b*x+a)*b^2*d*e^2-3/
2/c*ln(c*x^2+b*x+a)*b*d^2*e+ln(c*x^2+b*x+a)*d^3+4/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a^2*
e^3-5/c^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*b^2*e^3+12/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x
+b)/(4*a*c-b^2)^(1/2))*a*b*d*e^2-12/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*a*d^2*e+1/c^3/(4*a*c
-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b^4*e^3-3/c^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(
1/2))*b^3*d*e^2+3/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b^2*d^2*e

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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)^3/(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 = 2.00, size = 433, normalized size = 2.30 \begin {gather*} x^2\,\left (\frac {b\,e^3+6\,c\,d\,e^2}{2\,c}-\frac {b\,e^3}{c}\right )-x\,\left (\frac {b\,\left (\frac {b\,e^3+6\,c\,d\,e^2}{c}-\frac {2\,b\,e^3}{c}\right )}{c}+\frac {2\,a\,e^3}{c}-\frac {3\,d\,e\,\left (b\,e+2\,c\,d\right )}{c}\right )+\frac {2\,e^3\,x^3}{3}-\frac {\ln \left (b\,\sqrt {b^2-4\,a\,c}-4\,a\,c+b^2+2\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (b^3\,e^3-2\,c^3\,d^3+b^2\,e^3\,\sqrt {b^2-4\,a\,c}-3\,a\,b\,c\,e^3-a\,c\,e^3\,\sqrt {b^2-4\,a\,c}+6\,a\,c^2\,d\,e^2+3\,b\,c^2\,d^2\,e-3\,b^2\,c\,d\,e^2+3\,c^2\,d^2\,e\,\sqrt {b^2-4\,a\,c}-3\,b\,c\,d\,e^2\,\sqrt {b^2-4\,a\,c}\right )}{2\,c^3}-\frac {\ln \left (4\,a\,c+b\,\sqrt {b^2-4\,a\,c}-b^2+2\,c\,x\,\sqrt {b^2-4\,a\,c}\right )\,\left (b^3\,e^3-2\,c^3\,d^3-b^2\,e^3\,\sqrt {b^2-4\,a\,c}-3\,a\,b\,c\,e^3+a\,c\,e^3\,\sqrt {b^2-4\,a\,c}+6\,a\,c^2\,d\,e^2+3\,b\,c^2\,d^2\,e-3\,b^2\,c\,d\,e^2-3\,c^2\,d^2\,e\,\sqrt {b^2-4\,a\,c}+3\,b\,c\,d\,e^2\,\sqrt {b^2-4\,a\,c}\right )}{2\,c^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b + 2*c*x)*(d + e*x)^3)/(a + b*x + c*x^2),x)

[Out]

x^2*((b*e^3 + 6*c*d*e^2)/(2*c) - (b*e^3)/c) - x*((b*((b*e^3 + 6*c*d*e^2)/c - (2*b*e^3)/c))/c + (2*a*e^3)/c - (
3*d*e*(b*e + 2*c*d))/c) + (2*e^3*x^3)/3 - (log(b*(b^2 - 4*a*c)^(1/2) - 4*a*c + b^2 + 2*c*x*(b^2 - 4*a*c)^(1/2)
)*(b^3*e^3 - 2*c^3*d^3 + b^2*e^3*(b^2 - 4*a*c)^(1/2) - 3*a*b*c*e^3 - a*c*e^3*(b^2 - 4*a*c)^(1/2) + 6*a*c^2*d*e
^2 + 3*b*c^2*d^2*e - 3*b^2*c*d*e^2 + 3*c^2*d^2*e*(b^2 - 4*a*c)^(1/2) - 3*b*c*d*e^2*(b^2 - 4*a*c)^(1/2)))/(2*c^
3) - (log(4*a*c + b*(b^2 - 4*a*c)^(1/2) - b^2 + 2*c*x*(b^2 - 4*a*c)^(1/2))*(b^3*e^3 - 2*c^3*d^3 - b^2*e^3*(b^2
- 4*a*c)^(1/2) - 3*a*b*c*e^3 + a*c*e^3*(b^2 - 4*a*c)^(1/2) + 6*a*c^2*d*e^2 + 3*b*c^2*d^2*e - 3*b^2*c*d*e^2 -
3*c^2*d^2*e*(b^2 - 4*a*c)^(1/2) + 3*b*c*d*e^2*(b^2 - 4*a*c)^(1/2)))/(2*c^3)

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sympy [B]  time = 4.81, size = 561, normalized size = 2.98 \begin {gather*} \frac {2 e^{3} x^{3}}{3} + x^{2} \left (- \frac {b e^{3}}{2 c} + 3 d e^{2}\right ) + x \left (- \frac {2 a e^{3}}{c} + \frac {b^{2} e^{3}}{c^{2}} - \frac {3 b d e^{2}}{c} + 6 d^{2} e\right ) + \left (- \frac {e \sqrt {- 4 a c + b^{2}} \left (a c e^{2} - b^{2} e^{2} + 3 b c d e - 3 c^{2} d^{2}\right )}{2 c^{3}} + \frac {\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 )}{2 c^{3}}\right ) \log {\left (x + \frac {- a b e^{3} + 3 a c d e^{2} - c^{2} d^{3} + c^{2} \left (- \frac {e \sqrt {- 4 a c + b^{2}} \left (a c e^{2} - b^{2} e^{2} + 3 b c d e - 3 c^{2} d^{2}\right )}{2 c^{3}} + \frac {\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 )}{2 c^{3}}\right )}{a c e^{3} - b^{2} e^{3} + 3 b c d e^{2} - 3 c^{2} d^{2} e} \right )} + \left (\frac {e \sqrt {- 4 a c + b^{2}} \left (a c e^{2} - b^{2} e^{2} + 3 b c d e - 3 c^{2} d^{2}\right )}{2 c^{3}} + \frac {\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 )}{2 c^{3}}\right ) \log {\left (x + \frac {- a b e^{3} + 3 a c d e^{2} - c^{2} d^{3} + c^{2} \left (\frac {e \sqrt {- 4 a c + b^{2}} \left (a c e^{2} - b^{2} e^{2} + 3 b c d e - 3 c^{2} d^{2}\right )}{2 c^{3}} + \frac {\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 )}{2 c^{3}}\right )}{a c e^{3} - b^{2} e^{3} + 3 b c d e^{2} - 3 c^{2} d^{2} e} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(e*x+d)**3/(c*x**2+b*x+a),x)

[Out]

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

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