3.22.24 \(\int \frac {(A+B x) (d+e x)^2}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=205 \[ \frac {\log \left (a+b x+c x^2\right ) \left (B \left (-c e (a e+2 b d)+b^2 e^2+c^2 d^2\right )+A c e (2 c d-b e)\right )}{2 c^3}+\frac {\tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (b c \left (-3 a B e^2+2 A c d e+B c d^2\right )-2 c^2 \left (-a A e^2-2 a B d e+A c d^2\right )-b^2 c e (A e+2 B d)+b^3 B e^2\right )}{c^3 \sqrt {b^2-4 a c}}+\frac {e x (A c e-b B e+2 B c d)}{c^2}+\frac {B e^2 x^2}{2 c} \]

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Rubi [A]  time = 0.37, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {800, 634, 618, 206, 628} \begin {gather*} \frac {\log \left (a+b x+c x^2\right ) \left (B \left (-c e (a e+2 b d)+b^2 e^2+c^2 d^2\right )+A c e (2 c d-b e)\right )}{2 c^3}+\frac {\tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (b c \left (-3 a B e^2+2 A c d e+B c d^2\right )-2 c^2 \left (-a A e^2-2 a B d e+A c d^2\right )-b^2 c e (A e+2 B d)+b^3 B e^2\right )}{c^3 \sqrt {b^2-4 a c}}+\frac {e x (A c e-b B e+2 B c d)}{c^2}+\frac {B e^2 x^2}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(2*B*c*d - b*B*e + A*c*e)*x)/c^2 + (B*e^2*x^2)/(2*c) + ((b^3*B*e^2 - b^2*c*e*(2*B*d + A*e) - 2*c^2*(A*c*d^2
- 2*a*B*d*e - a*A*e^2) + b*c*(B*c*d^2 + 2*A*c*d*e - 3*a*B*e^2))*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(c^3*
Sqrt[b^2 - 4*a*c]) + ((A*c*e*(2*c*d - b*e) + B*(c^2*d^2 + b^2*e^2 - c*e*(2*b*d + 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 {(A+B x) (d+e x)^2}{a+b x+c x^2} \, dx &=\int \left (\frac {e (2 B c d-b B e+A c e)}{c^2}+\frac {B e^2 x}{c}+\frac {-a B e (2 c d-b e)+A c \left (c d^2-a e^2\right )+\left (A c e (2 c d-b e)+B \left (c^2 d^2+b^2 e^2-c e (2 b d+a e)\right )\right ) x}{c^2 \left (a+b x+c x^2\right )}\right ) \, dx\\ &=\frac {e (2 B c d-b B e+A c e) x}{c^2}+\frac {B e^2 x^2}{2 c}+\frac {\int \frac {-a B e (2 c d-b e)+A c \left (c d^2-a e^2\right )+\left (A c e (2 c d-b e)+B \left (c^2 d^2+b^2 e^2-c e (2 b d+a e)\right )\right ) x}{a+b x+c x^2} \, dx}{c^2}\\ &=\frac {e (2 B c d-b B e+A c e) x}{c^2}+\frac {B e^2 x^2}{2 c}-\frac {\left (b^3 B e^2-b^2 c e (2 B d+A e)-2 c^2 \left (A c d^2-2 a B d e-a A e^2\right )+b c \left (B c d^2+2 A c d e-3 a B e^2\right )\right ) \int \frac {1}{a+b x+c x^2} \, dx}{2 c^3}+\frac {\left (A c e (2 c d-b e)+B \left (c^2 d^2+b^2 e^2-c e (2 b d+a e)\right )\right ) \int \frac {b+2 c x}{a+b x+c x^2} \, dx}{2 c^3}\\ &=\frac {e (2 B c d-b B e+A c e) x}{c^2}+\frac {B e^2 x^2}{2 c}+\frac {\left (A c e (2 c d-b e)+B \left (c^2 d^2+b^2 e^2-c e (2 b d+a e)\right )\right ) \log \left (a+b x+c x^2\right )}{2 c^3}+\frac {\left (b^3 B e^2-b^2 c e (2 B d+A e)-2 c^2 \left (A c d^2-2 a B d e-a A e^2\right )+b c \left (B c d^2+2 A c d e-3 a B e^2\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 (2 B c d-b B e+A c e) x}{c^2}+\frac {B e^2 x^2}{2 c}+\frac {\left (b^3 B e^2-b^2 c e (2 B d+A e)-2 c^2 \left (A c d^2-2 a B d e-a A e^2\right )+b c \left (B c d^2+2 A c d e-3 a B e^2\right )\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 {\left (A c e (2 c d-b e)+B \left (c^2 d^2+b^2 e^2-c e (2 b d+a e)\right )\right ) \log \left (a+b x+c x^2\right )}{2 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 203, normalized size = 0.99 \begin {gather*} \frac {\log (a+x (b+c x)) \left (B \left (-c e (a e+2 b d)+b^2 e^2+c^2 d^2\right )+A c e (2 c d-b e)\right )-\frac {2 \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right ) \left (b c \left (-3 a B e^2+2 A c d e+B c d^2\right )+2 c^2 \left (a A e^2+2 a B d e-A c d^2\right )-b^2 c e (A e+2 B d)+b^3 B e^2\right )}{\sqrt {4 a c-b^2}}+2 c e x (A c e-b B e+2 B c d)+B c^2 e^2 x^2}{2 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 673, normalized size = 3.28 \begin {gather*} \left [\frac {{\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} e^{2} x^{2} + {\left ({\left (B b c^{2} - 2 \, A c^{3}\right )} d^{2} - 2 \, {\left (B b^{2} c - {\left (2 \, B a + A b\right )} c^{2}\right )} d e + {\left (B b^{3} + 2 \, A a c^{2} - {\left (3 \, B a b + A b^{2}\right )} c\right )} e^{2}\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 ) + 2 \, {\left (2 \, {\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} d e - {\left (B b^{3} c + 4 \, A a c^{3} - {\left (4 \, B a b + A b^{2}\right )} c^{2}\right )} e^{2}\right )} x + {\left ({\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} d^{2} - 2 \, {\left (B b^{3} c + 4 \, A a c^{3} - {\left (4 \, B a b + A b^{2}\right )} c^{2}\right )} d e + {\left (B b^{4} + 4 \, {\left (B a^{2} + A a b\right )} c^{2} - {\left (5 \, B a b^{2} + A b^{3}\right )} c\right )} e^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}, \frac {{\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} e^{2} x^{2} + 2 \, {\left ({\left (B b c^{2} - 2 \, A c^{3}\right )} d^{2} - 2 \, {\left (B b^{2} c - {\left (2 \, B a + A b\right )} c^{2}\right )} d e + {\left (B b^{3} + 2 \, A a c^{2} - {\left (3 \, B a b + A b^{2}\right )} c\right )} e^{2}\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 ) + 2 \, {\left (2 \, {\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} d e - {\left (B b^{3} c + 4 \, A a c^{3} - {\left (4 \, B a b + A b^{2}\right )} c^{2}\right )} e^{2}\right )} x + {\left ({\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} d^{2} - 2 \, {\left (B b^{3} c + 4 \, A a c^{3} - {\left (4 \, B a b + A b^{2}\right )} c^{2}\right )} d e + {\left (B b^{4} + 4 \, {\left (B a^{2} + A a b\right )} c^{2} - {\left (5 \, B a b^{2} + A b^{3}\right )} c\right )} e^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((B*b^2*c^2 - 4*B*a*c^3)*e^2*x^2 + ((B*b*c^2 - 2*A*c^3)*d^2 - 2*(B*b^2*c - (2*B*a + A*b)*c^2)*d*e + (B*b^
3 + 2*A*a*c^2 - (3*B*a*b + A*b^2)*c)*e^2)*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)) + 2*(2*(B*b^2*c^2 - 4*B*a*c^3)*d*e - (B*b^3*c + 4*A*a*c^3 - (4*B*a*b
+ A*b^2)*c^2)*e^2)*x + ((B*b^2*c^2 - 4*B*a*c^3)*d^2 - 2*(B*b^3*c + 4*A*a*c^3 - (4*B*a*b + A*b^2)*c^2)*d*e + (B
*b^4 + 4*(B*a^2 + A*a*b)*c^2 - (5*B*a*b^2 + A*b^3)*c)*e^2)*log(c*x^2 + b*x + a))/(b^2*c^3 - 4*a*c^4), 1/2*((B*
b^2*c^2 - 4*B*a*c^3)*e^2*x^2 + 2*((B*b*c^2 - 2*A*c^3)*d^2 - 2*(B*b^2*c - (2*B*a + A*b)*c^2)*d*e + (B*b^3 + 2*A
*a*c^2 - (3*B*a*b + A*b^2)*c)*e^2)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) +
2*(2*(B*b^2*c^2 - 4*B*a*c^3)*d*e - (B*b^3*c + 4*A*a*c^3 - (4*B*a*b + A*b^2)*c^2)*e^2)*x + ((B*b^2*c^2 - 4*B*a*
c^3)*d^2 - 2*(B*b^3*c + 4*A*a*c^3 - (4*B*a*b + A*b^2)*c^2)*d*e + (B*b^4 + 4*(B*a^2 + A*a*b)*c^2 - (5*B*a*b^2 +
A*b^3)*c)*e^2)*log(c*x^2 + b*x + a))/(b^2*c^3 - 4*a*c^4)]

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giac [A]  time = 0.16, size = 219, normalized size = 1.07 \begin {gather*} \frac {B c x^{2} e^{2} + 4 \, B c d x e - 2 \, B b x e^{2} + 2 \, A c x e^{2}}{2 \, c^{2}} + \frac {{\left (B c^{2} d^{2} - 2 \, B b c d e + 2 \, A c^{2} d e + B b^{2} e^{2} - B a c e^{2} - A b c e^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{3}} - \frac {{\left (B b c^{2} d^{2} - 2 \, A c^{3} d^{2} - 2 \, B b^{2} c d e + 4 \, B a c^{2} d e + 2 \, A b c^{2} d e + B b^{3} e^{2} - 3 \, B a b c e^{2} - A b^{2} c e^{2} + 2 \, A a c^{2} e^{2}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*c*x^2*e^2 + 4*B*c*d*x*e - 2*B*b*x*e^2 + 2*A*c*x*e^2)/c^2 + 1/2*(B*c^2*d^2 - 2*B*b*c*d*e + 2*A*c^2*d*e +
B*b^2*e^2 - B*a*c*e^2 - A*b*c*e^2)*log(c*x^2 + b*x + a)/c^3 - (B*b*c^2*d^2 - 2*A*c^3*d^2 - 2*B*b^2*c*d*e + 4*
B*a*c^2*d*e + 2*A*b*c^2*d*e + B*b^3*e^2 - 3*B*a*b*c*e^2 - A*b^2*c*e^2 + 2*A*a*c^2*e^2)*arctan((2*c*x + b)/sqrt
(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*c^3)

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maple [B]  time = 0.05, size = 543, normalized size = 2.65 \begin {gather*} -\frac {2 A a \,e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}+\frac {A \,b^{2} 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 {2 A b d e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}+\frac {2 A \,d^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}+\frac {3 B a b \,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 {4 B a d e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}-\frac {B \,b^{3} e^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{3}}+\frac {2 B \,b^{2} d e \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c^{2}}-\frac {B b \,d^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}+\frac {B \,e^{2} x^{2}}{2 c}-\frac {A b \,e^{2} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{2}}+\frac {A d e \ln \left (c \,x^{2}+b x +a \right )}{c}+\frac {A \,e^{2} x}{c}-\frac {B a \,e^{2} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{2}}+\frac {B \,b^{2} e^{2} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{3}}-\frac {B b d e \ln \left (c \,x^{2}+b x +a \right )}{c^{2}}-\frac {B b \,e^{2} x}{c^{2}}+\frac {B \,d^{2} \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 B d e x}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*B/c*e^2*x^2+1/c*e^2*A*x-1/c^2*e^2*B*b*x+2/c*e*B*d*x-1/2/c^2*ln(c*x^2+b*x+a)*A*b*e^2+1/c*ln(c*x^2+b*x+a)*A*
d*e-1/2/c^2*ln(c*x^2+b*x+a)*a*B*e^2+1/2/c^3*ln(c*x^2+b*x+a)*b^2*B*e^2-1/c^2*ln(c*x^2+b*x+a)*B*b*d*e+1/2/c*ln(c
*x^2+b*x+a)*B*d^2-2/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*A*a*e^2+2/(4*a*c-b^2)^(1/2)*arctan
((2*c*x+b)/(4*a*c-b^2)^(1/2))*A*d^2+3/c^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*B*a*b*e^2-4/c/
(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*B*a*d*e+1/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^2-2/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*A*b*d*e-1/c^3/(4*a*c-b^2)^(1/2
)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b^3*B*e^2+2/c^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*B*
b^2*d*e-1/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*B*b*d^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((B*x+A)*(e*x+d)^2/(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 = 2.78, size = 316, normalized size = 1.54 \begin {gather*} x\,\left (\frac {A\,e^2+2\,B\,d\,e}{c}-\frac {B\,b\,e^2}{c^2}\right )-\frac {\ln \left (c\,x^2+b\,x+a\right )\,\left (4\,B\,a^2\,c^2\,e^2-5\,B\,a\,b^2\,c\,e^2+8\,B\,a\,b\,c^2\,d\,e+4\,A\,a\,b\,c^2\,e^2-4\,B\,a\,c^3\,d^2-8\,A\,a\,c^3\,d\,e+B\,b^4\,e^2-2\,B\,b^3\,c\,d\,e-A\,b^3\,c\,e^2+B\,b^2\,c^2\,d^2+2\,A\,b^2\,c^2\,d\,e\right )}{2\,\left (4\,a\,c^4-b^2\,c^3\right )}-\frac {\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )\,\left (B\,b^3\,e^2-2\,B\,b^2\,c\,d\,e-A\,b^2\,c\,e^2+B\,b\,c^2\,d^2+2\,A\,b\,c^2\,d\,e-3\,B\,a\,b\,c\,e^2-2\,A\,c^3\,d^2+4\,B\,a\,c^2\,d\,e+2\,A\,a\,c^2\,e^2\right )}{c^3\,\sqrt {4\,a\,c-b^2}}+\frac {B\,e^2\,x^2}{2\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*((A*e^2 + 2*B*d*e)/c - (B*b*e^2)/c^2) - (log(a + b*x + c*x^2)*(B*b^4*e^2 - 4*B*a*c^3*d^2 - A*b^3*c*e^2 + 4*B
*a^2*c^2*e^2 + B*b^2*c^2*d^2 - 8*A*a*c^3*d*e - 2*B*b^3*c*d*e + 4*A*a*b*c^2*e^2 - 5*B*a*b^2*c*e^2 + 2*A*b^2*c^2
*d*e + 8*B*a*b*c^2*d*e))/(2*(4*a*c^4 - b^2*c^3)) - (atan(b/(4*a*c - b^2)^(1/2) + (2*c*x)/(4*a*c - b^2)^(1/2))*
(B*b^3*e^2 - 2*A*c^3*d^2 + 2*A*a*c^2*e^2 - A*b^2*c*e^2 + B*b*c^2*d^2 - 3*B*a*b*c*e^2 + 2*A*b*c^2*d*e + 4*B*a*c
^2*d*e - 2*B*b^2*c*d*e))/(c^3*(4*a*c - b^2)^(1/2)) + (B*e^2*x^2)/(2*c)

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sympy [B]  time = 11.09, size = 1532, normalized size = 7.47 \begin {gather*} \frac {B e^{2} x^{2}}{2 c} + x \left (\frac {A e^{2}}{c} - \frac {B b e^{2}}{c^{2}} + \frac {2 B d e}{c}\right ) + \left (- \frac {\sqrt {- 4 a c + b^{2}} \left (- 2 A a c^{2} e^{2} + A b^{2} c e^{2} - 2 A b c^{2} d e + 2 A c^{3} d^{2} + 3 B a b c e^{2} - 4 B a c^{2} d e - B b^{3} e^{2} + 2 B b^{2} c d e - B b c^{2} d^{2}\right )}{2 c^{3} \left (4 a c - b^{2}\right )} - \frac {A b c e^{2} - 2 A c^{2} d e + B a c e^{2} - B b^{2} e^{2} + 2 B b c d e - B c^{2} d^{2}}{2 c^{3}}\right ) \log {\left (x + \frac {A a b c e^{2} - 4 A a c^{2} d e + A b c^{2} d^{2} + 2 B a^{2} c e^{2} - B a b^{2} e^{2} + 2 B a b c d e - 2 B a c^{2} d^{2} + 4 a c^{3} \left (- \frac {\sqrt {- 4 a c + b^{2}} \left (- 2 A a c^{2} e^{2} + A b^{2} c e^{2} - 2 A b c^{2} d e + 2 A c^{3} d^{2} + 3 B a b c e^{2} - 4 B a c^{2} d e - B b^{3} e^{2} + 2 B b^{2} c d e - B b c^{2} d^{2}\right )}{2 c^{3} \left (4 a c - b^{2}\right )} - \frac {A b c e^{2} - 2 A c^{2} d e + B a c e^{2} - B b^{2} e^{2} + 2 B b c d e - B c^{2} d^{2}}{2 c^{3}}\right ) - b^{2} c^{2} \left (- \frac {\sqrt {- 4 a c + b^{2}} \left (- 2 A a c^{2} e^{2} + A b^{2} c e^{2} - 2 A b c^{2} d e + 2 A c^{3} d^{2} + 3 B a b c e^{2} - 4 B a c^{2} d e - B b^{3} e^{2} + 2 B b^{2} c d e - B b c^{2} d^{2}\right )}{2 c^{3} \left (4 a c - b^{2}\right )} - \frac {A b c e^{2} - 2 A c^{2} d e + B a c e^{2} - B b^{2} e^{2} + 2 B b c d e - B c^{2} d^{2}}{2 c^{3}}\right )}{- 2 A a c^{2} e^{2} + A b^{2} c e^{2} - 2 A b c^{2} d e + 2 A c^{3} d^{2} + 3 B a b c e^{2} - 4 B a c^{2} d e - B b^{3} e^{2} + 2 B b^{2} c d e - B b c^{2} d^{2}} \right )} + \left (\frac {\sqrt {- 4 a c + b^{2}} \left (- 2 A a c^{2} e^{2} + A b^{2} c e^{2} - 2 A b c^{2} d e + 2 A c^{3} d^{2} + 3 B a b c e^{2} - 4 B a c^{2} d e - B b^{3} e^{2} + 2 B b^{2} c d e - B b c^{2} d^{2}\right )}{2 c^{3} \left (4 a c - b^{2}\right )} - \frac {A b c e^{2} - 2 A c^{2} d e + B a c e^{2} - B b^{2} e^{2} + 2 B b c d e - B c^{2} d^{2}}{2 c^{3}}\right ) \log {\left (x + \frac {A a b c e^{2} - 4 A a c^{2} d e + A b c^{2} d^{2} + 2 B a^{2} c e^{2} - B a b^{2} e^{2} + 2 B a b c d e - 2 B a c^{2} d^{2} + 4 a c^{3} \left (\frac {\sqrt {- 4 a c + b^{2}} \left (- 2 A a c^{2} e^{2} + A b^{2} c e^{2} - 2 A b c^{2} d e + 2 A c^{3} d^{2} + 3 B a b c e^{2} - 4 B a c^{2} d e - B b^{3} e^{2} + 2 B b^{2} c d e - B b c^{2} d^{2}\right )}{2 c^{3} \left (4 a c - b^{2}\right )} - \frac {A b c e^{2} - 2 A c^{2} d e + B a c e^{2} - B b^{2} e^{2} + 2 B b c d e - B c^{2} d^{2}}{2 c^{3}}\right ) - b^{2} c^{2} \left (\frac {\sqrt {- 4 a c + b^{2}} \left (- 2 A a c^{2} e^{2} + A b^{2} c e^{2} - 2 A b c^{2} d e + 2 A c^{3} d^{2} + 3 B a b c e^{2} - 4 B a c^{2} d e - B b^{3} e^{2} + 2 B b^{2} c d e - B b c^{2} d^{2}\right )}{2 c^{3} \left (4 a c - b^{2}\right )} - \frac {A b c e^{2} - 2 A c^{2} d e + B a c e^{2} - B b^{2} e^{2} + 2 B b c d e - B c^{2} d^{2}}{2 c^{3}}\right )}{- 2 A a c^{2} e^{2} + A b^{2} c e^{2} - 2 A b c^{2} d e + 2 A c^{3} d^{2} + 3 B a b c e^{2} - 4 B a c^{2} d e - B b^{3} e^{2} + 2 B b^{2} c d e - B b c^{2} d^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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