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

Optimal. Leaf size=357 \[ \frac {e x \left (B \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )+A c e (3 c d-b e)\right )}{c^3}+\frac {\log \left (a+b x+c x^2\right ) \left (A c e \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )+B \left (-3 c^2 d e (a e+b d)+b c e^2 (2 a e+3 b d)-b^3 e^3+c^3 d^3\right )\right )}{2 c^4}-\frac {\tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (b^2 c e \left (-4 a B e^2+3 A c d e+3 B c d^2\right )-b c^2 \left (-3 a A e^3-9 a B d e^2+3 A c d^2 e+B c d^3\right )+2 c^2 \left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right )-b^3 c e^2 (A e+3 B d)+b^4 B e^3\right )}{c^4 \sqrt {b^2-4 a c}}+\frac {e^2 x^2 (A c e-b B e+3 B c d)}{2 c^2}+\frac {B e^3 x^3}{3 c} \]

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Rubi [A]  time = 0.65, antiderivative size = 357, 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 (A c e \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )+B \left (-3 c^2 d e (a e+b d)+b c e^2 (2 a e+3 b d)-b^3 e^3+c^3 d^3\right )\right )}{2 c^4}+\frac {e x \left (B \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )+A c e (3 c d-b e)\right )}{c^3}-\frac {\tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (b^2 c e \left (-4 a B e^2+3 A c d e+3 B c d^2\right )-b c^2 \left (-3 a A e^3-9 a B d e^2+3 A c d^2 e+B c d^3\right )+2 c^2 \left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right )-b^3 c e^2 (A e+3 B d)+b^4 B e^3\right )}{c^4 \sqrt {b^2-4 a c}}+\frac {e^2 x^2 (A c e-b B e+3 B c d)}{2 c^2}+\frac {B e^3 x^3}{3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(A*c*e*(3*c*d - b*e) + B*(3*c^2*d^2 + b^2*e^2 - c*e*(3*b*d + a*e)))*x)/c^3 + (e^2*(3*B*c*d - b*B*e + A*c*e)
*x^2)/(2*c^2) + (B*e^3*x^3)/(3*c) - ((b^4*B*e^3 - b^3*c*e^2*(3*B*d + A*e) + b^2*c*e*(3*B*c*d^2 + 3*A*c*d*e - 4
*a*B*e^2) - b*c^2*(B*c*d^3 + 3*A*c*d^2*e - 9*a*B*d*e^2 - 3*a*A*e^3) + 2*c^2*(A*c*d*(c*d^2 - 3*a*e^2) - a*B*e*(
3*c*d^2 - a*e^2)))*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(c^4*Sqrt[b^2 - 4*a*c]) + ((A*c*e*(3*c^2*d^2 + b^2*
e^2 - c*e*(3*b*d + a*e)) + B*(c^3*d^3 - b^3*e^3 - 3*c^2*d*e*(b*d + a*e) + b*c*e^2*(3*b*d + 2*a*e)))*Log[a + b*
x + c*x^2])/(2*c^4)

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)^3}{a+b x+c x^2} \, dx &=\int \left (\frac {e \left (A c e (3 c d-b e)+B \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )\right )}{c^3}+\frac {e^2 (3 B c d-b B e+A c e) x}{c^2}+\frac {B e^3 x^2}{c}+\frac {A c \left (c^2 d^3-3 a c d e^2+a b e^3\right )-a B e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )+\left (A c e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )+B \left (c^3 d^3-b^3 e^3-3 c^2 d e (b d+a e)+b c e^2 (3 b d+2 a e)\right )\right ) x}{c^3 \left (a+b x+c x^2\right )}\right ) \, dx\\ &=\frac {e \left (A c e (3 c d-b e)+B \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )\right ) x}{c^3}+\frac {e^2 (3 B c d-b B e+A c e) x^2}{2 c^2}+\frac {B e^3 x^3}{3 c}+\frac {\int \frac {A c \left (c^2 d^3-3 a c d e^2+a b e^3\right )-a B e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )+\left (A c e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )+B \left (c^3 d^3-b^3 e^3-3 c^2 d e (b d+a e)+b c e^2 (3 b d+2 a e)\right )\right ) x}{a+b x+c x^2} \, dx}{c^3}\\ &=\frac {e \left (A c e (3 c d-b e)+B \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )\right ) x}{c^3}+\frac {e^2 (3 B c d-b B e+A c e) x^2}{2 c^2}+\frac {B e^3 x^3}{3 c}+\frac {\left (A c e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )+B \left (c^3 d^3-b^3 e^3-3 c^2 d e (b d+a e)+b c e^2 (3 b d+2 a e)\right )\right ) \int \frac {b+2 c x}{a+b x+c x^2} \, dx}{2 c^4}+\frac {\left (b^4 B e^3-b^3 c e^2 (3 B d+A e)+b^2 c e \left (3 B c d^2+3 A c d e-4 a B e^2\right )-b c^2 \left (B c d^3+3 A c d^2 e-9 a B d e^2-3 a A e^3\right )+2 c^2 \left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right )\right ) \int \frac {1}{a+b x+c x^2} \, dx}{2 c^4}\\ &=\frac {e \left (A c e (3 c d-b e)+B \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )\right ) x}{c^3}+\frac {e^2 (3 B c d-b B e+A c e) x^2}{2 c^2}+\frac {B e^3 x^3}{3 c}+\frac {\left (A c e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )+B \left (c^3 d^3-b^3 e^3-3 c^2 d e (b d+a e)+b c e^2 (3 b d+2 a e)\right )\right ) \log \left (a+b x+c x^2\right )}{2 c^4}-\frac {\left (b^4 B e^3-b^3 c e^2 (3 B d+A e)+b^2 c e \left (3 B c d^2+3 A c d e-4 a B e^2\right )-b c^2 \left (B c d^3+3 A c d^2 e-9 a B d e^2-3 a A e^3\right )+2 c^2 \left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{c^4}\\ &=\frac {e \left (A c e (3 c d-b e)+B \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )\right ) x}{c^3}+\frac {e^2 (3 B c d-b B e+A c e) x^2}{2 c^2}+\frac {B e^3 x^3}{3 c}-\frac {\left (b^4 B e^3-b^3 c e^2 (3 B d+A e)+b^2 c e \left (3 B c d^2+3 A c d e-4 a B e^2\right )-b c^2 \left (B c d^3+3 A c d^2 e-9 a B d e^2-3 a A e^3\right )+2 c^2 \left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c^4 \sqrt {b^2-4 a c}}+\frac {\left (A c e \left (3 c^2 d^2+b^2 e^2-c e (3 b d+a e)\right )+B \left (c^3 d^3-b^3 e^3-3 c^2 d e (b d+a e)+b c e^2 (3 b d+2 a e)\right )\right ) \log \left (a+b x+c x^2\right )}{2 c^4}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 352, normalized size = 0.99 \begin {gather*} \frac {6 c e x \left (B \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )+A c e (3 c d-b e)\right )+3 \log (a+x (b+c x)) \left (A c e \left (-c e (a e+3 b d)+b^2 e^2+3 c^2 d^2\right )+B \left (-3 c^2 d e (a e+b d)+b c e^2 (2 a e+3 b d)-b^3 e^3+c^3 d^3\right )\right )+\frac {6 \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right ) \left (b^2 c e \left (-4 a B e^2+3 A c d e+3 B c d^2\right )+b c^2 \left (3 a A e^3+9 a B d e^2-3 A c d^2 e-B c d^3\right )+2 c^2 \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (a e^2-3 c d^2\right )\right )-b^3 c e^2 (A e+3 B d)+b^4 B e^3\right )}{\sqrt {4 a c-b^2}}+3 c^2 e^2 x^2 (A c e-b B e+3 B c d)+2 B c^3 e^3 x^3}{6 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*c*e*(A*c*e*(3*c*d - b*e) + B*(3*c^2*d^2 + b^2*e^2 - c*e*(3*b*d + a*e)))*x + 3*c^2*e^2*(3*B*c*d - b*B*e + A*
c*e)*x^2 + 2*B*c^3*e^3*x^3 + (6*(b^4*B*e^3 - b^3*c*e^2*(3*B*d + A*e) + b^2*c*e*(3*B*c*d^2 + 3*A*c*d*e - 4*a*B*
e^2) + b*c^2*(-(B*c*d^3) - 3*A*c*d^2*e + 9*a*B*d*e^2 + 3*a*A*e^3) + 2*c^2*(A*c*d*(c*d^2 - 3*a*e^2) + a*B*e*(-3
*c*d^2 + a*e^2)))*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] + 3*(A*c*e*(3*c^2*d^2 + b^2*e^2 -
c*e*(3*b*d + a*e)) + B*(c^3*d^3 - b^3*e^3 - 3*c^2*d*e*(b*d + a*e) + b*c*e^2*(3*b*d + 2*a*e)))*Log[a + x*(b +
c*x)])/(6*c^4)

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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)^3}{a+b x+c x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.54, size = 1139, normalized size = 3.19

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(2*(B*b^2*c^3 - 4*B*a*c^4)*e^3*x^3 + 3*(3*(B*b^2*c^3 - 4*B*a*c^4)*d*e^2 - (B*b^3*c^2 + 4*A*a*c^4 - (4*B*a
*b + A*b^2)*c^3)*e^3)*x^2 - 3*((B*b*c^3 - 2*A*c^4)*d^3 - 3*(B*b^2*c^2 - (2*B*a + A*b)*c^3)*d^2*e + 3*(B*b^3*c
+ 2*A*a*c^3 - (3*B*a*b + A*b^2)*c^2)*d*e^2 - (B*b^4 + (2*B*a^2 + 3*A*a*b)*c^2 - (4*B*a*b^2 + A*b^3)*c)*e^3)*sq
rt(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
*(3*(B*b^2*c^3 - 4*B*a*c^4)*d^2*e - 3*(B*b^3*c^2 + 4*A*a*c^4 - (4*B*a*b + A*b^2)*c^3)*d*e^2 + (B*b^4*c + 4*(B*
a^2 + A*a*b)*c^3 - (5*B*a*b^2 + A*b^3)*c^2)*e^3)*x + 3*((B*b^2*c^3 - 4*B*a*c^4)*d^3 - 3*(B*b^3*c^2 + 4*A*a*c^4
- (4*B*a*b + A*b^2)*c^3)*d^2*e + 3*(B*b^4*c + 4*(B*a^2 + A*a*b)*c^3 - (5*B*a*b^2 + A*b^3)*c^2)*d*e^2 - (B*b^5
- 4*A*a^2*c^3 + (8*B*a^2*b + 5*A*a*b^2)*c^2 - (6*B*a*b^3 + A*b^4)*c)*e^3)*log(c*x^2 + b*x + a))/(b^2*c^4 - 4*
a*c^5), 1/6*(2*(B*b^2*c^3 - 4*B*a*c^4)*e^3*x^3 + 3*(3*(B*b^2*c^3 - 4*B*a*c^4)*d*e^2 - (B*b^3*c^2 + 4*A*a*c^4 -
(4*B*a*b + A*b^2)*c^3)*e^3)*x^2 + 6*((B*b*c^3 - 2*A*c^4)*d^3 - 3*(B*b^2*c^2 - (2*B*a + A*b)*c^3)*d^2*e + 3*(B
*b^3*c + 2*A*a*c^3 - (3*B*a*b + A*b^2)*c^2)*d*e^2 - (B*b^4 + (2*B*a^2 + 3*A*a*b)*c^2 - (4*B*a*b^2 + A*b^3)*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*(3*(B*b^2*c^3 - 4*B*a*c^4)*d
^2*e - 3*(B*b^3*c^2 + 4*A*a*c^4 - (4*B*a*b + A*b^2)*c^3)*d*e^2 + (B*b^4*c + 4*(B*a^2 + A*a*b)*c^3 - (5*B*a*b^2
+ A*b^3)*c^2)*e^3)*x + 3*((B*b^2*c^3 - 4*B*a*c^4)*d^3 - 3*(B*b^3*c^2 + 4*A*a*c^4 - (4*B*a*b + A*b^2)*c^3)*d^2
*e + 3*(B*b^4*c + 4*(B*a^2 + A*a*b)*c^3 - (5*B*a*b^2 + A*b^3)*c^2)*d*e^2 - (B*b^5 - 4*A*a^2*c^3 + (8*B*a^2*b +
5*A*a*b^2)*c^2 - (6*B*a*b^3 + A*b^4)*c)*e^3)*log(c*x^2 + b*x + a))/(b^2*c^4 - 4*a*c^5)]

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giac [A]  time = 0.18, size = 402, normalized size = 1.13 \begin {gather*} \frac {2 \, B c^{2} x^{3} e^{3} + 9 \, B c^{2} d x^{2} e^{2} + 18 \, B c^{2} d^{2} x e - 3 \, B b c x^{2} e^{3} + 3 \, A c^{2} x^{2} e^{3} - 18 \, B b c d x e^{2} + 18 \, A c^{2} d x e^{2} + 6 \, B b^{2} x e^{3} - 6 \, B a c x e^{3} - 6 \, A b c x e^{3}}{6 \, c^{3}} + \frac {{\left (B c^{3} d^{3} - 3 \, B b c^{2} d^{2} e + 3 \, A c^{3} d^{2} e + 3 \, B b^{2} c d e^{2} - 3 \, B a c^{2} d e^{2} - 3 \, A b c^{2} d e^{2} - B b^{3} e^{3} + 2 \, B a b c e^{3} + A b^{2} c e^{3} - A a c^{2} e^{3}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, c^{4}} - \frac {{\left (B b c^{3} d^{3} - 2 \, A c^{4} d^{3} - 3 \, B b^{2} c^{2} d^{2} e + 6 \, B a c^{3} d^{2} e + 3 \, A b c^{3} d^{2} e + 3 \, B b^{3} c d e^{2} - 9 \, B a b c^{2} d e^{2} - 3 \, A b^{2} c^{2} d e^{2} + 6 \, A a c^{3} d e^{2} - B b^{4} e^{3} + 4 \, B a b^{2} c e^{3} + A b^{3} c e^{3} - 2 \, B a^{2} c^{2} e^{3} - 3 \, A a b 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^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*B*c^2*x^3*e^3 + 9*B*c^2*d*x^2*e^2 + 18*B*c^2*d^2*x*e - 3*B*b*c*x^2*e^3 + 3*A*c^2*x^2*e^3 - 18*B*b*c*d*x
*e^2 + 18*A*c^2*d*x*e^2 + 6*B*b^2*x*e^3 - 6*B*a*c*x*e^3 - 6*A*b*c*x*e^3)/c^3 + 1/2*(B*c^3*d^3 - 3*B*b*c^2*d^2*
e + 3*A*c^3*d^2*e + 3*B*b^2*c*d*e^2 - 3*B*a*c^2*d*e^2 - 3*A*b*c^2*d*e^2 - B*b^3*e^3 + 2*B*a*b*c*e^3 + A*b^2*c*
e^3 - A*a*c^2*e^3)*log(c*x^2 + b*x + a)/c^4 - (B*b*c^3*d^3 - 2*A*c^4*d^3 - 3*B*b^2*c^2*d^2*e + 6*B*a*c^3*d^2*e
+ 3*A*b*c^3*d^2*e + 3*B*b^3*c*d*e^2 - 9*B*a*b*c^2*d*e^2 - 3*A*b^2*c^2*d*e^2 + 6*A*a*c^3*d*e^2 - B*b^4*e^3 + 4
*B*a*b^2*c*e^3 + A*b^3*c*e^3 - 2*B*a^2*c^2*e^3 - 3*A*a*b*c^2*e^3)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/(sqrt
(-b^2 + 4*a*c)*c^4)

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maple [B]  time = 0.05, size = 946, normalized size = 2.65 \begin {gather*} \frac {B \,e^{3} x^{3}}{3 c}+\frac {3 A a b \,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 A a 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 {A \,b^{3} 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 A \,b^{2} 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 A b \,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 {A \,e^{3} x^{2}}{2 c}+\frac {2 A \,d^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}+\frac {2 B \,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^{2}}-\frac {4 B 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^{3}}+\frac {9 B 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^{2}}-\frac {6 B a \,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 \,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^{4}}-\frac {3 B \,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^{3}}+\frac {3 B \,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^{2}}-\frac {B b \,d^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}-\frac {B b \,e^{3} x^{2}}{2 c^{2}}+\frac {3 B d \,e^{2} x^{2}}{2 c}-\frac {A a \,e^{3} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{2}}+\frac {A \,b^{2} e^{3} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{3}}-\frac {3 A b d \,e^{2} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{2}}-\frac {A b \,e^{3} x}{c^{2}}+\frac {3 A \,d^{2} e \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {3 A d \,e^{2} x}{c}+\frac {B a b \,e^{3} \ln \left (c \,x^{2}+b x +a \right )}{c^{3}}-\frac {3 B a d \,e^{2} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{2}}-\frac {B a \,e^{3} x}{c^{2}}-\frac {B \,b^{3} e^{3} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{4}}+\frac {3 B \,b^{2} d \,e^{2} \ln \left (c \,x^{2}+b x +a \right )}{2 c^{3}}+\frac {B \,b^{2} e^{3} x}{c^{3}}-\frac {3 B b \,d^{2} e \ln \left (c \,x^{2}+b x +a \right )}{2 c^{2}}-\frac {3 B b d \,e^{2} x}{c^{2}}+\frac {B \,d^{3} \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {3 B \,d^{2} e x}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/c*ln(c*x^2+b*x+a)*B*d^3+2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*A*d^3+1/2*e^3/c*A*x^2+1/2/
c^3*ln(c*x^2+b*x+a)*A*b^2*e^3-1/c^3/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*A*b^3*e^3+1/c^4/(4*a
*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b^4*B*e^3-1/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)
^(1/2))*b*B*d^3-3/2/c^2*ln(c*x^2+b*x+a)*A*b*d*e^2+1/c^3*ln(c*x^2+b*x+a)*a*b*B*e^3-3/2/c^2*ln(c*x^2+b*x+a)*B*d*
a*e^2+3/2/c^3*ln(c*x^2+b*x+a)*B*d*b^2*e^2+9/c^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*B*a*b*d*
e^2+3*e^2/c*A*x*d+e^3/c^3*B*b^2*x+3*e/c*B*d^2*x-1/2/c^2*ln(c*x^2+b*x+a)*A*a*e^3-1/2/c^4*ln(c*x^2+b*x+a)*b^3*B*
e^3+3/2/c*ln(c*x^2+b*x+a)*A*d^2*e-1/2*e^3/c^2*B*x^2*b+3/2*e^2/c*B*x^2*d-e^3/c^2*B*x*a-e^3/c^2*A*b*x-3/c^3/(4*a
*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b^3*B*d*e^2+3/c^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-
b^2)^(1/2))*b^2*B*d^2*e-3*e^2/c^2*B*b*d*x-3/2/c^2*ln(c*x^2+b*x+a)*B*b*d^2*e+2/c^2/(4*a*c-b^2)^(1/2)*arctan((2*
c*x+b)/(4*a*c-b^2)^(1/2))*B*a^2*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^2*A*d*e^2+3/
c^2/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*A*a*b*e^3-3/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*
a*c-b^2)^(1/2))*b*A*d^2*e-6/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*A*a*d*e^2-4/c^3/(4*a*c-b^2
)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*B*a*b^2*e^3-6/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/
2))*B*a*d^2*e+1/3*B/c*e^3*x^3

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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)^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 = 3.11, size = 539, normalized size = 1.51 \begin {gather*} x^2\,\left (\frac {A\,e^3+3\,B\,d\,e^2}{2\,c}-\frac {B\,b\,e^3}{2\,c^2}\right )-x\,\left (\frac {b\,\left (\frac {A\,e^3+3\,B\,d\,e^2}{c}-\frac {B\,b\,e^3}{c^2}\right )}{c}-\frac {3\,d\,e\,\left (A\,e+B\,d\right )}{c}+\frac {B\,a\,e^3}{c^2}\right )-\frac {\ln \left (c\,x^2+b\,x+a\right )\,\left (-8\,B\,a^2\,b\,c^2\,e^3+12\,B\,a^2\,c^3\,d\,e^2+4\,A\,a^2\,c^3\,e^3+6\,B\,a\,b^3\,c\,e^3-15\,B\,a\,b^2\,c^2\,d\,e^2-5\,A\,a\,b^2\,c^2\,e^3+12\,B\,a\,b\,c^3\,d^2\,e+12\,A\,a\,b\,c^3\,d\,e^2-4\,B\,a\,c^4\,d^3-12\,A\,a\,c^4\,d^2\,e-B\,b^5\,e^3+3\,B\,b^4\,c\,d\,e^2+A\,b^4\,c\,e^3-3\,B\,b^3\,c^2\,d^2\,e-3\,A\,b^3\,c^2\,d\,e^2+B\,b^2\,c^3\,d^3+3\,A\,b^2\,c^3\,d^2\,e\right )}{2\,\left (4\,a\,c^5-b^2\,c^4\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 (2\,B\,a^2\,c^2\,e^3-4\,B\,a\,b^2\,c\,e^3+9\,B\,a\,b\,c^2\,d\,e^2+3\,A\,a\,b\,c^2\,e^3-6\,B\,a\,c^3\,d^2\,e-6\,A\,a\,c^3\,d\,e^2+B\,b^4\,e^3-3\,B\,b^3\,c\,d\,e^2-A\,b^3\,c\,e^3+3\,B\,b^2\,c^2\,d^2\,e+3\,A\,b^2\,c^2\,d\,e^2-B\,b\,c^3\,d^3-3\,A\,b\,c^3\,d^2\,e+2\,A\,c^4\,d^3\right )}{c^4\,\sqrt {4\,a\,c-b^2}}+\frac {B\,e^3\,x^3}{3\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2*((A*e^3 + 3*B*d*e^2)/(2*c) - (B*b*e^3)/(2*c^2)) - x*((b*((A*e^3 + 3*B*d*e^2)/c - (B*b*e^3)/c^2))/c - (3*d*
e*(A*e + B*d))/c + (B*a*e^3)/c^2) - (log(a + b*x + c*x^2)*(A*b^4*c*e^3 - 4*B*a*c^4*d^3 - B*b^5*e^3 + 4*A*a^2*c
^3*e^3 + B*b^2*c^3*d^3 - 5*A*a*b^2*c^2*e^3 - 8*B*a^2*b*c^2*e^3 + 3*A*b^2*c^3*d^2*e - 3*A*b^3*c^2*d*e^2 + 12*B*
a^2*c^3*d*e^2 - 3*B*b^3*c^2*d^2*e + 6*B*a*b^3*c*e^3 - 12*A*a*c^4*d^2*e + 3*B*b^4*c*d*e^2 + 12*A*a*b*c^3*d*e^2
+ 12*B*a*b*c^3*d^2*e - 15*B*a*b^2*c^2*d*e^2))/(2*(4*a*c^5 - b^2*c^4)) + (atan(b/(4*a*c - b^2)^(1/2) + (2*c*x)/
(4*a*c - b^2)^(1/2))*(2*A*c^4*d^3 + B*b^4*e^3 - A*b^3*c*e^3 - B*b*c^3*d^3 + 2*B*a^2*c^2*e^3 + 3*A*b^2*c^2*d*e^
2 + 3*B*b^2*c^2*d^2*e + 3*A*a*b*c^2*e^3 - 4*B*a*b^2*c*e^3 - 6*A*a*c^3*d*e^2 - 3*A*b*c^3*d^2*e - 6*B*a*c^3*d^2*
e - 3*B*b^3*c*d*e^2 + 9*B*a*b*c^2*d*e^2))/(c^4*(4*a*c - b^2)^(1/2)) + (B*e^3*x^3)/(3*c)

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sympy [B]  time = 25.81, size = 2759, normalized size = 7.73

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*e**3*x**3/(3*c) + x**2*(A*e**3/(2*c) - B*b*e**3/(2*c**2) + 3*B*d*e**2/(2*c)) + x*(-A*b*e**3/c**2 + 3*A*d*e**
2/c - B*a*e**3/c**2 + B*b**2*e**3/c**3 - 3*B*b*d*e**2/c**2 + 3*B*d**2*e/c) + (-sqrt(-4*a*c + b**2)*(3*A*a*b*c*
*2*e**3 - 6*A*a*c**3*d*e**2 - A*b**3*c*e**3 + 3*A*b**2*c**2*d*e**2 - 3*A*b*c**3*d**2*e + 2*A*c**4*d**3 + 2*B*a
**2*c**2*e**3 - 4*B*a*b**2*c*e**3 + 9*B*a*b*c**2*d*e**2 - 6*B*a*c**3*d**2*e + B*b**4*e**3 - 3*B*b**3*c*d*e**2
+ 3*B*b**2*c**2*d**2*e - B*b*c**3*d**3)/(2*c**4*(4*a*c - b**2)) + (-A*a*c**2*e**3 + A*b**2*c*e**3 - 3*A*b*c**2
*d*e**2 + 3*A*c**3*d**2*e + 2*B*a*b*c*e**3 - 3*B*a*c**2*d*e**2 - B*b**3*e**3 + 3*B*b**2*c*d*e**2 - 3*B*b*c**2*
d**2*e + B*c**3*d**3)/(2*c**4))*log(x + (2*A*a**2*c**2*e**3 - A*a*b**2*c*e**3 + 3*A*a*b*c**2*d*e**2 - 6*A*a*c*
*3*d**2*e + A*b*c**3*d**3 - 3*B*a**2*b*c*e**3 + 6*B*a**2*c**2*d*e**2 + B*a*b**3*e**3 - 3*B*a*b**2*c*d*e**2 + 3
*B*a*b*c**2*d**2*e - 2*B*a*c**3*d**3 + 4*a*c**4*(-sqrt(-4*a*c + b**2)*(3*A*a*b*c**2*e**3 - 6*A*a*c**3*d*e**2 -
A*b**3*c*e**3 + 3*A*b**2*c**2*d*e**2 - 3*A*b*c**3*d**2*e + 2*A*c**4*d**3 + 2*B*a**2*c**2*e**3 - 4*B*a*b**2*c*
e**3 + 9*B*a*b*c**2*d*e**2 - 6*B*a*c**3*d**2*e + B*b**4*e**3 - 3*B*b**3*c*d*e**2 + 3*B*b**2*c**2*d**2*e - B*b*
c**3*d**3)/(2*c**4*(4*a*c - b**2)) + (-A*a*c**2*e**3 + A*b**2*c*e**3 - 3*A*b*c**2*d*e**2 + 3*A*c**3*d**2*e + 2
*B*a*b*c*e**3 - 3*B*a*c**2*d*e**2 - B*b**3*e**3 + 3*B*b**2*c*d*e**2 - 3*B*b*c**2*d**2*e + B*c**3*d**3)/(2*c**4
)) - b**2*c**3*(-sqrt(-4*a*c + b**2)*(3*A*a*b*c**2*e**3 - 6*A*a*c**3*d*e**2 - A*b**3*c*e**3 + 3*A*b**2*c**2*d*
e**2 - 3*A*b*c**3*d**2*e + 2*A*c**4*d**3 + 2*B*a**2*c**2*e**3 - 4*B*a*b**2*c*e**3 + 9*B*a*b*c**2*d*e**2 - 6*B*
a*c**3*d**2*e + B*b**4*e**3 - 3*B*b**3*c*d*e**2 + 3*B*b**2*c**2*d**2*e - B*b*c**3*d**3)/(2*c**4*(4*a*c - b**2)
) + (-A*a*c**2*e**3 + A*b**2*c*e**3 - 3*A*b*c**2*d*e**2 + 3*A*c**3*d**2*e + 2*B*a*b*c*e**3 - 3*B*a*c**2*d*e**2
- B*b**3*e**3 + 3*B*b**2*c*d*e**2 - 3*B*b*c**2*d**2*e + B*c**3*d**3)/(2*c**4)))/(3*A*a*b*c**2*e**3 - 6*A*a*c*
*3*d*e**2 - A*b**3*c*e**3 + 3*A*b**2*c**2*d*e**2 - 3*A*b*c**3*d**2*e + 2*A*c**4*d**3 + 2*B*a**2*c**2*e**3 - 4*
B*a*b**2*c*e**3 + 9*B*a*b*c**2*d*e**2 - 6*B*a*c**3*d**2*e + B*b**4*e**3 - 3*B*b**3*c*d*e**2 + 3*B*b**2*c**2*d*
*2*e - B*b*c**3*d**3)) + (sqrt(-4*a*c + b**2)*(3*A*a*b*c**2*e**3 - 6*A*a*c**3*d*e**2 - A*b**3*c*e**3 + 3*A*b**
2*c**2*d*e**2 - 3*A*b*c**3*d**2*e + 2*A*c**4*d**3 + 2*B*a**2*c**2*e**3 - 4*B*a*b**2*c*e**3 + 9*B*a*b*c**2*d*e*
*2 - 6*B*a*c**3*d**2*e + B*b**4*e**3 - 3*B*b**3*c*d*e**2 + 3*B*b**2*c**2*d**2*e - B*b*c**3*d**3)/(2*c**4*(4*a*
c - b**2)) + (-A*a*c**2*e**3 + A*b**2*c*e**3 - 3*A*b*c**2*d*e**2 + 3*A*c**3*d**2*e + 2*B*a*b*c*e**3 - 3*B*a*c*
*2*d*e**2 - B*b**3*e**3 + 3*B*b**2*c*d*e**2 - 3*B*b*c**2*d**2*e + B*c**3*d**3)/(2*c**4))*log(x + (2*A*a**2*c**
2*e**3 - A*a*b**2*c*e**3 + 3*A*a*b*c**2*d*e**2 - 6*A*a*c**3*d**2*e + A*b*c**3*d**3 - 3*B*a**2*b*c*e**3 + 6*B*a
**2*c**2*d*e**2 + B*a*b**3*e**3 - 3*B*a*b**2*c*d*e**2 + 3*B*a*b*c**2*d**2*e - 2*B*a*c**3*d**3 + 4*a*c**4*(sqrt
(-4*a*c + b**2)*(3*A*a*b*c**2*e**3 - 6*A*a*c**3*d*e**2 - A*b**3*c*e**3 + 3*A*b**2*c**2*d*e**2 - 3*A*b*c**3*d**
2*e + 2*A*c**4*d**3 + 2*B*a**2*c**2*e**3 - 4*B*a*b**2*c*e**3 + 9*B*a*b*c**2*d*e**2 - 6*B*a*c**3*d**2*e + B*b**
4*e**3 - 3*B*b**3*c*d*e**2 + 3*B*b**2*c**2*d**2*e - B*b*c**3*d**3)/(2*c**4*(4*a*c - b**2)) + (-A*a*c**2*e**3 +
A*b**2*c*e**3 - 3*A*b*c**2*d*e**2 + 3*A*c**3*d**2*e + 2*B*a*b*c*e**3 - 3*B*a*c**2*d*e**2 - B*b**3*e**3 + 3*B*
b**2*c*d*e**2 - 3*B*b*c**2*d**2*e + B*c**3*d**3)/(2*c**4)) - b**2*c**3*(sqrt(-4*a*c + b**2)*(3*A*a*b*c**2*e**3
- 6*A*a*c**3*d*e**2 - A*b**3*c*e**3 + 3*A*b**2*c**2*d*e**2 - 3*A*b*c**3*d**2*e + 2*A*c**4*d**3 + 2*B*a**2*c**
2*e**3 - 4*B*a*b**2*c*e**3 + 9*B*a*b*c**2*d*e**2 - 6*B*a*c**3*d**2*e + B*b**4*e**3 - 3*B*b**3*c*d*e**2 + 3*B*b
**2*c**2*d**2*e - B*b*c**3*d**3)/(2*c**4*(4*a*c - b**2)) + (-A*a*c**2*e**3 + A*b**2*c*e**3 - 3*A*b*c**2*d*e**2
+ 3*A*c**3*d**2*e + 2*B*a*b*c*e**3 - 3*B*a*c**2*d*e**2 - B*b**3*e**3 + 3*B*b**2*c*d*e**2 - 3*B*b*c**2*d**2*e
+ B*c**3*d**3)/(2*c**4)))/(3*A*a*b*c**2*e**3 - 6*A*a*c**3*d*e**2 - A*b**3*c*e**3 + 3*A*b**2*c**2*d*e**2 - 3*A*
b*c**3*d**2*e + 2*A*c**4*d**3 + 2*B*a**2*c**2*e**3 - 4*B*a*b**2*c*e**3 + 9*B*a*b*c**2*d*e**2 - 6*B*a*c**3*d**2
*e + B*b**4*e**3 - 3*B*b**3*c*d*e**2 + 3*B*b**2*c**2*d**2*e - B*b*c**3*d**3))

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