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

Optimal. Leaf size=156 \[ -\frac {(b d-a e)^4 (B d-A e) \log (d+e x)}{e^6}+\frac {b x (b d-a e)^3 (B d-A e)}{e^5}-\frac {(a+b x)^2 (b d-a e)^2 (B d-A e)}{2 e^4}+\frac {(a+b x)^3 (b d-a e) (B d-A e)}{3 e^3}-\frac {(a+b x)^4 (B d-A e)}{4 e^2}+\frac {B (a+b x)^5}{5 b e} \]

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Rubi [A]  time = 0.12, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {27, 77} \begin {gather*} -\frac {(a+b x)^4 (B d-A e)}{4 e^2}+\frac {(a+b x)^3 (b d-a e) (B d-A e)}{3 e^3}-\frac {(a+b x)^2 (b d-a e)^2 (B d-A e)}{2 e^4}+\frac {b x (b d-a e)^3 (B d-A e)}{e^5}-\frac {(b d-a e)^4 (B d-A e) \log (d+e x)}{e^6}+\frac {B (a+b x)^5}{5 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(b*d - a*e)^3*(B*d - A*e)*x)/e^5 - ((b*d - a*e)^2*(B*d - A*e)*(a + b*x)^2)/(2*e^4) + ((b*d - a*e)*(B*d - A*
e)*(a + b*x)^3)/(3*e^3) - ((B*d - A*e)*(a + b*x)^4)/(4*e^2) + (B*(a + b*x)^5)/(5*b*e) - ((b*d - a*e)^4*(B*d -
A*e)*Log[d + e*x])/e^6

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2}{d+e x} \, dx &=\int \frac {(a+b x)^4 (A+B x)}{d+e x} \, dx\\ &=\int \left (-\frac {b (b d-a e)^3 (-B d+A e)}{e^5}+\frac {b (b d-a e)^2 (-B d+A e) (a+b x)}{e^4}-\frac {b (b d-a e) (-B d+A e) (a+b x)^2}{e^3}+\frac {b (-B d+A e) (a+b x)^3}{e^2}+\frac {B (a+b x)^4}{e}+\frac {(-b d+a e)^4 (-B d+A e)}{e^5 (d+e x)}\right ) \, dx\\ &=\frac {b (b d-a e)^3 (B d-A e) x}{e^5}-\frac {(b d-a e)^2 (B d-A e) (a+b x)^2}{2 e^4}+\frac {(b d-a e) (B d-A e) (a+b x)^3}{3 e^3}-\frac {(B d-A e) (a+b x)^4}{4 e^2}+\frac {B (a+b x)^5}{5 b e}-\frac {(b d-a e)^4 (B d-A e) \log (d+e x)}{e^6}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 258, normalized size = 1.65 \begin {gather*} \frac {e x \left (60 a^4 B e^4+120 a^3 b e^3 (2 A e-2 B d+B e x)+60 a^2 b^2 e^2 \left (3 A e (e x-2 d)+B \left (6 d^2-3 d e x+2 e^2 x^2\right )\right )+20 a b^3 e \left (2 A e \left (6 d^2-3 d e x+2 e^2 x^2\right )+B \left (-12 d^3+6 d^2 e x-4 d e^2 x^2+3 e^3 x^3\right )\right )+b^4 \left (5 A e \left (-12 d^3+6 d^2 e x-4 d e^2 x^2+3 e^3 x^3\right )+B \left (60 d^4-30 d^3 e x+20 d^2 e^2 x^2-15 d e^3 x^3+12 e^4 x^4\right )\right )\right )-60 (b d-a e)^4 (B d-A e) \log (d+e x)}{60 e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*x*(60*a^4*B*e^4 + 120*a^3*b*e^3*(-2*B*d + 2*A*e + B*e*x) + 60*a^2*b^2*e^2*(3*A*e*(-2*d + e*x) + B*(6*d^2 -
3*d*e*x + 2*e^2*x^2)) + 20*a*b^3*e*(2*A*e*(6*d^2 - 3*d*e*x + 2*e^2*x^2) + B*(-12*d^3 + 6*d^2*e*x - 4*d*e^2*x^2
+ 3*e^3*x^3)) + b^4*(5*A*e*(-12*d^3 + 6*d^2*e*x - 4*d*e^2*x^2 + 3*e^3*x^3) + B*(60*d^4 - 30*d^3*e*x + 20*d^2*
e^2*x^2 - 15*d*e^3*x^3 + 12*e^4*x^4))) - 60*(b*d - a*e)^4*(B*d - A*e)*Log[d + e*x])/(60*e^6)

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.41, size = 404, normalized size = 2.59 \begin {gather*} \frac {12 \, B b^{4} e^{5} x^{5} - 15 \, {\left (B b^{4} d e^{4} - {\left (4 \, B a b^{3} + A b^{4}\right )} e^{5}\right )} x^{4} + 20 \, {\left (B b^{4} d^{2} e^{3} - {\left (4 \, B a b^{3} + A b^{4}\right )} d e^{4} + 2 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} e^{5}\right )} x^{3} - 30 \, {\left (B b^{4} d^{3} e^{2} - {\left (4 \, B a b^{3} + A b^{4}\right )} d^{2} e^{3} + 2 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d e^{4} - 2 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} e^{5}\right )} x^{2} + 60 \, {\left (B b^{4} d^{4} e - {\left (4 \, B a b^{3} + A b^{4}\right )} d^{3} e^{2} + 2 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d^{2} e^{3} - 2 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} d e^{4} + {\left (B a^{4} + 4 \, A a^{3} b\right )} e^{5}\right )} x - 60 \, {\left (B b^{4} d^{5} - A a^{4} e^{5} - {\left (4 \, B a b^{3} + A b^{4}\right )} d^{4} e + 2 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d^{3} e^{2} - 2 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} d^{2} e^{3} + {\left (B a^{4} + 4 \, A a^{3} b\right )} d e^{4}\right )} \log \left (e x + d\right )}{60 \, e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*B*b^4*e^5*x^5 - 15*(B*b^4*d*e^4 - (4*B*a*b^3 + A*b^4)*e^5)*x^4 + 20*(B*b^4*d^2*e^3 - (4*B*a*b^3 + A*b
^4)*d*e^4 + 2*(3*B*a^2*b^2 + 2*A*a*b^3)*e^5)*x^3 - 30*(B*b^4*d^3*e^2 - (4*B*a*b^3 + A*b^4)*d^2*e^3 + 2*(3*B*a^
2*b^2 + 2*A*a*b^3)*d*e^4 - 2*(2*B*a^3*b + 3*A*a^2*b^2)*e^5)*x^2 + 60*(B*b^4*d^4*e - (4*B*a*b^3 + A*b^4)*d^3*e^
2 + 2*(3*B*a^2*b^2 + 2*A*a*b^3)*d^2*e^3 - 2*(2*B*a^3*b + 3*A*a^2*b^2)*d*e^4 + (B*a^4 + 4*A*a^3*b)*e^5)*x - 60*
(B*b^4*d^5 - A*a^4*e^5 - (4*B*a*b^3 + A*b^4)*d^4*e + 2*(3*B*a^2*b^2 + 2*A*a*b^3)*d^3*e^2 - 2*(2*B*a^3*b + 3*A*
a^2*b^2)*d^2*e^3 + (B*a^4 + 4*A*a^3*b)*d*e^4)*log(e*x + d))/e^6

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giac [B]  time = 0.20, size = 442, normalized size = 2.83 \begin {gather*} -{\left (B b^{4} d^{5} - 4 \, B a b^{3} d^{4} e - A b^{4} d^{4} e + 6 \, B a^{2} b^{2} d^{3} e^{2} + 4 \, A a b^{3} d^{3} e^{2} - 4 \, B a^{3} b d^{2} e^{3} - 6 \, A a^{2} b^{2} d^{2} e^{3} + B a^{4} d e^{4} + 4 \, A a^{3} b d e^{4} - A a^{4} e^{5}\right )} e^{\left (-6\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{60} \, {\left (12 \, B b^{4} x^{5} e^{4} - 15 \, B b^{4} d x^{4} e^{3} + 20 \, B b^{4} d^{2} x^{3} e^{2} - 30 \, B b^{4} d^{3} x^{2} e + 60 \, B b^{4} d^{4} x + 60 \, B a b^{3} x^{4} e^{4} + 15 \, A b^{4} x^{4} e^{4} - 80 \, B a b^{3} d x^{3} e^{3} - 20 \, A b^{4} d x^{3} e^{3} + 120 \, B a b^{3} d^{2} x^{2} e^{2} + 30 \, A b^{4} d^{2} x^{2} e^{2} - 240 \, B a b^{3} d^{3} x e - 60 \, A b^{4} d^{3} x e + 120 \, B a^{2} b^{2} x^{3} e^{4} + 80 \, A a b^{3} x^{3} e^{4} - 180 \, B a^{2} b^{2} d x^{2} e^{3} - 120 \, A a b^{3} d x^{2} e^{3} + 360 \, B a^{2} b^{2} d^{2} x e^{2} + 240 \, A a b^{3} d^{2} x e^{2} + 120 \, B a^{3} b x^{2} e^{4} + 180 \, A a^{2} b^{2} x^{2} e^{4} - 240 \, B a^{3} b d x e^{3} - 360 \, A a^{2} b^{2} d x e^{3} + 60 \, B a^{4} x e^{4} + 240 \, A a^{3} b x e^{4}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(B*b^4*d^5 - 4*B*a*b^3*d^4*e - A*b^4*d^4*e + 6*B*a^2*b^2*d^3*e^2 + 4*A*a*b^3*d^3*e^2 - 4*B*a^3*b*d^2*e^3 - 6*
A*a^2*b^2*d^2*e^3 + B*a^4*d*e^4 + 4*A*a^3*b*d*e^4 - A*a^4*e^5)*e^(-6)*log(abs(x*e + d)) + 1/60*(12*B*b^4*x^5*e
^4 - 15*B*b^4*d*x^4*e^3 + 20*B*b^4*d^2*x^3*e^2 - 30*B*b^4*d^3*x^2*e + 60*B*b^4*d^4*x + 60*B*a*b^3*x^4*e^4 + 15
*A*b^4*x^4*e^4 - 80*B*a*b^3*d*x^3*e^3 - 20*A*b^4*d*x^3*e^3 + 120*B*a*b^3*d^2*x^2*e^2 + 30*A*b^4*d^2*x^2*e^2 -
240*B*a*b^3*d^3*x*e - 60*A*b^4*d^3*x*e + 120*B*a^2*b^2*x^3*e^4 + 80*A*a*b^3*x^3*e^4 - 180*B*a^2*b^2*d*x^2*e^3
- 120*A*a*b^3*d*x^2*e^3 + 360*B*a^2*b^2*d^2*x*e^2 + 240*A*a*b^3*d^2*x*e^2 + 120*B*a^3*b*x^2*e^4 + 180*A*a^2*b^
2*x^2*e^4 - 240*B*a^3*b*d*x*e^3 - 360*A*a^2*b^2*d*x*e^3 + 60*B*a^4*x*e^4 + 240*A*a^3*b*x*e^4)*e^(-5)

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maple [B]  time = 0.05, size = 521, normalized size = 3.34 \begin {gather*} \frac {B \,b^{4} x^{5}}{5 e}+\frac {A \,b^{4} x^{4}}{4 e}+\frac {B a \,b^{3} x^{4}}{e}-\frac {B \,b^{4} d \,x^{4}}{4 e^{2}}+\frac {4 A a \,b^{3} x^{3}}{3 e}-\frac {A \,b^{4} d \,x^{3}}{3 e^{2}}+\frac {2 B \,a^{2} b^{2} x^{3}}{e}-\frac {4 B a \,b^{3} d \,x^{3}}{3 e^{2}}+\frac {B \,b^{4} d^{2} x^{3}}{3 e^{3}}+\frac {3 A \,a^{2} b^{2} x^{2}}{e}-\frac {2 A a \,b^{3} d \,x^{2}}{e^{2}}+\frac {A \,b^{4} d^{2} x^{2}}{2 e^{3}}+\frac {2 B \,a^{3} b \,x^{2}}{e}-\frac {3 B \,a^{2} b^{2} d \,x^{2}}{e^{2}}+\frac {2 B a \,b^{3} d^{2} x^{2}}{e^{3}}-\frac {B \,b^{4} d^{3} x^{2}}{2 e^{4}}+\frac {A \,a^{4} \ln \left (e x +d \right )}{e}-\frac {4 A \,a^{3} b d \ln \left (e x +d \right )}{e^{2}}+\frac {4 A \,a^{3} b x}{e}+\frac {6 A \,a^{2} b^{2} d^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {6 A \,a^{2} b^{2} d x}{e^{2}}-\frac {4 A a \,b^{3} d^{3} \ln \left (e x +d \right )}{e^{4}}+\frac {4 A a \,b^{3} d^{2} x}{e^{3}}+\frac {A \,b^{4} d^{4} \ln \left (e x +d \right )}{e^{5}}-\frac {A \,b^{4} d^{3} x}{e^{4}}-\frac {B \,a^{4} d \ln \left (e x +d \right )}{e^{2}}+\frac {B \,a^{4} x}{e}+\frac {4 B \,a^{3} b \,d^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {4 B \,a^{3} b d x}{e^{2}}-\frac {6 B \,a^{2} b^{2} d^{3} \ln \left (e x +d \right )}{e^{4}}+\frac {6 B \,a^{2} b^{2} d^{2} x}{e^{3}}+\frac {4 B a \,b^{3} d^{4} \ln \left (e x +d \right )}{e^{5}}-\frac {4 B a \,b^{3} d^{3} x}{e^{4}}-\frac {B \,b^{4} d^{5} \ln \left (e x +d \right )}{e^{6}}+\frac {B \,b^{4} d^{4} x}{e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/e^3*A*x^2*b^4*d^2+2/e*B*x^2*a^3*b-4/3/e^2*B*x^3*a*b^3*d+4/e^3*A*x*a*b^3*d^2-4/e^2*B*x*a^3*b*d-4/e^4*B*x*a*
b^3*d^3-6/e^2*A*x*a^2*b^2*d-3/e^2*B*x^2*a^2*b^2*d+2/e^3*B*x^2*a*b^3*d^2-2/e^2*A*x^2*a*b^3*d-4/e^2*ln(e*x+d)*A*
a^3*b*d+6/e^3*B*x*a^2*b^2*d^2-6/e^4*ln(e*x+d)*B*a^2*b^2*d^3+6/e^3*ln(e*x+d)*A*a^2*b^2*d^2+4/e^5*ln(e*x+d)*B*a*
b^3*d^4-4/e^4*ln(e*x+d)*A*a*b^3*d^3+4/e^3*ln(e*x+d)*B*a^3*b*d^2+1/5/e*b^4*B*x^5+1/4/e*A*x^4*b^4+1/e*B*x*a^4+1/
e*ln(e*x+d)*A*a^4+1/e^5*ln(e*x+d)*A*b^4*d^4+2/e*B*x^3*a^2*b^2+4/e*A*x*a^3*b-1/e^2*ln(e*x+d)*B*a^4*d-1/e^6*ln(e
*x+d)*B*b^4*d^5-1/2/e^4*B*x^2*b^4*d^3+4/3/e*A*x^3*a*b^3-1/4/e^2*B*x^4*b^4*d+1/e*B*x^4*a*b^3-1/e^4*A*x*b^4*d^3+
1/e^5*B*x*b^4*d^4+1/3/e^3*B*x^3*b^4*d^2+3/e*A*x^2*a^2*b^2-1/3/e^2*A*x^3*b^4*d

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maxima [B]  time = 0.72, size = 403, normalized size = 2.58 \begin {gather*} \frac {12 \, B b^{4} e^{4} x^{5} - 15 \, {\left (B b^{4} d e^{3} - {\left (4 \, B a b^{3} + A b^{4}\right )} e^{4}\right )} x^{4} + 20 \, {\left (B b^{4} d^{2} e^{2} - {\left (4 \, B a b^{3} + A b^{4}\right )} d e^{3} + 2 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} e^{4}\right )} x^{3} - 30 \, {\left (B b^{4} d^{3} e - {\left (4 \, B a b^{3} + A b^{4}\right )} d^{2} e^{2} + 2 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d e^{3} - 2 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} e^{4}\right )} x^{2} + 60 \, {\left (B b^{4} d^{4} - {\left (4 \, B a b^{3} + A b^{4}\right )} d^{3} e + 2 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d^{2} e^{2} - 2 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} d e^{3} + {\left (B a^{4} + 4 \, A a^{3} b\right )} e^{4}\right )} x}{60 \, e^{5}} - \frac {{\left (B b^{4} d^{5} - A a^{4} e^{5} - {\left (4 \, B a b^{3} + A b^{4}\right )} d^{4} e + 2 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d^{3} e^{2} - 2 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} d^{2} e^{3} + {\left (B a^{4} + 4 \, A a^{3} b\right )} d e^{4}\right )} \log \left (e x + d\right )}{e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*B*b^4*e^4*x^5 - 15*(B*b^4*d*e^3 - (4*B*a*b^3 + A*b^4)*e^4)*x^4 + 20*(B*b^4*d^2*e^2 - (4*B*a*b^3 + A*b
^4)*d*e^3 + 2*(3*B*a^2*b^2 + 2*A*a*b^3)*e^4)*x^3 - 30*(B*b^4*d^3*e - (4*B*a*b^3 + A*b^4)*d^2*e^2 + 2*(3*B*a^2*
b^2 + 2*A*a*b^3)*d*e^3 - 2*(2*B*a^3*b + 3*A*a^2*b^2)*e^4)*x^2 + 60*(B*b^4*d^4 - (4*B*a*b^3 + A*b^4)*d^3*e + 2*
(3*B*a^2*b^2 + 2*A*a*b^3)*d^2*e^2 - 2*(2*B*a^3*b + 3*A*a^2*b^2)*d*e^3 + (B*a^4 + 4*A*a^3*b)*e^4)*x)/e^5 - (B*b
^4*d^5 - A*a^4*e^5 - (4*B*a*b^3 + A*b^4)*d^4*e + 2*(3*B*a^2*b^2 + 2*A*a*b^3)*d^3*e^2 - 2*(2*B*a^3*b + 3*A*a^2*
b^2)*d^2*e^3 + (B*a^4 + 4*A*a^3*b)*d*e^4)*log(e*x + d)/e^6

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mupad [B]  time = 0.08, size = 411, normalized size = 2.63 \begin {gather*} x\,\left (\frac {B\,a^4+4\,A\,b\,a^3}{e}-\frac {d\,\left (\frac {d\,\left (\frac {d\,\left (\frac {A\,b^4+4\,B\,a\,b^3}{e}-\frac {B\,b^4\,d}{e^2}\right )}{e}-\frac {2\,a\,b^2\,\left (2\,A\,b+3\,B\,a\right )}{e}\right )}{e}+\frac {2\,a^2\,b\,\left (3\,A\,b+2\,B\,a\right )}{e}\right )}{e}\right )-x^3\,\left (\frac {d\,\left (\frac {A\,b^4+4\,B\,a\,b^3}{e}-\frac {B\,b^4\,d}{e^2}\right )}{3\,e}-\frac {2\,a\,b^2\,\left (2\,A\,b+3\,B\,a\right )}{3\,e}\right )+x^4\,\left (\frac {A\,b^4+4\,B\,a\,b^3}{4\,e}-\frac {B\,b^4\,d}{4\,e^2}\right )+x^2\,\left (\frac {d\,\left (\frac {d\,\left (\frac {A\,b^4+4\,B\,a\,b^3}{e}-\frac {B\,b^4\,d}{e^2}\right )}{e}-\frac {2\,a\,b^2\,\left (2\,A\,b+3\,B\,a\right )}{e}\right )}{2\,e}+\frac {a^2\,b\,\left (3\,A\,b+2\,B\,a\right )}{e}\right )+\frac {\ln \left (d+e\,x\right )\,\left (-B\,a^4\,d\,e^4+A\,a^4\,e^5+4\,B\,a^3\,b\,d^2\,e^3-4\,A\,a^3\,b\,d\,e^4-6\,B\,a^2\,b^2\,d^3\,e^2+6\,A\,a^2\,b^2\,d^2\,e^3+4\,B\,a\,b^3\,d^4\,e-4\,A\,a\,b^3\,d^3\,e^2-B\,b^4\,d^5+A\,b^4\,d^4\,e\right )}{e^6}+\frac {B\,b^4\,x^5}{5\,e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*((B*a^4 + 4*A*a^3*b)/e - (d*((d*((d*((A*b^4 + 4*B*a*b^3)/e - (B*b^4*d)/e^2))/e - (2*a*b^2*(2*A*b + 3*B*a))/e
))/e + (2*a^2*b*(3*A*b + 2*B*a))/e))/e) - x^3*((d*((A*b^4 + 4*B*a*b^3)/e - (B*b^4*d)/e^2))/(3*e) - (2*a*b^2*(2
*A*b + 3*B*a))/(3*e)) + x^4*((A*b^4 + 4*B*a*b^3)/(4*e) - (B*b^4*d)/(4*e^2)) + x^2*((d*((d*((A*b^4 + 4*B*a*b^3)
/e - (B*b^4*d)/e^2))/e - (2*a*b^2*(2*A*b + 3*B*a))/e))/(2*e) + (a^2*b*(3*A*b + 2*B*a))/e) + (log(d + e*x)*(A*a
^4*e^5 - B*b^4*d^5 + A*b^4*d^4*e - B*a^4*d*e^4 - 4*A*a*b^3*d^3*e^2 + 4*B*a^3*b*d^2*e^3 + 6*A*a^2*b^2*d^2*e^3 -
6*B*a^2*b^2*d^3*e^2 - 4*A*a^3*b*d*e^4 + 4*B*a*b^3*d^4*e))/e^6 + (B*b^4*x^5)/(5*e)

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sympy [B]  time = 0.92, size = 352, normalized size = 2.26 \begin {gather*} \frac {B b^{4} x^{5}}{5 e} + x^{4} \left (\frac {A b^{4}}{4 e} + \frac {B a b^{3}}{e} - \frac {B b^{4} d}{4 e^{2}}\right ) + x^{3} \left (\frac {4 A a b^{3}}{3 e} - \frac {A b^{4} d}{3 e^{2}} + \frac {2 B a^{2} b^{2}}{e} - \frac {4 B a b^{3} d}{3 e^{2}} + \frac {B b^{4} d^{2}}{3 e^{3}}\right ) + x^{2} \left (\frac {3 A a^{2} b^{2}}{e} - \frac {2 A a b^{3} d}{e^{2}} + \frac {A b^{4} d^{2}}{2 e^{3}} + \frac {2 B a^{3} b}{e} - \frac {3 B a^{2} b^{2} d}{e^{2}} + \frac {2 B a b^{3} d^{2}}{e^{3}} - \frac {B b^{4} d^{3}}{2 e^{4}}\right ) + x \left (\frac {4 A a^{3} b}{e} - \frac {6 A a^{2} b^{2} d}{e^{2}} + \frac {4 A a b^{3} d^{2}}{e^{3}} - \frac {A b^{4} d^{3}}{e^{4}} + \frac {B a^{4}}{e} - \frac {4 B a^{3} b d}{e^{2}} + \frac {6 B a^{2} b^{2} d^{2}}{e^{3}} - \frac {4 B a b^{3} d^{3}}{e^{4}} + \frac {B b^{4} d^{4}}{e^{5}}\right ) - \frac {\left (- A e + B d\right ) \left (a e - b d\right )^{4} \log {\left (d + e x \right )}}{e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*b**4*x**5/(5*e) + x**4*(A*b**4/(4*e) + B*a*b**3/e - B*b**4*d/(4*e**2)) + x**3*(4*A*a*b**3/(3*e) - A*b**4*d/(
3*e**2) + 2*B*a**2*b**2/e - 4*B*a*b**3*d/(3*e**2) + B*b**4*d**2/(3*e**3)) + x**2*(3*A*a**2*b**2/e - 2*A*a*b**3
*d/e**2 + A*b**4*d**2/(2*e**3) + 2*B*a**3*b/e - 3*B*a**2*b**2*d/e**2 + 2*B*a*b**3*d**2/e**3 - B*b**4*d**3/(2*e
**4)) + x*(4*A*a**3*b/e - 6*A*a**2*b**2*d/e**2 + 4*A*a*b**3*d**2/e**3 - A*b**4*d**3/e**4 + B*a**4/e - 4*B*a**3
*b*d/e**2 + 6*B*a**2*b**2*d**2/e**3 - 4*B*a*b**3*d**3/e**4 + B*b**4*d**4/e**5) - (-A*e + B*d)*(a*e - b*d)**4*l
og(d + e*x)/e**6

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