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

Optimal. Leaf size=92 \[ -\frac {(b d-a e)^2 (B d-A e) \log (d+e x)}{e^4}+\frac {b x (b d-a e) (B d-A e)}{e^3}-\frac {(a+b x)^2 (B d-A e)}{2 e^2}+\frac {B (a+b x)^3}{3 b e} \]

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Rubi [A]  time = 0.06, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {27, 77} \begin {gather*} -\frac {(a+b x)^2 (B d-A e)}{2 e^2}+\frac {b x (b d-a e) (B d-A e)}{e^3}-\frac {(b d-a e)^2 (B d-A e) \log (d+e x)}{e^4}+\frac {B (a+b x)^3}{3 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.06, size = 102, normalized size = 1.11 \begin {gather*} \frac {e x \left (6 a^2 B e^2+6 a b e (2 A e-2 B d+B e x)+b^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 )\right )-6 (b d-a e)^2 (B d-A e) \log (d+e x)}{6 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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 )}{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))/(d + e*x),x]

[Out]

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

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fricas [A]  time = 0.42, size = 153, normalized size = 1.66 \begin {gather*} \frac {2 \, B b^{2} e^{3} x^{3} - 3 \, {\left (B b^{2} d e^{2} - {\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + 6 \, {\left (B b^{2} d^{2} e - {\left (2 \, B a b + A b^{2}\right )} d e^{2} + {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x - 6 \, {\left (B b^{2} d^{3} - A a^{2} e^{3} - {\left (2 \, B a b + A b^{2}\right )} d^{2} e + {\left (B a^{2} + 2 \, A a b\right )} d e^{2}\right )} \log \left (e x + d\right )}{6 \, e^{4}} \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)/(e*x+d),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 162, normalized size = 1.76 \begin {gather*} -{\left (B b^{2} d^{3} - 2 \, B a b d^{2} e - A b^{2} d^{2} e + B a^{2} d e^{2} + 2 \, A a b d e^{2} - A a^{2} e^{3}\right )} e^{\left (-4\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{6} \, {\left (2 \, B b^{2} x^{3} e^{2} - 3 \, B b^{2} d x^{2} e + 6 \, B b^{2} d^{2} x + 6 \, B a b x^{2} e^{2} + 3 \, A b^{2} x^{2} e^{2} - 12 \, B a b d x e - 6 \, A b^{2} d x e + 6 \, B a^{2} x e^{2} + 12 \, A a b x e^{2}\right )} e^{\left (-3\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)/(e*x+d),x, algorithm="giac")

[Out]

-(B*b^2*d^3 - 2*B*a*b*d^2*e - A*b^2*d^2*e + B*a^2*d*e^2 + 2*A*a*b*d*e^2 - A*a^2*e^3)*e^(-4)*log(abs(x*e + d))
+ 1/6*(2*B*b^2*x^3*e^2 - 3*B*b^2*d*x^2*e + 6*B*b^2*d^2*x + 6*B*a*b*x^2*e^2 + 3*A*b^2*x^2*e^2 - 12*B*a*b*d*x*e
- 6*A*b^2*d*x*e + 6*B*a^2*x*e^2 + 12*A*a*b*x*e^2)*e^(-3)

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maple [B]  time = 0.05, size = 197, normalized size = 2.14 \begin {gather*} \frac {B \,b^{2} x^{3}}{3 e}+\frac {A \,b^{2} x^{2}}{2 e}+\frac {B a b \,x^{2}}{e}-\frac {B \,b^{2} d \,x^{2}}{2 e^{2}}+\frac {A \,a^{2} \ln \left (e x +d \right )}{e}-\frac {2 A a b d \ln \left (e x +d \right )}{e^{2}}+\frac {2 A a b x}{e}+\frac {A \,b^{2} d^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {A \,b^{2} d x}{e^{2}}-\frac {B \,a^{2} d \ln \left (e x +d \right )}{e^{2}}+\frac {B \,a^{2} x}{e}+\frac {2 B a b \,d^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {2 B a b d x}{e^{2}}-\frac {B \,b^{2} d^{3} \ln \left (e x +d \right )}{e^{4}}+\frac {B \,b^{2} d^{2} x}{e^{3}} \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)/(e*x+d),x)

[Out]

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

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maxima [A]  time = 0.54, size = 152, normalized size = 1.65 \begin {gather*} \frac {2 \, B b^{2} e^{2} x^{3} - 3 \, {\left (B b^{2} d e - {\left (2 \, B a b + A b^{2}\right )} e^{2}\right )} x^{2} + 6 \, {\left (B b^{2} d^{2} - {\left (2 \, B a b + A b^{2}\right )} d e + {\left (B a^{2} + 2 \, A a b\right )} e^{2}\right )} x}{6 \, e^{3}} - \frac {{\left (B b^{2} d^{3} - A a^{2} e^{3} - {\left (2 \, B a b + A b^{2}\right )} d^{2} e + {\left (B a^{2} + 2 \, A a b\right )} d e^{2}\right )} \log \left (e x + d\right )}{e^{4}} \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)/(e*x+d),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.97, size = 159, normalized size = 1.73 \begin {gather*} x\,\left (\frac {B\,a^2+2\,A\,b\,a}{e}-\frac {d\,\left (\frac {A\,b^2+2\,B\,a\,b}{e}-\frac {B\,b^2\,d}{e^2}\right )}{e}\right )+x^2\,\left (\frac {A\,b^2+2\,B\,a\,b}{2\,e}-\frac {B\,b^2\,d}{2\,e^2}\right )+\frac {\ln \left (d+e\,x\right )\,\left (-B\,a^2\,d\,e^2+A\,a^2\,e^3+2\,B\,a\,b\,d^2\,e-2\,A\,a\,b\,d\,e^2-B\,b^2\,d^3+A\,b^2\,d^2\,e\right )}{e^4}+\frac {B\,b^2\,x^3}{3\,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))/(d + e*x),x)

[Out]

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

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sympy [A]  time = 0.43, size = 117, normalized size = 1.27 \begin {gather*} \frac {B b^{2} x^{3}}{3 e} + x^{2} \left (\frac {A b^{2}}{2 e} + \frac {B a b}{e} - \frac {B b^{2} d}{2 e^{2}}\right ) + x \left (\frac {2 A a b}{e} - \frac {A b^{2} d}{e^{2}} + \frac {B a^{2}}{e} - \frac {2 B a b d}{e^{2}} + \frac {B b^{2} d^{2}}{e^{3}}\right ) - \frac {\left (- A e + B d\right ) \left (a e - b d\right )^{2} \log {\left (d + e x \right )}}{e^{4}} \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)/(e*x+d),x)

[Out]

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

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