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

Optimal. Leaf size=74 \[ -\frac {(b d-a e)^3 \log (d+e x)}{e^4}+\frac {b x (b d-a e)^2}{e^3}-\frac {(a+b x)^2 (b d-a e)}{2 e^2}+\frac {(a+b x)^3}{3 e} \]

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Rubi [A]  time = 0.03, antiderivative size = 74, 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, 43} \begin {gather*} \frac {b x (b d-a e)^2}{e^3}-\frac {(a+b x)^2 (b d-a e)}{2 e^2}-\frac {(b d-a e)^3 \log (d+e x)}{e^4}+\frac {(a+b x)^3}{3 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)^2*x)/e^3 - ((b*d - a*e)*(a + b*x)^2)/(2*e^2) + (a + b*x)^3/(3*e) - ((b*d - a*e)^3*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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

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Mathematica [A]  time = 0.03, size = 74, normalized size = 1.00 \begin {gather*} \frac {b e x \left (18 a^2 e^2+9 a b e (e x-2 d)+b^2 \left (6 d^2-3 d e x+2 e^2 x^2\right )\right )-6 (b d-a e)^3 \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]

(b*e*x*(18*a^2*e^2 + 9*a*b*e*(-2*d + e*x) + b^2*(6*d^2 - 3*d*e*x + 2*e^2*x^2)) - 6*(b*d - a*e)^3*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 = 0.41, size = 115, normalized size = 1.55 \begin {gather*} \frac {2 \, b^{3} e^{3} x^{3} - 3 \, {\left (b^{3} d e^{2} - 3 \, a b^{2} e^{3}\right )} x^{2} + 6 \, {\left (b^{3} d^{2} e - 3 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x - 6 \, {\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\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^3*e^3*x^3 - 3*(b^3*d*e^2 - 3*a*b^2*e^3)*x^2 + 6*(b^3*d^2*e - 3*a*b^2*d*e^2 + 3*a^2*b*e^3)*x - 6*(b^3*
d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*log(e*x + d))/e^4

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giac [A]  time = 0.18, size = 113, normalized size = 1.53 \begin {gather*} -{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} e^{\left (-4\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{6} \, {\left (2 \, b^{3} x^{3} e^{2} - 3 \, b^{3} d x^{2} e + 6 \, b^{3} d^{2} x + 9 \, a b^{2} x^{2} e^{2} - 18 \, a b^{2} d x e + 18 \, a^{2} 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^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*e^(-4)*log(abs(x*e + d)) + 1/6*(2*b^3*x^3*e^2 - 3*b^3*d*x
^2*e + 6*b^3*d^2*x + 9*a*b^2*x^2*e^2 - 18*a*b^2*d*x*e + 18*a^2*b*x*e^2)*e^(-3)

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maple [A]  time = 0.05, size = 133, normalized size = 1.80 \begin {gather*} \frac {b^{3} x^{3}}{3 e}+\frac {3 a \,b^{2} x^{2}}{2 e}-\frac {b^{3} d \,x^{2}}{2 e^{2}}+\frac {a^{3} \ln \left (e x +d \right )}{e}-\frac {3 a^{2} b d \ln \left (e x +d \right )}{e^{2}}+\frac {3 a^{2} b x}{e}+\frac {3 a \,b^{2} d^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {3 a \,b^{2} d x}{e^{2}}-\frac {b^{3} d^{3} \ln \left (e x +d \right )}{e^{4}}+\frac {b^{3} 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*b^3/e*x^3+3/2*b^2/e*x^2*a-1/2*b^3/e^2*x^2*d+3*b/e*a^2*x-3*b^2/e^2*a*d*x+b^3/e^3*d^2*x+1/e*ln(e*x+d)*a^3-3/
e^2*ln(e*x+d)*a^2*b*d+3/e^3*ln(e*x+d)*a*b^2*d^2-1/e^4*ln(e*x+d)*b^3*d^3

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maxima [A]  time = 0.68, size = 114, normalized size = 1.54 \begin {gather*} \frac {2 \, b^{3} e^{2} x^{3} - 3 \, {\left (b^{3} d e - 3 \, a b^{2} e^{2}\right )} x^{2} + 6 \, {\left (b^{3} d^{2} - 3 \, a b^{2} d e + 3 \, a^{2} b e^{2}\right )} x}{6 \, e^{3}} - \frac {{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\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^3*e^2*x^3 - 3*(b^3*d*e - 3*a*b^2*e^2)*x^2 + 6*(b^3*d^2 - 3*a*b^2*d*e + 3*a^2*b*e^2)*x)/e^3 - (b^3*d^3
- 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*log(e*x + d)/e^4

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mupad [B]  time = 2.03, size = 118, normalized size = 1.59 \begin {gather*} x^2\,\left (\frac {3\,a\,b^2}{2\,e}-\frac {b^3\,d}{2\,e^2}\right )+x\,\left (\frac {3\,a^2\,b}{e}-\frac {d\,\left (\frac {3\,a\,b^2}{e}-\frac {b^3\,d}{e^2}\right )}{e}\right )+\frac {\ln \left (d+e\,x\right )\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )}{e^4}+\frac {b^3\,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^2*((3*a*b^2)/(2*e) - (b^3*d)/(2*e^2)) + x*((3*a^2*b)/e - (d*((3*a*b^2)/e - (b^3*d)/e^2))/e) + (log(d + e*x)*
(a^3*e^3 - b^3*d^3 + 3*a*b^2*d^2*e - 3*a^2*b*d*e^2))/e^4 + (b^3*x^3)/(3*e)

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sympy [A]  time = 0.32, size = 83, normalized size = 1.12 \begin {gather*} \frac {b^{3} x^{3}}{3 e} + x^{2} \left (\frac {3 a b^{2}}{2 e} - \frac {b^{3} d}{2 e^{2}}\right ) + x \left (\frac {3 a^{2} b}{e} - \frac {3 a b^{2} d}{e^{2}} + \frac {b^{3} d^{2}}{e^{3}}\right ) + \frac {\left (a e - b d\right )^{3} \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**3*x**3/(3*e) + x**2*(3*a*b**2/(2*e) - b**3*d/(2*e**2)) + x*(3*a**2*b/e - 3*a*b**2*d/e**2 + b**3*d**2/e**3)
+ (a*e - b*d)**3*log(d + e*x)/e**4

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