3.17.89 \(\int \frac {(a+b x) (a^2+2 a b x+b^2 x^2)^2}{(d+e x)^3} \, dx\)

Optimal. Leaf size=133 \[ -\frac {5 b^4 (d+e x)^2 (b d-a e)}{2 e^6}+\frac {10 b^3 x (b d-a e)^2}{e^5}-\frac {10 b^2 (b d-a e)^3 \log (d+e x)}{e^6}-\frac {5 b (b d-a e)^4}{e^6 (d+e x)}+\frac {(b d-a e)^5}{2 e^6 (d+e x)^2}+\frac {b^5 (d+e x)^3}{3 e^6} \]

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Rubi [A]  time = 0.13, antiderivative size = 133, 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, 43} \begin {gather*} -\frac {5 b^4 (d+e x)^2 (b d-a e)}{2 e^6}+\frac {10 b^3 x (b d-a e)^2}{e^5}-\frac {10 b^2 (b d-a e)^3 \log (d+e x)}{e^6}-\frac {5 b (b d-a e)^4}{e^6 (d+e x)}+\frac {(b d-a e)^5}{2 e^6 (d+e x)^2}+\frac {b^5 (d+e x)^3}{3 e^6} \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)^3,x]

[Out]

(10*b^3*(b*d - a*e)^2*x)/e^5 + (b*d - a*e)^5/(2*e^6*(d + e*x)^2) - (5*b*(b*d - a*e)^4)/(e^6*(d + e*x)) - (5*b^
4*(b*d - a*e)*(d + e*x)^2)/(2*e^6) + (b^5*(d + e*x)^3)/(3*e^6) - (10*b^2*(b*d - a*e)^3*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 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 )^2}{(d+e x)^3} \, dx &=\int \frac {(a+b x)^5}{(d+e x)^3} \, dx\\ &=\int \left (\frac {10 b^3 (b d-a e)^2}{e^5}+\frac {(-b d+a e)^5}{e^5 (d+e x)^3}+\frac {5 b (b d-a e)^4}{e^5 (d+e x)^2}-\frac {10 b^2 (b d-a e)^3}{e^5 (d+e x)}-\frac {5 b^4 (b d-a e) (d+e x)}{e^5}+\frac {b^5 (d+e x)^2}{e^5}\right ) \, dx\\ &=\frac {10 b^3 (b d-a e)^2 x}{e^5}+\frac {(b d-a e)^5}{2 e^6 (d+e x)^2}-\frac {5 b (b d-a e)^4}{e^6 (d+e x)}-\frac {5 b^4 (b d-a e) (d+e x)^2}{2 e^6}+\frac {b^5 (d+e x)^3}{3 e^6}-\frac {10 b^2 (b d-a e)^3 \log (d+e x)}{e^6}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 230, normalized size = 1.73 \begin {gather*} \frac {-3 a^5 e^5-15 a^4 b e^4 (d+2 e x)+30 a^3 b^2 d e^3 (3 d+4 e x)+30 a^2 b^3 e^2 \left (-5 d^3-4 d^2 e x+4 d e^2 x^2+2 e^3 x^3\right )+15 a b^4 e \left (7 d^4+2 d^3 e x-11 d^2 e^2 x^2-4 d e^3 x^3+e^4 x^4\right )-60 b^2 (d+e x)^2 (b d-a e)^3 \log (d+e x)+b^5 \left (-27 d^5+6 d^4 e x+63 d^3 e^2 x^2+20 d^2 e^3 x^3-5 d e^4 x^4+2 e^5 x^5\right )}{6 e^6 (d+e x)^2} \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)^3,x]

[Out]

(-3*a^5*e^5 - 15*a^4*b*e^4*(d + 2*e*x) + 30*a^3*b^2*d*e^3*(3*d + 4*e*x) + 30*a^2*b^3*e^2*(-5*d^3 - 4*d^2*e*x +
4*d*e^2*x^2 + 2*e^3*x^3) + 15*a*b^4*e*(7*d^4 + 2*d^3*e*x - 11*d^2*e^2*x^2 - 4*d*e^3*x^3 + e^4*x^4) + b^5*(-27
*d^5 + 6*d^4*e*x + 63*d^3*e^2*x^2 + 20*d^2*e^3*x^3 - 5*d*e^4*x^4 + 2*e^5*x^5) - 60*b^2*(b*d - a*e)^3*(d + e*x)
^2*Log[d + e*x])/(6*e^6*(d + e*x)^2)

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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)^3} \, 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)^3,x]

[Out]

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

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fricas [B]  time = 0.42, size = 416, normalized size = 3.13 \begin {gather*} \frac {2 \, b^{5} e^{5} x^{5} - 27 \, b^{5} d^{5} + 105 \, a b^{4} d^{4} e - 150 \, a^{2} b^{3} d^{3} e^{2} + 90 \, a^{3} b^{2} d^{2} e^{3} - 15 \, a^{4} b d e^{4} - 3 \, a^{5} e^{5} - 5 \, {\left (b^{5} d e^{4} - 3 \, a b^{4} e^{5}\right )} x^{4} + 20 \, {\left (b^{5} d^{2} e^{3} - 3 \, a b^{4} d e^{4} + 3 \, a^{2} b^{3} e^{5}\right )} x^{3} + 3 \, {\left (21 \, b^{5} d^{3} e^{2} - 55 \, a b^{4} d^{2} e^{3} + 40 \, a^{2} b^{3} d e^{4}\right )} x^{2} + 6 \, {\left (b^{5} d^{4} e + 5 \, a b^{4} d^{3} e^{2} - 20 \, a^{2} b^{3} d^{2} e^{3} + 20 \, a^{3} b^{2} d e^{4} - 5 \, a^{4} b e^{5}\right )} x - 60 \, {\left (b^{5} d^{5} - 3 \, a b^{4} d^{4} e + 3 \, a^{2} b^{3} d^{3} e^{2} - a^{3} b^{2} d^{2} e^{3} + {\left (b^{5} d^{3} e^{2} - 3 \, a b^{4} d^{2} e^{3} + 3 \, a^{2} b^{3} d e^{4} - a^{3} b^{2} e^{5}\right )} x^{2} + 2 \, {\left (b^{5} d^{4} e - 3 \, a b^{4} d^{3} e^{2} + 3 \, a^{2} b^{3} d^{2} e^{3} - a^{3} b^{2} d e^{4}\right )} x\right )} \log \left (e x + d\right )}{6 \, {\left (e^{8} x^{2} + 2 \, d e^{7} x + d^{2} e^{6}\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)^3,x, algorithm="fricas")

[Out]

1/6*(2*b^5*e^5*x^5 - 27*b^5*d^5 + 105*a*b^4*d^4*e - 150*a^2*b^3*d^3*e^2 + 90*a^3*b^2*d^2*e^3 - 15*a^4*b*d*e^4
- 3*a^5*e^5 - 5*(b^5*d*e^4 - 3*a*b^4*e^5)*x^4 + 20*(b^5*d^2*e^3 - 3*a*b^4*d*e^4 + 3*a^2*b^3*e^5)*x^3 + 3*(21*b
^5*d^3*e^2 - 55*a*b^4*d^2*e^3 + 40*a^2*b^3*d*e^4)*x^2 + 6*(b^5*d^4*e + 5*a*b^4*d^3*e^2 - 20*a^2*b^3*d^2*e^3 +
20*a^3*b^2*d*e^4 - 5*a^4*b*e^5)*x - 60*(b^5*d^5 - 3*a*b^4*d^4*e + 3*a^2*b^3*d^3*e^2 - a^3*b^2*d^2*e^3 + (b^5*d
^3*e^2 - 3*a*b^4*d^2*e^3 + 3*a^2*b^3*d*e^4 - a^3*b^2*e^5)*x^2 + 2*(b^5*d^4*e - 3*a*b^4*d^3*e^2 + 3*a^2*b^3*d^2
*e^3 - a^3*b^2*d*e^4)*x)*log(e*x + d))/(e^8*x^2 + 2*d*e^7*x + d^2*e^6)

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giac [A]  time = 0.16, size = 250, normalized size = 1.88 \begin {gather*} -10 \, {\left (b^{5} d^{3} - 3 \, a b^{4} d^{2} e + 3 \, a^{2} b^{3} d e^{2} - a^{3} b^{2} e^{3}\right )} e^{\left (-6\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{6} \, {\left (2 \, b^{5} x^{3} e^{6} - 9 \, b^{5} d x^{2} e^{5} + 36 \, b^{5} d^{2} x e^{4} + 15 \, a b^{4} x^{2} e^{6} - 90 \, a b^{4} d x e^{5} + 60 \, a^{2} b^{3} x e^{6}\right )} e^{\left (-9\right )} - \frac {{\left (9 \, b^{5} d^{5} - 35 \, a b^{4} d^{4} e + 50 \, a^{2} b^{3} d^{3} e^{2} - 30 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} + a^{5} e^{5} + 10 \, {\left (b^{5} d^{4} e - 4 \, a b^{4} d^{3} e^{2} + 6 \, a^{2} b^{3} d^{2} e^{3} - 4 \, a^{3} b^{2} d e^{4} + a^{4} b e^{5}\right )} x\right )} e^{\left (-6\right )}}{2 \, {\left (x e + d\right )}^{2}} \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)^3,x, algorithm="giac")

[Out]

-10*(b^5*d^3 - 3*a*b^4*d^2*e + 3*a^2*b^3*d*e^2 - a^3*b^2*e^3)*e^(-6)*log(abs(x*e + d)) + 1/6*(2*b^5*x^3*e^6 -
9*b^5*d*x^2*e^5 + 36*b^5*d^2*x*e^4 + 15*a*b^4*x^2*e^6 - 90*a*b^4*d*x*e^5 + 60*a^2*b^3*x*e^6)*e^(-9) - 1/2*(9*b
^5*d^5 - 35*a*b^4*d^4*e + 50*a^2*b^3*d^3*e^2 - 30*a^3*b^2*d^2*e^3 + 5*a^4*b*d*e^4 + a^5*e^5 + 10*(b^5*d^4*e -
4*a*b^4*d^3*e^2 + 6*a^2*b^3*d^2*e^3 - 4*a^3*b^2*d*e^4 + a^4*b*e^5)*x)*e^(-6)/(x*e + d)^2

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maple [B]  time = 0.06, size = 346, normalized size = 2.60 \begin {gather*} \frac {b^{5} x^{3}}{3 e^{3}}-\frac {a^{5}}{2 \left (e x +d \right )^{2} e}+\frac {5 a^{4} b d}{2 \left (e x +d \right )^{2} e^{2}}-\frac {5 a^{3} b^{2} d^{2}}{\left (e x +d \right )^{2} e^{3}}+\frac {5 a^{2} b^{3} d^{3}}{\left (e x +d \right )^{2} e^{4}}-\frac {5 a \,b^{4} d^{4}}{2 \left (e x +d \right )^{2} e^{5}}+\frac {5 a \,b^{4} x^{2}}{2 e^{3}}+\frac {b^{5} d^{5}}{2 \left (e x +d \right )^{2} e^{6}}-\frac {3 b^{5} d \,x^{2}}{2 e^{4}}-\frac {5 a^{4} b}{\left (e x +d \right ) e^{2}}+\frac {20 a^{3} b^{2} d}{\left (e x +d \right ) e^{3}}+\frac {10 a^{3} b^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {30 a^{2} b^{3} d^{2}}{\left (e x +d \right ) e^{4}}-\frac {30 a^{2} b^{3} d \ln \left (e x +d \right )}{e^{4}}+\frac {10 a^{2} b^{3} x}{e^{3}}+\frac {20 a \,b^{4} d^{3}}{\left (e x +d \right ) e^{5}}+\frac {30 a \,b^{4} d^{2} \ln \left (e x +d \right )}{e^{5}}-\frac {15 a \,b^{4} d x}{e^{4}}-\frac {5 b^{5} d^{4}}{\left (e x +d \right ) e^{6}}-\frac {10 b^{5} d^{3} \ln \left (e x +d \right )}{e^{6}}+\frac {6 b^{5} d^{2} 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)^3,x)

[Out]

1/3*b^5/e^3*x^3+5/2*b^4/e^3*x^2*a-3/2*b^5/e^4*x^2*d+10*b^3/e^3*a^2*x-15*b^4/e^4*a*d*x+6*b^5/e^5*d^2*x-5*b/e^2/
(e*x+d)*a^4+20*b^2/e^3/(e*x+d)*a^3*d-30*b^3/e^4/(e*x+d)*a^2*d^2+20*b^4/e^5/(e*x+d)*a*d^3-5*b^5/e^6/(e*x+d)*d^4
-1/2/e/(e*x+d)^2*a^5+5/2/e^2/(e*x+d)^2*a^4*b*d-5/e^3/(e*x+d)^2*a^3*b^2*d^2+5/e^4/(e*x+d)^2*a^2*b^3*d^3-5/2/e^5
/(e*x+d)^2*a*b^4*d^4+1/2/e^6/(e*x+d)^2*b^5*d^5+10*b^2/e^3*ln(e*x+d)*a^3-30*b^3/e^4*ln(e*x+d)*a^2*d+30*b^4/e^5*
ln(e*x+d)*a*d^2-10*b^5/e^6*ln(e*x+d)*d^3

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maxima [B]  time = 0.72, size = 271, normalized size = 2.04 \begin {gather*} -\frac {9 \, b^{5} d^{5} - 35 \, a b^{4} d^{4} e + 50 \, a^{2} b^{3} d^{3} e^{2} - 30 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} + a^{5} e^{5} + 10 \, {\left (b^{5} d^{4} e - 4 \, a b^{4} d^{3} e^{2} + 6 \, a^{2} b^{3} d^{2} e^{3} - 4 \, a^{3} b^{2} d e^{4} + a^{4} b e^{5}\right )} x}{2 \, {\left (e^{8} x^{2} + 2 \, d e^{7} x + d^{2} e^{6}\right )}} + \frac {2 \, b^{5} e^{2} x^{3} - 3 \, {\left (3 \, b^{5} d e - 5 \, a b^{4} e^{2}\right )} x^{2} + 6 \, {\left (6 \, b^{5} d^{2} - 15 \, a b^{4} d e + 10 \, a^{2} b^{3} e^{2}\right )} x}{6 \, e^{5}} - \frac {10 \, {\left (b^{5} d^{3} - 3 \, a b^{4} d^{2} e + 3 \, a^{2} b^{3} d e^{2} - a^{3} b^{2} e^{3}\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)^3,x, algorithm="maxima")

[Out]

-1/2*(9*b^5*d^5 - 35*a*b^4*d^4*e + 50*a^2*b^3*d^3*e^2 - 30*a^3*b^2*d^2*e^3 + 5*a^4*b*d*e^4 + a^5*e^5 + 10*(b^5
*d^4*e - 4*a*b^4*d^3*e^2 + 6*a^2*b^3*d^2*e^3 - 4*a^3*b^2*d*e^4 + a^4*b*e^5)*x)/(e^8*x^2 + 2*d*e^7*x + d^2*e^6)
+ 1/6*(2*b^5*e^2*x^3 - 3*(3*b^5*d*e - 5*a*b^4*e^2)*x^2 + 6*(6*b^5*d^2 - 15*a*b^4*d*e + 10*a^2*b^3*e^2)*x)/e^5
- 10*(b^5*d^3 - 3*a*b^4*d^2*e + 3*a^2*b^3*d*e^2 - a^3*b^2*e^3)*log(e*x + d)/e^6

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mupad [B]  time = 0.09, size = 291, normalized size = 2.19 \begin {gather*} x^2\,\left (\frac {5\,a\,b^4}{2\,e^3}-\frac {3\,b^5\,d}{2\,e^4}\right )-\frac {\frac {a^5\,e^5+5\,a^4\,b\,d\,e^4-30\,a^3\,b^2\,d^2\,e^3+50\,a^2\,b^3\,d^3\,e^2-35\,a\,b^4\,d^4\,e+9\,b^5\,d^5}{2\,e}+x\,\left (5\,a^4\,b\,e^4-20\,a^3\,b^2\,d\,e^3+30\,a^2\,b^3\,d^2\,e^2-20\,a\,b^4\,d^3\,e+5\,b^5\,d^4\right )}{d^2\,e^5+2\,d\,e^6\,x+e^7\,x^2}-x\,\left (\frac {3\,d\,\left (\frac {5\,a\,b^4}{e^3}-\frac {3\,b^5\,d}{e^4}\right )}{e}-\frac {10\,a^2\,b^3}{e^3}+\frac {3\,b^5\,d^2}{e^5}\right )-\frac {\ln \left (d+e\,x\right )\,\left (-10\,a^3\,b^2\,e^3+30\,a^2\,b^3\,d\,e^2-30\,a\,b^4\,d^2\,e+10\,b^5\,d^3\right )}{e^6}+\frac {b^5\,x^3}{3\,e^3} \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)^3,x)

[Out]

x^2*((5*a*b^4)/(2*e^3) - (3*b^5*d)/(2*e^4)) - ((a^5*e^5 + 9*b^5*d^5 + 50*a^2*b^3*d^3*e^2 - 30*a^3*b^2*d^2*e^3
- 35*a*b^4*d^4*e + 5*a^4*b*d*e^4)/(2*e) + x*(5*b^5*d^4 + 5*a^4*b*e^4 - 20*a^3*b^2*d*e^3 + 30*a^2*b^3*d^2*e^2 -
20*a*b^4*d^3*e))/(d^2*e^5 + e^7*x^2 + 2*d*e^6*x) - x*((3*d*((5*a*b^4)/e^3 - (3*b^5*d)/e^4))/e - (10*a^2*b^3)/
e^3 + (3*b^5*d^2)/e^5) - (log(d + e*x)*(10*b^5*d^3 - 10*a^3*b^2*e^3 + 30*a^2*b^3*d*e^2 - 30*a*b^4*d^2*e))/e^6
+ (b^5*x^3)/(3*e^3)

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sympy [B]  time = 1.65, size = 258, normalized size = 1.94 \begin {gather*} \frac {b^{5} x^{3}}{3 e^{3}} + \frac {10 b^{2} \left (a e - b d\right )^{3} \log {\left (d + e x \right )}}{e^{6}} + x^{2} \left (\frac {5 a b^{4}}{2 e^{3}} - \frac {3 b^{5} d}{2 e^{4}}\right ) + x \left (\frac {10 a^{2} b^{3}}{e^{3}} - \frac {15 a b^{4} d}{e^{4}} + \frac {6 b^{5} d^{2}}{e^{5}}\right ) + \frac {- a^{5} e^{5} - 5 a^{4} b d e^{4} + 30 a^{3} b^{2} d^{2} e^{3} - 50 a^{2} b^{3} d^{3} e^{2} + 35 a b^{4} d^{4} e - 9 b^{5} d^{5} + x \left (- 10 a^{4} b e^{5} + 40 a^{3} b^{2} d e^{4} - 60 a^{2} b^{3} d^{2} e^{3} + 40 a b^{4} d^{3} e^{2} - 10 b^{5} d^{4} e\right )}{2 d^{2} e^{6} + 4 d e^{7} x + 2 e^{8} x^{2}} \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)**3,x)

[Out]

b**5*x**3/(3*e**3) + 10*b**2*(a*e - b*d)**3*log(d + e*x)/e**6 + x**2*(5*a*b**4/(2*e**3) - 3*b**5*d/(2*e**4)) +
x*(10*a**2*b**3/e**3 - 15*a*b**4*d/e**4 + 6*b**5*d**2/e**5) + (-a**5*e**5 - 5*a**4*b*d*e**4 + 30*a**3*b**2*d*
*2*e**3 - 50*a**2*b**3*d**3*e**2 + 35*a*b**4*d**4*e - 9*b**5*d**5 + x*(-10*a**4*b*e**5 + 40*a**3*b**2*d*e**4 -
60*a**2*b**3*d**2*e**3 + 40*a*b**4*d**3*e**2 - 10*b**5*d**4*e))/(2*d**2*e**6 + 4*d*e**7*x + 2*e**8*x**2)

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