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

Optimal. Leaf size=73 \[ \frac {(b d-a e)^3 \log (a+b x)}{b^4}+\frac {e x (b d-a e)^2}{b^3}+\frac {(d+e x)^2 (b d-a e)}{2 b^2}+\frac {(d+e x)^3}{3 b} \]

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Rubi [A]  time = 0.03, antiderivative size = 73, 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 {e x (b d-a e)^2}{b^3}+\frac {(d+e x)^2 (b d-a e)}{2 b^2}+\frac {(b d-a e)^3 \log (a+b x)}{b^4}+\frac {(d+e x)^3}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.03, size = 74, normalized size = 1.01 \begin {gather*} \frac {b e x \left (6 a^2 e^2-3 a b e (6 d+e x)+b^2 \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )+6 (b d-a e)^3 \log (a+b x)}{6 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*e*x*(6*a^2*e^2 - 3*a*b*e*(6*d + e*x) + b^2*(18*d^2 + 9*d*e*x + 2*e^2*x^2)) + 6*(b*d - a*e)^3*Log[a + b*x])/
(6*b^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^2+2 a b x+b^2 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.38, size = 116, normalized size = 1.59 \begin {gather*} \frac {2 \, b^{3} e^{3} x^{3} + 3 \, {\left (3 \, b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 6 \, {\left (3 \, b^{3} d^{2} e - 3 \, a b^{2} d e^{2} + 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 (b x + a\right )}{6 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 110, normalized size = 1.51 \begin {gather*} \frac {2 \, b^{2} x^{3} e^{3} + 9 \, b^{2} d x^{2} e^{2} + 18 \, b^{2} d^{2} x e - 3 \, a b x^{2} e^{3} - 18 \, a b d x e^{2} + 6 \, a^{2} x e^{3}}{6 \, b^{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 ({\left | b x + a \right |}\right )}{b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*b^2*x^3*e^3 + 9*b^2*d*x^2*e^2 + 18*b^2*d^2*x*e - 3*a*b*x^2*e^3 - 18*a*b*d*x*e^2 + 6*a^2*x*e^3)/b^3 + (b
^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*log(abs(b*x + a))/b^4

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maple [A]  time = 0.05, size = 133, normalized size = 1.82 \begin {gather*} \frac {e^{3} x^{3}}{3 b}-\frac {a \,e^{3} x^{2}}{2 b^{2}}+\frac {3 d \,e^{2} x^{2}}{2 b}-\frac {a^{3} e^{3} \ln \left (b x +a \right )}{b^{4}}+\frac {3 a^{2} d \,e^{2} \ln \left (b x +a \right )}{b^{3}}+\frac {a^{2} e^{3} x}{b^{3}}-\frac {3 a \,d^{2} e \ln \left (b x +a \right )}{b^{2}}-\frac {3 a d \,e^{2} x}{b^{2}}+\frac {d^{3} \ln \left (b x +a \right )}{b}+\frac {3 d^{2} e x}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.50, size = 114, normalized size = 1.56 \begin {gather*} \frac {2 \, b^{2} e^{3} x^{3} + 3 \, {\left (3 \, b^{2} d e^{2} - a b e^{3}\right )} x^{2} + 6 \, {\left (3 \, b^{2} d^{2} e - 3 \, a b d e^{2} + a^{2} e^{3}\right )} x}{6 \, b^{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 (b x + a\right )}{b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.99, size = 118, normalized size = 1.62 \begin {gather*} x\,\left (\frac {3\,d^2\,e}{b}+\frac {a\,\left (\frac {a\,e^3}{b^2}-\frac {3\,d\,e^2}{b}\right )}{b}\right )-x^2\,\left (\frac {a\,e^3}{2\,b^2}-\frac {3\,d\,e^2}{2\,b}\right )-\frac {\ln \left (a+b\,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 )}{b^4}+\frac {e^3\,x^3}{3\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.33, size = 83, normalized size = 1.14 \begin {gather*} x^{2} \left (- \frac {a e^{3}}{2 b^{2}} + \frac {3 d e^{2}}{2 b}\right ) + x \left (\frac {a^{2} e^{3}}{b^{3}} - \frac {3 a d e^{2}}{b^{2}} + \frac {3 d^{2} e}{b}\right ) + \frac {e^{3} x^{3}}{3 b} - \frac {\left (a e - b d\right )^{3} \log {\left (a + b x \right )}}{b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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