3.11.29 \(\int \frac {(a+b x)^2 (A+B x)}{(d+e x)^3} \, dx\) [1029]

Optimal. Leaf size=106 \[ \frac {b^2 B x}{e^3}+\frac {(b d-a e)^2 (B d-A e)}{2 e^4 (d+e x)^2}-\frac {(b d-a e) (3 b B d-2 A b e-a B e)}{e^4 (d+e x)}-\frac {b (3 b B d-A b e-2 a B e) \log (d+e x)}{e^4} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \begin {gather*} -\frac {(b d-a e) (-a B e-2 A b e+3 b B d)}{e^4 (d+e x)}+\frac {(b d-a e)^2 (B d-A e)}{2 e^4 (d+e x)^2}-\frac {b \log (d+e x) (-2 a B e-A b e+3 b B d)}{e^4}+\frac {b^2 B x}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 78

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

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Mathematica [A]
time = 0.06, size = 143, normalized size = 1.35 \begin {gather*} -\frac {a^2 e^2 (A e+B (d+2 e x))+2 a b e (A e (d+2 e x)-B d (3 d+4 e x))-b^2 \left (A d e (3 d+4 e x)+B \left (-5 d^3-4 d^2 e x+4 d e^2 x^2+2 e^3 x^3\right )\right )+2 b (3 b B d-A b e-2 a B e) (d+e x)^2 \log (d+e x)}{2 e^4 (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a^2*e^2*(A*e + B*(d + 2*e*x)) + 2*a*b*e*(A*e*(d + 2*e*x) - B*d*(3*d + 4*e*x)) - b^2*(A*d*e*(3*d + 4*e*x)
+ B*(-5*d^3 - 4*d^2*e*x + 4*d*e^2*x^2 + 2*e^3*x^3)) + 2*b*(3*b*B*d - A*b*e - 2*a*B*e)*(d + e*x)^2*Log[d + e*x
])/(e^4*(d + e*x)^2)

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Maple [A]
time = 0.07, size = 157, normalized size = 1.48

method	result	size
norman	\(\frac {\frac {b^{2} B \,x^{3}}{e}-\frac {a^{2} A \,e^{3}+2 A a b d \,e^{2}-3 A \,b^{2} d^{2} e +B \,a^{2} d \,e^{2}-6 B a b \,d^{2} e +9 b^{2} B \,d^{3}}{2 e^{4}}-\frac {\left (2 A a b \,e^{2}-2 A \,b^{2} d e +B \,a^{2} e^{2}-4 B a b d e +6 b^{2} B \,d^{2}\right ) x}{e^{3}}}{\left (e x +d \right )^{2}}+\frac {b \left (A b e +2 B a e -3 B b d \right ) \ln \left (e x +d \right )}{e^{4}}\)	\(155\)
default	\(\frac {b^{2} B x}{e^{3}}-\frac {a^{2} A \,e^{3}-2 A a b d \,e^{2}+A \,b^{2} d^{2} e -B \,a^{2} d \,e^{2}+2 B a b \,d^{2} e -b^{2} B \,d^{3}}{2 e^{4} \left (e x +d \right )^{2}}-\frac {2 A a b \,e^{2}-2 A \,b^{2} d e +B \,a^{2} e^{2}-4 B a b d e +3 b^{2} B \,d^{2}}{e^{4} \left (e x +d \right )}+\frac {b \left (A b e +2 B a e -3 B b d \right ) \ln \left (e x +d \right )}{e^{4}}\)	\(157\)
risch	\(\frac {b^{2} B x}{e^{3}}+\frac {\left (-2 A a b \,e^{2}+2 A \,b^{2} d e -B \,a^{2} e^{2}+4 B a b d e -3 b^{2} B \,d^{2}\right ) x -\frac {a^{2} A \,e^{3}+2 A a b d \,e^{2}-3 A \,b^{2} d^{2} e +B \,a^{2} d \,e^{2}-6 B a b \,d^{2} e +5 b^{2} B \,d^{3}}{2 e}}{e^{3} \left (e x +d \right )^{2}}+\frac {b^{2} \ln \left (e x +d \right ) A}{e^{3}}+\frac {2 b \ln \left (e x +d \right ) B a}{e^{3}}-\frac {3 b^{2} \ln \left (e x +d \right ) B d}{e^{4}}\)	\(171\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2*(B*x+A)/(e*x+d)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.31, size = 167, normalized size = 1.58 \begin {gather*} B b^{2} x e^{\left (-3\right )} - {\left (3 \, B b^{2} d - 2 \, B a b e - A b^{2} e\right )} e^{\left (-4\right )} \log \left (x e + d\right ) - \frac {5 \, B b^{2} d^{3} + A a^{2} e^{3} - 3 \, {\left (2 \, B a b e + A b^{2} e\right )} d^{2} + {\left (B a^{2} e^{2} + 2 \, A a b e^{2}\right )} d + 2 \, {\left (3 \, B b^{2} d^{2} e + B a^{2} e^{3} + 2 \, A a b e^{3} - 2 \, {\left (2 \, B a b e^{2} + A b^{2} e^{2}\right )} d\right )} x}{2 \, {\left (x^{2} e^{6} + 2 \, d x e^{5} + d^{2} e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 237 vs. \(2 (110) = 220\).
time = 0.97, size = 237, normalized size = 2.24 \begin {gather*} -\frac {5 \, B b^{2} d^{3} - {\left (2 \, B b^{2} x^{3} - A a^{2} - 2 \, {\left (B a^{2} + 2 \, A a b\right )} x\right )} e^{3} - {\left (4 \, B b^{2} d x^{2} + 4 \, {\left (2 \, B a b + A b^{2}\right )} d x - {\left (B a^{2} + 2 \, A a b\right )} d\right )} e^{2} + {\left (4 \, B b^{2} d^{2} x - 3 \, {\left (2 \, B a b + A b^{2}\right )} d^{2}\right )} e + 2 \, {\left (3 \, B b^{2} d^{3} - {\left (2 \, B a b + A b^{2}\right )} x^{2} e^{3} + {\left (3 \, B b^{2} d x^{2} - 2 \, {\left (2 \, B a b + A b^{2}\right )} d x\right )} e^{2} + {\left (6 \, B b^{2} d^{2} x - {\left (2 \, B a b + A b^{2}\right )} d^{2}\right )} e\right )} \log \left (x e + d\right )}{2 \, {\left (x^{2} e^{6} + 2 \, d x e^{5} + d^{2} e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 1.45, size = 187, normalized size = 1.76 \begin {gather*} \frac {B b^{2} x}{e^{3}} + \frac {b \left (A b e + 2 B a e - 3 B b d\right ) \log {\left (d + e x \right )}}{e^{4}} + \frac {- A a^{2} e^{3} - 2 A a b d e^{2} + 3 A b^{2} d^{2} e - B a^{2} d e^{2} + 6 B a b d^{2} e - 5 B b^{2} d^{3} + x \left (- 4 A a b e^{3} + 4 A b^{2} d e^{2} - 2 B a^{2} e^{3} + 8 B a b d e^{2} - 6 B b^{2} d^{2} e\right )}{2 d^{2} e^{4} + 4 d e^{5} x + 2 e^{6} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*b**2*x/e**3 + b*(A*b*e + 2*B*a*e - 3*B*b*d)*log(d + e*x)/e**4 + (-A*a**2*e**3 - 2*A*a*b*d*e**2 + 3*A*b**2*d*
*2*e - B*a**2*d*e**2 + 6*B*a*b*d**2*e - 5*B*b**2*d**3 + x*(-4*A*a*b*e**3 + 4*A*b**2*d*e**2 - 2*B*a**2*e**3 + 8
*B*a*b*d*e**2 - 6*B*b**2*d**2*e))/(2*d**2*e**4 + 4*d*e**5*x + 2*e**6*x**2)

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Giac [A]
time = 1.53, size = 156, normalized size = 1.47 \begin {gather*} B b^{2} x e^{\left (-3\right )} - {\left (3 \, B b^{2} d - 2 \, B a b e - A b^{2} e\right )} e^{\left (-4\right )} \log \left ({\left | x e + d \right |}\right ) - \frac {{\left (5 \, B b^{2} d^{3} - 6 \, B a b d^{2} e - 3 \, A b^{2} d^{2} e + B a^{2} d e^{2} + 2 \, A a b d e^{2} + A a^{2} e^{3} + 2 \, {\left (3 \, B b^{2} d^{2} e - 4 \, B a b d e^{2} - 2 \, A b^{2} d e^{2} + B a^{2} e^{3} + 2 \, A a b e^{3}\right )} x\right )} e^{\left (-4\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)^2*(B*x+A)/(e*x+d)^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.13, size = 170, normalized size = 1.60 \begin {gather*} \frac {\ln \left (d+e\,x\right )\,\left (A\,b^2\,e-3\,B\,b^2\,d+2\,B\,a\,b\,e\right )}{e^4}-\frac {x\,\left (B\,a^2\,e^2-4\,B\,a\,b\,d\,e+2\,A\,a\,b\,e^2+3\,B\,b^2\,d^2-2\,A\,b^2\,d\,e\right )+\frac {B\,a^2\,d\,e^2+A\,a^2\,e^3-6\,B\,a\,b\,d^2\,e+2\,A\,a\,b\,d\,e^2+5\,B\,b^2\,d^3-3\,A\,b^2\,d^2\,e}{2\,e}}{d^2\,e^3+2\,d\,e^4\,x+e^5\,x^2}+\frac {B\,b^2\,x}{e^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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