3.1136 \(\int \frac {(A+B x) (d+e x)^2}{(a+b x)^3} \, dx\)

Optimal. Leaf size=103 \[ -\frac {(b d-a e) (-3 a B e+2 A b e+b B d)}{b^4 (a+b x)}-\frac {(A b-a B) (b d-a e)^2}{2 b^4 (a+b x)^2}+\frac {e \log (a+b x) (-3 a B e+A b e+2 b B d)}{b^4}+\frac {B e^2 x}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 103, 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 = {77} \[ -\frac {(b d-a e) (-3 a B e+2 A b e+b B d)}{b^4 (a+b x)}-\frac {(A b-a B) (b d-a e)^2}{2 b^4 (a+b x)^2}+\frac {e \log (a+b x) (-3 a B e+A b e+2 b B d)}{b^4}+\frac {B e^2 x}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.09, size = 140, normalized size = 1.36 \[ \frac {B \left (-5 a^3 e^2+2 a^2 b e (3 d-2 e x)+a b^2 \left (-d^2+8 d e x+4 e^2 x^2\right )+2 b^3 x \left (e^2 x^2-d^2\right )\right )+2 e (a+b x)^2 \log (a+b x) (-3 a B e+A b e+2 b B d)-A b (b d-a e) (3 a e+b (d+4 e x))}{2 b^4 (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(A*b*(b*d - a*e)*(3*a*e + b*(d + 4*e*x))) + B*(-5*a^3*e^2 + 2*a^2*b*e*(3*d - 2*e*x) + 2*b^3*x*(-d^2 + e^2*x^
2) + a*b^2*(-d^2 + 8*d*e*x + 4*e^2*x^2)) + 2*e*(2*b*B*d + A*b*e - 3*a*B*e)*(a + b*x)^2*Log[a + b*x])/(2*b^4*(a
+ b*x)^2)

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fricas [B]  time = 0.83, size = 258, normalized size = 2.50 \[ \frac {2 \, B b^{3} e^{2} x^{3} + 4 \, B a b^{2} e^{2} x^{2} - {\left (B a b^{2} + A b^{3}\right )} d^{2} + 2 \, {\left (3 \, B a^{2} b - A a b^{2}\right )} d e - {\left (5 \, B a^{3} - 3 \, A a^{2} b\right )} e^{2} - 2 \, {\left (B b^{3} d^{2} - 2 \, {\left (2 \, B a b^{2} - A b^{3}\right )} d e + 2 \, {\left (B a^{2} b - A a b^{2}\right )} e^{2}\right )} x + 2 \, {\left (2 \, B a^{2} b d e - {\left (3 \, B a^{3} - A a^{2} b\right )} e^{2} + {\left (2 \, B b^{3} d e - {\left (3 \, B a b^{2} - A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (2 \, B a b^{2} d e - {\left (3 \, B a^{2} b - A a b^{2}\right )} e^{2}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.16, size = 155, normalized size = 1.50 \[ \frac {B x e^{2}}{b^{3}} + \frac {{\left (2 \, B b d e - 3 \, B a e^{2} + A b e^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4}} - \frac {B a b^{2} d^{2} + A b^{3} d^{2} - 6 \, B a^{2} b d e + 2 \, A a b^{2} d e + 5 \, B a^{3} e^{2} - 3 \, A a^{2} b e^{2} + 2 \, {\left (B b^{3} d^{2} - 4 \, B a b^{2} d e + 2 \, A b^{3} d e + 3 \, B a^{2} b e^{2} - 2 \, A a b^{2} e^{2}\right )} x}{2 \, {\left (b x + a\right )}^{2} b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 242, normalized size = 2.35 \[ -\frac {A \,a^{2} e^{2}}{2 \left (b x +a \right )^{2} b^{3}}+\frac {A a d e}{\left (b x +a \right )^{2} b^{2}}-\frac {A \,d^{2}}{2 \left (b x +a \right )^{2} b}+\frac {B \,a^{3} e^{2}}{2 \left (b x +a \right )^{2} b^{4}}-\frac {B \,a^{2} d e}{\left (b x +a \right )^{2} b^{3}}+\frac {B a \,d^{2}}{2 \left (b x +a \right )^{2} b^{2}}+\frac {2 A a \,e^{2}}{\left (b x +a \right ) b^{3}}-\frac {2 A d e}{\left (b x +a \right ) b^{2}}+\frac {A \,e^{2} \ln \left (b x +a \right )}{b^{3}}-\frac {3 B \,a^{2} e^{2}}{\left (b x +a \right ) b^{4}}+\frac {4 B a d e}{\left (b x +a \right ) b^{3}}-\frac {3 B a \,e^{2} \ln \left (b x +a \right )}{b^{4}}-\frac {B \,d^{2}}{\left (b x +a \right ) b^{2}}+\frac {2 B d e \ln \left (b x +a \right )}{b^{3}}+\frac {B \,e^{2} x}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.56, size = 170, normalized size = 1.65 \[ \frac {B e^{2} x}{b^{3}} - \frac {{\left (B a b^{2} + A b^{3}\right )} d^{2} - 2 \, {\left (3 \, B a^{2} b - A a b^{2}\right )} d e + {\left (5 \, B a^{3} - 3 \, A a^{2} b\right )} e^{2} + 2 \, {\left (B b^{3} d^{2} - 2 \, {\left (2 \, B a b^{2} - A b^{3}\right )} d e + {\left (3 \, B a^{2} b - 2 \, A a b^{2}\right )} e^{2}\right )} x}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} + \frac {{\left (2 \, B b d e - {\left (3 \, B a - A b\right )} e^{2}\right )} \log \left (b x + a\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.13, size = 170, normalized size = 1.65 \[ \frac {\ln \left (a+b\,x\right )\,\left (A\,b\,e^2-3\,B\,a\,e^2+2\,B\,b\,d\,e\right )}{b^4}-\frac {x\,\left (3\,B\,a^2\,e^2-4\,B\,a\,b\,d\,e-2\,A\,a\,b\,e^2+B\,b^2\,d^2+2\,A\,b^2\,d\,e\right )+\frac {5\,B\,a^3\,e^2-6\,B\,a^2\,b\,d\,e-3\,A\,a^2\,b\,e^2+B\,a\,b^2\,d^2+2\,A\,a\,b^2\,d\,e+A\,b^3\,d^2}{2\,b}}{a^2\,b^3+2\,a\,b^4\,x+b^5\,x^2}+\frac {B\,e^2\,x}{b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.69, size = 187, normalized size = 1.82 \[ \frac {B e^{2} x}{b^{3}} + \frac {3 A a^{2} b e^{2} - 2 A a b^{2} d e - A b^{3} d^{2} - 5 B a^{3} e^{2} + 6 B a^{2} b d e - B a b^{2} d^{2} + x \left (4 A a b^{2} e^{2} - 4 A b^{3} d e - 6 B a^{2} b e^{2} + 8 B a b^{2} d e - 2 B b^{3} d^{2}\right )}{2 a^{2} b^{4} + 4 a b^{5} x + 2 b^{6} x^{2}} - \frac {e \left (- A b e + 3 B a e - 2 B b d\right ) \log {\left (a + b x \right )}}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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