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

Optimal. Leaf size=99 \[ -\frac {(A b-a B) (b d-a e)^2}{b^4 (a+b x)}+\frac {(b d-a e) \log (a+b x) (-3 a B e+2 A b e+b B d)}{b^4}+\frac {e x (-2 a B e+A b e+2 b B d)}{b^3}+\frac {B e^2 x^2}{2 b^2} \]

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Rubi [A]  time = 0.10, antiderivative size = 99, 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, 77} \begin {gather*} -\frac {(A b-a B) (b d-a e)^2}{b^4 (a+b x)}+\frac {e x (-2 a B e+A b e+2 b B d)}{b^3}+\frac {(b d-a e) \log (a+b x) (-3 a B e+2 A b e+b B d)}{b^4}+\frac {B e^2 x^2}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.07, size = 153, normalized size = 1.55 \begin {gather*} \frac {\log (a+b x) \left (3 a^2 B e^2-2 a A b e^2-4 a b B d e+2 A b^2 d e+b^2 B d^2\right )}{b^4}+\frac {a^3 B e^2-a^2 A b e^2-2 a^2 b B d e+2 a A b^2 d e+a b^2 B d^2-A b^3 d^2}{b^4 (a+b x)}+\frac {e x (-2 a B e+A b e+2 b B d)}{b^3}+\frac {B e^2 x^2}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

[Out]

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

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fricas [B]  time = 0.42, size = 249, normalized size = 2.52 \begin {gather*} \frac {B b^{3} e^{2} x^{3} + 2 \, {\left (B a b^{2} - A b^{3}\right )} d^{2} - 4 \, {\left (B a^{2} b - A a b^{2}\right )} d e + 2 \, {\left (B a^{3} - A a^{2} b\right )} e^{2} + {\left (4 \, B b^{3} d e - {\left (3 \, B a b^{2} - 2 \, A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (2 \, B a b^{2} d e - {\left (2 \, B a^{2} b - A a b^{2}\right )} e^{2}\right )} x + 2 \, {\left (B a b^{2} d^{2} - 2 \, {\left (2 \, B a^{2} b - A a b^{2}\right )} d e + {\left (3 \, B a^{3} - 2 \, A a^{2} b\right )} e^{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\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x + a b^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 162, normalized size = 1.64 \begin {gather*} \frac {{\left (B b^{2} d^{2} - 4 \, B a b d e + 2 \, A b^{2} d e + 3 \, B a^{2} e^{2} - 2 \, A a b e^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4}} + \frac {B b^{2} x^{2} e^{2} + 4 \, B b^{2} d x e - 4 \, B a b x e^{2} + 2 \, A b^{2} x e^{2}}{2 \, b^{4}} + \frac {B a b^{2} d^{2} - A b^{3} d^{2} - 2 \, B a^{2} b d e + 2 \, A a b^{2} d e + B a^{3} e^{2} - A a^{2} b e^{2}}{{\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)^2/(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.06, size = 223, normalized size = 2.25 \begin {gather*} \frac {B \,e^{2} x^{2}}{2 b^{2}}-\frac {A \,a^{2} e^{2}}{\left (b x +a \right ) b^{3}}+\frac {2 A a d e}{\left (b x +a \right ) b^{2}}-\frac {2 A a \,e^{2} \ln \left (b x +a \right )}{b^{3}}-\frac {A \,d^{2}}{\left (b x +a \right ) b}+\frac {2 A d e \ln \left (b x +a \right )}{b^{2}}+\frac {A \,e^{2} x}{b^{2}}+\frac {B \,a^{3} e^{2}}{\left (b x +a \right ) b^{4}}-\frac {2 B \,a^{2} d e}{\left (b x +a \right ) b^{3}}+\frac {3 B \,a^{2} e^{2} \ln \left (b x +a \right )}{b^{4}}+\frac {B a \,d^{2}}{\left (b x +a \right ) b^{2}}-\frac {4 B a d e \ln \left (b x +a \right )}{b^{3}}-\frac {2 B a \,e^{2} x}{b^{3}}+\frac {B \,d^{2} \ln \left (b x +a \right )}{b^{2}}+\frac {2 B d e x}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.46, size = 158, normalized size = 1.60 \begin {gather*} \frac {{\left (B a b^{2} - A b^{3}\right )} d^{2} - 2 \, {\left (B a^{2} b - A a b^{2}\right )} d e + {\left (B a^{3} - A a^{2} b\right )} e^{2}}{b^{5} x + a b^{4}} + \frac {B b e^{2} x^{2} + 2 \, {\left (2 \, B b d e - {\left (2 \, B a - A b\right )} e^{2}\right )} x}{2 \, b^{3}} + \frac {{\left (B b^{2} d^{2} - 2 \, {\left (2 \, B a b - A b^{2}\right )} d e + {\left (3 \, B a^{2} - 2 \, A a b\right )} e^{2}\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)^2/(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.11, size = 165, normalized size = 1.67 \begin {gather*} x\,\left (\frac {A\,e^2+2\,B\,d\,e}{b^2}-\frac {2\,B\,a\,e^2}{b^3}\right )+\frac {\ln \left (a+b\,x\right )\,\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 )}{b^4}-\frac {-B\,a^3\,e^2+2\,B\,a^2\,b\,d\,e+A\,a^2\,b\,e^2-B\,a\,b^2\,d^2-2\,A\,a\,b^2\,d\,e+A\,b^3\,d^2}{b\,\left (x\,b^4+a\,b^3\right )}+\frac {B\,e^2\,x^2}{2\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.88, size = 151, normalized size = 1.53 \begin {gather*} \frac {B e^{2} x^{2}}{2 b^{2}} + x \left (\frac {A e^{2}}{b^{2}} - \frac {2 B a e^{2}}{b^{3}} + \frac {2 B d e}{b^{2}}\right ) + \frac {- A a^{2} b e^{2} + 2 A a b^{2} d e - A b^{3} d^{2} + B a^{3} e^{2} - 2 B a^{2} b d e + B a b^{2} d^{2}}{a b^{4} + b^{5} x} + \frac {\left (a e - b d\right ) \left (- 2 A b e + 3 B a e - B b d\right ) \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)**2/(b**2*x**2+2*a*b*x+a**2),x)

[Out]

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

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