3.1137 \(\int \frac {(A+B x) (d+e x)}{(a+b x)^3} \, dx\)

Optimal. Leaf size=69 \[ -\frac {(A b-a B) (b d-a e)}{2 b^3 (a+b x)^2}-\frac {-2 a B e+A b e+b B d}{b^3 (a+b x)}+\frac {B e \log (a+b x)}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {77} \[ -\frac {(A b-a B) (b d-a e)}{2 b^3 (a+b x)^2}-\frac {-2 a B e+A b e+b B d}{b^3 (a+b x)}+\frac {B e \log (a+b x)}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.04, size = 75, normalized size = 1.09 \[ \frac {B \left (3 a^2 e-a b d+4 a b e x-2 b^2 d x\right )-A b (a e+b d+2 b e x)+2 B e (a+b x)^2 \log (a+b x)}{2 b^3 (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.89, size = 110, normalized size = 1.59 \[ -\frac {{\left (B a b + A b^{2}\right )} d - {\left (3 \, B a^{2} - A a b\right )} e + 2 \, {\left (B b^{2} d - {\left (2 \, B a b - A b^{2}\right )} e\right )} x - 2 \, {\left (B b^{2} e x^{2} + 2 \, B a b e x + B a^{2} e\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.17, size = 77, normalized size = 1.12 \[ \frac {B e \log \left ({\left | b x + a \right |}\right )}{b^{3}} - \frac {2 \, {\left (B b d - 2 \, B a e + A b e\right )} x + \frac {B a b d + A b^{2} d - 3 \, B a^{2} e + A a b e}{b}}{2 \, {\left (b x + a\right )}^{2} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 118, normalized size = 1.71 \[ \frac {A a e}{2 \left (b x +a \right )^{2} b^{2}}-\frac {A d}{2 \left (b x +a \right )^{2} b}-\frac {B \,a^{2} e}{2 \left (b x +a \right )^{2} b^{3}}+\frac {B a d}{2 \left (b x +a \right )^{2} b^{2}}-\frac {A e}{\left (b x +a \right ) b^{2}}+\frac {2 B a e}{\left (b x +a \right ) b^{3}}-\frac {B d}{\left (b x +a \right ) b^{2}}+\frac {B e \ln \left (b x +a \right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.59, size = 92, normalized size = 1.33 \[ -\frac {{\left (B a b + A b^{2}\right )} d - {\left (3 \, B a^{2} - A a b\right )} e + 2 \, {\left (B b^{2} d - {\left (2 \, B a b - A b^{2}\right )} e\right )} x}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} + \frac {B e \log \left (b x + a\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.12, size = 82, normalized size = 1.19 \[ \frac {B\,e\,\ln \left (a+b\,x\right )}{b^3}-\frac {\frac {A\,b^2\,d-3\,B\,a^2\,e+A\,a\,b\,e+B\,a\,b\,d}{2\,b^3}+\frac {x\,\left (A\,b\,e-2\,B\,a\,e+B\,b\,d\right )}{b^2}}{a^2+2\,a\,b\,x+b^2\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.85, size = 94, normalized size = 1.36 \[ \frac {B e \log {\left (a + b x \right )}}{b^{3}} + \frac {- A a b e - A b^{2} d + 3 B a^{2} e - B a b d + x \left (- 2 A b^{2} e + 4 B a b e - 2 B b^{2} d\right )}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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