3.11.18 \(\int \frac {(a+b x) (A+B x)}{(d+e x)^3} \, dx\) [1018]

Optimal. Leaf size=70 \[ -\frac {(b d-a e) (B d-A e)}{2 e^3 (d+e x)^2}+\frac {2 b B d-A b e-a B e}{e^3 (d+e x)}+\frac {b B \log (d+e x)}{e^3} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 70, 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 = {78} \begin {gather*} -\frac {(b d-a e) (B d-A e)}{2 e^3 (d+e x)^2}+\frac {-a B e-A b e+2 b B d}{e^3 (d+e x)}+\frac {b B \log (d+e x)}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.03, size = 72, normalized size = 1.03 \begin {gather*} \frac {-a e (A e+B (d+2 e x))+b (-A e (d+2 e x)+B d (3 d+4 e x))+2 b B (d+e x)^2 \log (d+e x)}{2 e^3 (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.08, size = 77, normalized size = 1.10

method	result	size
norman	\(\frac {-\frac {A a \,e^{2}+A b d e +B a d e -3 B b \,d^{2}}{2 e^{3}}-\frac {\left (A b e +B a e -2 B b d \right ) x}{e^{2}}}{\left (e x +d \right )^{2}}+\frac {b B \ln \left (e x +d \right )}{e^{3}}\)	\(72\)
risch	\(\frac {-\frac {A a \,e^{2}+A b d e +B a d e -3 B b \,d^{2}}{2 e^{3}}-\frac {\left (A b e +B a e -2 B b d \right ) x}{e^{2}}}{\left (e x +d \right )^{2}}+\frac {b B \ln \left (e x +d \right )}{e^{3}}\)	\(72\)
default	\(-\frac {A a \,e^{2}-A b d e -B a d e +B b \,d^{2}}{2 e^{3} \left (e x +d \right )^{2}}-\frac {A b e +B a e -2 B b d}{e^{3} \left (e x +d \right )}+\frac {b B \ln \left (e x +d \right )}{e^{3}}\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 87, normalized size = 1.24 \begin {gather*} B b e^{\left (-3\right )} \log \left (x e + d\right ) + \frac {3 \, B b d^{2} - A a e^{2} - {\left (B a e + A b e\right )} d + 2 \, {\left (2 \, B b d e - B a e^{2} - A b e^{2}\right )} x}{2 \, {\left (x^{2} e^{5} + 2 \, d x e^{4} + d^{2} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.42, size = 101, normalized size = 1.44 \begin {gather*} \frac {3 \, B b d^{2} - {\left (A a + 2 \, {\left (B a + A b\right )} x\right )} e^{2} + {\left (4 \, B b d x - {\left (B a + A b\right )} d\right )} e + 2 \, {\left (B b x^{2} e^{2} + 2 \, B b d x e + B b d^{2}\right )} \log \left (x e + d\right )}{2 \, {\left (x^{2} e^{5} + 2 \, d x e^{4} + d^{2} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.45, size = 94, normalized size = 1.34 \begin {gather*} \frac {B b \log {\left (d + e x \right )}}{e^{3}} + \frac {- A a e^{2} - A b d e - B a d e + 3 B b d^{2} + x \left (- 2 A b e^{2} - 2 B a e^{2} + 4 B b d e\right )}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.72, size = 79, normalized size = 1.13 \begin {gather*} B b e^{\left (-3\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {{\left (2 \, {\left (2 \, B b d - B a e - A b e\right )} x + {\left (3 \, B b d^{2} - B a d e - A b d e - A a e^{2}\right )} e^{\left (-1\right )}\right )} e^{\left (-2\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)*(B*x+A)/(e*x+d)^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.09, size = 82, normalized size = 1.17 \begin {gather*} \frac {B\,b\,\ln \left (d+e\,x\right )}{e^3}-\frac {\frac {A\,a\,e^2-3\,B\,b\,d^2+A\,b\,d\,e+B\,a\,d\,e}{2\,e^3}+\frac {x\,\left (A\,b\,e+B\,a\,e-2\,B\,b\,d\right )}{e^2}}{d^2+2\,d\,e\,x+e^2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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