3.1016 \(\int \frac {(a+b x) (A+B x)}{d+e x} \, dx\)

Optimal. Leaf size=60 \[ \frac {(b d-a e) (B d-A e) \log (d+e x)}{e^3}+\frac {B (a+b x)^2}{2 b e}-\frac {b x (B d-A e)}{e^2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 60, 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} \[ \frac {(b d-a e) (B d-A e) \log (d+e x)}{e^3}+\frac {B (a+b x)^2}{2 b e}-\frac {b x (B d-A e)}{e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.03, size = 56, normalized size = 0.93 \[ \frac {e x (2 a B e+b (2 A e-2 B d+B e x))+2 (b d-a e) (B d-A e) \log (d+e x)}{2 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.85, size = 68, normalized size = 1.13 \[ \frac {B b e^{2} x^{2} - 2 \, {\left (B b d e - {\left (B a + A b\right )} e^{2}\right )} x + 2 \, {\left (B b d^{2} + A a e^{2} - {\left (B a + A b\right )} d e\right )} \log \left (e x + d\right )}{2 \, e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.18, size = 71, normalized size = 1.18 \[ {\left (B b d^{2} - B a d e - A b d e + A a e^{2}\right )} e^{\left (-3\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{2} \, {\left (B b x^{2} e - 2 \, B b d x + 2 \, B a x e + 2 \, A b x e\right )} e^{\left (-2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 90, normalized size = 1.50 \[ \frac {B b \,x^{2}}{2 e}+\frac {A a \ln \left (e x +d \right )}{e}-\frac {A b d \ln \left (e x +d \right )}{e^{2}}+\frac {A b x}{e}-\frac {B a d \ln \left (e x +d \right )}{e^{2}}+\frac {B a x}{e}+\frac {B b \,d^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {B b d x}{e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.52, size = 66, normalized size = 1.10 \[ \frac {B b e x^{2} - 2 \, {\left (B b d - {\left (B a + A b\right )} e\right )} x}{2 \, e^{2}} + \frac {{\left (B b d^{2} + A a e^{2} - {\left (B a + A b\right )} d e\right )} \log \left (e x + d\right )}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.09, size = 68, normalized size = 1.13 \[ x\,\left (\frac {A\,b+B\,a}{e}-\frac {B\,b\,d}{e^2}\right )+\frac {\ln \left (d+e\,x\right )\,\left (A\,a\,e^2+B\,b\,d^2-A\,b\,d\,e-B\,a\,d\,e\right )}{e^3}+\frac {B\,b\,x^2}{2\,e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.27, size = 53, normalized size = 0.88 \[ \frac {B b x^{2}}{2 e} + x \left (\frac {A b}{e} + \frac {B a}{e} - \frac {B b d}{e^{2}}\right ) - \frac {\left (- A e + B d\right ) \left (a e - b d\right ) \log {\left (d + e x \right )}}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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