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

Optimal. Leaf size=59 \[ \frac {(A b-a B) (b d-a e) \log (a+b x)}{b^3}+\frac {B x (b d-a e)}{b^2}+\frac {e (A+B x)^2}{2 b B} \]

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Rubi [A]  time = 0.04, antiderivative size = 59, 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} \begin {gather*} \frac {(A b-a B) (b d-a e) \log (a+b x)}{b^3}+\frac {B x (b d-a e)}{b^2}+\frac {e (A+B x)^2}{2 b B} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.04, size = 56, normalized size = 0.95 \begin {gather*} \frac {b x (b (2 A e+2 B d+B e x)-2 a B e)+2 (A b-a B) (b d-a e) \log (a+b x)}{2 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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)}{a+b x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.92, size = 75, normalized size = 1.27 \begin {gather*} \frac {B b^{2} e x^{2} + 2 \, {\left (B b^{2} d - {\left (B a b - A b^{2}\right )} e\right )} x - 2 \, {\left ({\left (B a b - A b^{2}\right )} d - {\left (B a^{2} - A a b\right )} e\right )} \log \left (b x + a\right )}{2 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.15, size = 74, normalized size = 1.25 \begin {gather*} \frac {B b x^{2} e + 2 \, B b d x - 2 \, B a x e + 2 \, A b x e}{2 \, b^{2}} - \frac {{\left (B a b d - A b^{2} d - B a^{2} e + A a b e\right )} \log \left ({\left | b x + a \right |}\right )}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 90, normalized size = 1.53 \begin {gather*} \frac {B e \,x^{2}}{2 b}-\frac {A a e \ln \left (b x +a \right )}{b^{2}}+\frac {A d \ln \left (b x +a \right )}{b}+\frac {A e x}{b}+\frac {B \,a^{2} e \ln \left (b x +a \right )}{b^{3}}-\frac {B a d \ln \left (b x +a \right )}{b^{2}}-\frac {B a e x}{b^{2}}+\frac {B d x}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.46, size = 72, normalized size = 1.22 \begin {gather*} \frac {B b e x^{2} + 2 \, {\left (B b d - {\left (B a - A b\right )} e\right )} x}{2 \, b^{2}} - \frac {{\left ({\left (B a b - A b^{2}\right )} d - {\left (B a^{2} - A a b\right )} e\right )} \log \left (b x + a\right )}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.10, size = 68, normalized size = 1.15 \begin {gather*} x\,\left (\frac {A\,e+B\,d}{b}-\frac {B\,a\,e}{b^2}\right )+\frac {\ln \left (a+b\,x\right )\,\left (A\,b^2\,d+B\,a^2\,e-A\,a\,b\,e-B\,a\,b\,d\right )}{b^3}+\frac {B\,e\,x^2}{2\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.28, size = 53, normalized size = 0.90 \begin {gather*} \frac {B e x^{2}}{2 b} + x \left (\frac {A e}{b} - \frac {B a e}{b^{2}} + \frac {B d}{b}\right ) + \frac {\left (- A b + B a\right ) \left (a e - b d\right ) \log {\left (a + b x \right )}}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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