3.1114 \(\int \frac {(A+B x) (d+e x)^2}{a+b x} \, dx\)

Optimal. Leaf size=91 \[ \frac {(A b-a B) (b d-a e)^2 \log (a+b x)}{b^4}+\frac {e x (A b-a B) (b d-a e)}{b^3}+\frac {(d+e x)^2 (A b-a B)}{2 b^2}+\frac {B (d+e x)^3}{3 b e} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {77} \[ \frac {(d+e x)^2 (A b-a B)}{2 b^2}+\frac {e x (A b-a B) (b d-a e)}{b^3}+\frac {(A b-a B) (b d-a e)^2 \log (a+b x)}{b^4}+\frac {B (d+e x)^3}{3 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.08, size = 102, normalized size = 1.12 \[ \frac {b x \left (6 a^2 B e^2-3 a b e (2 A e+4 B d+B e x)+b^2 \left (3 A e (4 d+e x)+2 B \left (3 d^2+3 d e x+e^2 x^2\right )\right )\right )+6 (A b-a B) (b d-a e)^2 \log (a+b x)}{6 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x*(6*a^2*B*e^2 - 3*a*b*e*(4*B*d + 2*A*e + B*e*x) + b^2*(3*A*e*(4*d + e*x) + 2*B*(3*d^2 + 3*d*e*x + e^2*x^2)
)) + 6*(A*b - a*B)*(b*d - a*e)^2*Log[a + b*x])/(6*b^4)

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fricas [A]  time = 0.87, size = 158, normalized size = 1.74 \[ \frac {2 \, B b^{3} e^{2} x^{3} + 3 \, {\left (2 \, B b^{3} d e - {\left (B a b^{2} - A b^{3}\right )} e^{2}\right )} x^{2} + 6 \, {\left (B b^{3} d^{2} - 2 \, {\left (B a b^{2} - A b^{3}\right )} d e + {\left (B a^{2} b - A a b^{2}\right )} e^{2}\right )} x - 6 \, {\left ({\left (B a b^{2} - A b^{3}\right )} d^{2} - 2 \, {\left (B a^{2} b - A a b^{2}\right )} d e + {\left (B a^{3} - A a^{2} b\right )} e^{2}\right )} \log \left (b x + a\right )}{6 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*B*b^3*e^2*x^3 + 3*(2*B*b^3*d*e - (B*a*b^2 - A*b^3)*e^2)*x^2 + 6*(B*b^3*d^2 - 2*(B*a*b^2 - A*b^3)*d*e +
(B*a^2*b - A*a*b^2)*e^2)*x - 6*((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)*log
(b*x + a))/b^4

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giac [A]  time = 1.24, size = 164, normalized size = 1.80 \[ \frac {2 \, B b^{2} x^{3} e^{2} + 6 \, B b^{2} d x^{2} e + 6 \, B b^{2} d^{2} x - 3 \, B a b x^{2} e^{2} + 3 \, A b^{2} x^{2} e^{2} - 12 \, B a b d x e + 12 \, A b^{2} d x e + 6 \, B a^{2} x e^{2} - 6 \, A a b x e^{2}}{6 \, b^{3}} - \frac {{\left (B a b^{2} d^{2} - A b^{3} d^{2} - 2 \, B a^{2} b d e + 2 \, A a b^{2} d e + B a^{3} e^{2} - A a^{2} b e^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*B*b^2*x^3*e^2 + 6*B*b^2*d*x^2*e + 6*B*b^2*d^2*x - 3*B*a*b*x^2*e^2 + 3*A*b^2*x^2*e^2 - 12*B*a*b*d*x*e +
12*A*b^2*d*x*e + 6*B*a^2*x*e^2 - 6*A*a*b*x*e^2)/b^3 - (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)*log(abs(b*x + a))/b^4

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maple [B]  time = 0.00, size = 197, normalized size = 2.16 \[ \frac {B \,e^{2} x^{3}}{3 b}+\frac {A \,e^{2} x^{2}}{2 b}-\frac {B a \,e^{2} x^{2}}{2 b^{2}}+\frac {B d e \,x^{2}}{b}+\frac {A \,a^{2} e^{2} \ln \left (b x +a \right )}{b^{3}}-\frac {2 A a d e \ln \left (b x +a \right )}{b^{2}}-\frac {A a \,e^{2} x}{b^{2}}+\frac {A \,d^{2} \ln \left (b x +a \right )}{b}+\frac {2 A d e x}{b}-\frac {B \,a^{3} e^{2} \ln \left (b x +a \right )}{b^{4}}+\frac {2 B \,a^{2} d e \ln \left (b x +a \right )}{b^{3}}+\frac {B \,a^{2} e^{2} x}{b^{3}}-\frac {B a \,d^{2} \ln \left (b x +a \right )}{b^{2}}-\frac {2 B a d e x}{b^{2}}+\frac {B \,d^{2} x}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.53, size = 155, normalized size = 1.70 \[ \frac {2 \, B b^{2} e^{2} x^{3} + 3 \, {\left (2 \, B b^{2} d e - {\left (B a b - A b^{2}\right )} e^{2}\right )} x^{2} + 6 \, {\left (B b^{2} d^{2} - 2 \, {\left (B a b - A b^{2}\right )} d e + {\left (B a^{2} - A a b\right )} e^{2}\right )} x}{6 \, b^{3}} - \frac {{\left ({\left (B a b^{2} - A b^{3}\right )} d^{2} - 2 \, {\left (B a^{2} b - A a b^{2}\right )} d e + {\left (B a^{3} - A a^{2} b\right )} e^{2}\right )} \log \left (b x + a\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*B*b^2*e^2*x^3 + 3*(2*B*b^2*d*e - (B*a*b - A*b^2)*e^2)*x^2 + 6*(B*b^2*d^2 - 2*(B*a*b - A*b^2)*d*e + (B*a
^2 - A*a*b)*e^2)*x)/b^3 - ((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)*log(b*x
+ a)/b^4

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mupad [B]  time = 0.07, size = 159, normalized size = 1.75 \[ x\,\left (\frac {B\,d^2+2\,A\,e\,d}{b}-\frac {a\,\left (\frac {A\,e^2+2\,B\,d\,e}{b}-\frac {B\,a\,e^2}{b^2}\right )}{b}\right )+x^2\,\left (\frac {A\,e^2+2\,B\,d\,e}{2\,b}-\frac {B\,a\,e^2}{2\,b^2}\right )+\frac {\ln \left (a+b\,x\right )\,\left (-B\,a^3\,e^2+2\,B\,a^2\,b\,d\,e+A\,a^2\,b\,e^2-B\,a\,b^2\,d^2-2\,A\,a\,b^2\,d\,e+A\,b^3\,d^2\right )}{b^4}+\frac {B\,e^2\,x^3}{3\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.49, size = 117, normalized size = 1.29 \[ \frac {B e^{2} x^{3}}{3 b} + x^{2} \left (\frac {A e^{2}}{2 b} - \frac {B a e^{2}}{2 b^{2}} + \frac {B d e}{b}\right ) + x \left (- \frac {A a e^{2}}{b^{2}} + \frac {2 A d e}{b} + \frac {B a^{2} e^{2}}{b^{3}} - \frac {2 B a d e}{b^{2}} + \frac {B d^{2}}{b}\right ) - \frac {\left (- A b + B a\right ) \left (a e - b d\right )^{2} \log {\left (a + b x \right )}}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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