3.1138 \(\int \frac {(A+B x) (d+e x)^3}{b x+c x^2} \, dx\)

Optimal. Leaf size=128 \[ \frac {e x \left (A c e (3 c d-b e)+B \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )\right )}{c^3}+\frac {(b B-A c) (c d-b e)^3 \log (b+c x)}{b c^4}+\frac {e^2 x^2 (A c e-b B e+3 B c d)}{2 c^2}+\frac {A d^3 \log (x)}{b}+\frac {B e^3 x^3}{3 c} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {771} \[ \frac {e x \left (A c e (3 c d-b e)+B \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )\right )}{c^3}+\frac {e^2 x^2 (A c e-b B e+3 B c d)}{2 c^2}+\frac {(b B-A c) (c d-b e)^3 \log (b+c x)}{b c^4}+\frac {A d^3 \log (x)}{b}+\frac {B e^3 x^3}{3 c} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^3)/(b*x + c*x^2),x]

[Out]

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

Rule 771

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 118, normalized size = 0.92 \[ \frac {b c e x \left (3 A c e (-2 b e+6 c d+c e x)+B \left (6 b^2 e^2-3 b c e (6 d+e x)+c^2 \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )\right )-6 (b B-A c) (b e-c d)^3 \log (b+c x)+6 A c^4 d^3 \log (x)}{6 b c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x)^3)/(b*x + c*x^2),x]

[Out]

(b*c*e*x*(3*A*c*e*(6*c*d - 2*b*e + c*e*x) + B*(6*b^2*e^2 - 3*b*c*e*(6*d + e*x) + c^2*(18*d^2 + 9*d*e*x + 2*e^2
*x^2))) + 6*A*c^4*d^3*Log[x] - 6*(b*B - A*c)*(-(c*d) + b*e)^3*Log[b + c*x])/(6*b*c^4)

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fricas [A]  time = 0.89, size = 216, normalized size = 1.69 \[ \frac {2 \, B b c^{3} e^{3} x^{3} + 6 \, A c^{4} d^{3} \log \relax (x) + 3 \, {\left (3 \, B b c^{3} d e^{2} - {\left (B b^{2} c^{2} - A b c^{3}\right )} e^{3}\right )} x^{2} + 6 \, {\left (3 \, B b c^{3} d^{2} e - 3 \, {\left (B b^{2} c^{2} - A b c^{3}\right )} d e^{2} + {\left (B b^{3} c - A b^{2} c^{2}\right )} e^{3}\right )} x + 6 \, {\left ({\left (B b c^{3} - A c^{4}\right )} d^{3} - 3 \, {\left (B b^{2} c^{2} - A b c^{3}\right )} d^{2} e + 3 \, {\left (B b^{3} c - A b^{2} c^{2}\right )} d e^{2} - {\left (B b^{4} - A b^{3} c\right )} e^{3}\right )} \log \left (c x + b\right )}{6 \, b c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^3/(c*x^2+b*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 207, normalized size = 1.62 \[ \frac {A d^{3} \log \left ({\left | x \right |}\right )}{b} + \frac {2 \, B c^{2} x^{3} e^{3} + 9 \, B c^{2} d x^{2} e^{2} + 18 \, B c^{2} d^{2} x e - 3 \, B b c x^{2} e^{3} + 3 \, A c^{2} x^{2} e^{3} - 18 \, B b c d x e^{2} + 18 \, A c^{2} d x e^{2} + 6 \, B b^{2} x e^{3} - 6 \, A b c x e^{3}}{6 \, c^{3}} + \frac {{\left (B b c^{3} d^{3} - A c^{4} d^{3} - 3 \, B b^{2} c^{2} d^{2} e + 3 \, A b c^{3} d^{2} e + 3 \, B b^{3} c d e^{2} - 3 \, A b^{2} c^{2} d e^{2} - B b^{4} e^{3} + A b^{3} c e^{3}\right )} \log \left ({\left | c x + b \right |}\right )}{b c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^3/(c*x^2+b*x),x, algorithm="giac")

[Out]

A*d^3*log(abs(x))/b + 1/6*(2*B*c^2*x^3*e^3 + 9*B*c^2*d*x^2*e^2 + 18*B*c^2*d^2*x*e - 3*B*b*c*x^2*e^3 + 3*A*c^2*
x^2*e^3 - 18*B*b*c*d*x*e^2 + 18*A*c^2*d*x*e^2 + 6*B*b^2*x*e^3 - 6*A*b*c*x*e^3)/c^3 + (B*b*c^3*d^3 - A*c^4*d^3
- 3*B*b^2*c^2*d^2*e + 3*A*b*c^3*d^2*e + 3*B*b^3*c*d*e^2 - 3*A*b^2*c^2*d*e^2 - B*b^4*e^3 + A*b^3*c*e^3)*log(abs
(c*x + b))/(b*c^4)

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maple [B]  time = 0.06, size = 252, normalized size = 1.97 \[ \frac {B \,e^{3} x^{3}}{3 c}+\frac {A \,e^{3} x^{2}}{2 c}-\frac {B b \,e^{3} x^{2}}{2 c^{2}}+\frac {3 B d \,e^{2} x^{2}}{2 c}+\frac {A \,b^{2} e^{3} \ln \left (c x +b \right )}{c^{3}}-\frac {3 A b d \,e^{2} \ln \left (c x +b \right )}{c^{2}}-\frac {A b \,e^{3} x}{c^{2}}+\frac {A \,d^{3} \ln \relax (x )}{b}-\frac {A \,d^{3} \ln \left (c x +b \right )}{b}+\frac {3 A \,d^{2} e \ln \left (c x +b \right )}{c}+\frac {3 A d \,e^{2} x}{c}-\frac {B \,b^{3} e^{3} \ln \left (c x +b \right )}{c^{4}}+\frac {3 B \,b^{2} d \,e^{2} \ln \left (c x +b \right )}{c^{3}}+\frac {B \,b^{2} e^{3} x}{c^{3}}-\frac {3 B b \,d^{2} e \ln \left (c x +b \right )}{c^{2}}-\frac {3 B b d \,e^{2} x}{c^{2}}+\frac {B \,d^{3} \ln \left (c x +b \right )}{c}+\frac {3 B \,d^{2} e x}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^3/(c*x^2+b*x),x)

[Out]

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

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maxima [A]  time = 0.61, size = 200, normalized size = 1.56 \[ \frac {A d^{3} \log \relax (x)}{b} + \frac {2 \, B c^{2} e^{3} x^{3} + 3 \, {\left (3 \, B c^{2} d e^{2} - {\left (B b c - A c^{2}\right )} e^{3}\right )} x^{2} + 6 \, {\left (3 \, B c^{2} d^{2} e - 3 \, {\left (B b c - A c^{2}\right )} d e^{2} + {\left (B b^{2} - A b c\right )} e^{3}\right )} x}{6 \, c^{3}} + \frac {{\left ({\left (B b c^{3} - A c^{4}\right )} d^{3} - 3 \, {\left (B b^{2} c^{2} - A b c^{3}\right )} d^{2} e + 3 \, {\left (B b^{3} c - A b^{2} c^{2}\right )} d e^{2} - {\left (B b^{4} - A b^{3} c\right )} e^{3}\right )} \log \left (c x + b\right )}{b c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^3/(c*x^2+b*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.58, size = 208, normalized size = 1.62 \[ x^2\,\left (\frac {A\,e^3+3\,B\,d\,e^2}{2\,c}-\frac {B\,b\,e^3}{2\,c^2}\right )-x\,\left (\frac {b\,\left (\frac {A\,e^3+3\,B\,d\,e^2}{c}-\frac {B\,b\,e^3}{c^2}\right )}{c}-\frac {3\,d\,e\,\left (A\,e+B\,d\right )}{c}\right )-\ln \left (b+c\,x\right )\,\left (\frac {A\,d^3}{b}-\frac {c^3\,\left (B\,b\,d^3+3\,A\,b\,e\,d^2\right )-c^2\,\left (3\,B\,b^2\,d^2\,e+3\,A\,b^2\,d\,e^2\right )+c\,\left (A\,b^3\,e^3+3\,B\,d\,b^3\,e^2\right )-B\,b^4\,e^3}{b\,c^4}\right )+\frac {A\,d^3\,\ln \relax (x)}{b}+\frac {B\,e^3\,x^3}{3\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(d + e*x)^3)/(b*x + c*x^2),x)

[Out]

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

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sympy [B]  time = 4.59, size = 264, normalized size = 2.06 \[ \frac {A d^{3} \log {\relax (x )}}{b} + \frac {B e^{3} x^{3}}{3 c} + x^{2} \left (\frac {A e^{3}}{2 c} - \frac {B b e^{3}}{2 c^{2}} + \frac {3 B d e^{2}}{2 c}\right ) + x \left (- \frac {A b e^{3}}{c^{2}} + \frac {3 A d e^{2}}{c} + \frac {B b^{2} e^{3}}{c^{3}} - \frac {3 B b d e^{2}}{c^{2}} + \frac {3 B d^{2} e}{c}\right ) - \frac {\left (- A c + B b\right ) \left (b e - c d\right )^{3} \log {\left (x + \frac {A b c^{3} d^{3} + \frac {b \left (- A c + B b\right ) \left (b e - c d\right )^{3}}{c}}{- A b^{3} c e^{3} + 3 A b^{2} c^{2} d e^{2} - 3 A b c^{3} d^{2} e + 2 A c^{4} d^{3} + B b^{4} e^{3} - 3 B b^{3} c d e^{2} + 3 B b^{2} c^{2} d^{2} e - B b c^{3} d^{3}} \right )}}{b c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**3/(c*x**2+b*x),x)

[Out]

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

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