3.1147 \(\int \frac {(A+B x) (d+e x)^3}{(b x+c x^2)^2} \, dx\)

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

[Out]

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

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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} \[ \frac {(b B-A c) (c d-b e)^3}{b^2 c^3 (b+c x)}+\frac {(c d-b e)^2 \log (b+c x) \left (-b c (B d-A e)+2 A c^2 d-2 b^2 B e\right )}{b^3 c^3}+\frac {d^2 \log (x) (3 A b e-2 A c d+b B d)}{b^3}-\frac {A d^3}{b^2 x}+\frac {B e^3 x}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.07, size = 128, normalized size = 1.00 \[ \frac {d^2 \log (x) (3 A b e-2 A c d+b B d)}{b^3}-\frac {(b B-A c) (b e-c d)^3}{b^2 c^3 (b+c x)}-\frac {A d^3}{b^2 x}+\frac {(c d-b e)^2 \log (b+c x) \left (A b c e+2 A c^2 d-2 b^2 B e-b B c d\right )}{b^3 c^3}+\frac {B e^3 x}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.24, size = 362, normalized size = 2.83 \[ \frac {B b^{3} c^{2} e^{3} x^{3} + B b^{4} c e^{3} x^{2} - A b^{2} c^{3} d^{3} + {\left ({\left (B b^{2} c^{3} - 2 \, A b c^{4}\right )} d^{3} - 3 \, {\left (B b^{3} c^{2} - A b^{2} c^{3}\right )} d^{2} e + 3 \, {\left (B b^{4} c - A b^{3} c^{2}\right )} d e^{2} - {\left (B b^{5} - A b^{4} c\right )} e^{3}\right )} x - {\left ({\left (3 \, A b c^{4} d^{2} e - 3 \, B b^{3} c^{2} d e^{2} + {\left (B b c^{4} - 2 \, A c^{5}\right )} d^{3} + {\left (2 \, B b^{4} c - A b^{3} c^{2}\right )} e^{3}\right )} x^{2} + {\left (3 \, A b^{2} c^{3} d^{2} e - 3 \, B b^{4} c d e^{2} + {\left (B b^{2} c^{3} - 2 \, A b c^{4}\right )} d^{3} + {\left (2 \, B b^{5} - A b^{4} c\right )} e^{3}\right )} x\right )} \log \left (c x + b\right ) + {\left ({\left (3 \, A b c^{4} d^{2} e + {\left (B b c^{4} - 2 \, A c^{5}\right )} d^{3}\right )} x^{2} + {\left (3 \, A b^{2} c^{3} d^{2} e + {\left (B b^{2} c^{3} - 2 \, A b c^{4}\right )} d^{3}\right )} x\right )} \log \relax (x)}{b^{3} c^{4} x^{2} + b^{4} c^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 229, normalized size = 1.79 \[ \frac {B x e^{3}}{c^{2}} + \frac {{\left (B b d^{3} - 2 \, A c d^{3} + 3 \, A b d^{2} e\right )} \log \left ({\left | x \right |}\right )}{b^{3}} - \frac {{\left (B b c^{3} d^{3} - 2 \, A c^{4} d^{3} + 3 \, A b c^{3} d^{2} e - 3 \, B b^{3} c d e^{2} + 2 \, B b^{4} e^{3} - A b^{3} c e^{3}\right )} \log \left ({\left | c x + b \right |}\right )}{b^{3} c^{3}} - \frac {A b c^{2} d^{3} - \frac {{\left (B b c^{3} d^{3} - 2 \, A c^{4} d^{3} - 3 \, B b^{2} c^{2} d^{2} e + 3 \, A b c^{3} d^{2} e + 3 \, B b^{3} c d e^{2} - 3 \, A b^{2} c^{2} d e^{2} - B b^{4} e^{3} + A b^{3} c e^{3}\right )} x}{c}}{{\left (c x + b\right )} b^{2} c^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 286, normalized size = 2.23 \[ \frac {A b \,e^{3}}{\left (c x +b \right ) c^{2}}+\frac {3 A \,d^{2} e}{\left (c x +b \right ) b}-\frac {A c \,d^{3}}{\left (c x +b \right ) b^{2}}+\frac {3 A \,d^{2} e \ln \relax (x )}{b^{2}}-\frac {3 A \,d^{2} e \ln \left (c x +b \right )}{b^{2}}-\frac {2 A c \,d^{3} \ln \relax (x )}{b^{3}}+\frac {2 A c \,d^{3} \ln \left (c x +b \right )}{b^{3}}-\frac {3 A d \,e^{2}}{\left (c x +b \right ) c}+\frac {A \,e^{3} \ln \left (c x +b \right )}{c^{2}}-\frac {B \,b^{2} e^{3}}{\left (c x +b \right ) c^{3}}+\frac {3 B b d \,e^{2}}{\left (c x +b \right ) c^{2}}-\frac {2 B b \,e^{3} \ln \left (c x +b \right )}{c^{3}}+\frac {B \,d^{3}}{\left (c x +b \right ) b}+\frac {B \,d^{3} \ln \relax (x )}{b^{2}}-\frac {B \,d^{3} \ln \left (c x +b \right )}{b^{2}}-\frac {3 B \,d^{2} e}{\left (c x +b \right ) c}+\frac {3 B d \,e^{2} \ln \left (c x +b \right )}{c^{2}}+\frac {B \,e^{3} x}{c^{2}}-\frac {A \,d^{3}}{b^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.51, size = 225, normalized size = 1.76 \[ \frac {B e^{3} x}{c^{2}} - \frac {A b c^{3} d^{3} - {\left ({\left (B b c^{3} - 2 \, A c^{4}\right )} d^{3} - 3 \, {\left (B b^{2} c^{2} - A b c^{3}\right )} d^{2} e + 3 \, {\left (B b^{3} c - A b^{2} c^{2}\right )} d e^{2} - {\left (B b^{4} - A b^{3} c\right )} e^{3}\right )} x}{b^{2} c^{4} x^{2} + b^{3} c^{3} x} + \frac {{\left (3 \, A b d^{2} e + {\left (B b - 2 \, A c\right )} d^{3}\right )} \log \relax (x)}{b^{3}} - \frac {{\left (3 \, A b c^{3} d^{2} e - 3 \, B b^{3} c d e^{2} + {\left (B b c^{3} - 2 \, A c^{4}\right )} d^{3} + {\left (2 \, B b^{4} - A b^{3} c\right )} e^{3}\right )} \log \left (c x + b\right )}{b^{3} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.59, size = 212, normalized size = 1.66 \[ \frac {\ln \relax (x)\,\left (b\,\left (B\,d^3+3\,A\,e\,d^2\right )-2\,A\,c\,d^3\right )}{b^3}-\frac {\frac {x\,\left (B\,b^4\,e^3-3\,B\,b^3\,c\,d\,e^2-A\,b^3\,c\,e^3+3\,B\,b^2\,c^2\,d^2\,e+3\,A\,b^2\,c^2\,d\,e^2-B\,b\,c^3\,d^3-3\,A\,b\,c^3\,d^2\,e+2\,A\,c^4\,d^3\right )}{b^2\,c}+\frac {A\,c^2\,d^3}{b}}{c^3\,x^2+b\,c^2\,x}+\frac {B\,e^3\,x}{c^2}+\frac {\ln \left (b+c\,x\right )\,{\left (b\,e-c\,d\right )}^2\,\left (2\,A\,c^2\,d-2\,B\,b^2\,e+A\,b\,c\,e-B\,b\,c\,d\right )}{b^3\,c^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 9.38, size = 502, normalized size = 3.92 \[ \frac {B e^{3} x}{c^{2}} + \frac {- A b c^{3} d^{3} + x \left (A b^{3} c e^{3} - 3 A b^{2} c^{2} d e^{2} + 3 A b c^{3} d^{2} e - 2 A c^{4} d^{3} - B b^{4} e^{3} + 3 B b^{3} c d e^{2} - 3 B b^{2} c^{2} d^{2} e + B b c^{3} d^{3}\right )}{b^{3} c^{3} x + b^{2} c^{4} x^{2}} + \frac {d^{2} \left (3 A b e - 2 A c d + B b d\right ) \log {\left (x + \frac {3 A b^{2} c^{2} d^{2} e - 2 A b c^{3} d^{3} + B b^{2} c^{2} d^{3} - b c^{2} d^{2} \left (3 A b e - 2 A c d + B b d\right )}{- A b^{3} c e^{3} + 6 A b c^{3} d^{2} e - 4 A c^{4} d^{3} + 2 B b^{4} e^{3} - 3 B b^{3} c d e^{2} + 2 B b c^{3} d^{3}} \right )}}{b^{3}} - \frac {\left (b e - c d\right )^{2} \left (- A b c e - 2 A c^{2} d + 2 B b^{2} e + B b c d\right ) \log {\left (x + \frac {3 A b^{2} c^{2} d^{2} e - 2 A b c^{3} d^{3} + B b^{2} c^{2} d^{3} + \frac {b \left (b e - c d\right )^{2} \left (- A b c e - 2 A c^{2} d + 2 B b^{2} e + B b c d\right )}{c}}{- A b^{3} c e^{3} + 6 A b c^{3} d^{2} e - 4 A c^{4} d^{3} + 2 B b^{4} e^{3} - 3 B b^{3} c d e^{2} + 2 B b c^{3} d^{3}} \right )}}{b^{3} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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