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

Optimal. Leaf size=171 \[ -\frac {\log (d+e x) \left (B c^2 d^3-A e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )\right )}{d^3 (c d-b e)^3}+\frac {c^2 (b B-A c) \log (b+c x)}{b (c d-b e)^3}+\frac {B c d^2-A e (2 c d-b e)}{d^2 (d+e x) (c d-b e)^2}+\frac {B d-A e}{2 d (d+e x)^2 (c d-b e)}+\frac {A \log (x)}{b d^3} \]

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Rubi [A]  time = 0.23, antiderivative size = 171, 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} \begin {gather*} -\frac {\log (d+e x) \left (B c^2 d^3-A e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )\right )}{d^3 (c d-b e)^3}+\frac {c^2 (b B-A c) \log (b+c x)}{b (c d-b e)^3}+\frac {B c d^2-A e (2 c d-b e)}{d^2 (d+e x) (c d-b e)^2}+\frac {B d-A e}{2 d (d+e x)^2 (c d-b e)}+\frac {A \log (x)}{b d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.20, size = 169, normalized size = 0.99 \begin {gather*} -\frac {\log (d+e x) \left (B c^2 d^3-A e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )\right )}{d^3 (c d-b e)^3}+\frac {c^2 (A c-b B) \log (b+c x)}{b (b e-c d)^3}+\frac {A e (b e-2 c d)+B c d^2}{d^2 (d+e x) (c d-b e)^2}+\frac {B d-A e}{2 d (d+e x)^2 (c d-b e)}+\frac {A \log (x)}{b d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(B*d - A*e)/(2*d*(c*d - b*e)*(d + e*x)^2) + (B*c*d^2 + A*e*(-2*c*d + b*e))/(d^2*(c*d - b*e)^2*(d + e*x)) + (A*
Log[x])/(b*d^3) + (c^2*(-(b*B) + A*c)*Log[b + c*x])/(b*(-(c*d) + b*e)^3) - ((B*c^2*d^3 - A*e*(3*c^2*d^2 - 3*b*
c*d*e + b^2*e^2))*Log[d + e*x])/(d^3*(c*d - b*e)^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)^3 \left (b x+c x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 37.44, size = 644, normalized size = 3.77 \begin {gather*} \frac {3 \, B b c^{2} d^{5} - 3 \, A b^{3} d^{2} e^{3} - {\left (4 \, B b^{2} c + 5 \, A b c^{2}\right )} d^{4} e + {\left (B b^{3} + 8 \, A b^{2} c\right )} d^{3} e^{2} + 2 \, {\left (B b c^{2} d^{4} e + 3 \, A b^{2} c d^{2} e^{3} - A b^{3} d e^{4} - {\left (B b^{2} c + 2 \, A b c^{2}\right )} d^{3} e^{2}\right )} x + 2 \, {\left ({\left (B b c^{2} - A c^{3}\right )} d^{3} e^{2} x^{2} + 2 \, {\left (B b c^{2} - A c^{3}\right )} d^{4} e x + {\left (B b c^{2} - A c^{3}\right )} d^{5}\right )} \log \left (c x + b\right ) - 2 \, {\left (B b c^{2} d^{5} - 3 \, A b c^{2} d^{4} e + 3 \, A b^{2} c d^{3} e^{2} - A b^{3} d^{2} e^{3} + {\left (B b c^{2} d^{3} e^{2} - 3 \, A b c^{2} d^{2} e^{3} + 3 \, A b^{2} c d e^{4} - A b^{3} e^{5}\right )} x^{2} + 2 \, {\left (B b c^{2} d^{4} e - 3 \, A b c^{2} d^{3} e^{2} + 3 \, A b^{2} c d^{2} e^{3} - A b^{3} d e^{4}\right )} x\right )} \log \left (e x + d\right ) + 2 \, {\left (A c^{3} d^{5} - 3 \, A b c^{2} d^{4} e + 3 \, A b^{2} c d^{3} e^{2} - A b^{3} d^{2} e^{3} + {\left (A c^{3} d^{3} e^{2} - 3 \, A b c^{2} d^{2} e^{3} + 3 \, A b^{2} c d e^{4} - A b^{3} e^{5}\right )} x^{2} + 2 \, {\left (A c^{3} d^{4} e - 3 \, A b c^{2} d^{3} e^{2} + 3 \, A b^{2} c d^{2} e^{3} - A b^{3} d e^{4}\right )} x\right )} \log \relax (x)}{2 \, {\left (b c^{3} d^{8} - 3 \, b^{2} c^{2} d^{7} e + 3 \, b^{3} c d^{6} e^{2} - b^{4} d^{5} e^{3} + {\left (b c^{3} d^{6} e^{2} - 3 \, b^{2} c^{2} d^{5} e^{3} + 3 \, b^{3} c d^{4} e^{4} - b^{4} d^{3} e^{5}\right )} x^{2} + 2 \, {\left (b c^{3} d^{7} e - 3 \, b^{2} c^{2} d^{6} e^{2} + 3 \, b^{3} c d^{5} e^{3} - b^{4} d^{4} e^{4}\right )} x\right )}} \end {gather*}

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]

1/2*(3*B*b*c^2*d^5 - 3*A*b^3*d^2*e^3 - (4*B*b^2*c + 5*A*b*c^2)*d^4*e + (B*b^3 + 8*A*b^2*c)*d^3*e^2 + 2*(B*b*c^
2*d^4*e + 3*A*b^2*c*d^2*e^3 - A*b^3*d*e^4 - (B*b^2*c + 2*A*b*c^2)*d^3*e^2)*x + 2*((B*b*c^2 - A*c^3)*d^3*e^2*x^
2 + 2*(B*b*c^2 - A*c^3)*d^4*e*x + (B*b*c^2 - A*c^3)*d^5)*log(c*x + b) - 2*(B*b*c^2*d^5 - 3*A*b*c^2*d^4*e + 3*A
*b^2*c*d^3*e^2 - A*b^3*d^2*e^3 + (B*b*c^2*d^3*e^2 - 3*A*b*c^2*d^2*e^3 + 3*A*b^2*c*d*e^4 - A*b^3*e^5)*x^2 + 2*(
B*b*c^2*d^4*e - 3*A*b*c^2*d^3*e^2 + 3*A*b^2*c*d^2*e^3 - A*b^3*d*e^4)*x)*log(e*x + d) + 2*(A*c^3*d^5 - 3*A*b*c^
2*d^4*e + 3*A*b^2*c*d^3*e^2 - A*b^3*d^2*e^3 + (A*c^3*d^3*e^2 - 3*A*b*c^2*d^2*e^3 + 3*A*b^2*c*d*e^4 - A*b^3*e^5
)*x^2 + 2*(A*c^3*d^4*e - 3*A*b*c^2*d^3*e^2 + 3*A*b^2*c*d^2*e^3 - A*b^3*d*e^4)*x)*log(x))/(b*c^3*d^8 - 3*b^2*c^
2*d^7*e + 3*b^3*c*d^6*e^2 - b^4*d^5*e^3 + (b*c^3*d^6*e^2 - 3*b^2*c^2*d^5*e^3 + 3*b^3*c*d^4*e^4 - b^4*d^3*e^5)*
x^2 + 2*(b*c^3*d^7*e - 3*b^2*c^2*d^6*e^2 + 3*b^3*c*d^5*e^3 - b^4*d^4*e^4)*x)

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giac [A]  time = 0.16, size = 308, normalized size = 1.80 \begin {gather*} \frac {{\left (B b c^{3} - A c^{4}\right )} \log \left ({\left | c x + b \right |}\right )}{b c^{4} d^{3} - 3 \, b^{2} c^{3} d^{2} e + 3 \, b^{3} c^{2} d e^{2} - b^{4} c e^{3}} - \frac {{\left (B c^{2} d^{3} e - 3 \, A c^{2} d^{2} e^{2} + 3 \, A b c d e^{3} - A b^{2} e^{4}\right )} \log \left ({\left | x e + d \right |}\right )}{c^{3} d^{6} e - 3 \, b c^{2} d^{5} e^{2} + 3 \, b^{2} c d^{4} e^{3} - b^{3} d^{3} e^{4}} + \frac {A \log \left ({\left | x \right |}\right )}{b d^{3}} + \frac {3 \, B c^{2} d^{5} - 4 \, B b c d^{4} e - 5 \, A c^{2} d^{4} e + B b^{2} d^{3} e^{2} + 8 \, A b c d^{3} e^{2} - 3 \, A b^{2} d^{2} e^{3} + 2 \, {\left (B c^{2} d^{4} e - B b c d^{3} e^{2} - 2 \, A c^{2} d^{3} e^{2} + 3 \, A b c d^{2} e^{3} - A b^{2} d e^{4}\right )} x}{2 \, {\left (c d - b e\right )}^{3} {\left (x e + d\right )}^{2} d^{3}} \end {gather*}

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]

(B*b*c^3 - A*c^4)*log(abs(c*x + b))/(b*c^4*d^3 - 3*b^2*c^3*d^2*e + 3*b^3*c^2*d*e^2 - b^4*c*e^3) - (B*c^2*d^3*e
- 3*A*c^2*d^2*e^2 + 3*A*b*c*d*e^3 - A*b^2*e^4)*log(abs(x*e + d))/(c^3*d^6*e - 3*b*c^2*d^5*e^2 + 3*b^2*c*d^4*e
^3 - b^3*d^3*e^4) + A*log(abs(x))/(b*d^3) + 1/2*(3*B*c^2*d^5 - 4*B*b*c*d^4*e - 5*A*c^2*d^4*e + B*b^2*d^3*e^2 +
8*A*b*c*d^3*e^2 - 3*A*b^2*d^2*e^3 + 2*(B*c^2*d^4*e - B*b*c*d^3*e^2 - 2*A*c^2*d^3*e^2 + 3*A*b*c*d^2*e^3 - A*b^
2*d*e^4)*x)/((c*d - b*e)^3*(x*e + d)^2*d^3)

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maple [A]  time = 0.05, size = 275, normalized size = 1.61 \begin {gather*} -\frac {A \,b^{2} e^{3} \ln \left (e x +d \right )}{\left (b e -c d \right )^{3} d^{3}}+\frac {3 A b c \,e^{2} \ln \left (e x +d \right )}{\left (b e -c d \right )^{3} d^{2}}+\frac {A \,c^{3} \ln \left (c x +b \right )}{\left (b e -c d \right )^{3} b}-\frac {3 A \,c^{2} e \ln \left (e x +d \right )}{\left (b e -c d \right )^{3} d}-\frac {B \,c^{2} \ln \left (c x +b \right )}{\left (b e -c d \right )^{3}}+\frac {B \,c^{2} \ln \left (e x +d \right )}{\left (b e -c d \right )^{3}}+\frac {A b \,e^{2}}{\left (b e -c d \right )^{2} \left (e x +d \right ) d^{2}}-\frac {2 A c e}{\left (b e -c d \right )^{2} \left (e x +d \right ) d}+\frac {B c}{\left (b e -c d \right )^{2} \left (e x +d \right )}+\frac {A e}{2 \left (b e -c d \right ) \left (e x +d \right )^{2} d}-\frac {B}{2 \left (b e -c d \right ) \left (e x +d \right )^{2}}+\frac {A \ln \relax (x )}{b \,d^{3}} \end {gather*}

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]

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

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maxima [A]  time = 0.71, size = 312, normalized size = 1.82 \begin {gather*} \frac {{\left (B b c^{2} - A c^{3}\right )} \log \left (c x + b\right )}{b c^{3} d^{3} - 3 \, b^{2} c^{2} d^{2} e + 3 \, b^{3} c d e^{2} - b^{4} e^{3}} - \frac {{\left (B c^{2} d^{3} - 3 \, A c^{2} d^{2} e + 3 \, A b c d e^{2} - A b^{2} e^{3}\right )} \log \left (e x + d\right )}{c^{3} d^{6} - 3 \, b c^{2} d^{5} e + 3 \, b^{2} c d^{4} e^{2} - b^{3} d^{3} e^{3}} + \frac {3 \, B c d^{3} + 3 \, A b d e^{2} - {\left (B b + 5 \, A c\right )} d^{2} e + 2 \, {\left (B c d^{2} e - 2 \, A c d e^{2} + A b e^{3}\right )} x}{2 \, {\left (c^{2} d^{6} - 2 \, b c d^{5} e + b^{2} d^{4} e^{2} + {\left (c^{2} d^{4} e^{2} - 2 \, b c d^{3} e^{3} + b^{2} d^{2} e^{4}\right )} x^{2} + 2 \, {\left (c^{2} d^{5} e - 2 \, b c d^{4} e^{2} + b^{2} d^{3} e^{3}\right )} x\right )}} + \frac {A \log \relax (x)}{b d^{3}} \end {gather*}

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]

(B*b*c^2 - A*c^3)*log(c*x + b)/(b*c^3*d^3 - 3*b^2*c^2*d^2*e + 3*b^3*c*d*e^2 - b^4*e^3) - (B*c^2*d^3 - 3*A*c^2*
d^2*e + 3*A*b*c*d*e^2 - A*b^2*e^3)*log(e*x + d)/(c^3*d^6 - 3*b*c^2*d^5*e + 3*b^2*c*d^4*e^2 - b^3*d^3*e^3) + 1/
2*(3*B*c*d^3 + 3*A*b*d*e^2 - (B*b + 5*A*c)*d^2*e + 2*(B*c*d^2*e - 2*A*c*d*e^2 + A*b*e^3)*x)/(c^2*d^6 - 2*b*c*d
^5*e + b^2*d^4*e^2 + (c^2*d^4*e^2 - 2*b*c*d^3*e^3 + b^2*d^2*e^4)*x^2 + 2*(c^2*d^5*e - 2*b*c*d^4*e^2 + b^2*d^3*
e^3)*x) + A*log(x)/(b*d^3)

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mupad [B]  time = 1.96, size = 284, normalized size = 1.66 \begin {gather*} \frac {\frac {3\,A\,b\,e^2+3\,B\,c\,d^2-5\,A\,c\,d\,e-B\,b\,d\,e}{2\,d\,\left (b^2\,e^2-2\,b\,c\,d\,e+c^2\,d^2\right )}+\frac {x\,\left (B\,c\,d^2\,e-2\,A\,c\,d\,e^2+A\,b\,e^3\right )}{d^2\,\left (b^2\,e^2-2\,b\,c\,d\,e+c^2\,d^2\right )}}{d^2+2\,d\,e\,x+e^2\,x^2}-\frac {\ln \left (d+e\,x\right )\,\left (c^2\,\left (B\,d^3-3\,A\,d^2\,e\right )-A\,b^2\,e^3+3\,A\,b\,c\,d\,e^2\right )}{-b^3\,d^3\,e^3+3\,b^2\,c\,d^4\,e^2-3\,b\,c^2\,d^5\,e+c^3\,d^6}+\frac {\ln \left (b+c\,x\right )\,\left (A\,c^3-B\,b\,c^2\right )}{b^4\,e^3-3\,b^3\,c\,d\,e^2+3\,b^2\,c^2\,d^2\,e-b\,c^3\,d^3}+\frac {A\,\ln \relax (x)}{b\,d^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((3*A*b*e^2 + 3*B*c*d^2 - 5*A*c*d*e - B*b*d*e)/(2*d*(b^2*e^2 + c^2*d^2 - 2*b*c*d*e)) + (x*(A*b*e^3 - 2*A*c*d*e
^2 + B*c*d^2*e))/(d^2*(b^2*e^2 + c^2*d^2 - 2*b*c*d*e)))/(d^2 + e^2*x^2 + 2*d*e*x) - (log(d + e*x)*(c^2*(B*d^3
- 3*A*d^2*e) - A*b^2*e^3 + 3*A*b*c*d*e^2))/(c^3*d^6 - b^3*d^3*e^3 + 3*b^2*c*d^4*e^2 - 3*b*c^2*d^5*e) + (log(b
+ c*x)*(A*c^3 - B*b*c^2))/(b^4*e^3 - b*c^3*d^3 + 3*b^2*c^2*d^2*e - 3*b^3*c*d*e^2) + (A*log(x))/(b*d^3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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]

Timed out

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