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

Optimal. Leaf size=140 \[ -\frac {2 c^2 \log (x) (5 c d-3 b e)}{b^6}+\frac {2 c^2 (5 c d-3 b e) \log (b+c x)}{b^6}-\frac {c^2 (4 c d-3 b e)}{b^5 (b+c x)}-\frac {3 c (2 c d-b e)}{b^5 x}-\frac {c^2 (c d-b e)}{2 b^4 (b+c x)^2}+\frac {3 c d-b e}{2 b^4 x^2}-\frac {d}{3 b^3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 140, 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 = {765} \[ -\frac {c^2 (4 c d-3 b e)}{b^5 (b+c x)}-\frac {c^2 (c d-b e)}{2 b^4 (b+c x)^2}-\frac {2 c^2 \log (x) (5 c d-3 b e)}{b^6}+\frac {2 c^2 (5 c d-3 b e) \log (b+c x)}{b^6}+\frac {3 c d-b e}{2 b^4 x^2}-\frac {3 c (2 c d-b e)}{b^5 x}-\frac {d}{3 b^3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)/(x*(b*x + c*x^2)^3),x]

[Out]

-d/(3*b^3*x^3) + (3*c*d - b*e)/(2*b^4*x^2) - (3*c*(2*c*d - b*e))/(b^5*x) - (c^2*(c*d - b*e))/(2*b^4*(b + c*x)^
2) - (c^2*(4*c*d - 3*b*e))/(b^5*(b + c*x)) - (2*c^2*(5*c*d - 3*b*e)*Log[x])/b^6 + (2*c^2*(5*c*d - 3*b*e)*Log[b
+ c*x])/b^6

Rule 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin {align*} \int \frac {d+e x}{x \left (b x+c x^2\right )^3} \, dx &=\int \left (\frac {d}{b^3 x^4}+\frac {-3 c d+b e}{b^4 x^3}-\frac {3 c (-2 c d+b e)}{b^5 x^2}+\frac {2 c^2 (-5 c d+3 b e)}{b^6 x}-\frac {c^3 (-c d+b e)}{b^4 (b+c x)^3}-\frac {c^3 (-4 c d+3 b e)}{b^5 (b+c x)^2}-\frac {2 c^3 (-5 c d+3 b e)}{b^6 (b+c x)}\right ) \, dx\\ &=-\frac {d}{3 b^3 x^3}+\frac {3 c d-b e}{2 b^4 x^2}-\frac {3 c (2 c d-b e)}{b^5 x}-\frac {c^2 (c d-b e)}{2 b^4 (b+c x)^2}-\frac {c^2 (4 c d-3 b e)}{b^5 (b+c x)}-\frac {2 c^2 (5 c d-3 b e) \log (x)}{b^6}+\frac {2 c^2 (5 c d-3 b e) \log (b+c x)}{b^6}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 129, normalized size = 0.92 \[ \frac {\frac {b \left (-\left (b^4 (2 d+3 e x)\right )+b^3 c x (5 d+12 e x)+2 b^2 c^2 x^2 (27 e x-10 d)+18 b c^3 x^3 (2 e x-5 d)-60 c^4 d x^4\right )}{x^3 (b+c x)^2}+12 c^2 \log (x) (3 b e-5 c d)+12 c^2 (5 c d-3 b e) \log (b+c x)}{6 b^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*(-60*c^4*d*x^4 + 18*b*c^3*x^3*(-5*d + 2*e*x) - b^4*(2*d + 3*e*x) + b^3*c*x*(5*d + 12*e*x) + 2*b^2*c^2*x^2*
(-10*d + 27*e*x)))/(x^3*(b + c*x)^2) + 12*c^2*(-5*c*d + 3*b*e)*Log[x] + 12*c^2*(5*c*d - 3*b*e)*Log[b + c*x])/(
6*b^6)

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fricas [A]  time = 0.84, size = 263, normalized size = 1.88 \[ -\frac {2 \, b^{5} d + 12 \, {\left (5 \, b c^{4} d - 3 \, b^{2} c^{3} e\right )} x^{4} + 18 \, {\left (5 \, b^{2} c^{3} d - 3 \, b^{3} c^{2} e\right )} x^{3} + 4 \, {\left (5 \, b^{3} c^{2} d - 3 \, b^{4} c e\right )} x^{2} - {\left (5 \, b^{4} c d - 3 \, b^{5} e\right )} x - 12 \, {\left ({\left (5 \, c^{5} d - 3 \, b c^{4} e\right )} x^{5} + 2 \, {\left (5 \, b c^{4} d - 3 \, b^{2} c^{3} e\right )} x^{4} + {\left (5 \, b^{2} c^{3} d - 3 \, b^{3} c^{2} e\right )} x^{3}\right )} \log \left (c x + b\right ) + 12 \, {\left ({\left (5 \, c^{5} d - 3 \, b c^{4} e\right )} x^{5} + 2 \, {\left (5 \, b c^{4} d - 3 \, b^{2} c^{3} e\right )} x^{4} + {\left (5 \, b^{2} c^{3} d - 3 \, b^{3} c^{2} e\right )} x^{3}\right )} \log \relax (x)}{6 \, {\left (b^{6} c^{2} x^{5} + 2 \, b^{7} c x^{4} + b^{8} x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*b^5*d + 12*(5*b*c^4*d - 3*b^2*c^3*e)*x^4 + 18*(5*b^2*c^3*d - 3*b^3*c^2*e)*x^3 + 4*(5*b^3*c^2*d - 3*b^4
*c*e)*x^2 - (5*b^4*c*d - 3*b^5*e)*x - 12*((5*c^5*d - 3*b*c^4*e)*x^5 + 2*(5*b*c^4*d - 3*b^2*c^3*e)*x^4 + (5*b^2
*c^3*d - 3*b^3*c^2*e)*x^3)*log(c*x + b) + 12*((5*c^5*d - 3*b*c^4*e)*x^5 + 2*(5*b*c^4*d - 3*b^2*c^3*e)*x^4 + (5
*b^2*c^3*d - 3*b^3*c^2*e)*x^3)*log(x))/(b^6*c^2*x^5 + 2*b^7*c*x^4 + b^8*x^3)

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giac [A]  time = 0.17, size = 165, normalized size = 1.18 \[ -\frac {2 \, {\left (5 \, c^{3} d - 3 \, b c^{2} e\right )} \log \left ({\left | x \right |}\right )}{b^{6}} + \frac {2 \, {\left (5 \, c^{4} d - 3 \, b c^{3} e\right )} \log \left ({\left | c x + b \right |}\right )}{b^{6} c} - \frac {2 \, b^{5} d + 12 \, {\left (5 \, b c^{4} d - 3 \, b^{2} c^{3} e\right )} x^{4} + 18 \, {\left (5 \, b^{2} c^{3} d - 3 \, b^{3} c^{2} e\right )} x^{3} + 4 \, {\left (5 \, b^{3} c^{2} d - 3 \, b^{4} c e\right )} x^{2} - {\left (5 \, b^{4} c d - 3 \, b^{5} e\right )} x}{6 \, {\left (c x + b\right )}^{2} b^{6} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(5*c^3*d - 3*b*c^2*e)*log(abs(x))/b^6 + 2*(5*c^4*d - 3*b*c^3*e)*log(abs(c*x + b))/(b^6*c) - 1/6*(2*b^5*d +
12*(5*b*c^4*d - 3*b^2*c^3*e)*x^4 + 18*(5*b^2*c^3*d - 3*b^3*c^2*e)*x^3 + 4*(5*b^3*c^2*d - 3*b^4*c*e)*x^2 - (5*b
^4*c*d - 3*b^5*e)*x)/((c*x + b)^2*b^6*x^3)

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maple [A]  time = 0.06, size = 168, normalized size = 1.20 \[ \frac {c^{2} e}{2 \left (c x +b \right )^{2} b^{3}}-\frac {c^{3} d}{2 \left (c x +b \right )^{2} b^{4}}+\frac {3 c^{2} e}{\left (c x +b \right ) b^{4}}-\frac {4 c^{3} d}{\left (c x +b \right ) b^{5}}+\frac {6 c^{2} e \ln \relax (x )}{b^{5}}-\frac {6 c^{2} e \ln \left (c x +b \right )}{b^{5}}-\frac {10 c^{3} d \ln \relax (x )}{b^{6}}+\frac {10 c^{3} d \ln \left (c x +b \right )}{b^{6}}+\frac {3 c e}{b^{4} x}-\frac {6 c^{2} d}{b^{5} x}-\frac {e}{2 b^{3} x^{2}}+\frac {3 c d}{2 b^{4} x^{2}}-\frac {d}{3 b^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*c^2/b^5*ln(c*x+b)*e+10*c^3/b^6*ln(c*x+b)*d+3*c^2/b^4/(c*x+b)*e-4*c^3/b^5/(c*x+b)*d+1/2*c^2/b^3/(c*x+b)^2*e-
1/2*c^3/b^4/(c*x+b)^2*d-1/3*d/b^3/x^3-1/2/b^3/x^2*e+3/2/b^4/x^2*c*d+3*c/b^4/x*e-6*c^2/b^5/x*d+6*c^2/b^5*ln(x)*
e-10*c^3/b^6*ln(x)*d

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maxima [A]  time = 0.99, size = 165, normalized size = 1.18 \[ -\frac {2 \, b^{4} d + 12 \, {\left (5 \, c^{4} d - 3 \, b c^{3} e\right )} x^{4} + 18 \, {\left (5 \, b c^{3} d - 3 \, b^{2} c^{2} e\right )} x^{3} + 4 \, {\left (5 \, b^{2} c^{2} d - 3 \, b^{3} c e\right )} x^{2} - {\left (5 \, b^{3} c d - 3 \, b^{4} e\right )} x}{6 \, {\left (b^{5} c^{2} x^{5} + 2 \, b^{6} c x^{4} + b^{7} x^{3}\right )}} + \frac {2 \, {\left (5 \, c^{3} d - 3 \, b c^{2} e\right )} \log \left (c x + b\right )}{b^{6}} - \frac {2 \, {\left (5 \, c^{3} d - 3 \, b c^{2} e\right )} \log \relax (x)}{b^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*b^4*d + 12*(5*c^4*d - 3*b*c^3*e)*x^4 + 18*(5*b*c^3*d - 3*b^2*c^2*e)*x^3 + 4*(5*b^2*c^2*d - 3*b^3*c*e)*
x^2 - (5*b^3*c*d - 3*b^4*e)*x)/(b^5*c^2*x^5 + 2*b^6*c*x^4 + b^7*x^3) + 2*(5*c^3*d - 3*b*c^2*e)*log(c*x + b)/b^
6 - 2*(5*c^3*d - 3*b*c^2*e)*log(x)/b^6

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mupad [B]  time = 1.09, size = 163, normalized size = 1.16 \[ \frac {\frac {2\,c\,x^2\,\left (3\,b\,e-5\,c\,d\right )}{3\,b^3}-\frac {x\,\left (3\,b\,e-5\,c\,d\right )}{6\,b^2}-\frac {d}{3\,b}+\frac {3\,c^2\,x^3\,\left (3\,b\,e-5\,c\,d\right )}{b^4}+\frac {2\,c^3\,x^4\,\left (3\,b\,e-5\,c\,d\right )}{b^5}}{b^2\,x^3+2\,b\,c\,x^4+c^2\,x^5}+\frac {4\,c^2\,\mathrm {atanh}\left (\frac {2\,c^2\,\left (3\,b\,e-5\,c\,d\right )\,\left (b+2\,c\,x\right )}{b\,\left (10\,c^3\,d-6\,b\,c^2\,e\right )}\right )\,\left (3\,b\,e-5\,c\,d\right )}{b^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)/(x*(b*x + c*x^2)^3),x)

[Out]

((2*c*x^2*(3*b*e - 5*c*d))/(3*b^3) - (x*(3*b*e - 5*c*d))/(6*b^2) - d/(3*b) + (3*c^2*x^3*(3*b*e - 5*c*d))/b^4 +
(2*c^3*x^4*(3*b*e - 5*c*d))/b^5)/(b^2*x^3 + c^2*x^5 + 2*b*c*x^4) + (4*c^2*atanh((2*c^2*(3*b*e - 5*c*d)*(b + 2
*c*x))/(b*(10*c^3*d - 6*b*c^2*e)))*(3*b*e - 5*c*d))/b^6

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sympy [A]  time = 0.80, size = 262, normalized size = 1.87 \[ \frac {- 2 b^{4} d + x^{4} \left (36 b c^{3} e - 60 c^{4} d\right ) + x^{3} \left (54 b^{2} c^{2} e - 90 b c^{3} d\right ) + x^{2} \left (12 b^{3} c e - 20 b^{2} c^{2} d\right ) + x \left (- 3 b^{4} e + 5 b^{3} c d\right )}{6 b^{7} x^{3} + 12 b^{6} c x^{4} + 6 b^{5} c^{2} x^{5}} + \frac {2 c^{2} \left (3 b e - 5 c d\right ) \log {\left (x + \frac {6 b^{2} c^{2} e - 10 b c^{3} d - 2 b c^{2} \left (3 b e - 5 c d\right )}{12 b c^{3} e - 20 c^{4} d} \right )}}{b^{6}} - \frac {2 c^{2} \left (3 b e - 5 c d\right ) \log {\left (x + \frac {6 b^{2} c^{2} e - 10 b c^{3} d + 2 b c^{2} \left (3 b e - 5 c d\right )}{12 b c^{3} e - 20 c^{4} d} \right )}}{b^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*b**4*d + x**4*(36*b*c**3*e - 60*c**4*d) + x**3*(54*b**2*c**2*e - 90*b*c**3*d) + x**2*(12*b**3*c*e - 20*b**
2*c**2*d) + x*(-3*b**4*e + 5*b**3*c*d))/(6*b**7*x**3 + 12*b**6*c*x**4 + 6*b**5*c**2*x**5) + 2*c**2*(3*b*e - 5*
c*d)*log(x + (6*b**2*c**2*e - 10*b*c**3*d - 2*b*c**2*(3*b*e - 5*c*d))/(12*b*c**3*e - 20*c**4*d))/b**6 - 2*c**2
*(3*b*e - 5*c*d)*log(x + (6*b**2*c**2*e - 10*b*c**3*d + 2*b*c**2*(3*b*e - 5*c*d))/(12*b*c**3*e - 20*c**4*d))/b
**6

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