3.3.66 \(\int \frac {1}{(d+e x)^3 (b x+c x^2)} \, dx\) [266]

Optimal. Leaf size=134 \[ -\frac {e}{2 d (c d-b e) (d+e x)^2}-\frac {e (2 c d-b e)}{d^2 (c d-b e)^2 (d+e x)}+\frac {\log (x)}{b d^3}-\frac {c^3 \log (b+c x)}{b (c d-b e)^3}+\frac {e \left (3 c^2 d^2-3 b c d e+b^2 e^2\right ) \log (d+e x)}{d^3 (c d-b e)^3} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {712} \begin {gather*} \frac {e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right ) \log (d+e x)}{d^3 (c d-b e)^3}-\frac {c^3 \log (b+c x)}{b (c d-b e)^3}-\frac {e (2 c d-b e)}{d^2 (d+e x) (c d-b e)^2}-\frac {e}{2 d (d+e x)^2 (c d-b e)}+\frac {\log (x)}{b d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^3*(b*x + c*x^2)),x]

[Out]

-1/2*e/(d*(c*d - b*e)*(d + e*x)^2) - (e*(2*c*d - b*e))/(d^2*(c*d - b*e)^2*(d + e*x)) + Log[x]/(b*d^3) - (c^3*L
og[b + c*x])/(b*(c*d - b*e)^3) + (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 712

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

Rubi steps

\begin {align*} \int \frac {1}{(d+e x)^3 \left (b x+c x^2\right )} \, dx &=\int \left (\frac {1}{b d^3 x}+\frac {c^4}{b (-c d+b e)^3 (b+c x)}+\frac {e^2}{d (c d-b e) (d+e x)^3}+\frac {e^2 (2 c d-b e)}{d^2 (c d-b e)^2 (d+e x)^2}+\frac {e^2 \left (3 c^2 d^2-3 b c d e+b^2 e^2\right )}{d^3 (c d-b e)^3 (d+e x)}\right ) \, dx\\ &=-\frac {e}{2 d (c d-b e) (d+e x)^2}-\frac {e (2 c d-b e)}{d^2 (c d-b e)^2 (d+e x)}+\frac {\log (x)}{b d^3}-\frac {c^3 \log (b+c x)}{b (c d-b e)^3}+\frac {e \left (3 c^2 d^2-3 b c d e+b^2 e^2\right ) \log (d+e x)}{d^3 (c d-b e)^3}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 116, normalized size = 0.87 \begin {gather*} \frac {\log (x)}{b d^3}+\frac {\frac {2 c^3 \log (b+c x)}{b}+\frac {e \left (\frac {d (c d-b e) (-b e (3 d+2 e x)+c d (5 d+4 e x))}{(d+e x)^2}-2 \left (3 c^2 d^2-3 b c d e+b^2 e^2\right ) \log (d+e x)\right )}{d^3}}{2 (-c d+b e)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^3*(b*x + c*x^2)),x]

[Out]

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

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Maple [A]
time = 0.52, size = 131, normalized size = 0.98

method	result	size
default	\(\frac {c^{3} \ln \left (c x +b \right )}{\left (b e -c d \right )^{3} b}+\frac {\ln \left (x \right )}{b \,d^{3}}+\frac {e}{2 d \left (b e -c d \right ) \left (e x +d \right )^{2}}+\frac {e \left (b e -2 c d \right )}{d^{2} \left (b e -c d \right )^{2} \left (e x +d \right )}-\frac {e \left (b^{2} e^{2}-3 b c d e +3 d^{2} c^{2}\right ) \ln \left (e x +d \right )}{d^{3} \left (b e -c d \right )^{3}}\)	\(131\)
norman	\(\frac {\frac {\left (-2 b \,e^{2}+3 d e c \right ) e x}{d^{2} \left (b^{2} e^{2}-2 b c d e +d^{2} c^{2}\right )}+\frac {\left (-3 b \,e^{2}+5 d e c \right ) e^{2} x^{2}}{2 d^{3} \left (b^{2} e^{2}-2 b c d e +d^{2} c^{2}\right )}}{\left (e x +d \right )^{2}}+\frac {\ln \left (x \right )}{b \,d^{3}}+\frac {c^{3} \ln \left (c x +b \right )}{b \left (b^{3} e^{3}-3 b^{2} d \,e^{2} c +3 b \,c^{2} d^{2} e -c^{3} d^{3}\right )}-\frac {e \left (b^{2} e^{2}-3 b c d e +3 d^{2} c^{2}\right ) \ln \left (e x +d \right )}{d^{3} \left (b^{3} e^{3}-3 b^{2} d \,e^{2} c +3 b \,c^{2} d^{2} e -c^{3} d^{3}\right )}\)	\(230\)
risch	\(\frac {\frac {e^{2} \left (b e -2 c d \right ) x}{d^{2} \left (b^{2} e^{2}-2 b c d e +d^{2} c^{2}\right )}+\frac {e \left (3 b e -5 c d \right )}{2 d \left (b^{2} e^{2}-2 b c d e +d^{2} c^{2}\right )}}{\left (e x +d \right )^{2}}+\frac {c^{3} \ln \left (c x +b \right )}{b \left (b^{3} e^{3}-3 b^{2} d \,e^{2} c +3 b \,c^{2} d^{2} e -c^{3} d^{3}\right )}-\frac {e^{3} \ln \left (-e x -d \right ) b^{2}}{d^{3} \left (b^{3} e^{3}-3 b^{2} d \,e^{2} c +3 b \,c^{2} d^{2} e -c^{3} d^{3}\right )}+\frac {3 e^{2} \ln \left (-e x -d \right ) b c}{d^{2} \left (b^{3} e^{3}-3 b^{2} d \,e^{2} c +3 b \,c^{2} d^{2} e -c^{3} d^{3}\right )}-\frac {3 e \ln \left (-e x -d \right ) c^{2}}{d \left (b^{3} e^{3}-3 b^{2} d \,e^{2} c +3 b \,c^{2} d^{2} e -c^{3} d^{3}\right )}+\frac {\ln \left (-x \right )}{d^{3} b}\)	\(321\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^3/(c*x^2+b*x),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 258, normalized size = 1.93 \begin {gather*} -\frac {c^{3} \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 (3 \, c^{2} d^{2} e - 3 \, b c d e^{2} + b^{2} e^{3}\right )} \log \left (x e + 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 {5 \, c d^{2} e - 3 \, b d e^{2} + 2 \, {\left (2 \, c d e^{2} - 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 {\log \left (x\right )}{b d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/(c*x^2+b*x),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 495 vs. \(2 (143) = 286\).
time = 12.76, size = 495, normalized size = 3.69 \begin {gather*} -\frac {5 \, b c^{2} d^{4} e + 2 \, b^{3} d x e^{4} - 3 \, {\left (2 \, b^{2} c d^{2} x - b^{3} d^{2}\right )} e^{3} + 4 \, {\left (b c^{2} d^{3} x - 2 \, b^{2} c d^{3}\right )} e^{2} + 2 \, {\left (c^{3} d^{3} x^{2} e^{2} + 2 \, c^{3} d^{4} x e + c^{3} d^{5}\right )} \log \left (c x + b\right ) - 2 \, {\left (3 \, b c^{2} d^{4} e + b^{3} x^{2} e^{5} - {\left (3 \, b^{2} c d x^{2} - 2 \, b^{3} d x\right )} e^{4} + {\left (3 \, b c^{2} d^{2} x^{2} - 6 \, b^{2} c d^{2} x + b^{3} d^{2}\right )} e^{3} + 3 \, {\left (2 \, b c^{2} d^{3} x - b^{2} c d^{3}\right )} e^{2}\right )} \log \left (x e + d\right ) - 2 \, {\left (c^{3} d^{5} - b^{3} x^{2} e^{5} + {\left (3 \, b^{2} c d x^{2} - 2 \, b^{3} d x\right )} e^{4} - {\left (3 \, b c^{2} d^{2} x^{2} - 6 \, b^{2} c d^{2} x + b^{3} d^{2}\right )} e^{3} + {\left (c^{3} d^{3} x^{2} - 6 \, b c^{2} d^{3} x + 3 \, b^{2} c d^{3}\right )} e^{2} + {\left (2 \, c^{3} d^{4} x - 3 \, b c^{2} d^{4}\right )} e\right )} \log \left (x\right )}{2 \, {\left (b c^{3} d^{8} - b^{4} d^{3} x^{2} e^{5} + {\left (3 \, b^{3} c d^{4} x^{2} - 2 \, b^{4} d^{4} x\right )} e^{4} - {\left (3 \, b^{2} c^{2} d^{5} x^{2} - 6 \, b^{3} c d^{5} x + b^{4} d^{5}\right )} e^{3} + {\left (b c^{3} d^{6} x^{2} - 6 \, b^{2} c^{2} d^{6} x + 3 \, b^{3} c d^{6}\right )} e^{2} + {\left (2 \, b c^{3} d^{7} x - 3 \, b^{2} c^{2} d^{7}\right )} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/(c*x^2+b*x),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
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(1/(e*x+d)**3/(c*x**2+b*x),x)

[Out]

Timed out

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Giac [A]
time = 1.11, size = 227, normalized size = 1.69 \begin {gather*} -\frac {c^{4} \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 (3 \, c^{2} d^{2} e^{2} - 3 \, b c d e^{3} + 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 {\log \left ({\left | x \right |}\right )}{b d^{3}} - \frac {5 \, c^{2} d^{4} e - 8 \, b c d^{3} e^{2} + 3 \, b^{2} d^{2} e^{3} + 2 \, {\left (2 \, c^{2} d^{3} e^{2} - 3 \, b c d^{2} e^{3} + 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(1/(e*x+d)^3/(c*x^2+b*x),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.66, size = 235, normalized size = 1.75 \begin {gather*} \frac {\frac {3\,b\,e^2-5\,c\,d\,e}{2\,d\,\left (b^2\,e^2-2\,b\,c\,d\,e+c^2\,d^2\right )}+\frac {e^2\,x\,\left (b\,e-2\,c\,d\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 {c^3\,\ln \left (b+c\,x\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 {\ln \left (d+e\,x\right )\,\left (b^2\,e^3-3\,b\,c\,d\,e^2+3\,c^2\,d^2\,e\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 (x\right )}{b\,d^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b*x + c*x^2)*(d + e*x)^3),x)

[Out]

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

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