3.3.73 \(\int \frac {1}{(d+e x) (b x+c x^2)^2} \, dx\) [273]

Optimal. Leaf size=110 \[ -\frac {1}{b^2 d x}-\frac {c^2}{b^2 (c d-b e) (b+c x)}-\frac {(2 c d+b e) \log (x)}{b^3 d^2}+\frac {c^2 (2 c d-3 b e) \log (b+c x)}{b^3 (c d-b e)^2}+\frac {e^3 \log (d+e x)}{d^2 (c d-b e)^2} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 110, 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 {c^2 (2 c d-3 b e) \log (b+c x)}{b^3 (c d-b e)^2}-\frac {\log (x) (b e+2 c d)}{b^3 d^2}-\frac {c^2}{b^2 (b+c x) (c d-b e)}-\frac {1}{b^2 d x}+\frac {e^3 \log (d+e x)}{d^2 (c d-b e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)*(b*x + c*x^2)^2),x]

[Out]

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

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

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Mathematica [A]
time = 0.06, size = 111, normalized size = 1.01 \begin {gather*} -\frac {1}{b^2 d x}+\frac {c^2}{b^2 (-c d+b e) (b+c x)}+\frac {(-2 c d-b e) \log (x)}{b^3 d^2}+\frac {\left (2 c^3 d-3 b c^2 e\right ) \log (b+c x)}{b^3 (-c d+b e)^2}+\frac {e^3 \log (d+e x)}{d^2 (c d-b e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.50, size = 111, normalized size = 1.01

method	result	size
default	\(\frac {c^{2}}{\left (b e -c d \right ) b^{2} \left (c x +b \right )}-\frac {c^{2} \left (3 b e -2 c d \right ) \ln \left (c x +b \right )}{\left (b e -c d \right )^{2} b^{3}}-\frac {1}{b^{2} d x}+\frac {\left (-b e -2 c d \right ) \ln \left (x \right )}{d^{2} b^{3}}+\frac {e^{3} \ln \left (e x +d \right )}{d^{2} \left (b e -c d \right )^{2}}\)	\(111\)
norman	\(\frac {\frac {\left (b c e -2 c^{2} d \right ) c \,x^{2}}{d \,b^{3} \left (b e -c d \right )}-\frac {1}{b d}}{x \left (c x +b \right )}+\frac {e^{3} \ln \left (e x +d \right )}{d^{2} \left (b^{2} e^{2}-2 b c d e +d^{2} c^{2}\right )}-\frac {\left (b e +2 c d \right ) \ln \left (x \right )}{b^{3} d^{2}}-\frac {c^{2} \left (3 b e -2 c d \right ) \ln \left (c x +b \right )}{b^{3} \left (b^{2} e^{2}-2 b c d e +d^{2} c^{2}\right )}\)	\(154\)
risch	\(\frac {-\frac {c \left (b e -2 c d \right ) x}{b^{2} d \left (b e -c d \right )}-\frac {1}{b d}}{x \left (c x +b \right )}-\frac {3 c^{2} \ln \left (c x +b \right ) e}{b^{2} \left (b^{2} e^{2}-2 b c d e +d^{2} c^{2}\right )}+\frac {2 c^{3} \ln \left (c x +b \right ) d}{b^{3} \left (b^{2} e^{2}-2 b c d e +d^{2} c^{2}\right )}+\frac {e^{3} \ln \left (-e x -d \right )}{d^{2} \left (b^{2} e^{2}-2 b c d e +d^{2} c^{2}\right )}-\frac {\ln \left (-x \right ) e}{d^{2} b^{2}}-\frac {2 \ln \left (-x \right ) c}{d \,b^{3}}\)	\(191\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 183, normalized size = 1.66 \begin {gather*} \frac {{\left (2 \, c^{3} d - 3 \, b c^{2} e\right )} \log \left (c x + b\right )}{b^{3} c^{2} d^{2} - 2 \, b^{4} c d e + b^{5} e^{2}} + \frac {e^{3} \log \left (x e + d\right )}{c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}} - \frac {b c d - b^{2} e + {\left (2 \, c^{2} d - b c e\right )} x}{{\left (b^{2} c^{2} d^{2} - b^{3} c d e\right )} x^{2} + {\left (b^{3} c d^{2} - b^{4} d e\right )} x} - \frac {{\left (2 \, c d + b e\right )} \log \left (x\right )}{b^{3} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 293 vs. \(2 (115) = 230\).
time = 7.97, size = 293, normalized size = 2.66 \begin {gather*} -\frac {2 \, b c^{3} d^{3} x + b^{2} c^{2} d^{3} - {\left (b^{3} c x^{2} + b^{4} x\right )} e^{3} \log \left (x e + d\right ) + {\left (b^{3} c d x + b^{4} d\right )} e^{2} - {\left (3 \, b^{2} c^{2} d^{2} x + 2 \, b^{3} c d^{2}\right )} e - {\left (2 \, c^{4} d^{3} x^{2} + 2 \, b c^{3} d^{3} x - 3 \, {\left (b c^{3} d^{2} x^{2} + b^{2} c^{2} d^{2} x\right )} e\right )} \log \left (c x + b\right ) + {\left (2 \, c^{4} d^{3} x^{2} + 2 \, b c^{3} d^{3} x + {\left (b^{3} c x^{2} + b^{4} x\right )} e^{3} - 3 \, {\left (b c^{3} d^{2} x^{2} + b^{2} c^{2} d^{2} x\right )} e\right )} \log \left (x\right )}{b^{3} c^{3} d^{4} x^{2} + b^{4} c^{2} d^{4} x + {\left (b^{5} c d^{2} x^{2} + b^{6} d^{2} x\right )} e^{2} - 2 \, {\left (b^{4} c^{2} d^{3} x^{2} + b^{5} c d^{3} x\right )} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]
time = 1.47, size = 202, normalized size = 1.84 \begin {gather*} \frac {{\left (2 \, c^{4} d - 3 \, b c^{3} e\right )} \log \left ({\left | c x + b \right |}\right )}{b^{3} c^{3} d^{2} - 2 \, b^{4} c^{2} d e + b^{5} c e^{2}} + \frac {e^{4} \log \left ({\left | x e + d \right |}\right )}{c^{2} d^{4} e - 2 \, b c d^{3} e^{2} + b^{2} d^{2} e^{3}} - \frac {{\left (2 \, c d + b e\right )} \log \left ({\left | x \right |}\right )}{b^{3} d^{2}} - \frac {b c^{2} d^{3} - 2 \, b^{2} c d^{2} e + b^{3} d e^{2} + {\left (2 \, c^{3} d^{3} - 3 \, b c^{2} d^{2} e + b^{2} c d e^{2}\right )} x}{{\left (c d - b e\right )}^{2} {\left (c x + b\right )} b^{2} d^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.52, size = 143, normalized size = 1.30 \begin {gather*} \frac {\ln \left (b+c\,x\right )\,\left (2\,c^3\,d-3\,b\,c^2\,e\right )}{b^5\,e^2-2\,b^4\,c\,d\,e+b^3\,c^2\,d^2}-\frac {\frac {1}{b\,d}-\frac {x\,\left (2\,c^2\,d-b\,c\,e\right )}{b^2\,d\,\left (b\,e-c\,d\right )}}{c\,x^2+b\,x}+\frac {e^3\,\ln \left (d+e\,x\right )}{d^2\,{\left (b\,e-c\,d\right )}^2}-\frac {\ln \left (x\right )\,\left (b\,e+2\,c\,d\right )}{b^3\,d^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b*x + c*x^2)^2*(d + e*x)),x)

[Out]

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

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