3.19.68 \(\int \frac {(d+e x)^2}{a d e+(c d^2+a e^2) x+c d e x^2} \, dx\) [1868]

Optimal. Leaf size=38 \[ \frac {e x}{c d}+\frac {\left (c d^2-a e^2\right ) \log (a e+c d x)}{c^2 d^2} \]

[Out]

e*x/c/d+(-a*e^2+c*d^2)*ln(c*d*x+a*e)/c^2/d^2

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Rubi [A]
time = 0.02, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {640, 45} \begin {gather*} \frac {\left (c d^2-a e^2\right ) \log (a e+c d x)}{c^2 d^2}+\frac {e x}{c d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2),x]

[Out]

(e*x)/(c*d) + ((c*d^2 - a*e^2)*Log[a*e + c*d*x])/(c^2*d^2)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 640

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

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 35, normalized size = 0.92 \begin {gather*} \frac {c d e x+\left (c d^2-a e^2\right ) \log (a e+c d x)}{c^2 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*d*e*x + (c*d^2 - a*e^2)*Log[a*e + c*d*x])/(c^2*d^2)

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Maple [A]
time = 0.72, size = 39, normalized size = 1.03

method	result	size
default	\(\frac {e x}{c d}+\frac {\left (-e^{2} a +c \,d^{2}\right ) \ln \left (c d x +a e \right )}{d^{2} c^{2}}\)	\(39\)
norman	\(\frac {e x}{c d}-\frac {\left (e^{2} a -c \,d^{2}\right ) \ln \left (c d x +a e \right )}{c^{2} d^{2}}\)	\(40\)
risch	\(\frac {e x}{c d}-\frac {\ln \left (c d x +a e \right ) e^{2} a}{c^{2} d^{2}}+\frac {\ln \left (c d x +a e \right )}{c}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*x/c/d+(-a*e^2+c*d^2)*ln(c*d*x+a*e)/d^2/c^2

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Maxima [A]
time = 0.29, size = 39, normalized size = 1.03 \begin {gather*} \frac {x e}{c d} + \frac {{\left (c d^{2} - a e^{2}\right )} \log \left (c d x + a e\right )}{c^{2} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*e/(c*d) + (c*d^2 - a*e^2)*log(c*d*x + a*e)/(c^2*d^2)

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Fricas [A]
time = 2.42, size = 36, normalized size = 0.95 \begin {gather*} \frac {c d x e + {\left (c d^{2} - a e^{2}\right )} \log \left (c d x + a e\right )}{c^{2} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*d*x*e + (c*d^2 - a*e^2)*log(c*d*x + a*e))/(c^2*d^2)

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Sympy [A]
time = 0.07, size = 32, normalized size = 0.84 \begin {gather*} \frac {e x}{c d} - \frac {\left (a e^{2} - c d^{2}\right ) \log {\left (a e + c d x \right )}}{c^{2} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*x/(c*d) - (a*e**2 - c*d**2)*log(a*e + c*d*x)/(c**2*d**2)

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Giac [A]
time = 0.71, size = 40, normalized size = 1.05 \begin {gather*} \frac {x e}{c d} + \frac {{\left (c d^{2} - a e^{2}\right )} \log \left ({\left | c d x + a e \right |}\right )}{c^{2} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*e/(c*d) + (c*d^2 - a*e^2)*log(abs(c*d*x + a*e))/(c^2*d^2)

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Mupad [B]
time = 0.06, size = 39, normalized size = 1.03 \begin {gather*} \frac {e\,x}{c\,d}-\frac {\ln \left (a\,e+c\,d\,x\right )\,\left (a\,e^2-c\,d^2\right )}{c^2\,d^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e*x)/(c*d) - (log(a*e + c*d*x)*(a*e^2 - c*d^2))/(c^2*d^2)

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