3.15.56 \(\int \frac {a^2+2 a b x+b^2 x^2}{d+e x} \, dx\) [1456]

Optimal. Leaf size=50 \[ -\frac {b (b d-a e) x}{e^2}+\frac {(a+b x)^2}{2 e}+\frac {(b d-a e)^2 \log (d+e x)}{e^3} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {27, 45} \begin {gather*} \frac {(b d-a e)^2 \log (d+e x)}{e^3}-\frac {b x (b d-a e)}{e^2}+\frac {(a+b x)^2}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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])

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 43, normalized size = 0.86 \begin {gather*} \frac {b e x (-2 b d+4 a e+b e x)+2 (b d-a e)^2 \log (d+e x)}{2 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.57, size = 56, normalized size = 1.12

method	result	size
default	\(\frac {b \left (\frac {1}{2} b e \,x^{2}+2 a e x -x b d \right )}{e^{2}}+\frac {\left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \ln \left (e x +d \right )}{e^{3}}\)	\(56\)
norman	\(\frac {b \left (2 a e -b d \right ) x}{e^{2}}+\frac {b^{2} x^{2}}{2 e}+\frac {\left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \ln \left (e x +d \right )}{e^{3}}\)	\(59\)
risch	\(\frac {b^{2} x^{2}}{2 e}+\frac {2 b a x}{e}-\frac {b^{2} x d}{e^{2}}+\frac {\ln \left (e x +d \right ) a^{2}}{e}-\frac {2 \ln \left (e x +d \right ) a b d}{e^{2}}+\frac {\ln \left (e x +d \right ) b^{2} d^{2}}{e^{3}}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 61, normalized size = 1.22 \begin {gather*} {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} e^{\left (-3\right )} \log \left (x e + d\right ) + \frac {1}{2} \, {\left (b^{2} x^{2} e - 2 \, {\left (b^{2} d - 2 \, a b e\right )} x\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*d^2 - 2*a*b*d*e + a^2*e^2)*e^(-3)*log(x*e + d) + 1/2*(b^2*x^2*e - 2*(b^2*d - 2*a*b*e)*x)*e^(-2)

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Fricas [A]
time = 2.07, size = 61, normalized size = 1.22 \begin {gather*} -\frac {1}{2} \, {\left (2 \, b^{2} d x e - {\left (b^{2} x^{2} + 4 \, a b x\right )} e^{2} - 2 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \log \left (x e + d\right )\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.10, size = 44, normalized size = 0.88 \begin {gather*} \frac {b^{2} x^{2}}{2 e} + x \left (\frac {2 a b}{e} - \frac {b^{2} d}{e^{2}}\right ) + \frac {\left (a e - b d\right )^{2} \log {\left (d + e x \right )}}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b**2*x**2/(2*e) + x*(2*a*b/e - b**2*d/e**2) + (a*e - b*d)**2*log(d + e*x)/e**3

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Giac [A]
time = 2.03, size = 61, normalized size = 1.22 \begin {gather*} {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} e^{\left (-3\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{2} \, {\left (b^{2} x^{2} e - 2 \, b^{2} d x + 4 \, a b x e\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.52, size = 62, normalized size = 1.24 \begin {gather*} \frac {\ln \left (d+e\,x\right )\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}{e^3}-x\,\left (\frac {b^2\,d}{e^2}-\frac {2\,a\,b}{e}\right )+\frac {b^2\,x^2}{2\,e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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