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

Optimal. Leaf size=62 \[ -\frac {c d x \left (c d^2-a e^2\right )}{e^2}+\frac {\left (c d^2-a e^2\right )^2 \log (d+e x)}{e^3}+\frac {(a e+c d x)^2}{2 e} \]

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Rubi [A]  time = 0.04, antiderivative size = 62, 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 = {626, 43} \begin {gather*} -\frac {c d x \left (c d^2-a e^2\right )}{e^2}+\frac {\left (c d^2-a e^2\right )^2 \log (d+e x)}{e^3}+\frac {(a e+c d x)^2}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2/(d + e*x)^3,x]

[Out]

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

Rule 43

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 626

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c*x)/e)^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 {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^3} \, dx &=\int \frac {(a e+c d x)^2}{d+e x} \, dx\\ &=\int \left (-\frac {c d \left (c d^2-a e^2\right )}{e^2}+\frac {c d (a e+c d x)}{e}+\frac {\left (-c d^2+a e^2\right )^2}{e^2 (d+e x)}\right ) \, dx\\ &=-\frac {c d \left (c d^2-a e^2\right ) x}{e^2}+\frac {(a e+c d x)^2}{2 e}+\frac {\left (c d^2-a e^2\right )^2 \log (d+e x)}{e^3}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 52, normalized size = 0.84 \begin {gather*} \frac {2 \left (c d^2-a e^2\right )^2 \log (d+e x)+c d e x \left (4 a e^2+c d (e x-2 d)\right )}{2 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2}{(d+e x)^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2/(d + e*x)^3,x]

[Out]

IntegrateAlgebraic[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2/(d + e*x)^3, x]

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fricas [A]  time = 0.40, size = 72, normalized size = 1.16 \begin {gather*} \frac {c^{2} d^{2} e^{2} x^{2} - 2 \, {\left (c^{2} d^{3} e - 2 \, a c d e^{3}\right )} x + 2 \, {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \log \left (e x + d\right )}{2 \, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 71, normalized size = 1.15 \begin {gather*} {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} e^{\left (-3\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{2} \, {\left (c^{2} d^{2} x^{2} e^{5} - 2 \, c^{2} d^{3} x e^{4} + 4 \, a c d x e^{6}\right )} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 77, normalized size = 1.24 \begin {gather*} \frac {c^{2} d^{2} x^{2}}{2 e}+a^{2} e \ln \left (e x +d \right )-\frac {2 a c \,d^{2} \ln \left (e x +d \right )}{e}+2 a c d x +\frac {c^{2} d^{4} \ln \left (e x +d \right )}{e^{3}}-\frac {c^{2} d^{3} x}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.96, size = 72, normalized size = 1.16 \begin {gather*} \frac {c^{2} d^{2} e x^{2} - 2 \, {\left (c^{2} d^{3} - 2 \, a c d e^{2}\right )} x}{2 \, e^{2}} + \frac {{\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \log \left (e x + d\right )}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.07, size = 69, normalized size = 1.11 \begin {gather*} x\,\left (2\,a\,c\,d-\frac {c^2\,d^3}{e^2}\right )+\frac {\ln \left (d+e\,x\right )\,\left (a^2\,e^4-2\,a\,c\,d^2\,e^2+c^2\,d^4\right )}{e^3}+\frac {c^2\,d^2\,x^2}{2\,e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.29, size = 53, normalized size = 0.85 \begin {gather*} \frac {c^{2} d^{2} x^{2}}{2 e} + x \left (2 a c d - \frac {c^{2} d^{3}}{e^{2}}\right ) + \frac {\left (a e^{2} - c d^{2}\right )^{2} \log {\left (d + e x \right )}}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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