3.10.99 \(\int \frac {(d+e x)^4}{c d^2+2 c d e x+c e^2 x^2} \, dx\) [999]

Optimal. Leaf size=17 \[ \frac {(d+e x)^3}{3 c e} \]

[Out]

1/3*(e*x+d)^3/c/e

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Rubi [A]
time = 0.00, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {27, 12, 32} \begin {gather*} \frac {(d+e x)^3}{3 c e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^4/(c*d^2 + 2*c*d*e*x + c*e^2*x^2),x]

[Out]

(d + e*x)^3/(3*c*e)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 17, normalized size = 1.00 \begin {gather*} \frac {(d+e x)^3}{3 c e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^4/(c*d^2 + 2*c*d*e*x + c*e^2*x^2),x]

[Out]

(d + e*x)^3/(3*c*e)

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Maple [A]
time = 0.60, size = 16, normalized size = 0.94

method	result	size
default	\(\frac {\left (e x +d \right )^{3}}{3 c e}\)	\(16\)
gosper	\(\frac {x \left (e^{2} x^{2}+3 d x e +3 d^{2}\right )}{3 c}\)	\(25\)
risch	\(\frac {e^{2} x^{3}}{3 c}+\frac {e d \,x^{2}}{c}+\frac {d^{2} x}{c}+\frac {d^{3}}{3 c e}\)	\(41\)
norman	\(\frac {-\frac {d^{4}}{c e}+\frac {e^{3} x^{4}}{3 c}+\frac {2 e \,d^{2} x^{2}}{c}+\frac {4 e^{2} d \,x^{3}}{3 c}}{e x +d}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^4/(c*e^2*x^2+2*c*d*e*x+c*d^2),x,method=_RETURNVERBOSE)

[Out]

1/3*(e*x+d)^3/c/e

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Maxima [A]
time = 0.28, size = 26, normalized size = 1.53 \begin {gather*} \frac {x^{3} e^{2} + 3 \, d x^{2} e + 3 \, d^{2} x}{3 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(c*e^2*x^2+2*c*d*e*x+c*d^2),x, algorithm="maxima")

[Out]

1/3*(x^3*e^2 + 3*d*x^2*e + 3*d^2*x)/c

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Fricas [A]
time = 2.12, size = 26, normalized size = 1.53 \begin {gather*} \frac {x^{3} e^{2} + 3 \, d x^{2} e + 3 \, d^{2} x}{3 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(c*e^2*x^2+2*c*d*e*x+c*d^2),x, algorithm="fricas")

[Out]

1/3*(x^3*e^2 + 3*d*x^2*e + 3*d^2*x)/c

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 24 vs. \(2 (10) = 20\).
time = 0.02, size = 24, normalized size = 1.41 \begin {gather*} \frac {d^{2} x}{c} + \frac {d e x^{2}}{c} + \frac {e^{2} x^{3}}{3 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4/(c*e**2*x**2+2*c*d*e*x+c*d**2),x)

[Out]

d**2*x/c + d*e*x**2/c + e**2*x**3/(3*c)

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Giac [A]
time = 2.35, size = 26, normalized size = 1.53 \begin {gather*} \frac {x^{3} e^{2} + 3 \, d x^{2} e + 3 \, d^{2} x}{3 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(c*e^2*x^2+2*c*d*e*x+c*d^2),x, algorithm="giac")

[Out]

1/3*(x^3*e^2 + 3*d*x^2*e + 3*d^2*x)/c

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Mupad [B]
time = 0.03, size = 24, normalized size = 1.41 \begin {gather*} \frac {x\,\left (3\,d^2+3\,d\,e\,x+e^2\,x^2\right )}{3\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^4/(c*d^2 + c*e^2*x^2 + 2*c*d*e*x),x)

[Out]

(x*(3*d^2 + e^2*x^2 + 3*d*e*x))/(3*c)

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