3.11.82 \(\int \frac {(d+e x)^3}{(c d^2+2 c d e x+c e^2 x^2)^{5/2}} \, dx\) [1082]

Optimal. Leaf size=32 \[ -\frac {1}{c^2 e \sqrt {c d^2+2 c d e x+c e^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {657, 643} \begin {gather*} -\frac {1}{c^2 e \sqrt {c d^2+2 c d e x+c e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2),x]

[Out]

-(1/(c^2*e*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2]))

Rule 643

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*((a + b*x + c*x^2)^(p +
1)/(b*(p + 1))), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 657

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[e^(m - 1)/c^((m - 1)/2
), Int[(d + e*x)*(a + b*x + c*x^2)^(p + (m - 1)/2), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[b^2 - 4*a*c,
0] &&  !IntegerQ[p] && EqQ[2*c*d - b*e, 0] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {(d+e x)^3}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx &=\frac {\int \frac {d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \, dx}{c}\\ &=-\frac {1}{c^2 e \sqrt {c d^2+2 c d e x+c e^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 21, normalized size = 0.66 \begin {gather*} -\frac {1}{c^2 e \sqrt {c (d+e x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3/(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2),x]

[Out]

-(1/(c^2*e*Sqrt[c*(d + e*x)^2]))

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Maple [A]
time = 0.57, size = 35, normalized size = 1.09

method	result	size
risch	\(-\frac {1}{c^{2} \sqrt {\left (e x +d \right )^{2} c}\, e}\)	\(20\)
gosper	\(-\frac {\left (e x +d \right )^{4}}{e \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )^{\frac {5}{2}}}\)	\(35\)
default	\(-\frac {\left (e x +d \right )^{4}}{e \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )^{\frac {5}{2}}}\)	\(35\)
trager	\(\frac {x \sqrt {x^{2} c \,e^{2}+2 c d e x +c \,d^{2}}}{c^{3} d \left (e x +d \right )^{2}}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (29) = 58\).
time = 0.30, size = 99, normalized size = 3.09 \begin {gather*} -\frac {x^{2} e}{{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{\frac {3}{2}} c} - \frac {5 \, d^{2} e^{\left (-1\right )}}{3 \, {\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{\frac {3}{2}} c} - \frac {2 \, d e^{\left (-3\right )}}{{\left (d e^{\left (-1\right )} + x\right )}^{2} c^{\frac {5}{2}}} + \frac {8 \, d^{2} e^{\left (-4\right )}}{3 \, {\left (d e^{\left (-1\right )} + x\right )}^{3} c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="maxima")

[Out]

-x^2*e/((c*x^2*e^2 + 2*c*d*x*e + c*d^2)^(3/2)*c) - 5/3*d^2*e^(-1)/((c*x^2*e^2 + 2*c*d*x*e + c*d^2)^(3/2)*c) -
2*d*e^(-3)/((d*e^(-1) + x)^2*c^(5/2)) + 8/3*d^2*e^(-4)/((d*e^(-1) + x)^3*c^(5/2))

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Fricas [A]
time = 2.58, size = 54, normalized size = 1.69 \begin {gather*} -\frac {\sqrt {c x^{2} e^{2} + 2 \, c d x e + c d^{2}}}{c^{3} x^{2} e^{3} + 2 \, c^{3} d x e^{2} + c^{3} d^{2} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="fricas")

[Out]

-sqrt(c*x^2*e^2 + 2*c*d*x*e + c*d^2)/(c^3*x^2*e^3 + 2*c^3*d*x*e^2 + c^3*d^2*e)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (32) = 64\).
time = 0.45, size = 70, normalized size = 2.19 \begin {gather*} \begin {cases} - \frac {\sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}}}{c^{3} d^{2} e + 2 c^{3} d e^{2} x + c^{3} e^{3} x^{2}} & \text {for}\: e \neq 0 \\\frac {d^{3} x}{\left (c d^{2}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3/(c*e**2*x**2+2*c*d*e*x+c*d**2)**(5/2),x)

[Out]

Piecewise((-sqrt(c*d**2 + 2*c*d*e*x + c*e**2*x**2)/(c**3*d**2*e + 2*c**3*d*e**2*x + c**3*e**3*x**2), Ne(e, 0))
, (d**3*x/(c*d**2)**(5/2), True))

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Giac [A]
time = 0.84, size = 24, normalized size = 0.75 \begin {gather*} -\frac {e^{\left (-1\right )}}{{\left (x e + d\right )} c^{\frac {5}{2}} \mathrm {sgn}\left (x e + d\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="giac")

[Out]

-e^(-1)/((x*e + d)*c^(5/2)*sgn(x*e + d))

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Mupad [B]
time = 0.48, size = 37, normalized size = 1.16 \begin {gather*} -\frac {\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}}{c^3\,e\,{\left (d+e\,x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3/(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(5/2),x)

[Out]

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

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