3.11.86 \(\int \frac {1}{(d+e x) (c d^2+2 c d e x+c e^2 x^2)^{5/2}} \, dx\) [1086]

Optimal. Leaf size=31 \[ -\frac {1}{5 e \left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 31, 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}{5 e \left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2)),x]

[Out]

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

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 {1}{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx &=c \int \frac {d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{7/2}} \, dx\\ &=-\frac {1}{5 e \left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 20, normalized size = 0.65 \begin {gather*} -\frac {1}{5 e \left (c (d+e x)^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2)),x]

[Out]

-1/5*1/(e*(c*(d + e*x)^2)^(5/2))

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Maple [A]
time = 0.59, size = 28, normalized size = 0.90

method	result	size
risch	\(-\frac {1}{5 c^{2} \left (e x +d \right )^{4} \sqrt {\left (e x +d \right )^{2} c}\, e}\)	\(27\)
gosper	\(-\frac {1}{5 e \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )^{\frac {5}{2}}}\)	\(28\)
default	\(-\frac {1}{5 e \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )^{\frac {5}{2}}}\)	\(28\)
trager	\(\frac {\left (e^{4} x^{4}+5 d \,e^{3} x^{3}+10 d^{2} e^{2} x^{2}+10 d^{3} e x +5 d^{4}\right ) x \sqrt {x^{2} c \,e^{2}+2 c d e x +c \,d^{2}}}{5 c^{3} d^{5} \left (e x +d \right )^{6}}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/5/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. 71 vs. \(2 (26) = 52\).
time = 0.28, size = 71, normalized size = 2.29 \begin {gather*} -\frac {1}{5 \, {\left (c^{\frac {5}{2}} x^{5} e^{6} + 5 \, c^{\frac {5}{2}} d x^{4} e^{5} + 10 \, c^{\frac {5}{2}} d^{2} x^{3} e^{4} + 10 \, c^{\frac {5}{2}} d^{3} x^{2} e^{3} + 5 \, c^{\frac {5}{2}} d^{4} x e^{2} + c^{\frac {5}{2}} d^{5} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="maxima")

[Out]

-1/5/(c^(5/2)*x^5*e^6 + 5*c^(5/2)*d*x^4*e^5 + 10*c^(5/2)*d^2*x^3*e^4 + 10*c^(5/2)*d^3*x^2*e^3 + 5*c^(5/2)*d^4*
x*e^2 + c^(5/2)*d^5*e)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (26) = 52\).
time = 2.22, size = 106, normalized size = 3.42 \begin {gather*} -\frac {\sqrt {c x^{2} e^{2} + 2 \, c d x e + c d^{2}}}{5 \, {\left (c^{3} x^{6} e^{7} + 6 \, c^{3} d x^{5} e^{6} + 15 \, c^{3} d^{2} x^{4} e^{5} + 20 \, c^{3} d^{3} x^{3} e^{4} + 15 \, c^{3} d^{4} x^{2} e^{3} + 6 \, c^{3} d^{5} x e^{2} + c^{3} d^{6} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="fricas")

[Out]

-1/5*sqrt(c*x^2*e^2 + 2*c*d*x*e + c*d^2)/(c^3*x^6*e^7 + 6*c^3*d*x^5*e^6 + 15*c^3*d^2*x^4*e^5 + 20*c^3*d^3*x^3*
e^4 + 15*c^3*d^4*x^2*e^3 + 6*c^3*d^5*x*e^2 + c^3*d^6*e)

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Sympy [A]
time = 2.56, size = 42, normalized size = 1.35 \begin {gather*} \begin {cases} - \frac {1}{5 e \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{\frac {5}{2}}} & \text {for}\: e \neq 0 \\\frac {x}{d \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(1/(e*x+d)/(c*e**2*x**2+2*c*d*e*x+c*d**2)**(5/2),x)

[Out]

Piecewise((-1/(5*e*(c*d**2 + 2*c*d*e*x + c*e**2*x**2)**(5/2)), Ne(e, 0)), (x/(d*(c*d**2)**(5/2)), True))

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Giac [A]
time = 1.27, size = 24, normalized size = 0.77 \begin {gather*} -\frac {e^{\left (-1\right )}}{5 \, {\left (x e + d\right )}^{5} 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(1/(e*x+d)/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.54, size = 37, normalized size = 1.19 \begin {gather*} -\frac {\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}}{5\,c^3\,e\,{\left (d+e\,x\right )}^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d + e*x)*(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)/(5*c^3*e*(d + e*x)^6)

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