3.11.70 \(\int \frac {1}{(d+e x)^4 \sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx\) [1070]

Optimal. Leaf size=39 \[ -\frac {c}{4 e (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 39, 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 = {656, 621} \begin {gather*} -\frac {c}{4 e (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^4*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2]),x]

[Out]

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

Rule 621

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[2*((a + b*x + c*x^2)^(p + 1)/((2*p + 1)*(b + 2*
c*x))), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && LtQ[p, -1]

Rule 656

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[e^m/c^(m/2), Int[(a +
b*x + c*x^2)^(p + m/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/2]

Rubi steps

\begin {align*} \int \frac {1}{(d+e x)^4 \sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx &=c^2 \int \frac {1}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx\\ &=-\frac {c}{4 e (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 27, normalized size = 0.69 \begin {gather*} -\frac {1}{4 e (d+e x)^3 \sqrt {c (d+e x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^4*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2]),x]

[Out]

-1/4*1/(e*(d + e*x)^3*Sqrt[c*(d + e*x)^2])

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Maple [A]
time = 0.68, size = 35, normalized size = 0.90

method	result	size
risch	\(-\frac {1}{4 \left (e x +d \right )^{3} \sqrt {\left (e x +d \right )^{2} c}\, e}\)	\(24\)
gosper	\(-\frac {1}{4 \left (e x +d \right )^{3} e \sqrt {x^{2} c \,e^{2}+2 c d e x +c \,d^{2}}}\)	\(35\)
default	\(-\frac {1}{4 \left (e x +d \right )^{3} e \sqrt {x^{2} c \,e^{2}+2 c d e x +c \,d^{2}}}\)	\(35\)
trager	\(\frac {\left (e^{3} x^{3}+4 d \,e^{2} x^{2}+6 d^{2} e x +4 d^{3}\right ) x \sqrt {x^{2} c \,e^{2}+2 c d e x +c \,d^{2}}}{4 d^{4} c \left (e x +d \right )^{5}}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^4/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 58, normalized size = 1.49 \begin {gather*} -\frac {1}{4 \, {\left (\sqrt {c} x^{4} e^{5} + 4 \, \sqrt {c} d x^{3} e^{4} + 6 \, \sqrt {c} d^{2} x^{2} e^{3} + 4 \, \sqrt {c} d^{3} x e^{2} + \sqrt {c} d^{4} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^4/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 81 vs. \(2 (35) = 70\).
time = 1.83, size = 81, normalized size = 2.08 \begin {gather*} -\frac {\sqrt {c x^{2} e^{2} + 2 \, c d x e + c d^{2}}}{4 \, {\left (c x^{5} e^{6} + 5 \, c d x^{4} e^{5} + 10 \, c d^{2} x^{3} e^{4} + 10 \, c d^{3} x^{2} e^{3} + 5 \, c d^{4} x e^{2} + c d^{5} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^4/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {c \left (d + e x\right )^{2}} \left (d + e x\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**4/(c*e**2*x**2+2*c*d*e*x+c*d**2)**(1/2),x)

[Out]

Integral(1/(sqrt(c*(d + e*x)**2)*(d + e*x)**4), x)

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Giac [A]
time = 1.73, size = 24, normalized size = 0.62 \begin {gather*} -\frac {e^{\left (-1\right )}}{4 \, {\left (x e + d\right )}^{4} \sqrt {c} \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)^4/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(1/2),x, algorithm="giac")

[Out]

-1/4*e^(-1)/((x*e + d)^4*sqrt(c)*sgn(x*e + d))

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Mupad [B]
time = 0.46, size = 37, normalized size = 0.95 \begin {gather*} -\frac {\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}}{4\,c\,e\,{\left (d+e\,x\right )}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d + e*x)^4*(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(1/2)),x)

[Out]

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

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