3.11.83 \(\int \frac {(d+e x)^2}{(c d^2+2 c d e x+c e^2 x^2)^{5/2}} \, dx\) [1083]

Optimal. Leaf size=41 \[ -\frac {1}{2 c^2 e (d+e x) \sqrt {c d^2+2 c d e x+c e^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 41, 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 {1}{2 c^2 e (d+e x) \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)^2/(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2),x]

[Out]

-1/2*1/(c^2*e*(d + e*x)*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^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 {(d+e x)^2}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx &=\frac {\int \frac {1}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \, dx}{c}\\ &=-\frac {1}{2 c^2 e (d+e x) \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 = 28, normalized size = 0.68 \begin {gather*} -\frac {d+e x}{2 c e \left (c (d+e x)^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.58, size = 35, normalized size = 0.85

method	result	size
risch	\(-\frac {1}{2 c^{2} \left (e x +d \right ) \sqrt {\left (e x +d \right )^{2} c}\, e}\)	\(27\)
gosper	\(-\frac {\left (e x +d \right )^{3}}{2 e \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )^{\frac {5}{2}}}\)	\(35\)
default	\(-\frac {\left (e x +d \right )^{3}}{2 e \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )^{\frac {5}{2}}}\)	\(35\)
trager	\(\frac {\left (e x +2 d \right ) x \sqrt {x^{2} c \,e^{2}+2 c d e x +c \,d^{2}}}{2 c^{3} d^{2} \left (e x +d \right )^{3}}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 62, normalized size = 1.51 \begin {gather*} -\frac {2 \, d 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 {e^{\left (-3\right )}}{2 \, {\left (d e^{\left (-1\right )} + x\right )}^{2} c^{\frac {5}{2}}} + \frac {2 \, d 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)^2/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.57, size = 67, normalized size = 1.63 \begin {gather*} -\frac {\sqrt {c x^{2} e^{2} + 2 \, c d x e + c d^{2}}}{2 \, {\left (c^{3} x^{3} e^{4} + 3 \, c^{3} d x^{2} e^{3} + 3 \, c^{3} d^{2} x e^{2} + c^{3} d^{3} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{2}}{\left (c \left (d + e x\right )^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 2.28, size = 24, normalized size = 0.59 \begin {gather*} -\frac {e^{\left (-1\right )}}{2 \, {\left (x e + d\right )}^{2} 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)^2/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.49, size = 37, normalized size = 0.90 \begin {gather*} -\frac {\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}}{2\,c^3\,e\,{\left (d+e\,x\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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