3.9.95 \(\int \frac {(d+e x)^2}{(c d^2+2 c d e x+c e^2 x^2)^{5/2}} \, dx\)

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

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Rubi [A]  time = 0.02, 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 = {642, 607} \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*c^2*e*(d + e*x)*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2])

Rule 607

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 642

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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IntegrateAlgebraic [B]  time = 0.62, size = 156, normalized size = 3.80 \begin {gather*} \frac {\sqrt {c e^2} (d-e x) \sqrt {c d^2+2 c d e x+c e^2 x^2}+c d^2 e+c e^3 x^2}{c e x^2 \sqrt {c e^2} \left (2 c^2 d^2 e^2+4 c^2 d e^3 x+2 c^2 e^4 x^2\right )+c e x^2 \left (-2 c^2 d e^3-2 c^2 e^4 x\right ) \sqrt {c d^2+2 c d e x+c e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.38, size = 69, normalized size = 1.68 \begin {gather*} -\frac {\sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{2 \, {\left (c^{3} e^{4} x^{3} + 3 \, c^{3} d e^{3} x^{2} + 3 \, c^{3} d^{2} e^{2} x + 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*e^2*x^2 + 2*c*d*e*x + c*d^2)/(c^3*e^4*x^3 + 3*c^3*d*e^3*x^2 + 3*c^3*d^2*e^2*x + c^3*d^3*e)

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giac [A]  time = 0.47, size = 70, normalized size = 1.71 \begin {gather*} \frac {4 \, C_{0} d^{3} e^{\left (-3\right )} + {\left (12 \, C_{0} d^{2} e^{\left (-2\right )} + 4 \, {\left (3 \, C_{0} d e^{\left (-1\right )} + C_{0} x\right )} x - \frac {1}{c}\right )} x - \frac {d e^{\left (-1\right )}}{c}}{2 \, {\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{\frac {3}{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="giac")

[Out]

1/2*(4*C_0*d^3*e^(-3) + (12*C_0*d^2*e^(-2) + 4*(3*C_0*d*e^(-1) + C_0*x)*x - 1/c)*x - d*e^(-1)/c)/(c*x^2*e^2 +
2*c*d*x*e + c*d^2)^(3/2)

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maple [A]  time = 0.06, size = 35, normalized size = 0.85 \begin {gather*} -\frac {\left (e x +d \right )^{3}}{2 \left (c \,e^{2} x^{2}+2 c d e x +c \,d^{2}\right )^{\frac {5}{2}} e} \end {gather*}

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)

[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 = 1.36, size = 67, normalized size = 1.63 \begin {gather*} -\frac {2 \, d}{3 \, {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {3}{2}} c e} - \frac {1}{2 \, c^{\frac {5}{2}} e^{3} {\left (x + \frac {d}{e}\right )}^{2}} + \frac {2 \, d}{3 \, c^{\frac {5}{2}} e^{4} {\left (x + \frac {d}{e}\right )}^{3}} \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/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c*e) - 1/2/(c^(5/2)*e^3*(x + d/e)^2) + 2/3*d/(c^(5/2)*e^4*(x + d
/e)^3)

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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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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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