3.1080 \(\int \frac {(d+e x)^5}{(c d^2+2 c d e x+c e^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=31 \[ \frac {\sqrt {c d^2+2 c d e x+c e^2 x^2}}{c^3 e} \]

[Out]

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

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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 = {643, 629} \[ \frac {\sqrt {c d^2+2 c d e x+c e^2 x^2}}{c^3 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 629

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 643

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

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Mathematica [A]  time = 0.00, size = 23, normalized size = 0.74 \[ \frac {x (d+e x)}{c^2 \sqrt {c (d+e x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.18, size = 38, normalized size = 1.23 \[ \frac {\sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}} x}{c^{3} e x + c^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.57, size = 93, normalized size = 3.00 \[ \frac {2 \, C_{0} d^{3} e^{\left (-3\right )} - \frac {3 \, d^{4} e^{\left (-1\right )}}{c} + {\left (6 \, C_{0} d^{2} e^{\left (-2\right )} - \frac {8 \, d^{3}}{c} + {\left (6 \, C_{0} d e^{\left (-1\right )} + {\left (2 \, C_{0} + \frac {x e^{3}}{c}\right )} x - \frac {6 \, d^{2} e}{c}\right )} x\right )} x}{{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*C_0*d^3*e^(-3) - 3*d^4*e^(-1)/c + (6*C_0*d^2*e^(-2) - 8*d^3/c + (6*C_0*d*e^(-1) + (2*C_0 + x*e^3/c)*x - 6*d
^2*e/c)*x)*x)/(c*x^2*e^2 + 2*c*d*x*e + c*d^2)^(3/2)

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maple [A]  time = 0.04, size = 32, normalized size = 1.03 \[ \frac {\left (e x +d \right )^{5} x}{\left (c \,e^{2} x^{2}+2 c d e x +c \,d^{2}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.55, size = 140, normalized size = 4.52 \[ \frac {e^{3} x^{4}}{{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {3}{2}} c} - \frac {6 \, d^{2} e x^{2}}{{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {3}{2}} c} - \frac {17 \, d^{4}}{3 \, {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {3}{2}} c e} - \frac {8 \, d^{3}}{c^{\frac {5}{2}} e^{3} {\left (x + \frac {d}{e}\right )}^{2}} + \frac {32 \, d^{4}}{3 \, c^{\frac {5}{2}} e^{4} {\left (x + \frac {d}{e}\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^3*x^4/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c) - 6*d^2*e*x^2/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c) - 17
/3*d^4/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c*e) - 8*d^3/(c^(5/2)*e^3*(x + d/e)^2) + 32/3*d^4/(c^(5/2)*e^4*(
x + d/e)^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {{\left (d+e\,x\right )}^5}{{\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.64, size = 42, normalized size = 1.35 \[ \begin {cases} \frac {\sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}}}{c^{3} e} & \text {for}\: e \neq 0 \\\frac {d^{5} x}{\left (c d^{2}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**5/(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*e), Ne(e, 0)), (d**5*x/(c*d**2)**(5/2), True))

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