3.11.80 \(\int \frac {(d+e x)^5}{(c d^2+2 c d e x+c e^2 x^2)^{5/2}} \, dx\) [1080]

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 = {657, 643} \begin {gather*} \frac {\sqrt {c d^2+2 c d e x+c e^2 x^2}}{c^3 e} \end {gather*}

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 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 {(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 \begin {gather*} \frac {x (d+e x)}{c^2 \sqrt {c (d+e x)^2}} \end {gather*}

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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Maple [A]
time = 0.58, size = 32, normalized size = 1.03

method	result	size
risch	\(\frac {\left (e x +d \right ) x}{c^{2} \sqrt {\left (e x +d \right )^{2} c}}\)	\(22\)
default	\(\frac {x \left (e x +d \right )^{5}}{\left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )^{\frac {5}{2}}}\)	\(32\)
trager	\(\frac {x \sqrt {x^{2} c \,e^{2}+2 c d e x +c \,d^{2}}}{c^{3} \left (e x +d \right )}\)	\(35\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 135 vs. \(2 (28) = 56\).
time = 0.29, size = 135, normalized size = 4.35 \begin {gather*} \frac {x^{4} e^{3}}{{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{\frac {3}{2}} c} - \frac {6 \, d^{2} x^{2} e}{{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{\frac {3}{2}} c} - \frac {17 \, d^{4} 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 {8 \, d^{3} e^{\left (-3\right )}}{{\left (d e^{\left (-1\right )} + x\right )}^{2} c^{\frac {5}{2}}} + \frac {32 \, d^{4} 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)^5/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="maxima")

[Out]

x^4*e^3/((c*x^2*e^2 + 2*c*d*x*e + c*d^2)^(3/2)*c) - 6*d^2*x^2*e/((c*x^2*e^2 + 2*c*d*x*e + c*d^2)^(3/2)*c) - 17
/3*d^4*e^(-1)/((c*x^2*e^2 + 2*c*d*x*e + c*d^2)^(3/2)*c) - 8*d^3*e^(-3)/((d*e^(-1) + x)^2*c^(5/2)) + 32/3*d^4*e
^(-4)/((d*e^(-1) + x)^3*c^(5/2))

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Fricas [A]
time = 1.99, size = 39, normalized size = 1.26 \begin {gather*} \frac {\sqrt {c x^{2} e^{2} + 2 \, c d x e + c d^{2}} x}{c^{3} x e + c^{3} d} \end {gather*}

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

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Sympy [A]
time = 0.46, size = 42, normalized size = 1.35 \begin {gather*} \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} \end {gather*}

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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Giac [A]
time = 0.91, size = 14, normalized size = 0.45 \begin {gather*} \frac {x}{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)^5/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="giac")

[Out]

x/(c^(5/2)*sgn(x*e + d))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \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 \end {gather*}

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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