3.11.55 \(\int \frac {(c d^2+2 c d e x+c e^2 x^2)^{5/2}}{(d+e x)^2} \, dx\) [1055]

Optimal. Leaf size=37 \[ \frac {c (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{4 e} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 37, 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, 623} \begin {gather*} \frac {c (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{4 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 623

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1)
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-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 {\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx &=c \int \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx\\ &=\frac {c (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{4 e}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 26, normalized size = 0.70 \begin {gather*} \frac {c (d+e x) \left (c (d+e x)^2\right )^{3/2}}{4 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.59, size = 35, normalized size = 0.95

method	result	size
risch	\(\frac {c^{2} \left (e x +d \right )^{3} \sqrt {\left (e x +d \right )^{2} c}}{4 e}\)	\(27\)
default	\(\frac {\left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )^{\frac {5}{2}}}{4 \left (e x +d \right ) e}\)	\(35\)
gosper	\(\frac {x \left (e^{3} x^{3}+4 d \,e^{2} x^{2}+6 d^{2} e x +4 d^{3}\right ) \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )^{\frac {5}{2}}}{4 \left (e x +d \right )^{5}}\)	\(62\)
trager	\(\frac {c^{2} x \left (e^{3} x^{3}+4 d \,e^{2} x^{2}+6 d^{2} e x +4 d^{3}\right ) \sqrt {x^{2} c \,e^{2}+2 c d e x +c \,d^{2}}}{4 e x +4 d}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 35, normalized size = 0.95 \begin {gather*} \frac {{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{\frac {5}{2}}}{4 \, {\left (x e^{2} + d e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (33) = 66\).
time = 3.11, size = 75, normalized size = 2.03 \begin {gather*} \frac {{\left (c^{2} x^{4} e^{3} + 4 \, c^{2} d x^{3} e^{2} + 6 \, c^{2} d^{2} x^{2} e + 4 \, c^{2} d^{3} x\right )} \sqrt {c x^{2} e^{2} + 2 \, c d x e + c d^{2}}}{4 \, {\left (x e + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (33) = 66\).
time = 1.04, size = 76, normalized size = 2.05 \begin {gather*} \frac {1}{4} \, {\left (c^{2} x^{4} e^{3} \mathrm {sgn}\left (x e + d\right ) + 4 \, c^{2} d x^{3} e^{2} \mathrm {sgn}\left (x e + d\right ) + 6 \, c^{2} d^{2} x^{2} e \mathrm {sgn}\left (x e + d\right ) + 4 \, c^{2} d^{3} x \mathrm {sgn}\left (x e + d\right )\right )} \sqrt {c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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