3.11.79 \(\int \frac {(d+e x)^6}{(c d^2+2 c d e x+c e^2 x^2)^{5/2}} \, dx\) [1079]

Optimal. Leaf size=39 \[ \frac {(d+e x) \sqrt {c d^2+2 c d e x+c e^2 x^2}}{2 c^3 e} \]

[Out]

1/2*(e*x+d)*(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 = 0.01, antiderivative size = 39, 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 {(d+e x) \sqrt {c d^2+2 c d e x+c e^2 x^2}}{2 c^3 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^6/(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2),x]

[Out]

((d + e*x)*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2])/(2*c^3*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 {(d+e x)^6}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx &=\frac {\int \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx}{c^3}\\ &=\frac {(d+e x) \sqrt {c d^2+2 c d e x+c e^2 x^2}}{2 c^3 e}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 33, normalized size = 0.85 \begin {gather*} \frac {x (d+e x) (2 d+e x)}{2 c^2 \sqrt {c (d+e x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^6/(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2),x]

[Out]

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

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Maple [A]
time = 0.68, size = 40, normalized size = 1.03

method	result	size
gosper	\(\frac {x \left (e x +2 d \right ) \left (e x +d \right )^{5}}{2 \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )^{\frac {5}{2}}}\)	\(40\)
default	\(\frac {x \left (e x +2 d \right ) \left (e x +d \right )^{5}}{2 \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )^{\frac {5}{2}}}\)	\(40\)
trager	\(\frac {x \left (e x +2 d \right ) \sqrt {x^{2} c \,e^{2}+2 c d e x +c \,d^{2}}}{2 c^{3} \left (e x +d \right )}\)	\(43\)
risch	\(\frac {\left (e x +d \right ) e \,x^{2}}{2 c^{2} \sqrt {\left (e x +d \right )^{2} c}}+\frac {\left (e x +d \right ) d x}{c^{2} \sqrt {\left (e x +d \right )^{2} c}}\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^6/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 169 vs. \(2 (35) = 70\).
time = 0.32, size = 169, normalized size = 4.33 \begin {gather*} \frac {x^{5} e^{4}}{2 \, {\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{\frac {3}{2}} c} + \frac {5 \, d x^{4} e^{3}}{2 \, {\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{\frac {3}{2}} c} - \frac {10 \, d^{3} x^{2} e}{{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{\frac {3}{2}} c} - \frac {26 \, d^{5} 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 {25 \, d^{4} e^{\left (-3\right )}}{2 \, {\left (d e^{\left (-1\right )} + x\right )}^{2} c^{\frac {5}{2}}} + \frac {50 \, d^{5} 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)^6/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="maxima")

[Out]

1/2*x^5*e^4/((c*x^2*e^2 + 2*c*d*x*e + c*d^2)^(3/2)*c) + 5/2*d*x^4*e^3/((c*x^2*e^2 + 2*c*d*x*e + c*d^2)^(3/2)*c
) - 10*d^3*x^2*e/((c*x^2*e^2 + 2*c*d*x*e + c*d^2)^(3/2)*c) - 26/3*d^5*e^(-1)/((c*x^2*e^2 + 2*c*d*x*e + c*d^2)^
(3/2)*c) - 25/2*d^4*e^(-3)/((d*e^(-1) + x)^2*c^(5/2)) + 50/3*d^5*e^(-4)/((d*e^(-1) + x)^3*c^(5/2))

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Fricas [A]
time = 2.36, size = 50, normalized size = 1.28 \begin {gather*} \frac {\sqrt {c x^{2} e^{2} + 2 \, c d x e + c d^{2}} {\left (x^{2} e + 2 \, d x\right )}}{2 \, {\left (c^{3} x e + c^{3} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^6/(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)*(x^2*e + 2*d*x)/(c^3*x*e + c^3*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{6}}{\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)**6/(c*e**2*x**2+2*c*d*e*x+c*d**2)**(5/2),x)

[Out]

Integral((d + e*x)**6/(c*(d + e*x)**2)**(5/2), x)

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Giac [A]
time = 1.16, size = 25, normalized size = 0.64 \begin {gather*} \frac {x^{2} e + 2 \, d x}{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)^6/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="giac")

[Out]

1/2*(x^2*e + 2*d*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 )}^6}{{\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)^6/(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(5/2),x)

[Out]

int((d + e*x)^6/(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(5/2), x)

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