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

Optimal result

Integrand size = 32, antiderivative size = 39 \[ \int \frac {(d+e x)^6}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=\frac {(d+e x) \sqrt {c d^2+2 c d e x+c e^2 x^2}}{2 c^3 e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {656, 623} \[ \int \frac {(d+e x)^6}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=\frac {(d+e x) \sqrt {c d^2+2 c d e x+c e^2 x^2}}{2 c^3 e} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.85 \[ \int \frac {(d+e x)^6}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=\frac {x (d+e x) (2 d+e x)}{2 c^2 \sqrt {c (d+e x)^2}} \]

[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])

Maple [A] (verified)

Time = 3.30 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.03

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.23 \[ \int \frac {(d+e x)^6}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}} {\left (e x^{2} + 2 \, d x\right )}}{2 \, {\left (c^{3} e x + c^{3} d\right )}} \]

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

Sympy [A] (verification not implemented)

Time = 4.71 (sec) , antiderivative size = 224, normalized size of antiderivative = 5.74 \[ \int \frac {(d+e x)^6}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=\frac {\begin {cases} \left (\frac {d}{2 c e} + \frac {x}{2 c}\right ) \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}} & \text {for}\: c e^{2} \neq 0 \\\frac {2 d^{2} \sqrt {c d^{2} + 2 c d e x} + \frac {2 \left (- c d^{2} \sqrt {c d^{2} + 2 c d e x} + \frac {\left (c d^{2} + 2 c d e x\right )^{\frac {3}{2}}}{3}\right )}{c} + \frac {c^{2} d^{4} \sqrt {c d^{2} + 2 c d e x} - \frac {2 c d^{2} \left (c d^{2} + 2 c d e x\right )^{\frac {3}{2}}}{3} + \frac {\left (c d^{2} + 2 c d e x\right )^{\frac {5}{2}}}{5}}{2 c^{2} d^{2}}}{2 c d e} & \text {for}\: c d e \neq 0 \\\frac {d^{2} x + d e x^{2} + \frac {e^{2} x^{3}}{3}}{\sqrt {c d^{2}}} & \text {otherwise} \end {cases}}{c^{2}} \]

[In]

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

[Out]

Piecewise(((d/(2*c*e) + x/(2*c))*sqrt(c*d**2 + 2*c*d*e*x + c*e**2*x**2), Ne(c*e**2, 0)), ((2*d**2*sqrt(c*d**2
+ 2*c*d*e*x) + 2*(-c*d**2*sqrt(c*d**2 + 2*c*d*e*x) + (c*d**2 + 2*c*d*e*x)**(3/2)/3)/c + (c**2*d**4*sqrt(c*d**2
+ 2*c*d*e*x) - 2*c*d**2*(c*d**2 + 2*c*d*e*x)**(3/2)/3 + (c*d**2 + 2*c*d*e*x)**(5/2)/5)/(2*c**2*d**2))/(2*c*d*
e), Ne(c*d*e, 0)), ((d**2*x + d*e*x**2 + e**2*x**3/3)/sqrt(c*d**2), True))/c**2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 232 vs. \(2 (35) = 70\).

Time = 0.21 (sec) , antiderivative size = 232, normalized size of antiderivative = 5.95 \[ \int \frac {(d+e x)^6}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=\frac {e^{4} x^{5}}{2 \, {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {3}{2}} c} + \frac {5 \, d e^{3} x^{4}}{2 \, {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {3}{2}} c} - \frac {25 \, c^{2} d^{6} e^{4}}{4 \, \left (c e^{2}\right )^{\frac {9}{2}} {\left (x + \frac {d}{e}\right )}^{4}} - \frac {10 \, d^{3} e x^{2}}{{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {3}{2}} c} + \frac {50 \, c d^{5} e^{3}}{3 \, \left (c e^{2}\right )^{\frac {7}{2}} {\left (x + \frac {d}{e}\right )}^{3}} - \frac {26 \, d^{5}}{3 \, {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {3}{2}} c e} - \frac {25 \, d^{4} e^{2}}{2 \, \left (c e^{2}\right )^{\frac {5}{2}} {\left (x + \frac {d}{e}\right )}^{2}} + \frac {25 \, d^{6}}{4 \, \left (c e^{2}\right )^{\frac {5}{2}} {\left (x + \frac {d}{e}\right )}^{4}} \]

[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*e^4*x^5/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c) + 5/2*d*e^3*x^4/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c
) - 25/4*c^2*d^6*e^4/((c*e^2)^(9/2)*(x + d/e)^4) - 10*d^3*e*x^2/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c) + 50
/3*c*d^5*e^3/((c*e^2)^(7/2)*(x + d/e)^3) - 26/3*d^5/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c*e) - 25/2*d^4*e^2
/((c*e^2)^(5/2)*(x + d/e)^2) + 25/4*d^6/((c*e^2)^(5/2)*(x + d/e)^4)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.59 \[ \int \frac {(d+e x)^6}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=\frac {e x^{2} + 2 \, d x}{2 \, c^{\frac {5}{2}} \mathrm {sgn}\left (e x + d\right )} \]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^6}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^6}{{\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^{5/2}} \,d x \]

[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)