\(\int (d+e x) (c d^2+2 c d e x+c e^2 x^2)^{5/2} \, dx\) [1052]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 30, antiderivative size = 34 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2} \, dx=\frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{7/2}}{7 c e} \]

[Out]

1/7*(c*e^2*x^2+2*c*d*e*x+c*d^2)^(7/2)/c/e

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {643} \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2} \, dx=\frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{7/2}}{7 c e} \]

[In]

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

[Out]

(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(7/2)/(7*c*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]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{7/2}}{7 c e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.68 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2} \, dx=\frac {\left (c (d+e x)^2\right )^{7/2}}{7 c e} \]

[In]

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

[Out]

(c*(d + e*x)^2)^(7/2)/(7*c*e)

Maple [A] (verified)

Time = 2.56 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79

method result size
risch \(\frac {c^{2} \left (e x +d \right )^{6} \sqrt {c \left (e x +d \right )^{2}}}{7 e}\) \(27\)
pseudoelliptic \(\frac {c^{2} \left (e x +d \right )^{6} \sqrt {c \left (e x +d \right )^{2}}}{7 e}\) \(27\)
default \(\frac {\left (e x +d \right )^{2} \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{\frac {5}{2}}}{7 e}\) \(35\)
gosper \(\frac {x \left (e^{6} x^{6}+7 d \,e^{5} x^{5}+21 d^{2} e^{4} x^{4}+35 x^{3} d^{3} e^{3}+35 d^{4} e^{2} x^{2}+21 d^{5} e x +7 d^{6}\right ) \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{\frac {5}{2}}}{7 \left (e x +d \right )^{5}}\) \(95\)
trager \(\frac {c^{2} x \left (e^{6} x^{6}+7 d \,e^{5} x^{5}+21 d^{2} e^{4} x^{4}+35 x^{3} d^{3} e^{3}+35 d^{4} e^{2} x^{2}+21 d^{5} e x +7 d^{6}\right ) \sqrt {c \,x^{2} e^{2}+2 x c d e +c \,d^{2}}}{7 e x +7 d}\) \(98\)

[In]

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

[Out]

1/7*c^2*(e*x+d)^6*(c*(e*x+d)^2)^(1/2)/e

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (30) = 60\).

Time = 0.30 (sec) , antiderivative size = 117, normalized size of antiderivative = 3.44 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2} \, dx=\frac {{\left (c^{2} e^{6} x^{7} + 7 \, c^{2} d e^{5} x^{6} + 21 \, c^{2} d^{2} e^{4} x^{5} + 35 \, c^{2} d^{3} e^{3} x^{4} + 35 \, c^{2} d^{4} e^{2} x^{3} + 21 \, c^{2} d^{5} e x^{2} + 7 \, c^{2} d^{6} x\right )} \sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{7 \, {\left (e x + d\right )}} \]

[In]

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

[Out]

1/7*(c^2*e^6*x^7 + 7*c^2*d*e^5*x^6 + 21*c^2*d^2*e^4*x^5 + 35*c^2*d^3*e^3*x^4 + 35*c^2*d^4*e^2*x^3 + 21*c^2*d^5
*e*x^2 + 7*c^2*d^6*x)*sqrt(c*e^2*x^2 + 2*c*d*e*x + c*d^2)/(e*x + d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (29) = 58\).

Time = 0.36 (sec) , antiderivative size = 287, normalized size of antiderivative = 8.44 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2} \, dx=\begin {cases} \frac {c^{2} d^{6} \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}}}{7 e} + \frac {6 c^{2} d^{5} x \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}}}{7} + \frac {15 c^{2} d^{4} e x^{2} \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}}}{7} + \frac {20 c^{2} d^{3} e^{2} x^{3} \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}}}{7} + \frac {15 c^{2} d^{2} e^{3} x^{4} \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}}}{7} + \frac {6 c^{2} d e^{4} x^{5} \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}}}{7} + \frac {c^{2} e^{5} x^{6} \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}}}{7} & \text {for}\: e \neq 0 \\d x \left (c d^{2}\right )^{\frac {5}{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((c**2*d**6*sqrt(c*d**2 + 2*c*d*e*x + c*e**2*x**2)/(7*e) + 6*c**2*d**5*x*sqrt(c*d**2 + 2*c*d*e*x + c*
e**2*x**2)/7 + 15*c**2*d**4*e*x**2*sqrt(c*d**2 + 2*c*d*e*x + c*e**2*x**2)/7 + 20*c**2*d**3*e**2*x**3*sqrt(c*d*
*2 + 2*c*d*e*x + c*e**2*x**2)/7 + 15*c**2*d**2*e**3*x**4*sqrt(c*d**2 + 2*c*d*e*x + c*e**2*x**2)/7 + 6*c**2*d*e
**4*x**5*sqrt(c*d**2 + 2*c*d*e*x + c*e**2*x**2)/7 + c**2*e**5*x**6*sqrt(c*d**2 + 2*c*d*e*x + c*e**2*x**2)/7, N
e(e, 0)), (d*x*(c*d**2)**(5/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2} \, dx=\frac {{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {7}{2}}}{7 \, c e} \]

[In]

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

[Out]

1/7*(c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(7/2)/(c*e)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (30) = 60\).

Time = 0.27 (sec) , antiderivative size = 149, normalized size of antiderivative = 4.38 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2} \, dx=\frac {1}{7} \, {\left (c^{2} e^{6} x^{7} \mathrm {sgn}\left (e x + d\right ) + 7 \, c^{2} d e^{5} x^{6} \mathrm {sgn}\left (e x + d\right ) + 21 \, c^{2} d^{2} e^{4} x^{5} \mathrm {sgn}\left (e x + d\right ) + 35 \, c^{2} d^{3} e^{3} x^{4} \mathrm {sgn}\left (e x + d\right ) + 35 \, c^{2} d^{4} e^{2} x^{3} \mathrm {sgn}\left (e x + d\right ) + 21 \, c^{2} d^{5} e x^{2} \mathrm {sgn}\left (e x + d\right ) + 7 \, c^{2} d^{6} x \mathrm {sgn}\left (e x + d\right ) + \frac {c^{2} d^{7} \mathrm {sgn}\left (e x + d\right )}{e}\right )} \sqrt {c} \]

[In]

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

[Out]

1/7*(c^2*e^6*x^7*sgn(e*x + d) + 7*c^2*d*e^5*x^6*sgn(e*x + d) + 21*c^2*d^2*e^4*x^5*sgn(e*x + d) + 35*c^2*d^3*e^
3*x^4*sgn(e*x + d) + 35*c^2*d^4*e^2*x^3*sgn(e*x + d) + 21*c^2*d^5*e*x^2*sgn(e*x + d) + 7*c^2*d^6*x*sgn(e*x + d
) + c^2*d^7*sgn(e*x + d)/e)*sqrt(c)

Mupad [B] (verification not implemented)

Time = 9.97 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.56 \[ \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2} \, dx=\frac {{\left (c\,{\left (d+e\,x\right )}^2\right )}^{7/2}}{7\,c\,e} \]

[In]

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

[Out]

(c*(d + e*x)^2)^(7/2)/(7*c*e)