\(\int \frac {(c d^2+2 c d e x+c e^2 x^2)^{5/2}}{d+e x} \, dx\) [1054]

Optimal result

Integrand size = 32, antiderivative size = 31 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{d+e x} \, dx=\frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{5 e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, 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 = {657, 643} \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{d+e x} \, dx=\frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{5 e} \]

[In]

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

[Out]

(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2)/(5*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*} \text {integral}& = c \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx \\ & = \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{5 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{d+e x} \, dx=\frac {\left (c (d+e x)^2\right )^{5/2}}{5 e} \]

[In]

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

[Out]

(c*(d + e*x)^2)^(5/2)/(5*e)

Maple [A] (verified)

Time = 2.69 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87

method	result	size
risch	\(\frac {c^{2} \left (e x +d \right )^{4} \sqrt {c \left (e x +d \right )^{2}}}{5 e}\)	\(27\)
pseudoelliptic	\(\frac {c^{2} \left (e x +d \right )^{4} \sqrt {c \left (e x +d \right )^{2}}}{5 e}\)	\(27\)
default	\(\frac {\left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{\frac {5}{2}}}{5 e}\)	\(28\)
gosper	\(\frac {x \left (e^{4} x^{4}+5 d \,e^{3} x^{3}+10 d^{2} e^{2} x^{2}+10 d^{3} e x +5 d^{4}\right ) \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{\frac {5}{2}}}{5 \left (e x +d \right )^{5}}\)	\(73\)
trager	\(\frac {c^{2} x \left (e^{4} x^{4}+5 d \,e^{3} x^{3}+10 d^{2} e^{2} x^{2}+10 d^{3} e x +5 d^{4}\right ) \sqrt {c \,x^{2} e^{2}+2 x c d e +c \,d^{2}}}{5 e x +5 d}\)	\(76\)

[In]

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

[Out]

1/5*c^2*(e*x+d)^4*(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. 89 vs. \(2 (27) = 54\).

Time = 0.27 (sec) , antiderivative size = 89, normalized size of antiderivative = 2.87 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{d+e x} \, dx=\frac {{\left (c^{2} e^{4} x^{5} + 5 \, c^{2} d e^{3} x^{4} + 10 \, c^{2} d^{2} e^{2} x^{3} + 10 \, c^{2} d^{3} e x^{2} + 5 \, c^{2} d^{4} x\right )} \sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{5 \, {\left (e x + d\right )}} \]

[In]

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

[Out]

1/5*(c^2*e^4*x^5 + 5*c^2*d*e^3*x^4 + 10*c^2*d^2*e^2*x^3 + 10*c^2*d^3*e*x^2 + 5*c^2*d^4*x)*sqrt(c*e^2*x^2 + 2*c
*d*e*x + c*d^2)/(e*x + d)

Sympy [A] (verification not implemented)

Time = 1.65 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{d+e x} \, dx=\begin {cases} \frac {\left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{\frac {5}{2}}}{5 e} & \text {for}\: e \neq 0 \\\frac {x \left (c d^{2}\right )^{\frac {5}{2}}}{d} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{d+e x} \, dx=\frac {{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {5}{2}}}{5 \, e} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (27) = 54\).

Time = 0.27 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.81 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{d+e x} \, dx=\frac {1}{5} \, {\left (\frac {c^{2} d^{5} \mathrm {sgn}\left (e x + d\right )}{e} + {\left (c^{2} e^{4} x^{5} + 5 \, c^{2} d e^{3} x^{4} + 10 \, c^{2} d^{2} e^{2} x^{3} + 10 \, c^{2} d^{3} e x^{2} + 5 \, c^{2} d^{4} x\right )} \mathrm {sgn}\left (e x + d\right )\right )} \sqrt {c} \]

[In]

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

[Out]

1/5*(c^2*d^5*sgn(e*x + d)/e + (c^2*e^4*x^5 + 5*c^2*d*e^3*x^4 + 10*c^2*d^2*e^2*x^3 + 10*c^2*d^3*e*x^2 + 5*c^2*d
^4*x)*sgn(e*x + d))*sqrt(c)

Mupad [B] (verification not implemented)

Time = 9.70 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.52 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{d+e x} \, dx=\frac {{\left (c\,{\left (d+e\,x\right )}^2\right )}^{5/2}}{5\,e} \]

[In]

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

[Out]

(c*(d + e*x)^2)^(5/2)/(5*e)