\(\int \frac {(d+e x)^4}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx\) [1062]

Optimal result

Integrand size = 32, antiderivative size = 39 \[ \int \frac {(d+e x)^4}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=\frac {(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{4 c^2 e} \]

[Out]

1/4*(e*x+d)*(c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2)/c^2/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} \[ \int \frac {(d+e x)^4}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=\frac {(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{4 c^2 e} \]

[In]

Int[(d + e*x)^4/Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2],x]

[Out]

((d + e*x)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2))/(4*c^2*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 \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx}{c^2} \\ & = \frac {(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{4 c^2 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.69 \[ \int \frac {(d+e x)^4}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=\frac {(d+e x)^5}{4 e \sqrt {c (d+e x)^2}} \]

[In]

Integrate[(d + e*x)^4/Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2],x]

[Out]

(d + e*x)^5/(4*e*Sqrt[c*(d + e*x)^2])

Maple [A] (verified)

Time = 2.56 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.62

method	result	size
risch	\(\frac {\left (e x +d \right )^{5}}{4 \sqrt {c \left (e x +d \right )^{2}}\, e}\)	\(24\)
default	\(\frac {\left (e x +d \right )^{5}}{4 \sqrt {c \,x^{2} e^{2}+2 x c d e +c \,d^{2}}\, e}\)	\(35\)
gosper	\(\frac {x \left (e^{3} x^{3}+4 d \,e^{2} x^{2}+6 d^{2} e x +4 d^{3}\right ) \left (e x +d \right )}{4 \sqrt {c \,x^{2} e^{2}+2 x c d e +c \,d^{2}}}\)	\(60\)
trager	\(\frac {x \left (e^{3} x^{3}+4 d \,e^{2} x^{2}+6 d^{2} e x +4 d^{3}\right ) \sqrt {c \,x^{2} e^{2}+2 x c d e +c \,d^{2}}}{4 c \left (e x +d \right )}\)	\(65\)

[In]

int((e*x+d)^4/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.69 \[ \int \frac {(d+e x)^4}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=\frac {{\left (e^{3} x^{4} + 4 \, d e^{2} x^{3} + 6 \, d^{2} e x^{2} + 4 \, d^{3} x\right )} \sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{4 \, {\left (c e x + c d\right )}} \]

[In]

integrate((e*x+d)^4/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (37) = 74\).

Time = 0.91 (sec) , antiderivative size = 219, normalized size of antiderivative = 5.62 \[ \int \frac {(d+e x)^4}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=\begin {cases} \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}} \left (\frac {d^{3}}{4 c e} + \frac {3 d^{2} x}{4 c} + \frac {3 d e x^{2}}{4 c} + \frac {e^{2} x^{3}}{4 c}\right ) & \text {for}\: c e^{2} \neq 0 \\\frac {\frac {d^{4} \sqrt {c d^{2} + 2 c d e x}}{16} + \frac {d^{2} \left (c d^{2} + 2 c d e x\right )^{\frac {3}{2}}}{12 c} + \frac {3 \left (c d^{2} + 2 c d e x\right )^{\frac {5}{2}}}{40 c^{2}} + \frac {\left (c d^{2} + 2 c d e x\right )^{\frac {7}{2}}}{28 c^{3} d^{2}} + \frac {\left (c d^{2} + 2 c d e x\right )^{\frac {9}{2}}}{144 c^{4} d^{4}}}{c d e} & \text {for}\: c d e \neq 0 \\\frac {\begin {cases} d^{4} x & \text {for}\: e = 0 \\\frac {\left (d + e x\right )^{5}}{5 e} & \text {otherwise} \end {cases}}{\sqrt {c d^{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate((e*x+d)**4/(c*e**2*x**2+2*c*d*e*x+c*d**2)**(1/2),x)

[Out]

Piecewise((sqrt(c*d**2 + 2*c*d*e*x + c*e**2*x**2)*(d**3/(4*c*e) + 3*d**2*x/(4*c) + 3*d*e*x**2/(4*c) + e**2*x**
3/(4*c)), Ne(c*e**2, 0)), ((d**4*sqrt(c*d**2 + 2*c*d*e*x)/16 + d**2*(c*d**2 + 2*c*d*e*x)**(3/2)/(12*c) + 3*(c*
d**2 + 2*c*d*e*x)**(5/2)/(40*c**2) + (c*d**2 + 2*c*d*e*x)**(7/2)/(28*c**3*d**2) + (c*d**2 + 2*c*d*e*x)**(9/2)/
(144*c**4*d**4))/(c*d*e), Ne(c*d*e, 0)), (Piecewise((d**4*x, Eq(e, 0)), ((d + e*x)**5/(5*e), True))/sqrt(c*d**
2), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (35) = 70\).

Time = 0.21 (sec) , antiderivative size = 182, normalized size of antiderivative = 4.67 \[ \int \frac {(d+e x)^4}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=\frac {3 \, c^{2} d^{4} e^{4} \log \left (x + \frac {d}{e}\right )}{2 \, \left (c e^{2}\right )^{\frac {5}{2}}} - \frac {3 \, c d^{3} e^{3} x}{2 \, \left (c e^{2}\right )^{\frac {3}{2}}} + \frac {3 \, d^{2} e^{2} x^{2}}{4 \, \sqrt {c e^{2}}} - \frac {3}{2} \, d^{4} \sqrt {\frac {1}{c e^{2}}} \log \left (x + \frac {d}{e}\right ) + \frac {\sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}} e^{2} x^{3}}{4 \, c} + \frac {3 \, \sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}} d e x^{2}}{4 \, c} + \frac {5 \, \sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}} d^{3}}{2 \, c e} \]

[In]

integrate((e*x+d)^4/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.46 \[ \int \frac {(d+e x)^4}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=\frac {\sqrt {c} e^{3} x^{4} + 4 \, \sqrt {c} d e^{2} x^{3} + 6 \, \sqrt {c} d^{2} e x^{2} + 4 \, \sqrt {c} d^{3} x}{4 \, c \mathrm {sgn}\left (e x + d\right )} \]

[In]

integrate((e*x+d)^4/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^4}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx=\int \frac {{\left (d+e\,x\right )}^4}{\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}} \,d x \]

[In]

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

[Out]

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