3.1029 \(\int (d+e x)^3 \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx\)

Optimal. Leaf size=34 \[ \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{5 c^2 e} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 34, 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 = {643, 629} \[ \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{5 c^2 e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2],x]

[Out]

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

Rule 629

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 643

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

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Mathematica [A]  time = 0.01, size = 27, normalized size = 0.79 \[ \frac {(d+e x)^4 \sqrt {c (d+e x)^2}}{5 e} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2],x]

[Out]

((d + e*x)^4*Sqrt[c*(d + e*x)^2])/(5*e)

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fricas [B]  time = 0.84, size = 74, normalized size = 2.18 \[ \frac {{\left (e^{4} x^{5} + 5 \, d e^{3} x^{4} + 10 \, d^{2} e^{2} x^{3} + 10 \, d^{3} e x^{2} + 5 \, d^{4} x\right )} \sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{5 \, {\left (e x + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(c*e^2*x^2+2*c*d*e*x+c*d^2)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.24, size = 61, normalized size = 1.79 \[ \frac {1}{5} \, {\left (d^{4} e^{\left (-1\right )} + {\left (4 \, d^{3} + {\left (6 \, d^{2} e + {\left (x e^{3} + 4 \, d e^{2}\right )} x\right )} x\right )} x\right )} \sqrt {c x^{2} e^{2} + 2 \, c d x e + c d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(c*e^2*x^2+2*c*d*e*x+c*d^2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.05, size = 73, normalized size = 2.15 \[ \frac {\left (e^{4} x^{4}+5 d \,e^{3} x^{3}+10 d^{2} e^{2} x^{2}+10 d^{3} e x +5 d^{4}\right ) \sqrt {c \,e^{2} x^{2}+2 c d e x +c \,d^{2}}\, x}{5 e x +5 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.48, size = 94, normalized size = 2.76 \[ \frac {{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {3}{2}} e x^{2}}{5 \, c} + \frac {2 \, {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {3}{2}} d x}{5 \, c} + \frac {{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {3}{2}} d^{2}}{5 \, c e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(c*e^2*x^2+2*c*d*e*x+c*d^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.91, size = 239, normalized size = 7.03 \[ \frac {d^3\,\left (\frac {x}{2}+\frac {d}{2\,e}\right )\,\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}}{4}+\frac {e\,x^2\,{\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^{3/2}}{5\,c}-\frac {23\,d^2\,\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}\,\left (8\,e^2\,\left (d^2+e^2\,x^2\right )-12\,d^2\,e^2+4\,d\,e^3\,x\right )}{480\,e^3}+\frac {3\,d\,x\,{\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^{3/2}}{4\,c}-\frac {7\,d\,\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}\,\left (c\,d^3+3\,e\,x\,\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )-4\,c\,d^2\,e\,x-5\,c\,d\,e^2\,x^2\right )}{60\,c\,e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d^3*(x/2 + d/(2*e))*(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(1/2))/4 + (e*x^2*(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(3/2))/
(5*c) - (23*d^2*(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(1/2)*(8*e^2*(d^2 + e^2*x^2) - 12*d^2*e^2 + 4*d*e^3*x))/(480*e
^3) + (3*d*x*(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(3/2))/(4*c) - (7*d*(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(1/2)*(c*d^3
+ 3*e*x*(c*d^2 + c*e^2*x^2 + 2*c*d*e*x) - 4*c*d^2*e*x - 5*c*d*e^2*x^2))/(60*c*e)

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sympy [A]  time = 0.55, size = 187, normalized size = 5.50 \[ \begin {cases} \frac {d^{4} \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}}}{5 e} + \frac {4 d^{3} x \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}}}{5} + \frac {6 d^{2} e x^{2} \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}}}{5} + \frac {4 d e^{2} x^{3} \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}}}{5} + \frac {e^{3} x^{4} \sqrt {c d^{2} + 2 c d e x + c e^{2} x^{2}}}{5} & \text {for}\: e \neq 0 \\d^{3} x \sqrt {c d^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((d**4*sqrt(c*d**2 + 2*c*d*e*x + c*e**2*x**2)/(5*e) + 4*d**3*x*sqrt(c*d**2 + 2*c*d*e*x + c*e**2*x**2)
/5 + 6*d**2*e*x**2*sqrt(c*d**2 + 2*c*d*e*x + c*e**2*x**2)/5 + 4*d*e**2*x**3*sqrt(c*d**2 + 2*c*d*e*x + c*e**2*x
**2)/5 + e**3*x**4*sqrt(c*d**2 + 2*c*d*e*x + c*e**2*x**2)/5, Ne(e, 0)), (d**3*x*sqrt(c*d**2), True))

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