### 3.1054 $$\int \frac{(c d^2+2 c d e x+c e^2 x^2)^{5/2}}{d+e x} \, dx$$

Optimal. Leaf size=31 $\frac{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{5 e}$

[Out]

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

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Rubi [A]  time = 0.0237312, antiderivative size = 31, normalized size of antiderivative = 1., 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} $\frac{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{5 e}$

Antiderivative was successfully veriﬁed.

[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_))^(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]

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]

Rubi steps

\begin{align*} \int \frac{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{d+e x} \, dx &=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]  time = 0.0054284, size = 20, normalized size = 0.65 $\frac{\left (c (d+e x)^2\right )^{5/2}}{5 e}$

Antiderivative was successfully veriﬁed.

[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)

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Maple [B]  time = 0.04, size = 73, normalized size = 2.4 \begin{align*}{\frac{x \left ({e}^{4}{x}^{4}+5\,d{e}^{3}{x}^{3}+10\,{d}^{2}{e}^{2}{x}^{2}+10\,{d}^{3}ex+5\,{d}^{4} \right ) }{5\, \left ( ex+d \right ) ^{5}} \left ( c{e}^{2}{x}^{2}+2\,cdex+c{d}^{2} \right ) ^{{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.37133, size = 185, normalized size = 5.97 \begin{align*} \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 )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 5.57784, size = 39, normalized size = 1.26 \begin{align*} \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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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))

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError