### 3.1058 $$\int \frac{(c d^2+2 c d e x+c e^2 x^2)^{5/2}}{(d+e x)^5} \, dx$$

Optimal. Leaf size=31 $\frac{c^2 \sqrt{c d^2+2 c d e x+c e^2 x^2}}{e}$

[Out]

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

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Rubi [A]  time = 0.0227217, 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} $\frac{c^2 \sqrt{c d^2+2 c d e x+c e^2 x^2}}{e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0063779, size = 23, normalized size = 0.74 $\frac{c^3 x (d+e x)}{\sqrt{c (d+e x)^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(c^3*x*(d + e*x))/Sqrt[c*(d + e*x)^2]

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Maple [A]  time = 0.041, size = 32, normalized size = 1. \begin{align*}{\frac{x}{ \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)^5,x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \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)^5,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.26425, size = 72, normalized size = 2.32 \begin{align*} \frac{\sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}} c^{2} x}{e x + d} \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)^5,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 8.02297, size = 41, normalized size = 1.32 \begin{align*} c^{2} \left (\begin{cases} \frac{x \sqrt{c d^{2}}}{d} & \text{for}\: e = 0 \\\frac{\sqrt{c d^{2} + 2 c d e x + c e^{2} x^{2}}}{e} & \text{otherwise} \end{cases}\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)**5,x)

[Out]

c**2*Piecewise((x*sqrt(c*d**2)/d, Eq(e, 0)), (sqrt(c*d**2 + 2*c*d*e*x + c*e**2*x**2)/e, True))

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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)^5,x, algorithm="giac")

[Out]

Exception raised: TypeError