3.1056 \(\int \frac{(c d^2+2 c d e x+c e^2 x^2)^{5/2}}{(d+e x)^3} \, dx\)

Optimal. Leaf size=32 \[ \frac{c \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{3 e} \]

[Out]

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

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Rubi [A]  time = 0.0227479, antiderivative size = 32, 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} \[ \frac{c \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{3 e} \]

Antiderivative was successfully verified.

[In]

Int[(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2)/(d + e*x)^3,x]

[Out]

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

Mathematica [A]  time = 0.0060843, size = 21, normalized size = 0.66 \[ \frac{c \left (c (d+e x)^2\right )^{3/2}}{3 e} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2)/(d + e*x)^3,x]

[Out]

(c*(c*(d + e*x)^2)^(3/2))/(3*e)

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Maple [A]  time = 0.04, size = 51, normalized size = 1.6 \begin{align*}{\frac{x \left ({e}^{2}{x}^{2}+3\,dex+3\,{d}^{2} \right ) }{3\, \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)^3,x)

[Out]

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

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.33436, size = 128, normalized size = 4. \begin{align*} \frac{{\left (c^{2} e^{2} x^{3} + 3 \, c^{2} d e x^{2} + 3 \, c^{2} d^{2} x\right )} \sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{3 \,{\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)^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c \left (d + e x\right )^{2}\right )^{\frac{5}{2}}}{\left (d + e x\right )^{3}}\, dx \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)**3,x)

[Out]

Integral((c*(d + e*x)**2)**(5/2)/(d + e*x)**3, x)

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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)^3,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError