3.1055 \(\int \frac{(c d^2+2 c d e x+c e^2 x^2)^{5/2}}{(d+e x)^2} \, dx\)

Optimal. Leaf size=37 \[ \frac{c (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{4 e} \]

[Out]

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

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Rubi [A]  time = 0.0209725, antiderivative size = 37, 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 = {642, 609} \[ \frac{c (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{4 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 642

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]

Rule 609

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)]

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

Mathematica [A]  time = 0.0048314, size = 26, normalized size = 0.7 \[ \frac{c (d+e x) \left (c (d+e x)^2\right )^{3/2}}{4 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.042, size = 62, normalized size = 1.7 \begin{align*}{\frac{x \left ({e}^{3}{x}^{3}+4\,d{e}^{2}{x}^{2}+6\,{d}^{2}ex+4\,{d}^{3} \right ) }{4\, \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)^2,x)

[Out]

1/4*x*(e^3*x^3+4*d*e^2*x^2+6*d^2*e*x+4*d^3)*(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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.25465, size = 155, normalized size = 4.19 \begin{align*} \frac{{\left (c^{2} e^{3} x^{4} + 4 \, c^{2} d e^{2} x^{3} + 6 \, c^{2} d^{2} e x^{2} + 4 \, c^{2} d^{3} x\right )} \sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{4 \,{\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)^2,x, algorithm="fricas")

[Out]

1/4*(c^2*e^3*x^4 + 4*c^2*d*e^2*x^3 + 6*c^2*d^2*e*x^2 + 4*c^2*d^3*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 )^{2}}\, 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)**2,x)

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)^2,x, algorithm="giac")

[Out]

Timed out