3.2066 $$\int \frac{(d+e x)^{5/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx$$

Optimal. Leaf size=105 $\frac{2 (d+e x)^{3/2}}{c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{4 \sqrt{d+e x} \left (c d^2-a e^2\right )}{c^2 d^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}$

[Out]

(-4*(c*d^2 - a*e^2)*Sqrt[d + e*x])/(c^2*d^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (2*(d + e*x)^(3/2))
/(c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])

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Rubi [A]  time = 0.0581765, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 39, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.051, Rules used = {656, 648} $\frac{2 (d+e x)^{3/2}}{c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{4 \sqrt{d+e x} \left (c d^2-a e^2\right )}{c^2 d^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-4*(c*d^2 - a*e^2)*Sqrt[d + e*x])/(c^2*d^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (2*(d + e*x)^(3/2))
/(c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])

Rule 656

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[(Simplify[m + p]*(2*c*d - b*e))/(c*(m + 2*p + 1)), In
t[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p], 0]

Rule 648

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c
*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0]

Rubi steps

\begin{align*} \int \frac{(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx &=\frac{2 (d+e x)^{3/2}}{c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{\left (2 \left (2 c d^2 e-e \left (c d^2+a e^2\right )\right )\right ) \int \frac{(d+e x)^{3/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{c d e}\\ &=-\frac{4 \left (c d^2-a e^2\right ) \sqrt{d+e x}}{c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{2 (d+e x)^{3/2}}{c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0339236, size = 51, normalized size = 0.49 $-\frac{2 \sqrt{d+e x} \left (c d (d-e x)-2 a e^2\right )}{c^2 d^2 \sqrt{(d+e x) (a e+c d x)}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.041, size = 68, normalized size = 0.7 \begin{align*} 2\,{\frac{ \left ( cdx+ae \right ) \left ( cdex+2\,a{e}^{2}-c{d}^{2} \right ) \left ( ex+d \right ) ^{3/2}}{{c}^{2}{d}^{2} \left ( cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade \right ) ^{3/2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.12429, size = 49, normalized size = 0.47 \begin{align*} \frac{2 \,{\left (c d e x - c d^{2} + 2 \, a e^{2}\right )}}{\sqrt{c d x + a e} c^{2} d^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.04646, size = 201, normalized size = 1.91 \begin{align*} \frac{2 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (c d e x - c d^{2} + 2 \, a e^{2}\right )} \sqrt{e x + d}}{c^{3} d^{3} e x^{2} + a c^{2} d^{3} e +{\left (c^{3} d^{4} + a c^{2} d^{2} e^{2}\right )} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [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((e*x+d)**(5/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x