### 3.2049 $$\int \frac{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{(d+e x)^{3/2}} \, dx$$

Optimal. Leaf size=109 $\frac{4 \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{7/2}}{63 c^2 d^2 (d+e x)^{7/2}}+\frac{2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{7/2}}{9 c d (d+e x)^{5/2}}$

[Out]

(4*(c*d^2 - a*e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(7/2))/(63*c^2*d^2*(d + e*x)^(7/2)) + (2*(a*d*e + (
c*d^2 + a*e^2)*x + c*d*e*x^2)^(7/2))/(9*c*d*(d + e*x)^(5/2))

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Rubi [A]  time = 0.0572526, antiderivative size = 109, 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} $\frac{4 \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{7/2}}{63 c^2 d^2 (d+e x)^{7/2}}+\frac{2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{7/2}}{9 c d (d+e x)^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(4*(c*d^2 - a*e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(7/2))/(63*c^2*d^2*(d + e*x)^(7/2)) + (2*(a*d*e + (
c*d^2 + a*e^2)*x + c*d*e*x^2)^(7/2))/(9*c*d*(d + e*x)^(5/2))

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{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{3/2}} \, dx &=\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{7/2}}{9 c d (d+e x)^{5/2}}+\frac{\left (2 \left (d^2-\frac{a e^2}{c}\right )\right ) \int \frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2}} \, dx}{9 d}\\ &=\frac{4 \left (c d^2-a e^2\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{7/2}}{63 c^2 d^2 (d+e x)^{7/2}}+\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{7/2}}{9 c d (d+e x)^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0561066, size = 65, normalized size = 0.6 $\frac{2 (a e+c d x)^3 \sqrt{(d+e x) (a e+c d x)} \left (c d (9 d+7 e x)-2 a e^2\right )}{63 c^2 d^2 \sqrt{d+e x}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(a*e + c*d*x)^3*Sqrt[(a*e + c*d*x)*(d + e*x)]*(-2*a*e^2 + c*d*(9*d + 7*e*x)))/(63*c^2*d^2*Sqrt[d + e*x])

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Maple [A]  time = 0.053, size = 69, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2\,cdx+2\,ae \right ) \left ( -7\,cdex+2\,a{e}^{2}-9\,c{d}^{2} \right ) }{63\,{c}^{2}{d}^{2}} \left ( cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade \right ) ^{{\frac{5}{2}}} \left ( ex+d \right ) ^{-{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/63*(c*d*x+a*e)*(-7*c*d*e*x+2*a*e^2-9*c*d^2)*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(5/2)/c^2/d^2/(e*x+d)^(5/2)

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Maxima [A]  time = 1.09768, size = 178, normalized size = 1.63 \begin{align*} \frac{2 \,{\left (7 \, c^{4} d^{4} e x^{4} + 9 \, a^{3} c d^{2} e^{3} - 2 \, a^{4} e^{5} +{\left (9 \, c^{4} d^{5} + 19 \, a c^{3} d^{3} e^{2}\right )} x^{3} + 3 \,{\left (9 \, a c^{3} d^{4} e + 5 \, a^{2} c^{2} d^{2} e^{3}\right )} x^{2} +{\left (27 \, a^{2} c^{2} d^{3} e^{2} + a^{3} c d e^{4}\right )} x\right )} \sqrt{c d x + a e}}{63 \, c^{2} d^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/63*(7*c^4*d^4*e*x^4 + 9*a^3*c*d^2*e^3 - 2*a^4*e^5 + (9*c^4*d^5 + 19*a*c^3*d^3*e^2)*x^3 + 3*(9*a*c^3*d^4*e +
5*a^2*c^2*d^2*e^3)*x^2 + (27*a^2*c^2*d^3*e^2 + a^3*c*d*e^4)*x)*sqrt(c*d*x + a*e)/(c^2*d^2)

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Fricas [A]  time = 1.82335, size = 346, normalized size = 3.17 \begin{align*} \frac{2 \,{\left (7 \, c^{4} d^{4} e x^{4} + 9 \, a^{3} c d^{2} e^{3} - 2 \, a^{4} e^{5} +{\left (9 \, c^{4} d^{5} + 19 \, a c^{3} d^{3} e^{2}\right )} x^{3} + 3 \,{\left (9 \, a c^{3} d^{4} e + 5 \, a^{2} c^{2} d^{2} e^{3}\right )} x^{2} +{\left (27 \, a^{2} c^{2} d^{3} e^{2} + a^{3} c d e^{4}\right )} x\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{e x + d}}{63 \,{\left (c^{2} d^{2} e x + c^{2} d^{3}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/63*(7*c^4*d^4*e*x^4 + 9*a^3*c*d^2*e^3 - 2*a^4*e^5 + (9*c^4*d^5 + 19*a*c^3*d^3*e^2)*x^3 + 3*(9*a*c^3*d^4*e +
5*a^2*c^2*d^2*e^3)*x^2 + (27*a^2*c^2*d^3*e^2 + a^3*c*d*e^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqr
t(e*x + d)/(c^2*d^2*e*x + c^2*d^3)

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

[Out]

Timed out

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

[Out]

Timed out