### 3.2038 $$\int \sqrt{d+e x} (a d e+(c d^2+a e^2) x+c d e x^2)^{3/2} \, dx$$

Optimal. Leaf size=171 $\frac{8 \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{63 c^2 d^2 (d+e x)^{3/2}}+\frac{16 \left (c d^2-a e^2\right )^2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{315 c^3 d^3 (d+e x)^{5/2}}+\frac{2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{9 c d \sqrt{d+e x}}$

[Out]

(16*(c*d^2 - a*e^2)^2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(315*c^3*d^3*(d + e*x)^(5/2)) + (8*(c*d^2
- a*e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(63*c^2*d^2*(d + e*x)^(3/2)) + (2*(a*d*e + (c*d^2 + a
*e^2)*x + c*d*e*x^2)^(5/2))/(9*c*d*Sqrt[d + e*x])

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Rubi [A]  time = 0.121992, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 39, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.051, Rules used = {656, 648} $\frac{8 \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{63 c^2 d^2 (d+e x)^{3/2}}+\frac{16 \left (c d^2-a e^2\right )^2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{315 c^3 d^3 (d+e x)^{5/2}}+\frac{2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{9 c d \sqrt{d+e x}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(16*(c*d^2 - a*e^2)^2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(315*c^3*d^3*(d + e*x)^(5/2)) + (8*(c*d^2
- a*e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(63*c^2*d^2*(d + e*x)^(3/2)) + (2*(a*d*e + (c*d^2 + a
*e^2)*x + c*d*e*x^2)^(5/2))/(9*c*d*Sqrt[d + e*x])

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 \sqrt{d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2} \, dx &=\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{9 c d \sqrt{d+e x}}+\frac{\left (4 \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 )^{3/2}}{\sqrt{d+e x}} \, dx}{9 d}\\ &=\frac{8 \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 )^{5/2}}{63 c^2 d^2 (d+e x)^{3/2}}+\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{9 c d \sqrt{d+e x}}+\frac{\left (8 \left (d^2-\frac{a e^2}{c}\right )^2\right ) \int \frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx}{63 d^2}\\ &=\frac{16 \left (c d^2-a e^2\right )^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{315 c^3 d^3 (d+e x)^{5/2}}+\frac{8 \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 )^{5/2}}{63 c^2 d^2 (d+e x)^{3/2}}+\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{9 c d \sqrt{d+e x}}\\ \end{align*}

Mathematica [A]  time = 0.0921158, size = 88, normalized size = 0.51 $\frac{2 ((d+e x) (a e+c d x))^{5/2} \left (8 a^2 e^4-4 a c d e^2 (9 d+5 e x)+c^2 d^2 \left (63 d^2+90 d e x+35 e^2 x^2\right )\right )}{315 c^3 d^3 (d+e x)^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*((a*e + c*d*x)*(d + e*x))^(5/2)*(8*a^2*e^4 - 4*a*c*d*e^2*(9*d + 5*e*x) + c^2*d^2*(63*d^2 + 90*d*e*x + 35*e^
2*x^2)))/(315*c^3*d^3*(d + e*x)^(5/2))

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Maple [A]  time = 0.046, size = 110, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2\,cdx+2\,ae \right ) \left ( 35\,{e}^{2}{x}^{2}{c}^{2}{d}^{2}-20\,acd{e}^{3}x+90\,{c}^{2}{d}^{3}ex+8\,{a}^{2}{e}^{4}-36\,ac{d}^{2}{e}^{2}+63\,{c}^{2}{d}^{4} \right ) }{315\,{c}^{3}{d}^{3}} \left ( cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade \right ) ^{{\frac{3}{2}}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x)

[Out]

2/315*(c*d*x+a*e)*(35*c^2*d^2*e^2*x^2-20*a*c*d*e^3*x+90*c^2*d^3*e*x+8*a^2*e^4-36*a*c*d^2*e^2+63*c^2*d^4)*(c*d*
e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(3/2)/c^3/d^3/(e*x+d)^(3/2)

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Maxima [A]  time = 1.09035, size = 255, normalized size = 1.49 \begin{align*} \frac{2 \,{\left (35 \, c^{4} d^{4} e^{2} x^{4} + 63 \, a^{2} c^{2} d^{4} e^{2} - 36 \, a^{3} c d^{2} e^{4} + 8 \, a^{4} e^{6} + 10 \,{\left (9 \, c^{4} d^{5} e + 5 \, a c^{3} d^{3} e^{3}\right )} x^{3} + 3 \,{\left (21 \, c^{4} d^{6} + 48 \, a c^{3} d^{4} e^{2} + a^{2} c^{2} d^{2} e^{4}\right )} x^{2} + 2 \,{\left (63 \, a c^{3} d^{5} e + 9 \, a^{2} c^{2} d^{3} e^{3} - 2 \, a^{3} c d e^{5}\right )} x\right )} \sqrt{c d x + a e}{\left (e x + d\right )}}{315 \,{\left (c^{3} d^{3} e x + c^{3} d^{4}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*c^4*d^4*e^2*x^4 + 63*a^2*c^2*d^4*e^2 - 36*a^3*c*d^2*e^4 + 8*a^4*e^6 + 10*(9*c^4*d^5*e + 5*a*c^3*d^3*
e^3)*x^3 + 3*(21*c^4*d^6 + 48*a*c^3*d^4*e^2 + a^2*c^2*d^2*e^4)*x^2 + 2*(63*a*c^3*d^5*e + 9*a^2*c^2*d^3*e^3 - 2
*a^3*c*d*e^5)*x)*sqrt(c*d*x + a*e)*(e*x + d)/(c^3*d^3*e*x + c^3*d^4)

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Fricas [A]  time = 1.88297, size = 432, normalized size = 2.53 \begin{align*} \frac{2 \,{\left (35 \, c^{4} d^{4} e^{2} x^{4} + 63 \, a^{2} c^{2} d^{4} e^{2} - 36 \, a^{3} c d^{2} e^{4} + 8 \, a^{4} e^{6} + 10 \,{\left (9 \, c^{4} d^{5} e + 5 \, a c^{3} d^{3} e^{3}\right )} x^{3} + 3 \,{\left (21 \, c^{4} d^{6} + 48 \, a c^{3} d^{4} e^{2} + a^{2} c^{2} d^{2} e^{4}\right )} x^{2} + 2 \,{\left (63 \, a c^{3} d^{5} e + 9 \, a^{2} c^{2} d^{3} e^{3} - 2 \, a^{3} c d e^{5}\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}}{315 \,{\left (c^{3} d^{3} e x + c^{3} d^{4}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="fricas")

[Out]

2/315*(35*c^4*d^4*e^2*x^4 + 63*a^2*c^2*d^4*e^2 - 36*a^3*c*d^2*e^4 + 8*a^4*e^6 + 10*(9*c^4*d^5*e + 5*a*c^3*d^3*
e^3)*x^3 + 3*(21*c^4*d^6 + 48*a*c^3*d^4*e^2 + a^2*c^2*d^2*e^4)*x^2 + 2*(63*a*c^3*d^5*e + 9*a^2*c^2*d^3*e^3 - 2
*a^3*c*d*e^5)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d)/(c^3*d^3*e*x + c^3*d^4)

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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)**(1/2)*(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(-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)^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="giac")

[Out]

Timed out