### 3.2028 $$\int (d+e x)^{5/2} \sqrt{a d e+(c d^2+a e^2) x+c d e x^2} \, dx$$

Optimal. Leaf size=233 $\frac{32 \left (c d^2-a e^2\right )^3 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{315 c^4 d^4 (d+e x)^{3/2}}+\frac{16 \left (c d^2-a e^2\right )^2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{105 c^3 d^3 \sqrt{d+e x}}+\frac{4 \sqrt{d+e x} \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{21 c^2 d^2}+\frac{2 (d+e x)^{3/2} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{9 c d}$

[Out]

(32*(c*d^2 - a*e^2)^3*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(315*c^4*d^4*(d + e*x)^(3/2)) + (16*(c*d^
2 - a*e^2)^2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(105*c^3*d^3*Sqrt[d + e*x]) + (4*(c*d^2 - a*e^2)*S
qrt[d + e*x]*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(21*c^2*d^2) + (2*(d + e*x)^(3/2)*(a*d*e + (c*d^2
+ a*e^2)*x + c*d*e*x^2)^(3/2))/(9*c*d)

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Rubi [A]  time = 0.188651, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 39, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.051, Rules used = {656, 648} $\frac{32 \left (c d^2-a e^2\right )^3 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{315 c^4 d^4 (d+e x)^{3/2}}+\frac{16 \left (c d^2-a e^2\right )^2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{105 c^3 d^3 \sqrt{d+e x}}+\frac{4 \sqrt{d+e x} \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{21 c^2 d^2}+\frac{2 (d+e x)^{3/2} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{9 c d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(32*(c*d^2 - a*e^2)^3*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(315*c^4*d^4*(d + e*x)^(3/2)) + (16*(c*d^
2 - a*e^2)^2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(105*c^3*d^3*Sqrt[d + e*x]) + (4*(c*d^2 - a*e^2)*S
qrt[d + e*x]*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(21*c^2*d^2) + (2*(d + e*x)^(3/2)*(a*d*e + (c*d^2
+ a*e^2)*x + c*d*e*x^2)^(3/2))/(9*c*d)

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 (d+e x)^{5/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx &=\frac{2 (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}}{9 c d}+\frac{\left (2 \left (d^2-\frac{a e^2}{c}\right )\right ) \int (d+e x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx}{3 d}\\ &=\frac{4 \left (c d^2-a e^2\right ) \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}}{21 c^2 d^2}+\frac{2 (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}}{9 c d}+\frac{\left (8 \left (d^2-\frac{a e^2}{c}\right )^2\right ) \int \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx}{21 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 )^{3/2}}{105 c^3 d^3 \sqrt{d+e x}}+\frac{4 \left (c d^2-a e^2\right ) \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}}{21 c^2 d^2}+\frac{2 (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}}{9 c d}+\frac{\left (16 \left (d^2-\frac{a e^2}{c}\right )^3\right ) \int \frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt{d+e x}} \, dx}{105 d^3}\\ &=\frac{32 \left (c d^2-a e^2\right )^3 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{315 c^4 d^4 (d+e x)^{3/2}}+\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 )^{3/2}}{105 c^3 d^3 \sqrt{d+e x}}+\frac{4 \left (c d^2-a e^2\right ) \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}}{21 c^2 d^2}+\frac{2 (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}}{9 c d}\\ \end{align*}

Mathematica [A]  time = 0.105582, size = 131, normalized size = 0.56 $\frac{2 ((d+e x) (a e+c d x))^{3/2} \left (24 a^2 c d e^4 (3 d+e x)-16 a^3 e^6-6 a c^2 d^2 e^2 \left (21 d^2+18 d e x+5 e^2 x^2\right )+c^3 d^3 \left (189 d^2 e x+105 d^3+135 d e^2 x^2+35 e^3 x^3\right )\right )}{315 c^4 d^4 (d+e x)^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*((a*e + c*d*x)*(d + e*x))^(3/2)*(-16*a^3*e^6 + 24*a^2*c*d*e^4*(3*d + e*x) - 6*a*c^2*d^2*e^2*(21*d^2 + 18*d*
e*x + 5*e^2*x^2) + c^3*d^3*(105*d^3 + 189*d^2*e*x + 135*d*e^2*x^2 + 35*e^3*x^3)))/(315*c^4*d^4*(d + e*x)^(3/2)
)

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Maple [A]  time = 0.044, size = 168, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,cdx+2\,ae \right ) \left ( -35\,{e}^{3}{x}^{3}{c}^{3}{d}^{3}+30\,a{c}^{2}{d}^{2}{e}^{4}{x}^{2}-135\,{c}^{3}{d}^{4}{e}^{2}{x}^{2}-24\,{a}^{2}cd{e}^{5}x+108\,a{c}^{2}{d}^{3}{e}^{3}x-189\,{c}^{3}{d}^{5}ex+16\,{a}^{3}{e}^{6}-72\,{a}^{2}c{d}^{2}{e}^{4}+126\,a{c}^{2}{d}^{4}{e}^{2}-105\,{c}^{3}{d}^{6} \right ) }{315\,{c}^{4}{d}^{4}}\sqrt{cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade}{\frac{1}{\sqrt{ex+d}}}} \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)^(1/2),x)

[Out]

-2/315*(c*d*x+a*e)*(-35*c^3*d^3*e^3*x^3+30*a*c^2*d^2*e^4*x^2-135*c^3*d^4*e^2*x^2-24*a^2*c*d*e^5*x+108*a*c^2*d^
3*e^3*x-189*c^3*d^5*e*x+16*a^3*e^6-72*a^2*c*d^2*e^4+126*a*c^2*d^4*e^2-105*c^3*d^6)*(c*d*e*x^2+a*e^2*x+c*d^2*x+
a*d*e)^(1/2)/c^4/d^4/(e*x+d)^(1/2)

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Maxima [A]  time = 1.12121, size = 285, normalized size = 1.22 \begin{align*} \frac{2 \,{\left (35 \, c^{4} d^{4} e^{3} x^{4} + 105 \, a c^{3} d^{6} e - 126 \, a^{2} c^{2} d^{4} e^{3} + 72 \, a^{3} c d^{2} e^{5} - 16 \, a^{4} e^{7} + 5 \,{\left (27 \, c^{4} d^{5} e^{2} + a c^{3} d^{3} e^{4}\right )} x^{3} + 3 \,{\left (63 \, c^{4} d^{6} e + 9 \, a c^{3} d^{4} e^{3} - 2 \, a^{2} c^{2} d^{2} e^{5}\right )} x^{2} +{\left (105 \, c^{4} d^{7} + 63 \, a c^{3} d^{5} e^{2} - 36 \, a^{2} c^{2} d^{3} e^{4} + 8 \, a^{3} c d e^{6}\right )} x\right )} \sqrt{c d x + a e}{\left (e x + d\right )}}{315 \,{\left (c^{4} d^{4} e x + c^{4} d^{5}\right )}} \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)^(1/2),x, algorithm="maxima")

[Out]

2/315*(35*c^4*d^4*e^3*x^4 + 105*a*c^3*d^6*e - 126*a^2*c^2*d^4*e^3 + 72*a^3*c*d^2*e^5 - 16*a^4*e^7 + 5*(27*c^4*
d^5*e^2 + a*c^3*d^3*e^4)*x^3 + 3*(63*c^4*d^6*e + 9*a*c^3*d^4*e^3 - 2*a^2*c^2*d^2*e^5)*x^2 + (105*c^4*d^7 + 63*
a*c^3*d^5*e^2 - 36*a^2*c^2*d^3*e^4 + 8*a^3*c*d*e^6)*x)*sqrt(c*d*x + a*e)*(e*x + d)/(c^4*d^4*e*x + c^4*d^5)

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Fricas [A]  time = 1.86278, size = 483, normalized size = 2.07 \begin{align*} \frac{2 \,{\left (35 \, c^{4} d^{4} e^{3} x^{4} + 105 \, a c^{3} d^{6} e - 126 \, a^{2} c^{2} d^{4} e^{3} + 72 \, a^{3} c d^{2} e^{5} - 16 \, a^{4} e^{7} + 5 \,{\left (27 \, c^{4} d^{5} e^{2} + a c^{3} d^{3} e^{4}\right )} x^{3} + 3 \,{\left (63 \, c^{4} d^{6} e + 9 \, a c^{3} d^{4} e^{3} - 2 \, a^{2} c^{2} d^{2} e^{5}\right )} x^{2} +{\left (105 \, c^{4} d^{7} + 63 \, a c^{3} d^{5} e^{2} - 36 \, a^{2} c^{2} d^{3} e^{4} + 8 \, a^{3} c d e^{6}\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^{4} d^{4} e x + c^{4} d^{5}\right )}} \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)^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*c^4*d^4*e^3*x^4 + 105*a*c^3*d^6*e - 126*a^2*c^2*d^4*e^3 + 72*a^3*c*d^2*e^5 - 16*a^4*e^7 + 5*(27*c^4*
d^5*e^2 + a*c^3*d^3*e^4)*x^3 + 3*(63*c^4*d^6*e + 9*a*c^3*d^4*e^3 - 2*a^2*c^2*d^2*e^5)*x^2 + (105*c^4*d^7 + 63*
a*c^3*d^5*e^2 - 36*a^2*c^2*d^3*e^4 + 8*a^3*c*d*e^6)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x +
d)/(c^4*d^4*e*x + c^4*d^5)

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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)**(1/2),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)^(5/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="giac")

[Out]

Timed out