3.2029 \(\int (d+e x)^{3/2} \sqrt{a d e+(c d^2+a e^2) x+c d e x^2} \, dx\)

Optimal. Leaf size=171 \[ \frac{16 \left (c d^2-a e^2\right )^2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{105 c^3 d^3 (d+e x)^{3/2}}+\frac{8 \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{35 c^2 d^2 \sqrt{d+e x}}+\frac{2 \sqrt{d+e x} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{7 c d} \]

[Out]

(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*(d + e*x)^(3/2)) + (8*(c*d^2
- a*e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(35*c^2*d^2*Sqrt[d + e*x]) + (2*Sqrt[d + e*x]*(a*d*e
+ (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(7*c*d)

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Rubi [A]  time = 0.112851, 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} \[ \frac{16 \left (c d^2-a e^2\right )^2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{105 c^3 d^3 (d+e x)^{3/2}}+\frac{8 \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{35 c^2 d^2 \sqrt{d+e x}}+\frac{2 \sqrt{d+e x} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{7 c d} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(3/2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^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)^(3/2))/(105*c^3*d^3*(d + e*x)^(3/2)) + (8*(c*d^2
- a*e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(35*c^2*d^2*Sqrt[d + e*x]) + (2*Sqrt[d + e*x]*(a*d*e
+ (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(7*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)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx &=\frac{2 \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}}{7 c d}+\frac{\left (4 \left (d^2-\frac{a e^2}{c}\right )\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}{7 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 )^{3/2}}{35 c^2 d^2 \sqrt{d+e x}}+\frac{2 \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}}{7 c d}+\frac{\left (8 \left (d^2-\frac{a e^2}{c}\right )^2\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}{35 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 (d+e x)^{3/2}}+\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 )^{3/2}}{35 c^2 d^2 \sqrt{d+e x}}+\frac{2 \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}}{7 c d}\\ \end{align*}

Mathematica [A]  time = 0.0736661, size = 88, normalized size = 0.51 \[ \frac{2 ((d+e x) (a e+c d x))^{3/2} \left (8 a^2 e^4-4 a c d e^2 (7 d+3 e x)+c^2 d^2 \left (35 d^2+42 d e x+15 e^2 x^2\right )\right )}{105 c^3 d^3 (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(3/2)*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)*(8*a^2*e^4 - 4*a*c*d*e^2*(7*d + 3*e*x) + c^2*d^2*(35*d^2 + 42*d*e*x + 15*e^
2*x^2)))/(105*c^3*d^3*(d + e*x)^(3/2))

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Maple [A]  time = 0.044, size = 110, normalized size = 0.6 \begin{align*}{\frac{ \left ( 2\,cdx+2\,ae \right ) \left ( 15\,{e}^{2}{x}^{2}{c}^{2}{d}^{2}-12\,acd{e}^{3}x+42\,{c}^{2}{d}^{3}ex+8\,{a}^{2}{e}^{4}-28\,ac{d}^{2}{e}^{2}+35\,{c}^{2}{d}^{4} \right ) }{105\,{c}^{3}{d}^{3}}\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)^(3/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x)

[Out]

2/105*(c*d*x+a*e)*(15*c^2*d^2*e^2*x^2-12*a*c*d*e^3*x+42*c^2*d^3*e*x+8*a^2*e^4-28*a*c*d^2*e^2+35*c^2*d^4)*(c*d*
e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)/c^3/d^3/(e*x+d)^(1/2)

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Maxima [A]  time = 1.08813, size = 189, normalized size = 1.11 \begin{align*} \frac{2 \,{\left (15 \, c^{3} d^{3} e^{2} x^{3} + 35 \, a c^{2} d^{4} e - 28 \, a^{2} c d^{2} e^{3} + 8 \, a^{3} e^{5} + 3 \,{\left (14 \, c^{3} d^{4} e + a c^{2} d^{2} e^{3}\right )} x^{2} +{\left (35 \, c^{3} d^{5} + 14 \, a c^{2} d^{3} e^{2} - 4 \, a^{2} c d e^{4}\right )} x\right )} \sqrt{c d x + a e}{\left (e x + d\right )}}{105 \,{\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)^(3/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="maxima")

[Out]

2/105*(15*c^3*d^3*e^2*x^3 + 35*a*c^2*d^4*e - 28*a^2*c*d^2*e^3 + 8*a^3*e^5 + 3*(14*c^3*d^4*e + a*c^2*d^2*e^3)*x
^2 + (35*c^3*d^5 + 14*a*c^2*d^3*e^2 - 4*a^2*c*d*e^4)*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 = 1.84812, size = 336, normalized size = 1.96 \begin{align*} \frac{2 \,{\left (15 \, c^{3} d^{3} e^{2} x^{3} + 35 \, a c^{2} d^{4} e - 28 \, a^{2} c d^{2} e^{3} + 8 \, a^{3} e^{5} + 3 \,{\left (14 \, c^{3} d^{4} e + a c^{2} d^{2} e^{3}\right )} x^{2} +{\left (35 \, c^{3} d^{5} + 14 \, a c^{2} d^{3} e^{2} - 4 \, a^{2} 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}}{105 \,{\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)^(3/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="fricas")

[Out]

2/105*(15*c^3*d^3*e^2*x^3 + 35*a*c^2*d^4*e - 28*a^2*c*d^2*e^3 + 8*a^3*e^5 + 3*(14*c^3*d^4*e + a*c^2*d^2*e^3)*x
^2 + (35*c^3*d^5 + 14*a*c^2*d^3*e^2 - 4*a^2*c*d*e^4)*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)**(3/2)*(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)^(3/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="giac")

[Out]

Timed out