3.2048 \(\int \frac{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}}{\sqrt{d+e x}} \, dx\)

Optimal. Leaf size=171 \[ \frac{8 \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{7/2}}{99 c^2 d^2 (d+e x)^{5/2}}+\frac{16 \left (c d^2-a e^2\right )^2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{7/2}}{693 c^3 d^3 (d+e x)^{7/2}}+\frac{2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{7/2}}{11 c d (d+e x)^{3/2}} \]

[Out]

(16*(c*d^2 - a*e^2)^2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(7/2))/(693*c^3*d^3*(d + e*x)^(7/2)) + (8*(c*d^2
- a*e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(7/2))/(99*c^2*d^2*(d + e*x)^(5/2)) + (2*(a*d*e + (c*d^2 + a
*e^2)*x + c*d*e*x^2)^(7/2))/(11*c*d*(d + e*x)^(3/2))

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Rubi [A]  time = 0.114086, 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} \[ \frac{8 \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{7/2}}{99 c^2 d^2 (d+e x)^{5/2}}+\frac{16 \left (c d^2-a e^2\right )^2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{7/2}}{693 c^3 d^3 (d+e x)^{7/2}}+\frac{2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{7/2}}{11 c d (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/Sqrt[d + e*x],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)^(7/2))/(693*c^3*d^3*(d + e*x)^(7/2)) + (8*(c*d^2
- a*e^2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(7/2))/(99*c^2*d^2*(d + e*x)^(5/2)) + (2*(a*d*e + (c*d^2 + a
*e^2)*x + c*d*e*x^2)^(7/2))/(11*c*d*(d + e*x)^(3/2))

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}}{\sqrt{d+e x}} \, dx &=\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{7/2}}{11 c d (d+e x)^{3/2}}+\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 )^{5/2}}{(d+e x)^{3/2}} \, dx}{11 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 )^{7/2}}{99 c^2 d^2 (d+e x)^{5/2}}+\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{7/2}}{11 c d (d+e x)^{3/2}}+\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 )^{5/2}}{(d+e x)^{5/2}} \, dx}{99 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 )^{7/2}}{693 c^3 d^3 (d+e x)^{7/2}}+\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 )^{7/2}}{99 c^2 d^2 (d+e x)^{5/2}}+\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{7/2}}{11 c d (d+e x)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0844179, size = 98, normalized size = 0.57 \[ \frac{2 (a e+c d x)^3 \sqrt{(d+e x) (a e+c d x)} \left (8 a^2 e^4-4 a c d e^2 (11 d+7 e x)+c^2 d^2 \left (99 d^2+154 d e x+63 e^2 x^2\right )\right )}{693 c^3 d^3 \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a*e + c*d*x)^3*Sqrt[(a*e + c*d*x)*(d + e*x)]*(8*a^2*e^4 - 4*a*c*d*e^2*(11*d + 7*e*x) + c^2*d^2*(99*d^2 + 1
54*d*e*x + 63*e^2*x^2)))/(693*c^3*d^3*Sqrt[d + e*x])

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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 ( 63\,{e}^{2}{x}^{2}{c}^{2}{d}^{2}-28\,acd{e}^{3}x+154\,{c}^{2}{d}^{3}ex+8\,{a}^{2}{e}^{4}-44\,ac{d}^{2}{e}^{2}+99\,{c}^{2}{d}^{4} \right ) }{693\,{c}^{3}{d}^{3}} \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)^(1/2),x)

[Out]

2/693*(c*d*x+a*e)*(63*c^2*d^2*e^2*x^2-28*a*c*d*e^3*x+154*c^2*d^3*e*x+8*a^2*e^4-44*a*c*d^2*e^2+99*c^2*d^4)*(c*d
*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(5/2)/c^3/d^3/(e*x+d)^(5/2)

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Maxima [A]  time = 1.04736, size = 293, normalized size = 1.71 \begin{align*} \frac{2 \,{\left (63 \, c^{5} d^{5} e^{2} x^{5} + 99 \, a^{3} c^{2} d^{4} e^{3} - 44 \, a^{4} c d^{2} e^{5} + 8 \, a^{5} e^{7} + 7 \,{\left (22 \, c^{5} d^{6} e + 23 \, a c^{4} d^{4} e^{3}\right )} x^{4} +{\left (99 \, c^{5} d^{7} + 418 \, a c^{4} d^{5} e^{2} + 113 \, a^{2} c^{3} d^{3} e^{4}\right )} x^{3} + 3 \,{\left (99 \, a c^{4} d^{6} e + 110 \, a^{2} c^{3} d^{4} e^{3} + a^{3} c^{2} d^{2} e^{5}\right )} x^{2} +{\left (297 \, a^{2} c^{3} d^{5} e^{2} + 22 \, a^{3} c^{2} d^{3} e^{4} - 4 \, a^{4} c d e^{6}\right )} x\right )} \sqrt{c d x + a e}}{693 \, c^{3} d^{3}} \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)^(1/2),x, algorithm="maxima")

[Out]

2/693*(63*c^5*d^5*e^2*x^5 + 99*a^3*c^2*d^4*e^3 - 44*a^4*c*d^2*e^5 + 8*a^5*e^7 + 7*(22*c^5*d^6*e + 23*a*c^4*d^4
*e^3)*x^4 + (99*c^5*d^7 + 418*a*c^4*d^5*e^2 + 113*a^2*c^3*d^3*e^4)*x^3 + 3*(99*a*c^4*d^6*e + 110*a^2*c^3*d^4*e
^3 + a^3*c^2*d^2*e^5)*x^2 + (297*a^2*c^3*d^5*e^2 + 22*a^3*c^2*d^3*e^4 - 4*a^4*c*d*e^6)*x)*sqrt(c*d*x + a*e)/(c
^3*d^3)

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Fricas [A]  time = 1.88095, size = 531, normalized size = 3.11 \begin{align*} \frac{2 \,{\left (63 \, c^{5} d^{5} e^{2} x^{5} + 99 \, a^{3} c^{2} d^{4} e^{3} - 44 \, a^{4} c d^{2} e^{5} + 8 \, a^{5} e^{7} + 7 \,{\left (22 \, c^{5} d^{6} e + 23 \, a c^{4} d^{4} e^{3}\right )} x^{4} +{\left (99 \, c^{5} d^{7} + 418 \, a c^{4} d^{5} e^{2} + 113 \, a^{2} c^{3} d^{3} e^{4}\right )} x^{3} + 3 \,{\left (99 \, a c^{4} d^{6} e + 110 \, a^{2} c^{3} d^{4} e^{3} + a^{3} c^{2} d^{2} e^{5}\right )} x^{2} +{\left (297 \, a^{2} c^{3} d^{5} e^{2} + 22 \, a^{3} c^{2} d^{3} e^{4} - 4 \, a^{4} 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}}{693 \,{\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((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

2/693*(63*c^5*d^5*e^2*x^5 + 99*a^3*c^2*d^4*e^3 - 44*a^4*c*d^2*e^5 + 8*a^5*e^7 + 7*(22*c^5*d^6*e + 23*a*c^4*d^4
*e^3)*x^4 + (99*c^5*d^7 + 418*a*c^4*d^5*e^2 + 113*a^2*c^3*d^3*e^4)*x^3 + 3*(99*a*c^4*d^6*e + 110*a^2*c^3*d^4*e
^3 + a^3*c^2*d^2*e^5)*x^2 + (297*a^2*c^3*d^5*e^2 + 22*a^3*c^2*d^3*e^4 - 4*a^4*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^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((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/2)/(e*x+d)**(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((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

Timed out