3.2065 \(\int \frac{(d+e x)^{7/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.116254, 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} \[ \frac{8 (d+e x)^{3/2} \left (c d^2-a e^2\right )}{3 c^2 d^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{16 \sqrt{d+e x} \left (c d^2-a e^2\right )^2}{3 c^3 d^3 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{2 (d+e x)^{5/2}}{3 c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(7/2)/(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*Sqrt[d + e*x])/(3*c^3*d^3*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (8*(c*d^2 - a*
e^2)*(d + e*x)^(3/2))/(3*c^2*d^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (2*(d + e*x)^(5/2))/(3*c*d*Sqr
t[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^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{(d+e x)^{7/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx &=\frac{2 (d+e x)^{5/2}}{3 c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{\left (4 \left (d^2-\frac{a e^2}{c}\right )\right ) \int \frac{(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{3 d}\\ &=\frac{8 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{3 c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{2 (d+e x)^{5/2}}{3 c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{\left (8 \left (c d^2-a e^2\right )^2\right ) \int \frac{(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}} \, dx}{3 c^2 d^2}\\ &=-\frac{16 \left (c d^2-a e^2\right )^2 \sqrt{d+e x}}{3 c^3 d^3 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{8 \left (c d^2-a e^2\right ) (d+e x)^{3/2}}{3 c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{2 (d+e x)^{5/2}}{3 c d \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0599896, size = 87, normalized size = 0.51 \[ -\frac{2 \sqrt{d+e x} \left (8 a^2 e^4+4 a c d e^2 (e x-3 d)+c^2 d^2 \left (3 d^2-6 d e x-e^2 x^2\right )\right )}{3 c^3 d^3 \sqrt{(d+e x) (a e+c d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(7/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2),x]

[Out]

(-2*Sqrt[d + e*x]*(8*a^2*e^4 + 4*a*c*d*e^2*(-3*d + e*x) + c^2*d^2*(3*d^2 - 6*d*e*x - e^2*x^2)))/(3*c^3*d^3*Sqr
t[(a*e + c*d*x)*(d + e*x)])

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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 ( -{e}^{2}{x}^{2}{c}^{2}{d}^{2}+4\,acd{e}^{3}x-6\,{c}^{2}{d}^{3}ex+8\,{a}^{2}{e}^{4}-12\,ac{d}^{2}{e}^{2}+3\,{c}^{2}{d}^{4} \right ) }{3\,{c}^{3}{d}^{3}} \left ( ex+d \right ) ^{{\frac{3}{2}}} \left ( cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x)

[Out]

-2/3*(c*d*x+a*e)*(-c^2*d^2*e^2*x^2+4*a*c*d*e^3*x-6*c^2*d^3*e*x+8*a^2*e^4-12*a*c*d^2*e^2+3*c^2*d^4)*(e*x+d)^(3/
2)/c^3/d^3/(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(3/2)

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Maxima [A]  time = 1.05948, size = 107, normalized size = 0.63 \begin{align*} \frac{2 \,{\left (c^{2} d^{2} e^{2} x^{2} - 3 \, c^{2} d^{4} + 12 \, a c d^{2} e^{2} - 8 \, a^{2} e^{4} + 2 \,{\left (3 \, c^{2} d^{3} e - 2 \, a c d e^{3}\right )} x\right )}}{3 \, \sqrt{c d x + a e} c^{3} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="maxima")

[Out]

2/3*(c^2*d^2*e^2*x^2 - 3*c^2*d^4 + 12*a*c*d^2*e^2 - 8*a^2*e^4 + 2*(3*c^2*d^3*e - 2*a*c*d*e^3)*x)/(sqrt(c*d*x +
a*e)*c^3*d^3)

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Fricas [A]  time = 2.18425, size = 292, normalized size = 1.71 \begin{align*} \frac{2 \,{\left (c^{2} d^{2} e^{2} x^{2} - 3 \, c^{2} d^{4} + 12 \, a c d^{2} e^{2} - 8 \, a^{2} e^{4} + 2 \,{\left (3 \, c^{2} d^{3} e - 2 \, a c d e^{3}\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}}{3 \,{\left (c^{4} d^{4} e x^{2} + a c^{3} d^{4} e +{\left (c^{4} d^{5} + a c^{3} d^{3} e^{2}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="fricas")

[Out]

2/3*(c^2*d^2*e^2*x^2 - 3*c^2*d^4 + 12*a*c*d^2*e^2 - 8*a^2*e^4 + 2*(3*c^2*d^3*e - 2*a*c*d*e^3)*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^2 + a*c^3*d^4*e + (c^4*d^5 + a*c^3*d^3*e^2)*x)

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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)**(7/2)/(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(7/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x