3.1928 \(\int \frac{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}}{(d+e x)^6} \, dx\)

Optimal. Leaf size=111 \[ \frac{4 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{35 (d+e x)^5 \left (c d^2-a e^2\right )^2}+\frac{2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{7 (d+e x)^6 \left (c d^2-a e^2\right )} \]

[Out]

(2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(7*(c*d^2 - a*e^2)*(d + e*x)^6) + (4*c*d*(a*d*e + (c*d^2 + a
*e^2)*x + c*d*e*x^2)^(5/2))/(35*(c*d^2 - a*e^2)^2*(d + e*x)^5)

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Rubi [A]  time = 0.0479102, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.054, Rules used = {658, 650} \[ \frac{4 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{35 (d+e x)^5 \left (c d^2-a e^2\right )^2}+\frac{2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{7 (d+e x)^6 \left (c d^2-a e^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)/(d + e*x)^6,x]

[Out]

(2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(7*(c*d^2 - a*e^2)*(d + e*x)^6) + (4*c*d*(a*d*e + (c*d^2 + a
*e^2)*x + c*d*e*x^2)^(5/2))/(35*(c*d^2 - a*e^2)^2*(d + e*x)^5)

Rule 658

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*Simplify[m + 2*p + 2])/((m + p + 1)*(2*c*d -
b*e)), Int[(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] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2], 0]

Rule 650

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)), 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 + 2*p + 2, 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 )^{3/2}}{(d+e x)^6} \, dx &=\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{7 \left (c d^2-a e^2\right ) (d+e x)^6}+\frac{(2 c d) \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)^5} \, dx}{7 \left (c d^2-a e^2\right )}\\ &=\frac{2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{7 \left (c d^2-a e^2\right ) (d+e x)^6}+\frac{4 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{35 \left (c d^2-a e^2\right )^2 (d+e x)^5}\\ \end{align*}

Mathematica [A]  time = 0.047925, size = 61, normalized size = 0.55 \[ \frac{2 ((d+e x) (a e+c d x))^{5/2} \left (c d (7 d+2 e x)-5 a e^2\right )}{35 (d+e x)^6 \left (c d^2-a e^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)/(d + e*x)^6,x]

[Out]

(2*((a*e + c*d*x)*(d + e*x))^(5/2)*(-5*a*e^2 + c*d*(7*d + 2*e*x)))/(35*(c*d^2 - a*e^2)^2*(d + e*x)^6)

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Maple [A]  time = 0.045, size = 90, normalized size = 0.8 \begin{align*} -{\frac{ \left ( 2\,cdx+2\,ae \right ) \left ( -2\,cdex+5\,a{e}^{2}-7\,c{d}^{2} \right ) }{35\, \left ( ex+d \right ) ^{5} \left ({a}^{2}{e}^{4}-2\,ac{d}^{2}{e}^{2}+{c}^{2}{d}^{4} \right ) } \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((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d)^6,x)

[Out]

-2/35*(c*d*x+a*e)*(-2*c*d*e*x+5*a*e^2-7*c*d^2)*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(3/2)/(e*x+d)^5/(a^2*e^4-2*a*
c*d^2*e^2+c^2*d^4)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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)^(3/2)/(e*x+d)^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 22.4999, size = 548, normalized size = 4.94 \begin{align*} \frac{2 \,{\left (2 \, c^{3} d^{3} e x^{3} + 7 \, a^{2} c d^{2} e^{2} - 5 \, a^{3} e^{4} +{\left (7 \, c^{3} d^{4} - a c^{2} d^{2} e^{2}\right )} x^{2} + 2 \,{\left (7 \, a c^{2} d^{3} e - 4 \, a^{2} 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}}{35 \,{\left (c^{2} d^{8} - 2 \, a c d^{6} e^{2} + a^{2} d^{4} e^{4} +{\left (c^{2} d^{4} e^{4} - 2 \, a c d^{2} e^{6} + a^{2} e^{8}\right )} x^{4} + 4 \,{\left (c^{2} d^{5} e^{3} - 2 \, a c d^{3} e^{5} + a^{2} d e^{7}\right )} x^{3} + 6 \,{\left (c^{2} d^{6} e^{2} - 2 \, a c d^{4} e^{4} + a^{2} d^{2} e^{6}\right )} x^{2} + 4 \,{\left (c^{2} d^{7} e - 2 \, a c d^{5} e^{3} + a^{2} d^{3} e^{5}\right )} x\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)^(3/2)/(e*x+d)^6,x, algorithm="fricas")

[Out]

2/35*(2*c^3*d^3*e*x^3 + 7*a^2*c*d^2*e^2 - 5*a^3*e^4 + (7*c^3*d^4 - a*c^2*d^2*e^2)*x^2 + 2*(7*a*c^2*d^3*e - 4*a
^2*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)/(c^2*d^8 - 2*a*c*d^6*e^2 + a^2*d^4*e^4 + (c^2*d^4*e
^4 - 2*a*c*d^2*e^6 + a^2*e^8)*x^4 + 4*(c^2*d^5*e^3 - 2*a*c*d^3*e^5 + a^2*d*e^7)*x^3 + 6*(c^2*d^6*e^2 - 2*a*c*d
^4*e^4 + a^2*d^2*e^6)*x^2 + 4*(c^2*d^7*e - 2*a*c*d^5*e^3 + a^2*d^3*e^5)*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((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2)/(e*x+d)**6,x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \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)^(3/2)/(e*x+d)^6,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError