3.1086 \(\int \frac{1}{(d+e x) (c d^2+2 c d e x+c e^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=31 \[ -\frac{1}{5 e \left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \]

[Out]

-1/(5*e*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0242698, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {643, 629} \[ -\frac{1}{5 e \left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2)),x]

[Out]

-1/(5*e*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2))

Rule 643

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[e^(m - 1)/c^((m - 1)/2
), Int[(d + e*x)*(a + b*x + c*x^2)^(p + (m - 1)/2), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[b^2 - 4*a*c,
 0] &&  !IntegerQ[p] && EqQ[2*c*d - b*e, 0] && IntegerQ[(m - 1)/2]

Rule 629

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d*(a + b*x + c*x^2)^(p +
 1))/(b*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rubi steps

\begin{align*} \int \frac{1}{(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx &=c \int \frac{d+e x}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{7/2}} \, dx\\ &=-\frac{1}{5 e \left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0266593, size = 20, normalized size = 0.65 \[ -\frac{1}{5 e \left (c (d+e x)^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2)),x]

[Out]

-1/(5*e*(c*(d + e*x)^2)^(5/2))

________________________________________________________________________________________

Maple [A]  time = 0.042, size = 28, normalized size = 0.9 \begin{align*} -{\frac{1}{5\,e} \left ( c{e}^{2}{x}^{2}+2\,cdex+c{d}^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x)

[Out]

-1/5/e/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2)

________________________________________________________________________________________

Maxima [B]  time = 1.21331, size = 101, normalized size = 3.26 \begin{align*} -\frac{1}{5 \,{\left (c^{\frac{5}{2}} e^{6} x^{5} + 5 \, c^{\frac{5}{2}} d e^{5} x^{4} + 10 \, c^{\frac{5}{2}} d^{2} e^{4} x^{3} + 10 \, c^{\frac{5}{2}} d^{3} e^{3} x^{2} + 5 \, c^{\frac{5}{2}} d^{4} e^{2} x + c^{\frac{5}{2}} d^{5} e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5/(c^(5/2)*e^6*x^5 + 5*c^(5/2)*d*e^5*x^4 + 10*c^(5/2)*d^2*e^4*x^3 + 10*c^(5/2)*d^3*e^3*x^2 + 5*c^(5/2)*d^4*
e^2*x + c^(5/2)*d^5*e)

________________________________________________________________________________________

Fricas [B]  time = 2.32166, size = 225, normalized size = 7.26 \begin{align*} -\frac{\sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{5 \,{\left (c^{3} e^{7} x^{6} + 6 \, c^{3} d e^{6} x^{5} + 15 \, c^{3} d^{2} e^{5} x^{4} + 20 \, c^{3} d^{3} e^{4} x^{3} + 15 \, c^{3} d^{4} e^{3} x^{2} + 6 \, c^{3} d^{5} e^{2} x + c^{3} d^{6} e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*sqrt(c*e^2*x^2 + 2*c*d*e*x + c*d^2)/(c^3*e^7*x^6 + 6*c^3*d*e^6*x^5 + 15*c^3*d^2*e^5*x^4 + 20*c^3*d^3*e^4*
x^3 + 15*c^3*d^4*e^3*x^2 + 6*c^3*d^5*e^2*x + c^3*d^6*e)

________________________________________________________________________________________

Sympy [A]  time = 3.66332, size = 42, normalized size = 1.35 \begin{align*} \begin{cases} - \frac{1}{5 e \left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{\frac{5}{2}}} & \text{for}\: e \neq 0 \\\frac{x}{d \left (c d^{2}\right )^{\frac{5}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*e**2*x**2+2*c*d*e*x+c*d**2)**(5/2),x)

[Out]

Piecewise((-1/(5*e*(c*d**2 + 2*c*d*e*x + c*e**2*x**2)**(5/2)), Ne(e, 0)), (x/(d*(c*d**2)**(5/2)), True))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \left [\mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, 1\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[undef, undef, undef, undef, undef, 1]