### 3.1077 $$\int \frac{1}{(d+e x)^2 (c d^2+2 c d e x+c e^2 x^2)^{3/2}} \, dx$$

Optimal. Leaf size=38 $-\frac{1}{4 e (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}$

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0202065, antiderivative size = 38, 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 = {642, 607} $-\frac{1}{4 e (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((d + e*x)^2*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2)),x]

[Out]

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

Rule 642

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[e^m/c^(m/2), Int[(a +
b*x + c*x^2)^(p + m/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/2]

Rule 607

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(2*(a + b*x + c*x^2)^(p + 1))/((2*p + 1)*(b + 2
*c*x)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && LtQ[p, -1]

Rubi steps

\begin{align*} \int \frac{1}{(d+e x)^2 \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \, dx &=c \int \frac{1}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx\\ &=-\frac{1}{4 e (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0177119, size = 26, normalized size = 0.68 $-\frac{c (d+e x)}{4 e \left (c (d+e x)^2\right )^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((d + e*x)^2*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2)),x]

[Out]

-(c*(d + e*x))/(4*e*(c*(d + e*x)^2)^(5/2))

________________________________________________________________________________________

Maple [A]  time = 0.041, size = 35, normalized size = 0.9 \begin{align*} -{\frac{1}{4\,e \left ( ex+d \right ) } \left ( c{e}^{2}{x}^{2}+2\,cdex+c{d}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^2/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.20297, size = 82, normalized size = 2.16 \begin{align*} -\frac{1}{4 \,{\left (c^{\frac{3}{2}} e^{5} x^{4} + 4 \, c^{\frac{3}{2}} d e^{4} x^{3} + 6 \, c^{\frac{3}{2}} d^{2} e^{3} x^{2} + 4 \, c^{\frac{3}{2}} d^{3} e^{2} x + c^{\frac{3}{2}} d^{4} e\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^2/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 2.42569, size = 197, normalized size = 5.18 \begin{align*} -\frac{\sqrt{c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{4 \,{\left (c^{2} e^{6} x^{5} + 5 \, c^{2} d e^{5} x^{4} + 10 \, c^{2} d^{2} e^{4} x^{3} + 10 \, c^{2} d^{3} e^{3} x^{2} + 5 \, c^{2} d^{4} e^{2} x + c^{2} d^{5} e\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^2/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (c \left (d + e x\right )^{2}\right )^{\frac{3}{2}} \left (d + e x\right )^{2}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**2/(c*e**2*x**2+2*c*d*e*x+c*d**2)**(3/2),x)

[Out]

Integral(1/((c*(d + e*x)**2)**(3/2)*(d + e*x)**2), x)

________________________________________________________________________________________

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

[Out]

Timed out