### 3.332 $$\int \frac{d+e x}{(b x+c x^2)^{5/2}} \, dx$$

Optimal. Leaf size=71 $\frac{8 (b+2 c x) (2 c d-b e)}{3 b^4 \sqrt{b x+c x^2}}-\frac{2 (x (2 c d-b e)+b d)}{3 b^2 \left (b x+c x^2\right )^{3/2}}$

[Out]

(-2*(b*d + (2*c*d - b*e)*x))/(3*b^2*(b*x + c*x^2)^(3/2)) + (8*(2*c*d - b*e)*(b + 2*c*x))/(3*b^4*Sqrt[b*x + c*x
^2])

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Rubi [A]  time = 0.0181793, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.105, Rules used = {638, 613} $\frac{8 (b+2 c x) (2 c d-b e)}{3 b^4 \sqrt{b x+c x^2}}-\frac{2 (x (2 c d-b e)+b d)}{3 b^2 \left (b x+c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)/(b*x + c*x^2)^(5/2),x]

[Out]

(-2*(b*d + (2*c*d - b*e)*x))/(3*b^2*(b*x + c*x^2)^(3/2)) + (8*(2*c*d - b*e)*(b + 2*c*x))/(3*b^4*Sqrt[b*x + c*x
^2])

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin{align*} \int \frac{d+e x}{\left (b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}-\frac{(4 (2 c d-b e)) \int \frac{1}{\left (b x+c x^2\right )^{3/2}} \, dx}{3 b^2}\\ &=-\frac{2 (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac{8 (2 c d-b e) (b+2 c x)}{3 b^4 \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0244812, size = 67, normalized size = 0.94 $-\frac{2 \left (-6 b^2 c x (d-2 e x)+b^3 (d+3 e x)+8 b c^2 x^2 (e x-3 d)-16 c^3 d x^3\right )}{3 b^4 (x (b+c x))^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*(-16*c^3*d*x^3 - 6*b^2*c*x*(d - 2*e*x) + 8*b*c^2*x^2*(-3*d + e*x) + b^3*(d + 3*e*x)))/(3*b^4*(x*(b + c*x))
^(3/2))

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Maple [A]  time = 0.046, size = 83, normalized size = 1.2 \begin{align*} -{\frac{2\,x \left ( cx+b \right ) \left ( 8\,b{c}^{2}e{x}^{3}-16\,{c}^{3}d{x}^{3}+12\,{b}^{2}ce{x}^{2}-24\,b{c}^{2}d{x}^{2}+3\,{b}^{3}ex-6\,{b}^{2}cdx+d{b}^{3} \right ) }{3\,{b}^{4}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*x*(c*x+b)*(8*b*c^2*e*x^3-16*c^3*d*x^3+12*b^2*c*e*x^2-24*b*c^2*d*x^2+3*b^3*e*x-6*b^2*c*d*x+b^3*d)/b^4/(c*x
^2+b*x)^(5/2)

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Maxima [B]  time = 1.14867, size = 176, normalized size = 2.48 \begin{align*} -\frac{4 \, c d x}{3 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}} b^{2}} + \frac{32 \, c^{2} d x}{3 \, \sqrt{c x^{2} + b x} b^{4}} + \frac{2 \, e x}{3 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}} b} - \frac{16 \, c e x}{3 \, \sqrt{c x^{2} + b x} b^{3}} - \frac{2 \, d}{3 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}} b} + \frac{16 \, c d}{3 \, \sqrt{c x^{2} + b x} b^{3}} - \frac{8 \, e}{3 \, \sqrt{c x^{2} + b x} b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3*c*d*x/((c*x^2 + b*x)^(3/2)*b^2) + 32/3*c^2*d*x/(sqrt(c*x^2 + b*x)*b^4) + 2/3*e*x/((c*x^2 + b*x)^(3/2)*b)
- 16/3*c*e*x/(sqrt(c*x^2 + b*x)*b^3) - 2/3*d/((c*x^2 + b*x)^(3/2)*b) + 16/3*c*d/(sqrt(c*x^2 + b*x)*b^3) - 8/3*
e/(sqrt(c*x^2 + b*x)*b^2)

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Fricas [A]  time = 1.96854, size = 209, normalized size = 2.94 \begin{align*} -\frac{2 \,{\left (b^{3} d - 8 \,{\left (2 \, c^{3} d - b c^{2} e\right )} x^{3} - 12 \,{\left (2 \, b c^{2} d - b^{2} c e\right )} x^{2} - 3 \,{\left (2 \, b^{2} c d - b^{3} e\right )} x\right )} \sqrt{c x^{2} + b x}}{3 \,{\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(b^3*d - 8*(2*c^3*d - b*c^2*e)*x^3 - 12*(2*b*c^2*d - b^2*c*e)*x^2 - 3*(2*b^2*c*d - b^3*e)*x)*sqrt(c*x^2 +
b*x)/(b^4*c^2*x^4 + 2*b^5*c*x^3 + b^6*x^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d + e x}{\left (x \left (b + c x\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d + e*x)/(x*(b + c*x))**(5/2), x)

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Giac [A]  time = 1.39542, size = 136, normalized size = 1.92 \begin{align*} \frac{{\left (4 \, x{\left (\frac{2 \,{\left (2 \, c^{3} d - b c^{2} e\right )} x}{b^{4} c^{2}} + \frac{3 \,{\left (2 \, b c^{2} d - b^{2} c e\right )}}{b^{4} c^{2}}\right )} + \frac{3 \,{\left (2 \, b^{2} c d - b^{3} e\right )}}{b^{4} c^{2}}\right )} x - \frac{d}{b c^{2}}}{3 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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