### 3.330 $$\int \frac{(d+e x)^3}{(b x+c x^2)^{5/2}} \, dx$$

Optimal. Leaf size=87 $\frac{16 d (c d-b e) (x (2 c d-b e)+b d)}{3 b^4 \sqrt{b x+c x^2}}-\frac{2 (d+e x)^2 (x (2 c d-b e)+b d)}{3 b^2 \left (b x+c x^2\right )^{3/2}}$

[Out]

(-2*(d + e*x)^2*(b*d + (2*c*d - b*e)*x))/(3*b^2*(b*x + c*x^2)^(3/2)) + (16*d*(c*d - b*e)*(b*d + (2*c*d - b*e)*
x))/(3*b^4*Sqrt[b*x + c*x^2])

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Rubi [A]  time = 0.0364209, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.095, Rules used = {722, 636} $\frac{16 d (c d-b e) (x (2 c d-b e)+b d)}{3 b^4 \sqrt{b x+c x^2}}-\frac{2 (d+e x)^2 (x (2 c d-b e)+b d)}{3 b^2 \left (b x+c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^3/(b*x + c*x^2)^(5/2),x]

[Out]

(-2*(d + e*x)^2*(b*d + (2*c*d - b*e)*x))/(3*b^2*(b*x + c*x^2)^(3/2)) + (16*d*(c*d - b*e)*(b*d + (2*c*d - b*e)*
x))/(3*b^4*Sqrt[b*x + c*x^2])

Rule 722

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

Rule 636

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

Rubi steps

\begin{align*} \int \frac{(d+e x)^3}{\left (b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 (d+e x)^2 (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}-\frac{(8 d (c d-b e)) \int \frac{d+e x}{\left (b x+c x^2\right )^{3/2}} \, dx}{3 b^2}\\ &=-\frac{2 (d+e x)^2 (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac{16 d (c d-b e) (b d+(2 c d-b e) x)}{3 b^4 \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0482613, size = 105, normalized size = 1.21 $\frac{2 \left (6 b^2 c d x \left (d^2-6 d e x+e^2 x^2\right )+b^3 \left (-9 d^2 e x-d^3+9 d e^2 x^2+e^3 x^3\right )+24 b c^2 d^2 x^2 (d-e x)+16 c^3 d^3 x^3\right )}{3 b^4 (x (b+c x))^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^3/(b*x + c*x^2)^(5/2),x]

[Out]

(2*(16*c^3*d^3*x^3 + 24*b*c^2*d^2*x^2*(d - e*x) + 6*b^2*c*d*x*(d^2 - 6*d*e*x + e^2*x^2) + b^3*(-d^3 - 9*d^2*e*
x + 9*d*e^2*x^2 + e^3*x^3)))/(3*b^4*(x*(b + c*x))^(3/2))

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Maple [A]  time = 0.047, size = 136, normalized size = 1.6 \begin{align*} -{\frac{2\,x \left ( cx+b \right ) \left ( -{b}^{3}{e}^{3}{x}^{3}-6\,{b}^{2}cd{e}^{2}{x}^{3}+24\,b{c}^{2}{d}^{2}e{x}^{3}-16\,{c}^{3}{d}^{3}{x}^{3}-9\,{b}^{3}d{e}^{2}{x}^{2}+36\,{b}^{2}c{d}^{2}e{x}^{2}-24\,b{c}^{2}{d}^{3}{x}^{2}+9\,{b}^{3}{d}^{2}ex-6\,{b}^{2}c{d}^{3}x+{d}^{3}{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)^3/(c*x^2+b*x)^(5/2),x)

[Out]

-2/3*x*(c*x+b)*(-b^3*e^3*x^3-6*b^2*c*d*e^2*x^3+24*b*c^2*d^2*e*x^3-16*c^3*d^3*x^3-9*b^3*d*e^2*x^2+36*b^2*c*d^2*
e*x^2-24*b*c^2*d^3*x^2+9*b^3*d^2*e*x-6*b^2*c*d^3*x+b^3*d^3)/b^4/(c*x^2+b*x)^(5/2)

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Maxima [B]  time = 1.17347, size = 401, normalized size = 4.61 \begin{align*} -\frac{e^{3} x^{2}}{{\left (c x^{2} + b x\right )}^{\frac{3}{2}} c} - \frac{4 \, c d^{3} x}{3 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}} b^{2}} + \frac{32 \, c^{2} d^{3} x}{3 \, \sqrt{c x^{2} + b x} b^{4}} + \frac{2 \, d^{2} e x}{{\left (c x^{2} + b x\right )}^{\frac{3}{2}} b} - \frac{16 \, c d^{2} e x}{\sqrt{c x^{2} + b x} b^{3}} + \frac{4 \, d e^{2} x}{\sqrt{c x^{2} + b x} b^{2}} - \frac{2 \, d e^{2} x}{{\left (c x^{2} + b x\right )}^{\frac{3}{2}} c} - \frac{b e^{3} x}{3 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}} c^{2}} + \frac{2 \, e^{3} x}{3 \, \sqrt{c x^{2} + b x} b c} - \frac{2 \, d^{3}}{3 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}} b} + \frac{16 \, c d^{3}}{3 \, \sqrt{c x^{2} + b x} b^{3}} - \frac{8 \, d^{2} e}{\sqrt{c x^{2} + b x} b^{2}} + \frac{2 \, d e^{2}}{\sqrt{c x^{2} + b x} b c} + \frac{e^{3}}{3 \, \sqrt{c x^{2} + b x} c^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

-e^3*x^2/((c*x^2 + b*x)^(3/2)*c) - 4/3*c*d^3*x/((c*x^2 + b*x)^(3/2)*b^2) + 32/3*c^2*d^3*x/(sqrt(c*x^2 + b*x)*b
^4) + 2*d^2*e*x/((c*x^2 + b*x)^(3/2)*b) - 16*c*d^2*e*x/(sqrt(c*x^2 + b*x)*b^3) + 4*d*e^2*x/(sqrt(c*x^2 + b*x)*
b^2) - 2*d*e^2*x/((c*x^2 + b*x)^(3/2)*c) - 1/3*b*e^3*x/((c*x^2 + b*x)^(3/2)*c^2) + 2/3*e^3*x/(sqrt(c*x^2 + b*x
)*b*c) - 2/3*d^3/((c*x^2 + b*x)^(3/2)*b) + 16/3*c*d^3/(sqrt(c*x^2 + b*x)*b^3) - 8*d^2*e/(sqrt(c*x^2 + b*x)*b^2
) + 2*d*e^2/(sqrt(c*x^2 + b*x)*b*c) + 1/3*e^3/(sqrt(c*x^2 + b*x)*c^2)

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Fricas [A]  time = 1.94477, size = 298, normalized size = 3.43 \begin{align*} -\frac{2 \,{\left (b^{3} d^{3} -{\left (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 6 \, b^{2} c d e^{2} + b^{3} e^{3}\right )} x^{3} - 3 \,{\left (8 \, b c^{2} d^{3} - 12 \, b^{2} c d^{2} e + 3 \, b^{3} d e^{2}\right )} x^{2} - 3 \,{\left (2 \, b^{2} c d^{3} - 3 \, b^{3} d^{2} 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)^3/(c*x^2+b*x)^(5/2),x, algorithm="fricas")

[Out]

-2/3*(b^3*d^3 - (16*c^3*d^3 - 24*b*c^2*d^2*e + 6*b^2*c*d*e^2 + b^3*e^3)*x^3 - 3*(8*b*c^2*d^3 - 12*b^2*c*d^2*e
+ 3*b^3*d*e^2)*x^2 - 3*(2*b^2*c*d^3 - 3*b^3*d^2*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{\left (d + e x\right )^{3}}{\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)**3/(c*x**2+b*x)**(5/2),x)

[Out]

Integral((d + e*x)**3/(x*(b + c*x))**(5/2), x)

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Giac [A]  time = 1.28533, size = 188, normalized size = 2.16 \begin{align*} \frac{{\left (x{\left (\frac{{\left (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 6 \, b^{2} c d e^{2} + b^{3} e^{3}\right )} x}{b^{4} c^{2}} + \frac{3 \,{\left (8 \, b c^{2} d^{3} - 12 \, b^{2} c d^{2} e + 3 \, b^{3} d e^{2}\right )}}{b^{4} c^{2}}\right )} + \frac{3 \,{\left (2 \, b^{2} c d^{3} - 3 \, b^{3} d^{2} e\right )}}{b^{4} c^{2}}\right )} x - \frac{d^{3}}{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)^3/(c*x^2+b*x)^(5/2),x, algorithm="giac")

[Out]

1/3*((x*((16*c^3*d^3 - 24*b*c^2*d^2*e + 6*b^2*c*d*e^2 + b^3*e^3)*x/(b^4*c^2) + 3*(8*b*c^2*d^3 - 12*b^2*c*d^2*e
+ 3*b^3*d*e^2)/(b^4*c^2)) + 3*(2*b^2*c*d^3 - 3*b^3*d^2*e)/(b^4*c^2))*x - d^3/(b*c^2))/(c*x^2 + b*x)^(3/2)