### 3.2393 $$\int \frac{(d+e x)^3}{(a+b x+c x^2)^{5/2}} \, dx$$

Optimal. Leaf size=118 $\frac{16 \left (a e^2-b d e+c d^2\right ) (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}-\frac{2 (d+e x)^2 (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}$

[Out]

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

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Rubi [A]  time = 0.0416106, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.091, Rules used = {722, 636} $\frac{16 \left (a e^2-b d e+c d^2\right ) (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}}-\frac{2 (d+e x)^2 (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Mathematica [A]  time = 0.816332, size = 190, normalized size = 1.61 $\frac{2 \left (12 b (d-e x) \left (2 a^2 e^2+a c (d-e x)^2+2 c^2 d^2 x^2\right )-8 \left (3 a^2 c e \left (d^2+e^2 x^2\right )+2 a^3 e^3-3 a c^2 d x \left (d^2+e^2 x^2\right )-2 c^3 d^3 x^3\right )-6 b^2 \left (d^2-6 d e x+e^2 x^2\right ) (a e-c d x)+b^3 \left (-9 d^2 e x-d^3+9 d e^2 x^2+e^3 x^3\right )\right )}{3 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.047, size = 296, normalized size = 2.5 \begin{align*} -{\frac{24\,abc{e}^{3}{x}^{3}-48\,a{c}^{2}d{e}^{2}{x}^{3}-2\,{b}^{3}{e}^{3}{x}^{3}-12\,{b}^{2}cd{e}^{2}{x}^{3}+48\,b{c}^{2}{d}^{2}e{x}^{3}-32\,{c}^{3}{d}^{3}{x}^{3}+48\,{a}^{2}c{e}^{3}{x}^{2}+12\,a{b}^{2}{e}^{3}{x}^{2}-72\,abcd{e}^{2}{x}^{2}-18\,{b}^{3}d{e}^{2}{x}^{2}+72\,{b}^{2}c{d}^{2}e{x}^{2}-48\,b{c}^{2}{d}^{3}{x}^{2}+48\,{a}^{2}b{e}^{3}x-72\,a{b}^{2}d{e}^{2}x+72\,abc{d}^{2}ex-48\,a{c}^{2}{d}^{3}x+18\,{b}^{3}{d}^{2}ex-12\,{b}^{2}c{d}^{3}x+32\,{a}^{3}{e}^{3}-48\,d{e}^{2}{a}^{2}b+48\,{a}^{2}c{d}^{2}e+12\,a{b}^{2}{d}^{2}e-24\,abc{d}^{3}+2\,{b}^{3}{d}^{3}}{48\,{a}^{2}{c}^{2}-24\,ac{b}^{2}+3\,{b}^{4}} \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/(c*x^2+b*x+a)^(5/2),x)

[Out]

-2/3/(c*x^2+b*x+a)^(3/2)*(12*a*b*c*e^3*x^3-24*a*c^2*d*e^2*x^3-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+24*a^2*c*e^3*x^2+6*a*b^2*e^3*x^2-36*a*b*c*d*e^2*x^2-9*b^3*d*e^2*x^2+36*b^2*c*d^2*e*x^2-24*b*c^
2*d^3*x^2+24*a^2*b*e^3*x-36*a*b^2*d*e^2*x+36*a*b*c*d^2*e*x-24*a*c^2*d^3*x+9*b^3*d^2*e*x-6*b^2*c*d^3*x+16*a^3*e
^3-24*a^2*b*d*e^2+24*a^2*c*d^2*e+6*a*b^2*d^2*e-12*a*b*c*d^3+b^3*d^3)/(16*a^2*c^2-8*a*b^2*c+b^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((e*x+d)^3/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 8.75005, size = 783, normalized size = 6.64 \begin{align*} \frac{2 \,{\left (24 \, a^{2} b d e^{2} - 16 \, a^{3} e^{3} -{\left (b^{3} - 12 \, a b c\right )} d^{3} - 6 \,{\left (a b^{2} + 4 \, a^{2} c\right )} d^{2} e +{\left (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 6 \,{\left (b^{2} c + 4 \, a c^{2}\right )} d e^{2} +{\left (b^{3} - 12 \, a b c\right )} e^{3}\right )} x^{3} + 3 \,{\left (8 \, b c^{2} d^{3} - 12 \, b^{2} c d^{2} e + 3 \,{\left (b^{3} + 4 \, a b c\right )} d e^{2} - 2 \,{\left (a b^{2} + 4 \, a^{2} c\right )} e^{3}\right )} x^{2} + 3 \,{\left (12 \, a b^{2} d e^{2} - 8 \, a^{2} b e^{3} + 2 \,{\left (b^{2} c + 4 \, a c^{2}\right )} d^{3} - 3 \,{\left (b^{3} + 4 \, a b c\right )} d^{2} e\right )} x\right )} \sqrt{c x^{2} + b x + a}}{3 \,{\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} +{\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{4} + 2 \,{\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{3} +{\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} x^{2} + 2 \,{\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(24*a^2*b*d*e^2 - 16*a^3*e^3 - (b^3 - 12*a*b*c)*d^3 - 6*(a*b^2 + 4*a^2*c)*d^2*e + (16*c^3*d^3 - 24*b*c^2*d
^2*e + 6*(b^2*c + 4*a*c^2)*d*e^2 + (b^3 - 12*a*b*c)*e^3)*x^3 + 3*(8*b*c^2*d^3 - 12*b^2*c*d^2*e + 3*(b^3 + 4*a*
b*c)*d*e^2 - 2*(a*b^2 + 4*a^2*c)*e^3)*x^2 + 3*(12*a*b^2*d*e^2 - 8*a^2*b*e^3 + 2*(b^2*c + 4*a*c^2)*d^3 - 3*(b^3
+ 4*a*b*c)*d^2*e)*x)*sqrt(c*x^2 + b*x + a)/(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 16*
a^2*c^4)*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 - 8*
a^2*b^3*c + 16*a^3*b*c^2)*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((e*x+d)**3/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.12298, size = 474, normalized size = 4.02 \begin{align*} \frac{{\left ({\left (\frac{{\left (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 6 \, b^{2} c d e^{2} + 24 \, a c^{2} d e^{2} + b^{3} e^{3} - 12 \, a b c e^{3}\right )} x}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}} + \frac{3 \,{\left (8 \, b c^{2} d^{3} - 12 \, b^{2} c d^{2} e + 3 \, b^{3} d e^{2} + 12 \, a b c d e^{2} - 2 \, a b^{2} e^{3} - 8 \, a^{2} c e^{3}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac{3 \,{\left (2 \, b^{2} c d^{3} + 8 \, a c^{2} d^{3} - 3 \, b^{3} d^{2} e - 12 \, a b c d^{2} e + 12 \, a b^{2} d e^{2} - 8 \, a^{2} b e^{3}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x - \frac{b^{3} d^{3} - 12 \, a b c d^{3} + 6 \, a b^{2} d^{2} e + 24 \, a^{2} c d^{2} e - 24 \, a^{2} b d e^{2} + 16 \, a^{3} e^{3}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}}{3 \,{\left (c x^{2} + b x + a\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+a)^(5/2),x, algorithm="giac")

[Out]

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