### 3.1273 $$\int \frac{(a+b x+c x^2)^2}{(b d+2 c d x)^{5/2}} \, dx$$

Optimal. Leaf size=88 $-\frac{\left (b^2-4 a c\right ) \sqrt{b d+2 c d x}}{8 c^3 d^3}-\frac{\left (b^2-4 a c\right )^2}{48 c^3 d (b d+2 c d x)^{3/2}}+\frac{(b d+2 c d x)^{5/2}}{80 c^3 d^5}$

[Out]

-(b^2 - 4*a*c)^2/(48*c^3*d*(b*d + 2*c*d*x)^(3/2)) - ((b^2 - 4*a*c)*Sqrt[b*d + 2*c*d*x])/(8*c^3*d^3) + (b*d + 2
*c*d*x)^(5/2)/(80*c^3*d^5)

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Rubi [A]  time = 0.0403534, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.038, Rules used = {683} $-\frac{\left (b^2-4 a c\right ) \sqrt{b d+2 c d x}}{8 c^3 d^3}-\frac{\left (b^2-4 a c\right )^2}{48 c^3 d (b d+2 c d x)^{3/2}}+\frac{(b d+2 c d x)^{5/2}}{80 c^3 d^5}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(b^2 - 4*a*c)^2/(48*c^3*d*(b*d + 2*c*d*x)^(3/2)) - ((b^2 - 4*a*c)*Sqrt[b*d + 2*c*d*x])/(8*c^3*d^3) + (b*d + 2
*c*d*x)^(5/2)/(80*c^3*d^5)

Rule 683

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

Rubi steps

\begin{align*} \int \frac{\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{5/2}} \, dx &=\int \left (\frac{\left (-b^2+4 a c\right )^2}{16 c^2 (b d+2 c d x)^{5/2}}+\frac{-b^2+4 a c}{8 c^2 d^2 \sqrt{b d+2 c d x}}+\frac{(b d+2 c d x)^{3/2}}{16 c^2 d^4}\right ) \, dx\\ &=-\frac{\left (b^2-4 a c\right )^2}{48 c^3 d (b d+2 c d x)^{3/2}}-\frac{\left (b^2-4 a c\right ) \sqrt{b d+2 c d x}}{8 c^3 d^3}+\frac{(b d+2 c d x)^{5/2}}{80 c^3 d^5}\\ \end{align*}

Mathematica [A]  time = 0.0398138, size = 91, normalized size = 1.03 $\frac{c^2 \left (-5 a^2+30 a c x^2+3 c^2 x^4\right )+b^2 c \left (10 a-3 c x^2\right )+6 b c^2 x \left (5 a+c x^2\right )-6 b^3 c x-2 b^4}{15 c^3 d (d (b+2 c x))^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*b^4 - 6*b^3*c*x + b^2*c*(10*a - 3*c*x^2) + 6*b*c^2*x*(5*a + c*x^2) + c^2*(-5*a^2 + 30*a*c*x^2 + 3*c^2*x^4)
)/(15*c^3*d*(d*(b + 2*c*x))^(3/2))

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Maple [A]  time = 0.044, size = 96, normalized size = 1.1 \begin{align*} -{\frac{ \left ( 2\,cx+b \right ) \left ( -3\,{c}^{4}{x}^{4}-6\,b{x}^{3}{c}^{3}-30\,a{c}^{3}{x}^{2}+3\,{b}^{2}{c}^{2}{x}^{2}-30\,ab{c}^{2}x+6\,{b}^{3}cx+5\,{a}^{2}{c}^{2}-10\,ac{b}^{2}+2\,{b}^{4} \right ) }{15\,{c}^{3}} \left ( 2\,cdx+bd \right ) ^{-{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(2*c*x+b)*(-3*c^4*x^4-6*b*c^3*x^3-30*a*c^3*x^2+3*b^2*c^2*x^2-30*a*b*c^2*x+6*b^3*c*x+5*a^2*c^2-10*a*b^2*c
+2*b^4)/c^3/(2*c*d*x+b*d)^(5/2)

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Maxima [A]  time = 1.10376, size = 122, normalized size = 1.39 \begin{align*} -\frac{\frac{5 \,{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )}}{{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} c^{2}} + \frac{3 \,{\left (10 \, \sqrt{2 \, c d x + b d}{\left (b^{2} - 4 \, a c\right )} d^{2} -{\left (2 \, c d x + b d\right )}^{\frac{5}{2}}\right )}}{c^{2} d^{4}}}{240 \, c d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(5*(b^4 - 8*a*b^2*c + 16*a^2*c^2)/((2*c*d*x + b*d)^(3/2)*c^2) + 3*(10*sqrt(2*c*d*x + b*d)*(b^2 - 4*a*c)
*d^2 - (2*c*d*x + b*d)^(5/2))/(c^2*d^4))/(c*d)

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Fricas [A]  time = 2.01993, size = 251, normalized size = 2.85 \begin{align*} \frac{{\left (3 \, c^{4} x^{4} + 6 \, b c^{3} x^{3} - 2 \, b^{4} + 10 \, a b^{2} c - 5 \, a^{2} c^{2} - 3 \,{\left (b^{2} c^{2} - 10 \, a c^{3}\right )} x^{2} - 6 \,{\left (b^{3} c - 5 \, a b c^{2}\right )} x\right )} \sqrt{2 \, c d x + b d}}{15 \,{\left (4 \, c^{5} d^{3} x^{2} + 4 \, b c^{4} d^{3} x + b^{2} c^{3} d^{3}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*c^4*x^4 + 6*b*c^3*x^3 - 2*b^4 + 10*a*b^2*c - 5*a^2*c^2 - 3*(b^2*c^2 - 10*a*c^3)*x^2 - 6*(b^3*c - 5*a*b
*c^2)*x)*sqrt(2*c*d*x + b*d)/(4*c^5*d^3*x^2 + 4*b*c^4*d^3*x + b^2*c^3*d^3)

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Sympy [A]  time = 49.9339, size = 82, normalized size = 0.93 \begin{align*} - \frac{\left (4 a c - b^{2}\right )^{2}}{48 c^{3} d \left (b d + 2 c d x\right )^{\frac{3}{2}}} + \frac{\left (4 a c - b^{2}\right ) \sqrt{b d + 2 c d x}}{8 c^{3} d^{3}} + \frac{\left (b d + 2 c d x\right )^{\frac{5}{2}}}{80 c^{3} d^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*a*c - b**2)**2/(48*c**3*d*(b*d + 2*c*d*x)**(3/2)) + (4*a*c - b**2)*sqrt(b*d + 2*c*d*x)/(8*c**3*d**3) + (b*
d + 2*c*d*x)**(5/2)/(80*c**3*d**5)

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Giac [A]  time = 1.18923, size = 147, normalized size = 1.67 \begin{align*} -\frac{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}{48 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} c^{3} d} - \frac{10 \, \sqrt{2 \, c d x + b d} b^{2} c^{12} d^{22} - 40 \, \sqrt{2 \, c d x + b d} a c^{13} d^{22} -{\left (2 \, c d x + b d\right )}^{\frac{5}{2}} c^{12} d^{20}}{80 \, c^{15} d^{25}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(b^4 - 8*a*b^2*c + 16*a^2*c^2)/((2*c*d*x + b*d)^(3/2)*c^3*d) - 1/80*(10*sqrt(2*c*d*x + b*d)*b^2*c^12*d^2
2 - 40*sqrt(2*c*d*x + b*d)*a*c^13*d^22 - (2*c*d*x + b*d)^(5/2)*c^12*d^20)/(c^15*d^25)