3.1272 \(\int \frac{(a+b x+c x^2)^2}{(b d+2 c d x)^{3/2}} \, dx\)

Optimal. Leaf size=88 \[ -\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}{24 c^3 d^3}-\frac{\left (b^2-4 a c\right )^2}{16 c^3 d \sqrt{b d+2 c d x}}+\frac{(b d+2 c d x)^{7/2}}{112 c^3 d^5} \]

[Out]

-(b^2 - 4*a*c)^2/(16*c^3*d*Sqrt[b*d + 2*c*d*x]) - ((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(3/2))/(24*c^3*d^3) + (b*d +
2*c*d*x)^(7/2)/(112*c^3*d^5)

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Rubi [A]  time = 0.0382175, 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} \[ -\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}{24 c^3 d^3}-\frac{\left (b^2-4 a c\right )^2}{16 c^3 d \sqrt{b d+2 c d x}}+\frac{(b d+2 c d x)^{7/2}}{112 c^3 d^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^2 - 4*a*c)^2/(16*c^3*d*Sqrt[b*d + 2*c*d*x]) - ((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(3/2))/(24*c^3*d^3) + (b*d +
2*c*d*x)^(7/2)/(112*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)^{3/2}} \, dx &=\int \left (\frac{\left (-b^2+4 a c\right )^2}{16 c^2 (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^2 d^2}+\frac{(b d+2 c d x)^{5/2}}{16 c^2 d^4}\right ) \, dx\\ &=-\frac{\left (b^2-4 a c\right )^2}{16 c^3 d \sqrt{b d+2 c d x}}-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}{24 c^3 d^3}+\frac{(b d+2 c d x)^{7/2}}{112 c^3 d^5}\\ \end{align*}

Mathematica [A]  time = 0.0382741, size = 91, normalized size = 1.03 \[ \frac{c^2 \left (-21 a^2+14 a c x^2+3 c^2 x^4\right )+b^2 c \left (14 a+c x^2\right )+2 b c^2 x \left (7 a+3 c x^2\right )-2 b^3 c x-2 b^4}{21 c^3 d \sqrt{d (b+2 c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^4 - 2*b^3*c*x + b^2*c*(14*a + c*x^2) + 2*b*c^2*x*(7*a + 3*c*x^2) + c^2*(-21*a^2 + 14*a*c*x^2 + 3*c^2*x^4
))/(21*c^3*d*Sqrt[d*(b + 2*c*x)])

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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}-14\,a{c}^{3}{x}^{2}-{b}^{2}{c}^{2}{x}^{2}-14\,ab{c}^{2}x+2\,{b}^{3}cx+21\,{a}^{2}{c}^{2}-14\,ac{b}^{2}+2\,{b}^{4} \right ) }{21\,{c}^{3}} \left ( 2\,cdx+bd \right ) ^{-{\frac{3}{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)^(3/2),x)

[Out]

-1/21*(2*c*x+b)*(-3*c^4*x^4-6*b*c^3*x^3-14*a*c^3*x^2-b^2*c^2*x^2-14*a*b*c^2*x+2*b^3*c*x+21*a^2*c^2-14*a*b^2*c+
2*b^4)/c^3/(2*c*d*x+b*d)^(3/2)

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Maxima [A]  time = 1.2528, size = 120, normalized size = 1.36 \begin{align*} -\frac{\frac{21 \,{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )}}{\sqrt{2 \, c d x + b d} c^{2}} + \frac{14 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}}{\left (b^{2} - 4 \, a c\right )} d^{2} - 3 \,{\left (2 \, c d x + b d\right )}^{\frac{7}{2}}}{c^{2} d^{4}}}{336 \, 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)^(3/2),x, algorithm="maxima")

[Out]

-1/336*(21*(b^4 - 8*a*b^2*c + 16*a^2*c^2)/(sqrt(2*c*d*x + b*d)*c^2) + (14*(2*c*d*x + b*d)^(3/2)*(b^2 - 4*a*c)*
d^2 - 3*(2*c*d*x + b*d)^(7/2))/(c^2*d^4))/(c*d)

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Fricas [A]  time = 2.01115, size = 223, normalized size = 2.53 \begin{align*} \frac{{\left (3 \, c^{4} x^{4} + 6 \, b c^{3} x^{3} - 2 \, b^{4} + 14 \, a b^{2} c - 21 \, a^{2} c^{2} +{\left (b^{2} c^{2} + 14 \, a c^{3}\right )} x^{2} - 2 \,{\left (b^{3} c - 7 \, a b c^{2}\right )} x\right )} \sqrt{2 \, c d x + b d}}{21 \,{\left (2 \, c^{4} d^{2} x + b c^{3} d^{2}\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)^(3/2),x, algorithm="fricas")

[Out]

1/21*(3*c^4*x^4 + 6*b*c^3*x^3 - 2*b^4 + 14*a*b^2*c - 21*a^2*c^2 + (b^2*c^2 + 14*a*c^3)*x^2 - 2*(b^3*c - 7*a*b*
c^2)*x)*sqrt(2*c*d*x + b*d)/(2*c^4*d^2*x + b*c^3*d^2)

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Sympy [A]  time = 27.4823, size = 82, normalized size = 0.93 \begin{align*} - \frac{\left (4 a c - b^{2}\right )^{2}}{16 c^{3} d \sqrt{b d + 2 c d x}} + \frac{\left (4 a c - b^{2}\right ) \left (b d + 2 c d x\right )^{\frac{3}{2}}}{24 c^{3} d^{3}} + \frac{\left (b d + 2 c d x\right )^{\frac{7}{2}}}{112 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)**(3/2),x)

[Out]

-(4*a*c - b**2)**2/(16*c**3*d*sqrt(b*d + 2*c*d*x)) + (4*a*c - b**2)*(b*d + 2*c*d*x)**(3/2)/(24*c**3*d**3) + (b
*d + 2*c*d*x)**(7/2)/(112*c**3*d**5)

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Giac [A]  time = 1.19754, size = 147, normalized size = 1.67 \begin{align*} -\frac{b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}{16 \, \sqrt{2 \, c d x + b d} c^{3} d} - \frac{14 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} b^{2} c^{18} d^{32} - 56 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} a c^{19} d^{32} - 3 \,{\left (2 \, c d x + b d\right )}^{\frac{7}{2}} c^{18} d^{30}}{336 \, c^{21} d^{35}} \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)^(3/2),x, algorithm="giac")

[Out]

-1/16*(b^4 - 8*a*b^2*c + 16*a^2*c^2)/(sqrt(2*c*d*x + b*d)*c^3*d) - 1/336*(14*(2*c*d*x + b*d)^(3/2)*b^2*c^18*d^
32 - 56*(2*c*d*x + b*d)^(3/2)*a*c^19*d^32 - 3*(2*c*d*x + b*d)^(7/2)*c^18*d^30)/(c^21*d^35)