### 3.1270 $$\int \sqrt{b d+2 c d x} (a+b x+c x^2)^2 \, dx$$

Optimal. Leaf size=88 $-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}{56 c^3 d^3}+\frac{\left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}}{48 c^3 d}+\frac{(b d+2 c d x)^{11/2}}{176 c^3 d^5}$

[Out]

((b^2 - 4*a*c)^2*(b*d + 2*c*d*x)^(3/2))/(48*c^3*d) - ((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(7/2))/(56*c^3*d^3) + (b*d
+ 2*c*d*x)^(11/2)/(176*c^3*d^5)

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Rubi [A]  time = 0.039476, 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} $-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}{56 c^3 d^3}+\frac{\left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}}{48 c^3 d}+\frac{(b d+2 c d x)^{11/2}}{176 c^3 d^5}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((b^2 - 4*a*c)^2*(b*d + 2*c*d*x)^(3/2))/(48*c^3*d) - ((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(7/2))/(56*c^3*d^3) + (b*d
+ 2*c*d*x)^(11/2)/(176*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 \sqrt{b d+2 c d x} \left (a+b x+c x^2\right )^2 \, dx &=\int \left (\frac{\left (-b^2+4 a c\right )^2 \sqrt{b d+2 c d x}}{16 c^2}+\frac{\left (-b^2+4 a c\right ) (b d+2 c d x)^{5/2}}{8 c^2 d^2}+\frac{(b d+2 c d x)^{9/2}}{16 c^2 d^4}\right ) \, dx\\ &=\frac{\left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}}{48 c^3 d}-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}{56 c^3 d^3}+\frac{(b d+2 c d x)^{11/2}}{176 c^3 d^5}\\ \end{align*}

Mathematica [A]  time = 0.0439803, size = 92, normalized size = 1.05 $\frac{\left (c^2 \left (77 a^2+66 a c x^2+21 c^2 x^4\right )+b^2 c \left (15 c x^2-22 a\right )+6 b c^2 x \left (11 a+7 c x^2\right )-6 b^3 c x+2 b^4\right ) (d (b+2 c x))^{3/2}}{231 c^3 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((d*(b + 2*c*x))^(3/2)*(2*b^4 - 6*b^3*c*x + 6*b*c^2*x*(11*a + 7*c*x^2) + b^2*c*(-22*a + 15*c*x^2) + c^2*(77*a^
2 + 66*a*c*x^2 + 21*c^2*x^4)))/(231*c^3*d)

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Maple [A]  time = 0.044, size = 96, normalized size = 1.1 \begin{align*}{\frac{ \left ( 2\,cx+b \right ) \left ( 21\,{c}^{4}{x}^{4}+42\,b{x}^{3}{c}^{3}+66\,a{c}^{3}{x}^{2}+15\,{b}^{2}{c}^{2}{x}^{2}+66\,ab{c}^{2}x-6\,{b}^{3}cx+77\,{a}^{2}{c}^{2}-22\,ac{b}^{2}+2\,{b}^{4} \right ) }{231\,{c}^{3}}\sqrt{2\,cdx+bd}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(1/2)*(c*x^2+b*x+a)^2,x)

[Out]

1/231*(2*c*x+b)*(21*c^4*x^4+42*b*c^3*x^3+66*a*c^3*x^2+15*b^2*c^2*x^2+66*a*b*c^2*x-6*b^3*c*x+77*a^2*c^2-22*a*b^
2*c+2*b^4)*(2*c*d*x+b*d)^(1/2)/c^3

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Maxima [A]  time = 1.19286, size = 109, normalized size = 1.24 \begin{align*} -\frac{66 \,{\left (2 \, c d x + b d\right )}^{\frac{7}{2}}{\left (b^{2} - 4 \, a c\right )} d^{2} - 77 \,{\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}} d^{4} - 21 \,{\left (2 \, c d x + b d\right )}^{\frac{11}{2}}}{3696 \, c^{3} d^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(1/2)*(c*x^2+b*x+a)^2,x, algorithm="maxima")

[Out]

-1/3696*(66*(2*c*d*x + b*d)^(7/2)*(b^2 - 4*a*c)*d^2 - 77*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*(2*c*d*x + b*d)^(3/2)*
d^4 - 21*(2*c*d*x + b*d)^(11/2))/(c^3*d^5)

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Fricas [A]  time = 1.99735, size = 271, normalized size = 3.08 \begin{align*} \frac{{\left (42 \, c^{5} x^{5} + 105 \, b c^{4} x^{4} + 2 \, b^{5} - 22 \, a b^{3} c + 77 \, a^{2} b c^{2} + 12 \,{\left (6 \, b^{2} c^{3} + 11 \, a c^{4}\right )} x^{3} + 3 \,{\left (b^{3} c^{2} + 66 \, a b c^{3}\right )} x^{2} - 2 \,{\left (b^{4} c - 11 \, a b^{2} c^{2} - 77 \, a^{2} c^{3}\right )} x\right )} \sqrt{2 \, c d x + b d}}{231 \, c^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(1/2)*(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

1/231*(42*c^5*x^5 + 105*b*c^4*x^4 + 2*b^5 - 22*a*b^3*c + 77*a^2*b*c^2 + 12*(6*b^2*c^3 + 11*a*c^4)*x^3 + 3*(b^3
*c^2 + 66*a*b*c^3)*x^2 - 2*(b^4*c - 11*a*b^2*c^2 - 77*a^2*c^3)*x)*sqrt(2*c*d*x + b*d)/c^3

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Sympy [A]  time = 3.29871, size = 94, normalized size = 1.07 \begin{align*} \frac{\frac{\left (b d + 2 c d x\right )^{\frac{3}{2}} \left (16 a^{2} c^{2} - 8 a b^{2} c + b^{4}\right )}{48 c^{2}} + \frac{\left (4 a c - b^{2}\right ) \left (b d + 2 c d x\right )^{\frac{7}{2}}}{56 c^{2} d^{2}} + \frac{\left (b d + 2 c d x\right )^{\frac{11}{2}}}{176 c^{2} d^{4}}}{c d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)**(1/2)*(c*x**2+b*x+a)**2,x)

[Out]

((b*d + 2*c*d*x)**(3/2)*(16*a**2*c**2 - 8*a*b**2*c + b**4)/(48*c**2) + (4*a*c - b**2)*(b*d + 2*c*d*x)**(7/2)/(
56*c**2*d**2) + (b*d + 2*c*d*x)**(11/2)/(176*c**2*d**4))/(c*d)

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Giac [B]  time = 1.14695, size = 471, normalized size = 5.35 \begin{align*} \frac{18480 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} a^{2} - \frac{3696 \,{\left (5 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} b d - 3 \,{\left (2 \, c d x + b d\right )}^{\frac{5}{2}}\right )} a b}{c d} + \frac{132 \,{\left (35 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} b^{2} d^{2} - 42 \,{\left (2 \, c d x + b d\right )}^{\frac{5}{2}} b d + 15 \,{\left (2 \, c d x + b d\right )}^{\frac{7}{2}}\right )} b^{2}}{c^{2} d^{2}} + \frac{264 \,{\left (35 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} b^{2} d^{2} - 42 \,{\left (2 \, c d x + b d\right )}^{\frac{5}{2}} b d + 15 \,{\left (2 \, c d x + b d\right )}^{\frac{7}{2}}\right )} a}{c d^{2}} - \frac{44 \,{\left (105 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} b^{3} d^{3} - 189 \,{\left (2 \, c d x + b d\right )}^{\frac{5}{2}} b^{2} d^{2} + 135 \,{\left (2 \, c d x + b d\right )}^{\frac{7}{2}} b d - 35 \,{\left (2 \, c d x + b d\right )}^{\frac{9}{2}}\right )} b}{c^{2} d^{3}} + \frac{1155 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} b^{4} d^{4} - 2772 \,{\left (2 \, c d x + b d\right )}^{\frac{5}{2}} b^{3} d^{3} + 2970 \,{\left (2 \, c d x + b d\right )}^{\frac{7}{2}} b^{2} d^{2} - 1540 \,{\left (2 \, c d x + b d\right )}^{\frac{9}{2}} b d + 315 \,{\left (2 \, c d x + b d\right )}^{\frac{11}{2}}}{c^{2} d^{4}}}{55440 \, c d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(1/2)*(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

1/55440*(18480*(2*c*d*x + b*d)^(3/2)*a^2 - 3696*(5*(2*c*d*x + b*d)^(3/2)*b*d - 3*(2*c*d*x + b*d)^(5/2))*a*b/(c
*d) + 132*(35*(2*c*d*x + b*d)^(3/2)*b^2*d^2 - 42*(2*c*d*x + b*d)^(5/2)*b*d + 15*(2*c*d*x + b*d)^(7/2))*b^2/(c^
2*d^2) + 264*(35*(2*c*d*x + b*d)^(3/2)*b^2*d^2 - 42*(2*c*d*x + b*d)^(5/2)*b*d + 15*(2*c*d*x + b*d)^(7/2))*a/(c
*d^2) - 44*(105*(2*c*d*x + b*d)^(3/2)*b^3*d^3 - 189*(2*c*d*x + b*d)^(5/2)*b^2*d^2 + 135*(2*c*d*x + b*d)^(7/2)*
b*d - 35*(2*c*d*x + b*d)^(9/2))*b/(c^2*d^3) + (1155*(2*c*d*x + b*d)^(3/2)*b^4*d^4 - 2772*(2*c*d*x + b*d)^(5/2)
*b^3*d^3 + 2970*(2*c*d*x + b*d)^(7/2)*b^2*d^2 - 1540*(2*c*d*x + b*d)^(9/2)*b*d + 315*(2*c*d*x + b*d)^(11/2))/(
c^2*d^4))/(c*d)