### 3.1269 $$\int (b d+2 c d x)^{3/2} (a+b x+c x^2)^2 \, dx$$

Optimal. Leaf size=88 $-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{9/2}}{72 c^3 d^3}+\frac{\left (b^2-4 a c\right )^2 (b d+2 c d x)^{5/2}}{80 c^3 d}+\frac{(b d+2 c d x)^{13/2}}{208 c^3 d^5}$

[Out]

((b^2 - 4*a*c)^2*(b*d + 2*c*d*x)^(5/2))/(80*c^3*d) - ((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(9/2))/(72*c^3*d^3) + (b*d
+ 2*c*d*x)^(13/2)/(208*c^3*d^5)

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Rubi [A]  time = 0.0549081, 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} $-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{9/2}}{72 c^3 d^3}+\frac{\left (b^2-4 a c\right )^2 (b d+2 c d x)^{5/2}}{80 c^3 d}+\frac{(b d+2 c d x)^{13/2}}{208 c^3 d^5}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((b^2 - 4*a*c)^2*(b*d + 2*c*d*x)^(5/2))/(80*c^3*d) - ((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(9/2))/(72*c^3*d^3) + (b*d
+ 2*c*d*x)^(13/2)/(208*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 (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )^2 \, dx &=\int \left (\frac{\left (-b^2+4 a c\right )^2 (b d+2 c d x)^{3/2}}{16 c^2}+\frac{\left (-b^2+4 a c\right ) (b d+2 c d x)^{7/2}}{8 c^2 d^2}+\frac{(b d+2 c d x)^{11/2}}{16 c^2 d^4}\right ) \, dx\\ &=\frac{\left (b^2-4 a c\right )^2 (b d+2 c d x)^{5/2}}{80 c^3 d}-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{9/2}}{72 c^3 d^3}+\frac{(b d+2 c d x)^{13/2}}{208 c^3 d^5}\\ \end{align*}

Mathematica [A]  time = 0.0608572, size = 63, normalized size = 0.72 $\frac{\left (-130 \left (b^2-4 a c\right ) (b+2 c x)^2+117 \left (b^2-4 a c\right )^2+45 (b+2 c x)^4\right ) (d (b+2 c x))^{5/2}}{9360 c^3 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((d*(b + 2*c*x))^(5/2)*(117*(b^2 - 4*a*c)^2 - 130*(b^2 - 4*a*c)*(b + 2*c*x)^2 + 45*(b + 2*c*x)^4))/(9360*c^3*d
)

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Maple [A]  time = 0.043, size = 96, normalized size = 1.1 \begin{align*}{\frac{ \left ( 2\,cx+b \right ) \left ( 45\,{c}^{4}{x}^{4}+90\,b{x}^{3}{c}^{3}+130\,a{c}^{3}{x}^{2}+35\,{b}^{2}{c}^{2}{x}^{2}+130\,ab{c}^{2}x-10\,{b}^{3}cx+117\,{a}^{2}{c}^{2}-26\,ac{b}^{2}+2\,{b}^{4} \right ) }{585\,{c}^{3}} \left ( 2\,cdx+bd \right ) ^{{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/585*(2*c*x+b)*(45*c^4*x^4+90*b*c^3*x^3+130*a*c^3*x^2+35*b^2*c^2*x^2+130*a*b*c^2*x-10*b^3*c*x+117*a^2*c^2-26*
a*b^2*c+2*b^4)*(2*c*d*x+b*d)^(3/2)/c^3

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Maxima [A]  time = 1.05443, size = 109, normalized size = 1.24 \begin{align*} -\frac{130 \,{\left (2 \, c d x + b d\right )}^{\frac{9}{2}}{\left (b^{2} - 4 \, a c\right )} d^{2} - 117 \,{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )}{\left (2 \, c d x + b d\right )}^{\frac{5}{2}} d^{4} - 45 \,{\left (2 \, c d x + b d\right )}^{\frac{13}{2}}}{9360 \, c^{3} d^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9360*(130*(2*c*d*x + b*d)^(9/2)*(b^2 - 4*a*c)*d^2 - 117*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*(2*c*d*x + b*d)^(5/2
)*d^4 - 45*(2*c*d*x + b*d)^(13/2))/(c^3*d^5)

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Fricas [B]  time = 2.02091, size = 377, normalized size = 4.28 \begin{align*} \frac{{\left (180 \, c^{6} d x^{6} + 540 \, b c^{5} d x^{5} + 5 \,{\left (109 \, b^{2} c^{4} + 104 \, a c^{5}\right )} d x^{4} + 10 \,{\left (19 \, b^{3} c^{3} + 104 \, a b c^{4}\right )} d x^{3} + 3 \,{\left (b^{4} c^{2} + 182 \, a b^{2} c^{3} + 156 \, a^{2} c^{4}\right )} d x^{2} - 2 \,{\left (b^{5} c - 13 \, a b^{3} c^{2} - 234 \, a^{2} b c^{3}\right )} d x +{\left (2 \, b^{6} - 26 \, a b^{4} c + 117 \, a^{2} b^{2} c^{2}\right )} d\right )} \sqrt{2 \, c d x + b d}}{585 \, c^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/585*(180*c^6*d*x^6 + 540*b*c^5*d*x^5 + 5*(109*b^2*c^4 + 104*a*c^5)*d*x^4 + 10*(19*b^3*c^3 + 104*a*b*c^4)*d*x
^3 + 3*(b^4*c^2 + 182*a*b^2*c^3 + 156*a^2*c^4)*d*x^2 - 2*(b^5*c - 13*a*b^3*c^2 - 234*a^2*b*c^3)*d*x + (2*b^6 -
26*a*b^4*c + 117*a^2*b^2*c^2)*d)*sqrt(2*c*d*x + b*d)/c^3

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Sympy [A]  time = 21.6616, size = 695, normalized size = 7.9 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)**(3/2)*(c*x**2+b*x+a)**2,x)

[Out]

a**2*b*d*Piecewise((x*sqrt(b*d), Eq(c, 0)), (0, Eq(d, 0)), ((b*d + 2*c*d*x)**(3/2)/(3*c*d), True)) + a**2*(-b*
d*(b*d + 2*c*d*x)**(3/2)/3 + (b*d + 2*c*d*x)**(5/2)/5)/(c*d) + a*b**2*(-b*d*(b*d + 2*c*d*x)**(3/2)/3 + (b*d +
2*c*d*x)**(5/2)/5)/(c**2*d) + 3*a*b*(b**2*d**2*(b*d + 2*c*d*x)**(3/2)/3 - 2*b*d*(b*d + 2*c*d*x)**(5/2)/5 + (b*
d + 2*c*d*x)**(7/2)/7)/(2*c**2*d**2) + a*(-b**3*d**3*(b*d + 2*c*d*x)**(3/2)/3 + 3*b**2*d**2*(b*d + 2*c*d*x)**(
5/2)/5 - 3*b*d*(b*d + 2*c*d*x)**(7/2)/7 + (b*d + 2*c*d*x)**(9/2)/9)/(2*c**2*d**3) + b**3*(b**2*d**2*(b*d + 2*c
*d*x)**(3/2)/3 - 2*b*d*(b*d + 2*c*d*x)**(5/2)/5 + (b*d + 2*c*d*x)**(7/2)/7)/(4*c**3*d**2) + b**2*(-b**3*d**3*(
b*d + 2*c*d*x)**(3/2)/3 + 3*b**2*d**2*(b*d + 2*c*d*x)**(5/2)/5 - 3*b*d*(b*d + 2*c*d*x)**(7/2)/7 + (b*d + 2*c*d
*x)**(9/2)/9)/(2*c**3*d**3) + 5*b*(b**4*d**4*(b*d + 2*c*d*x)**(3/2)/3 - 4*b**3*d**3*(b*d + 2*c*d*x)**(5/2)/5 +
6*b**2*d**2*(b*d + 2*c*d*x)**(7/2)/7 - 4*b*d*(b*d + 2*c*d*x)**(9/2)/9 + (b*d + 2*c*d*x)**(11/2)/11)/(16*c**3*
d**4) + (-b**5*d**5*(b*d + 2*c*d*x)**(3/2)/3 + b**4*d**4*(b*d + 2*c*d*x)**(5/2) - 10*b**3*d**3*(b*d + 2*c*d*x)
**(7/2)/7 + 10*b**2*d**2*(b*d + 2*c*d*x)**(9/2)/9 - 5*b*d*(b*d + 2*c*d*x)**(11/2)/11 + (b*d + 2*c*d*x)**(13/2)
/13)/(16*c**3*d**5)

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Giac [B]  time = 1.16506, size = 783, normalized size = 8.9 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/720720*(240240*(2*c*d*x + b*d)^(3/2)*a^2*b - 48048*(5*(2*c*d*x + b*d)^(3/2)*b*d - 3*(2*c*d*x + b*d)^(5/2))*a
^2/d - 48048*(5*(2*c*d*x + b*d)^(3/2)*b*d - 3*(2*c*d*x + b*d)^(5/2))*a*b^2/(c*d) + 1716*(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^3/(c^2*d^2) + 10296*(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*b/(c*d^2) - 1144*(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^2/(c^2*d^3) - 1144*(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))*a/(c*d^3) + 65*(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))*b/(c^2*d^4) - 5*(3003*(2*c*d*x + b*d)^(3/2)*b^5*d^5 - 9009*(2*c*d*x + b*d)^(5/2)*b^4*d^4 + 128
70*(2*c*d*x + b*d)^(7/2)*b^3*d^3 - 10010*(2*c*d*x + b*d)^(9/2)*b^2*d^2 + 4095*(2*c*d*x + b*d)^(11/2)*b*d - 693
*(2*c*d*x + b*d)^(13/2))/(c^2*d^5))/c