\(\int \frac {(a+b x+c x^2)^2}{\sqrt {b d+2 c d x}} \, dx\) [1271]

Optimal result

Integrand size = 26, antiderivative size = 88 \[ \int \frac {\left (a+b x+c x^2\right )^2}{\sqrt {b d+2 c d x}} \, dx=\frac {\left (b^2-4 a c\right )^2 \sqrt {b d+2 c d x}}{16 c^3 d}-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}{40 c^3 d^3}+\frac {(b d+2 c d x)^{9/2}}{144 c^3 d^5} \]

[Out]

-1/40*(-4*a*c+b^2)*(2*c*d*x+b*d)^(5/2)/c^3/d^3+1/144*(2*c*d*x+b*d)^(9/2)/c^3/d^5+1/16*(-4*a*c+b^2)^2*(2*c*d*x+
b*d)^(1/2)/c^3/d

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {697} \[ \int \frac {\left (a+b x+c x^2\right )^2}{\sqrt {b d+2 c d x}} \, dx=-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}{40 c^3 d^3}+\frac {\left (b^2-4 a c\right )^2 \sqrt {b d+2 c d x}}{16 c^3 d}+\frac {(b d+2 c d x)^{9/2}}{144 c^3 d^5} \]

[In]

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

[Out]

((b^2 - 4*a*c)^2*Sqrt[b*d + 2*c*d*x])/(16*c^3*d) - ((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(5/2))/(40*c^3*d^3) + (b*d +
2*c*d*x)^(9/2)/(144*c^3*d^5)

Rule 697

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*} \text {integral}& = \int \left (\frac {\left (-b^2+4 a c\right )^2}{16 c^2 \sqrt {b d+2 c d x}}+\frac {\left (-b^2+4 a c\right ) (b d+2 c d x)^{3/2}}{8 c^2 d^2}+\frac {(b d+2 c d x)^{7/2}}{16 c^2 d^4}\right ) \, dx \\ & = \frac {\left (b^2-4 a c\right )^2 \sqrt {b d+2 c d x}}{16 c^3 d}-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}{40 c^3 d^3}+\frac {(b d+2 c d x)^{9/2}}{144 c^3 d^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x+c x^2\right )^2}{\sqrt {b d+2 c d x}} \, dx=\frac {\sqrt {d (b+2 c x)} \left (45 b^4-360 a b^2 c+720 a^2 c^2-18 b^2 (b+2 c x)^2+72 a c (b+2 c x)^2+5 (b+2 c x)^4\right )}{720 c^3 d} \]

[In]

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

[Out]

(Sqrt[d*(b + 2*c*x)]*(45*b^4 - 360*a*b^2*c + 720*a^2*c^2 - 18*b^2*(b + 2*c*x)^2 + 72*a*c*(b + 2*c*x)^2 + 5*(b
+ 2*c*x)^4))/(720*c^3*d)

Maple [A] (verified)

Time = 2.91 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(\frac {\frac {\left (2 c d x +b d \right )^{\frac {9}{2}}}{9}+\frac {\left (8 a c \,d^{2}-2 b^{2} d^{2}\right ) \left (2 c d x +b d \right )^{\frac {5}{2}}}{5}+\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{2} \sqrt {2 c d x +b d}}{16 d^{5} c^{3}}\)	\(82\)
default	\(\frac {\frac {\left (2 c d x +b d \right )^{\frac {9}{2}}}{9}+\frac {\left (8 a c \,d^{2}-2 b^{2} d^{2}\right ) \left (2 c d x +b d \right )^{\frac {5}{2}}}{5}+\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{2} \sqrt {2 c d x +b d}}{16 d^{5} c^{3}}\)	\(82\)
pseudoelliptic	\(\frac {\left (5 c^{4} x^{4}+10 b \,c^{3} x^{3}+18 x^{2} c^{3} a +3 b^{2} c^{2} x^{2}+18 a b \,c^{2} x -2 b^{3} c x +45 a^{2} c^{2}-18 a \,b^{2} c +2 b^{4}\right ) \sqrt {d \left (2 c x +b \right )}}{45 d \,c^{3}}\)	\(92\)
trager	\(\frac {\left (5 c^{4} x^{4}+10 b \,c^{3} x^{3}+18 x^{2} c^{3} a +3 b^{2} c^{2} x^{2}+18 a b \,c^{2} x -2 b^{3} c x +45 a^{2} c^{2}-18 a \,b^{2} c +2 b^{4}\right ) \sqrt {2 c d x +b d}}{45 d \,c^{3}}\)	\(93\)
gosper	\(\frac {\left (2 c x +b \right ) \left (5 c^{4} x^{4}+10 b \,c^{3} x^{3}+18 x^{2} c^{3} a +3 b^{2} c^{2} x^{2}+18 a b \,c^{2} x -2 b^{3} c x +45 a^{2} c^{2}-18 a \,b^{2} c +2 b^{4}\right )}{45 c^{3} \sqrt {2 c d x +b d}}\)	\(96\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+b x+c x^2\right )^2}{\sqrt {b d+2 c d x}} \, dx=\frac {{\left (5 \, c^{4} x^{4} + 10 \, b c^{3} x^{3} + 2 \, b^{4} - 18 \, a b^{2} c + 45 \, a^{2} c^{2} + 3 \, {\left (b^{2} c^{2} + 6 \, a c^{3}\right )} x^{2} - 2 \, {\left (b^{3} c - 9 \, a b c^{2}\right )} x\right )} \sqrt {2 \, c d x + b d}}{45 \, c^{3} d} \]

[In]

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

[Out]

1/45*(5*c^4*x^4 + 10*b*c^3*x^3 + 2*b^4 - 18*a*b^2*c + 45*a^2*c^2 + 3*(b^2*c^2 + 6*a*c^3)*x^2 - 2*(b^3*c - 9*a*
b*c^2)*x)*sqrt(2*c*d*x + b*d)/(c^3*d)

Sympy [A] (verification not implemented)

Time = 0.86 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.64 \[ \int \frac {\left (a+b x+c x^2\right )^2}{\sqrt {b d+2 c d x}} \, dx=\begin {cases} \frac {\frac {\sqrt {b d + 2 c d x} \left (16 a^{2} c^{2} - 8 a b^{2} c + b^{4}\right )}{16 c^{2}} + \frac {\left (4 a c - b^{2}\right ) \left (b d + 2 c d x\right )^{\frac {5}{2}}}{40 c^{2} d^{2}} + \frac {\left (b d + 2 c d x\right )^{\frac {9}{2}}}{144 c^{2} d^{4}}}{c d} & \text {for}\: c d \neq 0 \\\frac {a^{2} x + a b x^{2} + \frac {b c x^{4}}{2} + \frac {c^{2} x^{5}}{5} + \frac {x^{3} \cdot \left (2 a c + b^{2}\right )}{3}}{\sqrt {b d}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise(((sqrt(b*d + 2*c*d*x)*(16*a**2*c**2 - 8*a*b**2*c + b**4)/(16*c**2) + (4*a*c - b**2)*(b*d + 2*c*d*x)*
*(5/2)/(40*c**2*d**2) + (b*d + 2*c*d*x)**(9/2)/(144*c**2*d**4))/(c*d), Ne(c*d, 0)), ((a**2*x + a*b*x**2 + b*c*
x**4/2 + c**2*x**5/5 + x**3*(2*a*c + b**2)/3)/sqrt(b*d), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 351 vs. \(2 (76) = 152\).

Time = 0.20 (sec) , antiderivative size = 351, normalized size of antiderivative = 3.99 \[ \int \frac {\left (a+b x+c x^2\right )^2}{\sqrt {b d+2 c d x}} \, dx=\frac {5040 \, \sqrt {2 \, c d x + b d} a^{2} - 168 \, a {\left (\frac {10 \, {\left (3 \, \sqrt {2 \, c d x + b d} b d - {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}\right )} b}{c d} - \frac {15 \, \sqrt {2 \, c d x + b d} b^{2} d^{2} - 10 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b d + 3 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}}{c d^{2}}\right )} + \frac {84 \, {\left (15 \, \sqrt {2 \, c d x + b d} b^{2} d^{2} - 10 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b d + 3 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}\right )} b^{2}}{c^{2} d^{2}} - \frac {36 \, {\left (35 \, \sqrt {2 \, c d x + b d} b^{3} d^{3} - 35 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} d^{2} + 21 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} b d - 5 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}}\right )} b}{c^{2} d^{3}} + \frac {315 \, \sqrt {2 \, c d x + b d} b^{4} d^{4} - 420 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{3} d^{3} + 378 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} b^{2} d^{2} - 180 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} b d + 35 \, {\left (2 \, c d x + b d\right )}^{\frac {9}{2}}}{c^{2} d^{4}}}{5040 \, c d} \]

[In]

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

[Out]

1/5040*(5040*sqrt(2*c*d*x + b*d)*a^2 - 168*a*(10*(3*sqrt(2*c*d*x + b*d)*b*d - (2*c*d*x + b*d)^(3/2))*b/(c*d) -
(15*sqrt(2*c*d*x + b*d)*b^2*d^2 - 10*(2*c*d*x + b*d)^(3/2)*b*d + 3*(2*c*d*x + b*d)^(5/2))/(c*d^2)) + 84*(15*s
qrt(2*c*d*x + b*d)*b^2*d^2 - 10*(2*c*d*x + b*d)^(3/2)*b*d + 3*(2*c*d*x + b*d)^(5/2))*b^2/(c^2*d^2) - 36*(35*sq
rt(2*c*d*x + b*d)*b^3*d^3 - 35*(2*c*d*x + b*d)^(3/2)*b^2*d^2 + 21*(2*c*d*x + b*d)^(5/2)*b*d - 5*(2*c*d*x + b*d
)^(7/2))*b/(c^2*d^3) + (315*sqrt(2*c*d*x + b*d)*b^4*d^4 - 420*(2*c*d*x + b*d)^(3/2)*b^3*d^3 + 378*(2*c*d*x + b
*d)^(5/2)*b^2*d^2 - 180*(2*c*d*x + b*d)^(7/2)*b*d + 35*(2*c*d*x + b*d)^(9/2))/(c^2*d^4))/(c*d)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 349 vs. \(2 (76) = 152\).

Time = 0.28 (sec) , antiderivative size = 349, normalized size of antiderivative = 3.97 \[ \int \frac {\left (a+b x+c x^2\right )^2}{\sqrt {b d+2 c d x}} \, dx=\frac {5040 \, \sqrt {2 \, c d x + b d} a^{2} - \frac {1680 \, {\left (3 \, \sqrt {2 \, c d x + b d} b d - {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}\right )} a b}{c d} + \frac {84 \, {\left (15 \, \sqrt {2 \, c d x + b d} b^{2} d^{2} - 10 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b d + 3 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}\right )} b^{2}}{c^{2} d^{2}} + \frac {168 \, {\left (15 \, \sqrt {2 \, c d x + b d} b^{2} d^{2} - 10 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b d + 3 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}\right )} a}{c d^{2}} - \frac {36 \, {\left (35 \, \sqrt {2 \, c d x + b d} b^{3} d^{3} - 35 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} d^{2} + 21 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} b d - 5 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}}\right )} b}{c^{2} d^{3}} + \frac {315 \, \sqrt {2 \, c d x + b d} b^{4} d^{4} - 420 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{3} d^{3} + 378 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} b^{2} d^{2} - 180 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} b d + 35 \, {\left (2 \, c d x + b d\right )}^{\frac {9}{2}}}{c^{2} d^{4}}}{5040 \, c d} \]

[In]

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

[Out]

1/5040*(5040*sqrt(2*c*d*x + b*d)*a^2 - 1680*(3*sqrt(2*c*d*x + b*d)*b*d - (2*c*d*x + b*d)^(3/2))*a*b/(c*d) + 84
*(15*sqrt(2*c*d*x + b*d)*b^2*d^2 - 10*(2*c*d*x + b*d)^(3/2)*b*d + 3*(2*c*d*x + b*d)^(5/2))*b^2/(c^2*d^2) + 168
*(15*sqrt(2*c*d*x + b*d)*b^2*d^2 - 10*(2*c*d*x + b*d)^(3/2)*b*d + 3*(2*c*d*x + b*d)^(5/2))*a/(c*d^2) - 36*(35*
sqrt(2*c*d*x + b*d)*b^3*d^3 - 35*(2*c*d*x + b*d)^(3/2)*b^2*d^2 + 21*(2*c*d*x + b*d)^(5/2)*b*d - 5*(2*c*d*x + b
*d)^(7/2))*b/(c^2*d^3) + (315*sqrt(2*c*d*x + b*d)*b^4*d^4 - 420*(2*c*d*x + b*d)^(3/2)*b^3*d^3 + 378*(2*c*d*x +
b*d)^(5/2)*b^2*d^2 - 180*(2*c*d*x + b*d)^(7/2)*b*d + 35*(2*c*d*x + b*d)^(9/2))/(c^2*d^4))/(c*d)

Mupad [B] (verification not implemented)

Time = 9.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x+c x^2\right )^2}{\sqrt {b d+2 c d x}} \, dx=\frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (5\,{\left (b\,d+2\,c\,d\,x\right )}^4+45\,b^4\,d^4-18\,b^2\,d^2\,{\left (b\,d+2\,c\,d\,x\right )}^2+720\,a^2\,c^2\,d^4+72\,a\,c\,d^2\,{\left (b\,d+2\,c\,d\,x\right )}^2-360\,a\,b^2\,c\,d^4\right )}{720\,c^3\,d^5} \]

[In]

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

[Out]

((b*d + 2*c*d*x)^(1/2)*(5*(b*d + 2*c*d*x)^4 + 45*b^4*d^4 - 18*b^2*d^2*(b*d + 2*c*d*x)^2 + 720*a^2*c^2*d^4 + 72
*a*c*d^2*(b*d + 2*c*d*x)^2 - 360*a*b^2*c*d^4))/(720*c^3*d^5)