\(\int \frac {(a+b x+c x^2)^2}{(b d+2 c d x)^{5/2}} \, dx\) [1273]

Optimal result

Integrand size = 26, antiderivative size = 88 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{5/2}} \, dx=-\frac {\left (b^2-4 a c\right )^2}{48 c^3 d (b d+2 c d x)^{3/2}}-\frac {\left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}{8 c^3 d^3}+\frac {(b d+2 c d x)^{5/2}}{80 c^3 d^5} \]

[Out]

-1/48*(-4*a*c+b^2)^2/c^3/d/(2*c*d*x+b*d)^(3/2)+1/80*(2*c*d*x+b*d)^(5/2)/c^3/d^5-1/8*(-4*a*c+b^2)*(2*c*d*x+b*d)
^(1/2)/c^3/d^3

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} \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{5/2}} \, dx=-\frac {\left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}{8 c^3 d^3}-\frac {\left (b^2-4 a c\right )^2}{48 c^3 d (b d+2 c d x)^{3/2}}+\frac {(b d+2 c d x)^{5/2}}{80 c^3 d^5} \]

[In]

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

[Out]

-1/48*(b^2 - 4*a*c)^2/(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 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 (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] (verified)

Time = 0.06 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{5/2}} \, dx=\frac {-5 b^4+40 a b^2 c-80 a^2 c^2-30 b^2 (b+2 c x)^2+120 a c (b+2 c x)^2+3 (b+2 c x)^4}{240 c^3 d (d (b+2 c x))^{3/2}} \]

[In]

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

[Out]

(-5*b^4 + 40*a*b^2*c - 80*a^2*c^2 - 30*b^2*(b + 2*c*x)^2 + 120*a*c*(b + 2*c*x)^2 + 3*(b + 2*c*x)^4)/(240*c^3*d
*(d*(b + 2*c*x))^(3/2))

Maple [A] (verified)

Time = 2.52 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(\frac {8 a c \,d^{2} \sqrt {2 c d x +b d}-2 b^{2} d^{2} \sqrt {2 c d x +b d}+\frac {\left (2 c d x +b d \right )^{\frac {5}{2}}}{5}-\frac {d^{4} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}{3 \left (2 c d x +b d \right )^{\frac {3}{2}}}}{16 c^{3} d^{5}}\)	\(95\)
default	\(\frac {8 a c \,d^{2} \sqrt {2 c d x +b d}-2 b^{2} d^{2} \sqrt {2 c d x +b d}+\frac {\left (2 c d x +b d \right )^{\frac {5}{2}}}{5}-\frac {d^{4} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}{3 \left (2 c d x +b d \right )^{\frac {3}{2}}}}{16 c^{3} d^{5}}\)	\(95\)
gosper	\(-\frac {\left (2 c x +b \right ) \left (-3 c^{4} x^{4}-6 b \,c^{3} x^{3}-30 x^{2} c^{3} a +3 b^{2} c^{2} x^{2}-30 a b \,c^{2} x +6 b^{3} c x +5 a^{2} c^{2}-10 a \,b^{2} c +2 b^{4}\right )}{15 c^{3} \left (2 c d x +b d \right )^{\frac {5}{2}}}\)	\(96\)
pseudoelliptic	\(\frac {3 c^{4} x^{4}+6 b \,c^{3} x^{3}+30 x^{2} c^{3} a -3 b^{2} c^{2} x^{2}+30 a b \,c^{2} x -6 b^{3} c x -5 a^{2} c^{2}+10 a \,b^{2} c -2 b^{4}}{15 d^{2} \left (2 c x +b \right ) \sqrt {d \left (2 c x +b \right )}\, c^{3}}\)	\(100\)
trager	\(-\frac {\left (-3 c^{4} x^{4}-6 b \,c^{3} x^{3}-30 x^{2} c^{3} a +3 b^{2} c^{2} x^{2}-30 a b \,c^{2} x +6 b^{3} c x +5 a^{2} c^{2}-10 a \,b^{2} c +2 b^{4}\right ) \sqrt {2 c d x +b d}}{15 d^{3} c^{3} \left (2 c x +b \right )^{2}}\)	\(101\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.36 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{5/2}} \, dx=\frac {{\left (3 \, c^{4} x^{4} + 6 \, b c^{3} x^{3} - 2 \, b^{4} + 10 \, a b^{2} c - 5 \, a^{2} c^{2} - 3 \, {\left (b^{2} c^{2} - 10 \, a c^{3}\right )} x^{2} - 6 \, {\left (b^{3} c - 5 \, a b c^{2}\right )} x\right )} \sqrt {2 \, c d x + b d}}{15 \, {\left (4 \, c^{5} d^{3} x^{2} + 4 \, b c^{4} d^{3} x + b^{2} c^{3} d^{3}\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 1.38 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.52 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{5/2}} \, dx=\begin {cases} \frac {- \frac {\left (4 a c - b^{2}\right )^{2}}{48 c^{2} \left (b d + 2 c d x\right )^{\frac {3}{2}}} + \frac {\left (4 a c - b^{2}\right ) \sqrt {b d + 2 c d x}}{8 c^{2} d^{2}} + \frac {\left (b d + 2 c d x\right )^{\frac {5}{2}}}{80 c^{2} d^{4}}}{c d} & \text {for}\: c d \neq 0 \\\frac {a^{2} x + a b x^{2} + \frac {b c x^{4}}{2} + \frac {c^{2} x^{5}}{5} + \frac {x^{3} \cdot \left (2 a c + b^{2}\right )}{3}}{\left (b d\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{5/2}} \, dx=-\frac {\frac {5 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )}}{{\left (2 \, c d x + b d\right )}^{\frac {3}{2}} c^{2}} + \frac {3 \, {\left (10 \, \sqrt {2 \, c d x + b d} {\left (b^{2} - 4 \, a c\right )} d^{2} - {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}\right )}}{c^{2} d^{4}}}{240 \, c d} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{5/2}} \, dx=-\frac {b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}{48 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} c^{3} d} - \frac {10 \, \sqrt {2 \, c d x + b d} b^{2} c^{12} d^{22} - 40 \, \sqrt {2 \, c d x + b d} a c^{13} d^{22} - {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} c^{12} d^{20}}{80 \, c^{15} d^{25}} \]

[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)

Mupad [B] (verification not implemented)

Time = 9.22 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^{5/2}} \, dx=\frac {3\,{\left (b\,d+2\,c\,d\,x\right )}^4-5\,b^4\,d^4-30\,b^2\,d^2\,{\left (b\,d+2\,c\,d\,x\right )}^2-80\,a^2\,c^2\,d^4+120\,a\,c\,d^2\,{\left (b\,d+2\,c\,d\,x\right )}^2+40\,a\,b^2\,c\,d^4}{240\,c^3\,d^5\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}} \]

[In]

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

[Out]

(3*(b*d + 2*c*d*x)^4 - 5*b^4*d^4 - 30*b^2*d^2*(b*d + 2*c*d*x)^2 - 80*a^2*c^2*d^4 + 120*a*c*d^2*(b*d + 2*c*d*x)
^2 + 40*a*b^2*c*d^4)/(240*c^3*d^5*(b*d + 2*c*d*x)^(3/2))