\(\int \frac {(a+b x+c x^2)^2}{(d+e x)^{5/2}} \, dx\) [2280]

Optimal result

Integrand size = 22, antiderivative size = 162 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=-\frac {2 \left (c d^2-b d e+a e^2\right )^2}{3 e^5 (d+e x)^{3/2}}+\frac {4 (2 c d-b e) \left (c d^2-b d e+a e^2\right )}{e^5 \sqrt {d+e x}}+\frac {2 \left (6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)\right ) \sqrt {d+e x}}{e^5}-\frac {4 c (2 c d-b e) (d+e x)^{3/2}}{3 e^5}+\frac {2 c^2 (d+e x)^{5/2}}{5 e^5} \]

[Out]

-2/3*(a*e^2-b*d*e+c*d^2)^2/e^5/(e*x+d)^(3/2)-4/3*c*(-b*e+2*c*d)*(e*x+d)^(3/2)/e^5+2/5*c^2*(e*x+d)^(5/2)/e^5+4*
(-b*e+2*c*d)*(a*e^2-b*d*e+c*d^2)/e^5/(e*x+d)^(1/2)+2*(6*c^2*d^2+b^2*e^2-2*c*e*(-a*e+3*b*d))*(e*x+d)^(1/2)/e^5

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {712} \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=\frac {2 \sqrt {d+e x} \left (-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2\right )}{e^5}+\frac {4 (2 c d-b e) \left (a e^2-b d e+c d^2\right )}{e^5 \sqrt {d+e x}}-\frac {2 \left (a e^2-b d e+c d^2\right )^2}{3 e^5 (d+e x)^{3/2}}-\frac {4 c (d+e x)^{3/2} (2 c d-b e)}{3 e^5}+\frac {2 c^2 (d+e x)^{5/2}}{5 e^5} \]

[In]

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

[Out]

(-2*(c*d^2 - b*d*e + a*e^2)^2)/(3*e^5*(d + e*x)^(3/2)) + (4*(2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2))/(e^5*Sqrt[d
+ e*x]) + (2*(6*c^2*d^2 + b^2*e^2 - 2*c*e*(3*b*d - a*e))*Sqrt[d + e*x])/e^5 - (4*c*(2*c*d - b*e)*(d + e*x)^(3
/2))/(3*e^5) + (2*c^2*(d + e*x)^(5/2))/(5*e^5)

Rule 712

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] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\left (c d^2-b d e+a e^2\right )^2}{e^4 (d+e x)^{5/2}}+\frac {2 (-2 c d+b e) \left (c d^2-b d e+a e^2\right )}{e^4 (d+e x)^{3/2}}+\frac {6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)}{e^4 \sqrt {d+e x}}-\frac {2 c (2 c d-b e) \sqrt {d+e x}}{e^4}+\frac {c^2 (d+e x)^{3/2}}{e^4}\right ) \, dx \\ & = -\frac {2 \left (c d^2-b d e+a e^2\right )^2}{3 e^5 (d+e x)^{3/2}}+\frac {4 (2 c d-b e) \left (c d^2-b d e+a e^2\right )}{e^5 \sqrt {d+e x}}+\frac {2 \left (6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)\right ) \sqrt {d+e x}}{e^5}-\frac {4 c (2 c d-b e) (d+e x)^{3/2}}{3 e^5}+\frac {2 c^2 (d+e x)^{5/2}}{5 e^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=\frac {2 \left (c^2 \left (128 d^4+192 d^3 e x+48 d^2 e^2 x^2-8 d e^3 x^3+3 e^4 x^4\right )-5 e^2 \left (a^2 e^2+2 a b e (2 d+3 e x)-b^2 \left (8 d^2+12 d e x+3 e^2 x^2\right )\right )+10 c e \left (a e \left (8 d^2+12 d e x+3 e^2 x^2\right )+b \left (-16 d^3-24 d^2 e x-6 d e^2 x^2+e^3 x^3\right )\right )\right )}{15 e^5 (d+e x)^{3/2}} \]

[In]

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

[Out]

(2*(c^2*(128*d^4 + 192*d^3*e*x + 48*d^2*e^2*x^2 - 8*d*e^3*x^3 + 3*e^4*x^4) - 5*e^2*(a^2*e^2 + 2*a*b*e*(2*d + 3
*e*x) - b^2*(8*d^2 + 12*d*e*x + 3*e^2*x^2)) + 10*c*e*(a*e*(8*d^2 + 12*d*e*x + 3*e^2*x^2) + b*(-16*d^3 - 24*d^2
*e*x - 6*d*e^2*x^2 + e^3*x^3))))/(15*e^5*(d + e*x)^(3/2))

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.80

method	result	size
risch	\(\frac {2 \left (3 c^{2} e^{2} x^{2}+10 b c \,e^{2} x -14 c^{2} d e x +30 a c \,e^{2}+15 b^{2} e^{2}-80 b c d e +73 c^{2} d^{2}\right ) \sqrt {e x +d}}{15 e^{5}}-\frac {2 \left (6 b \,e^{2} x -12 c d e x +a \,e^{2}+5 b d e -11 c \,d^{2}\right ) \left (a \,e^{2}-b d e +c \,d^{2}\right )}{3 e^{5} \left (e x +d \right )^{\frac {3}{2}}}\)	\(129\)
pseudoelliptic	\(-\frac {2 \left (\left (-\frac {3 c^{2} x^{4}}{5}+\left (-2 b \,x^{3}-6 x^{2} a \right ) c -3 b^{2} x^{2}+6 a b x +a^{2}\right ) e^{4}+4 \left (\frac {2 c^{2} x^{3}}{5}+\left (3 b \,x^{2}-6 a x \right ) c +b \left (-3 b x +a \right )\right ) d \,e^{3}-16 \left (\frac {3 c^{2} x^{2}}{5}+\left (-3 b x +a \right ) c +\frac {b^{2}}{2}\right ) d^{2} e^{2}+32 c \,d^{3} \left (-\frac {6 c x}{5}+b \right ) e -\frac {128 c^{2} d^{4}}{5}\right )}{3 \left (e x +d \right )^{\frac {3}{2}} e^{5}}\)	\(145\)
gosper	\(-\frac {2 \left (-3 c^{2} x^{4} e^{4}-10 b c \,e^{4} x^{3}+8 c^{2} d \,e^{3} x^{3}-30 a c \,e^{4} x^{2}-15 b^{2} e^{4} x^{2}+60 b c d \,e^{3} x^{2}-48 c^{2} d^{2} e^{2} x^{2}+30 a b \,e^{4} x -120 a c d \,e^{3} x -60 b^{2} d \,e^{3} x +240 b c \,d^{2} e^{2} x -192 c^{2} d^{3} e x +5 a^{2} e^{4}+20 a b d \,e^{3}-80 a c \,d^{2} e^{2}-40 b^{2} d^{2} e^{2}+160 b c \,d^{3} e -128 c^{2} d^{4}\right )}{15 \left (e x +d \right )^{\frac {3}{2}} e^{5}}\)	\(194\)
trager	\(-\frac {2 \left (-3 c^{2} x^{4} e^{4}-10 b c \,e^{4} x^{3}+8 c^{2} d \,e^{3} x^{3}-30 a c \,e^{4} x^{2}-15 b^{2} e^{4} x^{2}+60 b c d \,e^{3} x^{2}-48 c^{2} d^{2} e^{2} x^{2}+30 a b \,e^{4} x -120 a c d \,e^{3} x -60 b^{2} d \,e^{3} x +240 b c \,d^{2} e^{2} x -192 c^{2} d^{3} e x +5 a^{2} e^{4}+20 a b d \,e^{3}-80 a c \,d^{2} e^{2}-40 b^{2} d^{2} e^{2}+160 b c \,d^{3} e -128 c^{2} d^{4}\right )}{15 \left (e x +d \right )^{\frac {3}{2}} e^{5}}\)	\(194\)
derivativedivides	\(\frac {\frac {2 c^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {4 b c e \left (e x +d \right )^{\frac {3}{2}}}{3}-\frac {8 c^{2} d \left (e x +d \right )^{\frac {3}{2}}}{3}+4 a c \,e^{2} \sqrt {e x +d}+2 b^{2} e^{2} \sqrt {e x +d}-12 b c d e \sqrt {e x +d}+12 c^{2} d^{2} \sqrt {e x +d}-\frac {2 \left (2 a b \,e^{3}-4 a c d \,e^{2}-2 b^{2} d \,e^{2}+6 b c \,d^{2} e -4 c^{2} d^{3}\right )}{\sqrt {e x +d}}-\frac {2 \left (a^{2} e^{4}-2 a b d \,e^{3}+2 a c \,d^{2} e^{2}+b^{2} d^{2} e^{2}-2 b c \,d^{3} e +c^{2} d^{4}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{5}}\)	\(210\)
default	\(\frac {\frac {2 c^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {4 b c e \left (e x +d \right )^{\frac {3}{2}}}{3}-\frac {8 c^{2} d \left (e x +d \right )^{\frac {3}{2}}}{3}+4 a c \,e^{2} \sqrt {e x +d}+2 b^{2} e^{2} \sqrt {e x +d}-12 b c d e \sqrt {e x +d}+12 c^{2} d^{2} \sqrt {e x +d}-\frac {2 \left (2 a b \,e^{3}-4 a c d \,e^{2}-2 b^{2} d \,e^{2}+6 b c \,d^{2} e -4 c^{2} d^{3}\right )}{\sqrt {e x +d}}-\frac {2 \left (a^{2} e^{4}-2 a b d \,e^{3}+2 a c \,d^{2} e^{2}+b^{2} d^{2} e^{2}-2 b c \,d^{3} e +c^{2} d^{4}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{5}}\)	\(210\)

[In]

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

[Out]

2/15*(3*c^2*e^2*x^2+10*b*c*e^2*x-14*c^2*d*e*x+30*a*c*e^2+15*b^2*e^2-80*b*c*d*e+73*c^2*d^2)*(e*x+d)^(1/2)/e^5-2
/3*(6*b*e^2*x-12*c*d*e*x+a*e^2+5*b*d*e-11*c*d^2)*(a*e^2-b*d*e+c*d^2)/e^5/(e*x+d)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.22 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (3 \, c^{2} e^{4} x^{4} + 128 \, c^{2} d^{4} - 160 \, b c d^{3} e - 20 \, a b d e^{3} - 5 \, a^{2} e^{4} + 40 \, {\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} - 2 \, {\left (4 \, c^{2} d e^{3} - 5 \, b c e^{4}\right )} x^{3} + 3 \, {\left (16 \, c^{2} d^{2} e^{2} - 20 \, b c d e^{3} + 5 \, {\left (b^{2} + 2 \, a c\right )} e^{4}\right )} x^{2} + 6 \, {\left (32 \, c^{2} d^{3} e - 40 \, b c d^{2} e^{2} - 5 \, a b e^{4} + 10 \, {\left (b^{2} + 2 \, a c\right )} d e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} \]

[In]

integrate((c*x^2+b*x+a)^2/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

2/15*(3*c^2*e^4*x^4 + 128*c^2*d^4 - 160*b*c*d^3*e - 20*a*b*d*e^3 - 5*a^2*e^4 + 40*(b^2 + 2*a*c)*d^2*e^2 - 2*(4
*c^2*d*e^3 - 5*b*c*e^4)*x^3 + 3*(16*c^2*d^2*e^2 - 20*b*c*d*e^3 + 5*(b^2 + 2*a*c)*e^4)*x^2 + 6*(32*c^2*d^3*e -
40*b*c*d^2*e^2 - 5*a*b*e^4 + 10*(b^2 + 2*a*c)*d*e^3)*x)*sqrt(e*x + d)/(e^7*x^2 + 2*d*e^6*x + d^2*e^5)

Sympy [A] (verification not implemented)

Time = 4.64 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=\begin {cases} \frac {2 \left (\frac {c^{2} \left (d + e x\right )^{\frac {5}{2}}}{5 e^{4}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \cdot \left (2 b c e - 4 c^{2} d\right )}{3 e^{4}} + \frac {\sqrt {d + e x} \left (2 a c e^{2} + b^{2} e^{2} - 6 b c d e + 6 c^{2} d^{2}\right )}{e^{4}} - \frac {2 \left (b e - 2 c d\right ) \left (a e^{2} - b d e + c d^{2}\right )}{e^{4} \sqrt {d + e x}} - \frac {\left (a e^{2} - b d e + c d^{2}\right )^{2}}{3 e^{4} \left (d + e x\right )^{\frac {3}{2}}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {a^{2} x + a b x^{2} + \frac {b c x^{4}}{2} + \frac {c^{2} x^{5}}{5} + \frac {x^{3} \cdot \left (2 a c + b^{2}\right )}{3}}{d^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(c**2*(d + e*x)**(5/2)/(5*e**4) + (d + e*x)**(3/2)*(2*b*c*e - 4*c**2*d)/(3*e**4) + sqrt(d + e*x)*
(2*a*c*e**2 + b**2*e**2 - 6*b*c*d*e + 6*c**2*d**2)/e**4 - 2*(b*e - 2*c*d)*(a*e**2 - b*d*e + c*d**2)/(e**4*sqrt
(d + e*x)) - (a*e**2 - b*d*e + c*d**2)**2/(3*e**4*(d + e*x)**(3/2)))/e, Ne(e, 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)/d**(5/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}} c^{2} - 10 \, {\left (2 \, c^{2} d - b c e\right )} {\left (e x + d\right )}^{\frac {3}{2}} + 15 \, {\left (6 \, c^{2} d^{2} - 6 \, b c d e + {\left (b^{2} + 2 \, a c\right )} e^{2}\right )} \sqrt {e x + d}}{e^{4}} - \frac {5 \, {\left (c^{2} d^{4} - 2 \, b c d^{3} e - 2 \, a b d e^{3} + a^{2} e^{4} + {\left (b^{2} + 2 \, a c\right )} d^{2} e^{2} - 6 \, {\left (2 \, c^{2} d^{3} - 3 \, b c d^{2} e - a b e^{3} + {\left (b^{2} + 2 \, a c\right )} d e^{2}\right )} {\left (e x + d\right )}\right )}}{{\left (e x + d\right )}^{\frac {3}{2}} e^{4}}\right )}}{15 \, e} \]

[In]

integrate((c*x^2+b*x+a)^2/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

2/15*((3*(e*x + d)^(5/2)*c^2 - 10*(2*c^2*d - b*c*e)*(e*x + d)^(3/2) + 15*(6*c^2*d^2 - 6*b*c*d*e + (b^2 + 2*a*c
)*e^2)*sqrt(e*x + d))/e^4 - 5*(c^2*d^4 - 2*b*c*d^3*e - 2*a*b*d*e^3 + a^2*e^4 + (b^2 + 2*a*c)*d^2*e^2 - 6*(2*c^
2*d^3 - 3*b*c*d^2*e - a*b*e^3 + (b^2 + 2*a*c)*d*e^2)*(e*x + d))/((e*x + d)^(3/2)*e^4))/e

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.51 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (12 \, {\left (e x + d\right )} c^{2} d^{3} - c^{2} d^{4} - 18 \, {\left (e x + d\right )} b c d^{2} e + 2 \, b c d^{3} e + 6 \, {\left (e x + d\right )} b^{2} d e^{2} + 12 \, {\left (e x + d\right )} a c d e^{2} - b^{2} d^{2} e^{2} - 2 \, a c d^{2} e^{2} - 6 \, {\left (e x + d\right )} a b e^{3} + 2 \, a b d e^{3} - a^{2} e^{4}\right )}}{3 \, {\left (e x + d\right )}^{\frac {3}{2}} e^{5}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} c^{2} e^{20} - 20 \, {\left (e x + d\right )}^{\frac {3}{2}} c^{2} d e^{20} + 90 \, \sqrt {e x + d} c^{2} d^{2} e^{20} + 10 \, {\left (e x + d\right )}^{\frac {3}{2}} b c e^{21} - 90 \, \sqrt {e x + d} b c d e^{21} + 15 \, \sqrt {e x + d} b^{2} e^{22} + 30 \, \sqrt {e x + d} a c e^{22}\right )}}{15 \, e^{25}} \]

[In]

integrate((c*x^2+b*x+a)^2/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

2/3*(12*(e*x + d)*c^2*d^3 - c^2*d^4 - 18*(e*x + d)*b*c*d^2*e + 2*b*c*d^3*e + 6*(e*x + d)*b^2*d*e^2 + 12*(e*x +
d)*a*c*d*e^2 - b^2*d^2*e^2 - 2*a*c*d^2*e^2 - 6*(e*x + d)*a*b*e^3 + 2*a*b*d*e^3 - a^2*e^4)/((e*x + d)^(3/2)*e^
5) + 2/15*(3*(e*x + d)^(5/2)*c^2*e^20 - 20*(e*x + d)^(3/2)*c^2*d*e^20 + 90*sqrt(e*x + d)*c^2*d^2*e^20 + 10*(e*
x + d)^(3/2)*b*c*e^21 - 90*sqrt(e*x + d)*b*c*d*e^21 + 15*sqrt(e*x + d)*b^2*e^22 + 30*sqrt(e*x + d)*a*c*e^22)/e
^25

Mupad [B] (verification not implemented)

Time = 9.81 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.20 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(d+e x)^{5/2}} \, dx=\frac {2\,c^2\,{\left (d+e\,x\right )}^{5/2}}{5\,e^5}-\frac {\frac {2\,a^2\,e^4}{3}-\left (d+e\,x\right )\,\left (4\,b^2\,d\,e^2-12\,b\,c\,d^2\,e-4\,a\,b\,e^3+8\,c^2\,d^3+8\,a\,c\,d\,e^2\right )+\frac {2\,c^2\,d^4}{3}+\frac {2\,b^2\,d^2\,e^2}{3}-\frac {4\,a\,b\,d\,e^3}{3}-\frac {4\,b\,c\,d^3\,e}{3}+\frac {4\,a\,c\,d^2\,e^2}{3}}{e^5\,{\left (d+e\,x\right )}^{3/2}}+\frac {\sqrt {d+e\,x}\,\left (2\,b^2\,e^2-12\,b\,c\,d\,e+12\,c^2\,d^2+4\,a\,c\,e^2\right )}{e^5}-\frac {\left (8\,c^2\,d-4\,b\,c\,e\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,e^5} \]

[In]

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

[Out]

(2*c^2*(d + e*x)^(5/2))/(5*e^5) - ((2*a^2*e^4)/3 - (d + e*x)*(8*c^2*d^3 + 4*b^2*d*e^2 - 4*a*b*e^3 + 8*a*c*d*e^
2 - 12*b*c*d^2*e) + (2*c^2*d^4)/3 + (2*b^2*d^2*e^2)/3 - (4*a*b*d*e^3)/3 - (4*b*c*d^3*e)/3 + (4*a*c*d^2*e^2)/3)
/(e^5*(d + e*x)^(3/2)) + ((d + e*x)^(1/2)*(2*b^2*e^2 + 12*c^2*d^2 + 4*a*c*e^2 - 12*b*c*d*e))/e^5 - ((8*c^2*d -
4*b*c*e)*(d + e*x)^(3/2))/(3*e^5)