\(\int \frac {(a^2+2 a b x+b^2 x^2)^2}{\sqrt {d+e x}} \, dx\) [1633]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [B] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 28, antiderivative size = 127 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{\sqrt {d+e x}} \, dx=\frac {2 (b d-a e)^4 \sqrt {d+e x}}{e^5}-\frac {8 b (b d-a e)^3 (d+e x)^{3/2}}{3 e^5}+\frac {12 b^2 (b d-a e)^2 (d+e x)^{5/2}}{5 e^5}-\frac {8 b^3 (b d-a e) (d+e x)^{7/2}}{7 e^5}+\frac {2 b^4 (d+e x)^{9/2}}{9 e^5} \]

[Out]

-8/3*b*(-a*e+b*d)^3*(e*x+d)^(3/2)/e^5+12/5*b^2*(-a*e+b*d)^2*(e*x+d)^(5/2)/e^5-8/7*b^3*(-a*e+b*d)*(e*x+d)^(7/2)
/e^5+2/9*b^4*(e*x+d)^(9/2)/e^5+2*(-a*e+b*d)^4*(e*x+d)^(1/2)/e^5

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {27, 45} \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{\sqrt {d+e x}} \, dx=-\frac {8 b^3 (d+e x)^{7/2} (b d-a e)}{7 e^5}+\frac {12 b^2 (d+e x)^{5/2} (b d-a e)^2}{5 e^5}-\frac {8 b (d+e x)^{3/2} (b d-a e)^3}{3 e^5}+\frac {2 \sqrt {d+e x} (b d-a e)^4}{e^5}+\frac {2 b^4 (d+e x)^{9/2}}{9 e^5} \]

[In]

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

[Out]

(2*(b*d - a*e)^4*Sqrt[d + e*x])/e^5 - (8*b*(b*d - a*e)^3*(d + e*x)^(3/2))/(3*e^5) + (12*b^2*(b*d - a*e)^2*(d +
 e*x)^(5/2))/(5*e^5) - (8*b^3*(b*d - a*e)*(d + e*x)^(7/2))/(7*e^5) + (2*b^4*(d + e*x)^(9/2))/(9*e^5)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \frac {(a+b x)^4}{\sqrt {d+e x}} \, dx \\ & = \int \left (\frac {(-b d+a e)^4}{e^4 \sqrt {d+e x}}-\frac {4 b (b d-a e)^3 \sqrt {d+e x}}{e^4}+\frac {6 b^2 (b d-a e)^2 (d+e x)^{3/2}}{e^4}-\frac {4 b^3 (b d-a e) (d+e x)^{5/2}}{e^4}+\frac {b^4 (d+e x)^{7/2}}{e^4}\right ) \, dx \\ & = \frac {2 (b d-a e)^4 \sqrt {d+e x}}{e^5}-\frac {8 b (b d-a e)^3 (d+e x)^{3/2}}{3 e^5}+\frac {12 b^2 (b d-a e)^2 (d+e x)^{5/2}}{5 e^5}-\frac {8 b^3 (b d-a e) (d+e x)^{7/2}}{7 e^5}+\frac {2 b^4 (d+e x)^{9/2}}{9 e^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.20 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {d+e x} \left (315 a^4 e^4+420 a^3 b e^3 (-2 d+e x)+126 a^2 b^2 e^2 \left (8 d^2-4 d e x+3 e^2 x^2\right )+36 a b^3 e \left (-16 d^3+8 d^2 e x-6 d e^2 x^2+5 e^3 x^3\right )+b^4 \left (128 d^4-64 d^3 e x+48 d^2 e^2 x^2-40 d e^3 x^3+35 e^4 x^4\right )\right )}{315 e^5} \]

[In]

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

[Out]

(2*Sqrt[d + e*x]*(315*a^4*e^4 + 420*a^3*b*e^3*(-2*d + e*x) + 126*a^2*b^2*e^2*(8*d^2 - 4*d*e*x + 3*e^2*x^2) + 3
6*a*b^3*e*(-16*d^3 + 8*d^2*e*x - 6*d*e^2*x^2 + 5*e^3*x^3) + b^4*(128*d^4 - 64*d^3*e*x + 48*d^2*e^2*x^2 - 40*d*
e^3*x^3 + 35*e^4*x^4)))/(315*e^5)

Maple [A] (verified)

Time = 2.29 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.13

method result size
pseudoelliptic \(\frac {2 \sqrt {e x +d}\, \left (\left (\frac {1}{9} b^{4} x^{4}+\frac {4}{7} a \,b^{3} x^{3}+\frac {6}{5} a^{2} b^{2} x^{2}+\frac {4}{3} a^{3} b x +a^{4}\right ) e^{4}-\frac {8 \left (\frac {1}{21} b^{3} x^{3}+\frac {9}{35} a \,b^{2} x^{2}+\frac {3}{5} a^{2} b x +a^{3}\right ) b d \,e^{3}}{3}+\frac {16 \left (\frac {1}{21} b^{2} x^{2}+\frac {2}{7} a b x +a^{2}\right ) b^{2} d^{2} e^{2}}{5}-\frac {64 b^{3} d^{3} \left (\frac {b x}{9}+a \right ) e}{35}+\frac {128 b^{4} d^{4}}{315}\right )}{e^{5}}\) \(143\)
derivativedivides \(\frac {\frac {2 b^{4} \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {4 \left (2 a e b -2 b^{2} d \right ) b^{2} \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (2 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) b^{2}+\left (2 a e b -2 b^{2} d \right )^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {4 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \left (2 a e b -2 b^{2} d \right ) \left (e x +d \right )^{\frac {3}{2}}}{3}+2 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )^{2} \sqrt {e x +d}}{e^{5}}\) \(166\)
default \(\frac {\frac {2 b^{4} \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {4 \left (2 a e b -2 b^{2} d \right ) b^{2} \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (2 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) b^{2}+\left (2 a e b -2 b^{2} d \right )^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {4 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \left (2 a e b -2 b^{2} d \right ) \left (e x +d \right )^{\frac {3}{2}}}{3}+2 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )^{2} \sqrt {e x +d}}{e^{5}}\) \(166\)
gosper \(\frac {2 \left (35 b^{4} x^{4} e^{4}+180 x^{3} a \,b^{3} e^{4}-40 x^{3} b^{4} d \,e^{3}+378 x^{2} a^{2} b^{2} e^{4}-216 x^{2} a \,b^{3} d \,e^{3}+48 x^{2} b^{4} d^{2} e^{2}+420 x \,a^{3} b \,e^{4}-504 x \,a^{2} b^{2} d \,e^{3}+288 x a \,b^{3} d^{2} e^{2}-64 x \,b^{4} d^{3} e +315 e^{4} a^{4}-840 b \,e^{3} d \,a^{3}+1008 b^{2} e^{2} d^{2} a^{2}-576 a \,b^{3} d^{3} e +128 b^{4} d^{4}\right ) \sqrt {e x +d}}{315 e^{5}}\) \(186\)
trager \(\frac {2 \left (35 b^{4} x^{4} e^{4}+180 x^{3} a \,b^{3} e^{4}-40 x^{3} b^{4} d \,e^{3}+378 x^{2} a^{2} b^{2} e^{4}-216 x^{2} a \,b^{3} d \,e^{3}+48 x^{2} b^{4} d^{2} e^{2}+420 x \,a^{3} b \,e^{4}-504 x \,a^{2} b^{2} d \,e^{3}+288 x a \,b^{3} d^{2} e^{2}-64 x \,b^{4} d^{3} e +315 e^{4} a^{4}-840 b \,e^{3} d \,a^{3}+1008 b^{2} e^{2} d^{2} a^{2}-576 a \,b^{3} d^{3} e +128 b^{4} d^{4}\right ) \sqrt {e x +d}}{315 e^{5}}\) \(186\)
risch \(\frac {2 \left (35 b^{4} x^{4} e^{4}+180 x^{3} a \,b^{3} e^{4}-40 x^{3} b^{4} d \,e^{3}+378 x^{2} a^{2} b^{2} e^{4}-216 x^{2} a \,b^{3} d \,e^{3}+48 x^{2} b^{4} d^{2} e^{2}+420 x \,a^{3} b \,e^{4}-504 x \,a^{2} b^{2} d \,e^{3}+288 x a \,b^{3} d^{2} e^{2}-64 x \,b^{4} d^{3} e +315 e^{4} a^{4}-840 b \,e^{3} d \,a^{3}+1008 b^{2} e^{2} d^{2} a^{2}-576 a \,b^{3} d^{3} e +128 b^{4} d^{4}\right ) \sqrt {e x +d}}{315 e^{5}}\) \(186\)

[In]

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

[Out]

2*(e*x+d)^(1/2)*((1/9*b^4*x^4+4/7*a*b^3*x^3+6/5*a^2*b^2*x^2+4/3*a^3*b*x+a^4)*e^4-8/3*(1/21*b^3*x^3+9/35*a*b^2*
x^2+3/5*a^2*b*x+a^3)*b*d*e^3+16/5*(1/21*b^2*x^2+2/7*a*b*x+a^2)*b^2*d^2*e^2-64/35*b^3*d^3*(1/9*b*x+a)*e+128/315
*b^4*d^4)/e^5

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.43 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (35 \, b^{4} e^{4} x^{4} + 128 \, b^{4} d^{4} - 576 \, a b^{3} d^{3} e + 1008 \, a^{2} b^{2} d^{2} e^{2} - 840 \, a^{3} b d e^{3} + 315 \, a^{4} e^{4} - 20 \, {\left (2 \, b^{4} d e^{3} - 9 \, a b^{3} e^{4}\right )} x^{3} + 6 \, {\left (8 \, b^{4} d^{2} e^{2} - 36 \, a b^{3} d e^{3} + 63 \, a^{2} b^{2} e^{4}\right )} x^{2} - 4 \, {\left (16 \, b^{4} d^{3} e - 72 \, a b^{3} d^{2} e^{2} + 126 \, a^{2} b^{2} d e^{3} - 105 \, a^{3} b e^{4}\right )} x\right )} \sqrt {e x + d}}{315 \, e^{5}} \]

[In]

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

[Out]

2/315*(35*b^4*e^4*x^4 + 128*b^4*d^4 - 576*a*b^3*d^3*e + 1008*a^2*b^2*d^2*e^2 - 840*a^3*b*d*e^3 + 315*a^4*e^4 -
 20*(2*b^4*d*e^3 - 9*a*b^3*e^4)*x^3 + 6*(8*b^4*d^2*e^2 - 36*a*b^3*d*e^3 + 63*a^2*b^2*e^4)*x^2 - 4*(16*b^4*d^3*
e - 72*a*b^3*d^2*e^2 + 126*a^2*b^2*d*e^3 - 105*a^3*b*e^4)*x)*sqrt(e*x + d)/e^5

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 270 vs. \(2 (117) = 234\).

Time = 1.01 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.13 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{\sqrt {d+e x}} \, dx=\begin {cases} \frac {2 \left (\frac {b^{4} \left (d + e x\right )^{\frac {9}{2}}}{9 e^{4}} + \frac {\left (d + e x\right )^{\frac {7}{2}} \cdot \left (4 a b^{3} e - 4 b^{4} d\right )}{7 e^{4}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \cdot \left (6 a^{2} b^{2} e^{2} - 12 a b^{3} d e + 6 b^{4} d^{2}\right )}{5 e^{4}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \cdot \left (4 a^{3} b e^{3} - 12 a^{2} b^{2} d e^{2} + 12 a b^{3} d^{2} e - 4 b^{4} d^{3}\right )}{3 e^{4}} + \frac {\sqrt {d + e x} \left (a^{4} e^{4} - 4 a^{3} b d e^{3} + 6 a^{2} b^{2} d^{2} e^{2} - 4 a b^{3} d^{3} e + b^{4} d^{4}\right )}{e^{4}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {a^{4} x + 2 a^{3} b x^{2} + 2 a^{2} b^{2} x^{3} + a b^{3} x^{4} + \frac {b^{4} x^{5}}{5}}{\sqrt {d}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(b**4*(d + e*x)**(9/2)/(9*e**4) + (d + e*x)**(7/2)*(4*a*b**3*e - 4*b**4*d)/(7*e**4) + (d + e*x)**
(5/2)*(6*a**2*b**2*e**2 - 12*a*b**3*d*e + 6*b**4*d**2)/(5*e**4) + (d + e*x)**(3/2)*(4*a**3*b*e**3 - 12*a**2*b*
*2*d*e**2 + 12*a*b**3*d**2*e - 4*b**4*d**3)/(3*e**4) + sqrt(d + e*x)*(a**4*e**4 - 4*a**3*b*d*e**3 + 6*a**2*b**
2*d**2*e**2 - 4*a*b**3*d**3*e + b**4*d**4)/e**4)/e, Ne(e, 0)), ((a**4*x + 2*a**3*b*x**2 + 2*a**2*b**2*x**3 + a
*b**3*x**4 + b**4*x**5/5)/sqrt(d), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 247 vs. \(2 (109) = 218\).

Time = 0.20 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.94 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (315 \, \sqrt {e x + d} a^{4} + 42 \, {\left (\frac {10 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a b}{e} + \frac {{\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} b^{2}}{e^{2}}\right )} a^{2} + \frac {84 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a^{2} b^{2}}{e^{2}} + \frac {36 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} a b^{3}}{e^{3}} + \frac {{\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} b^{4}}{e^{4}}\right )}}{315 \, e} \]

[In]

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

[Out]

2/315*(315*sqrt(e*x + d)*a^4 + 42*(10*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a*b/e + (3*(e*x + d)^(5/2) - 10*(e
*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*b^2/e^2)*a^2 + 84*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(
e*x + d)*d^2)*a^2*b^2/e^2 + 36*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*
x + d)*d^3)*a*b^3/e^3 + (35*(e*x + d)^(9/2) - 180*(e*x + d)^(7/2)*d + 378*(e*x + d)^(5/2)*d^2 - 420*(e*x + d)^
(3/2)*d^3 + 315*sqrt(e*x + d)*d^4)*b^4/e^4)/e

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.61 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (315 \, \sqrt {e x + d} a^{4} + \frac {420 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a^{3} b}{e} + \frac {126 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a^{2} b^{2}}{e^{2}} + \frac {36 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} a b^{3}}{e^{3}} + \frac {{\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} b^{4}}{e^{4}}\right )}}{315 \, e} \]

[In]

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

[Out]

2/315*(315*sqrt(e*x + d)*a^4 + 420*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a^3*b/e + 126*(3*(e*x + d)^(5/2) - 10
*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a^2*b^2/e^2 + 36*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*
x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*a*b^3/e^3 + (35*(e*x + d)^(9/2) - 180*(e*x + d)^(7/2)*d + 378*(e*x +
d)^(5/2)*d^2 - 420*(e*x + d)^(3/2)*d^3 + 315*sqrt(e*x + d)*d^4)*b^4/e^4)/e

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^2}{\sqrt {d+e x}} \, dx=\frac {2\,b^4\,{\left (d+e\,x\right )}^{9/2}}{9\,e^5}-\frac {\left (8\,b^4\,d-8\,a\,b^3\,e\right )\,{\left (d+e\,x\right )}^{7/2}}{7\,e^5}+\frac {2\,{\left (a\,e-b\,d\right )}^4\,\sqrt {d+e\,x}}{e^5}+\frac {12\,b^2\,{\left (a\,e-b\,d\right )}^2\,{\left (d+e\,x\right )}^{5/2}}{5\,e^5}+\frac {8\,b\,{\left (a\,e-b\,d\right )}^3\,{\left (d+e\,x\right )}^{3/2}}{3\,e^5} \]

[In]

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

[Out]

(2*b^4*(d + e*x)^(9/2))/(9*e^5) - ((8*b^4*d - 8*a*b^3*e)*(d + e*x)^(7/2))/(7*e^5) + (2*(a*e - b*d)^4*(d + e*x)
^(1/2))/e^5 + (12*b^2*(a*e - b*d)^2*(d + e*x)^(5/2))/(5*e^5) + (8*b*(a*e - b*d)^3*(d + e*x)^(3/2))/(3*e^5)