\(\int \frac {(a+b x+c x^2)^3}{(b d+2 c d x)^{11/2}} \, dx\) [1283]

Optimal result

Integrand size = 26, antiderivative size = 121 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{11/2}} \, dx=\frac {\left (b^2-4 a c\right )^3}{576 c^4 d (b d+2 c d x)^{9/2}}-\frac {3 \left (b^2-4 a c\right )^2}{320 c^4 d^3 (b d+2 c d x)^{5/2}}+\frac {3 \left (b^2-4 a c\right )}{64 c^4 d^5 \sqrt {b d+2 c d x}}+\frac {(b d+2 c d x)^{3/2}}{192 c^4 d^7} \]

[Out]

1/576*(-4*a*c+b^2)^3/c^4/d/(2*c*d*x+b*d)^(9/2)-3/320*(-4*a*c+b^2)^2/c^4/d^3/(2*c*d*x+b*d)^(5/2)+1/192*(2*c*d*x
+b*d)^(3/2)/c^4/d^7+3/64*(-4*a*c+b^2)/c^4/d^5/(2*c*d*x+b*d)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 121, 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} \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{11/2}} \, dx=\frac {3 \left (b^2-4 a c\right )}{64 c^4 d^5 \sqrt {b d+2 c d x}}-\frac {3 \left (b^2-4 a c\right )^2}{320 c^4 d^3 (b d+2 c d x)^{5/2}}+\frac {\left (b^2-4 a c\right )^3}{576 c^4 d (b d+2 c d x)^{9/2}}+\frac {(b d+2 c d x)^{3/2}}{192 c^4 d^7} \]

[In]

Int[(a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^(11/2),x]

[Out]

(b^2 - 4*a*c)^3/(576*c^4*d*(b*d + 2*c*d*x)^(9/2)) - (3*(b^2 - 4*a*c)^2)/(320*c^4*d^3*(b*d + 2*c*d*x)^(5/2)) +
(3*(b^2 - 4*a*c))/(64*c^4*d^5*Sqrt[b*d + 2*c*d*x]) + (b*d + 2*c*d*x)^(3/2)/(192*c^4*d^7)

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 )^3}{64 c^3 (b d+2 c d x)^{11/2}}+\frac {3 \left (-b^2+4 a c\right )^2}{64 c^3 d^2 (b d+2 c d x)^{7/2}}+\frac {3 \left (-b^2+4 a c\right )}{64 c^3 d^4 (b d+2 c d x)^{3/2}}+\frac {\sqrt {b d+2 c d x}}{64 c^3 d^6}\right ) \, dx \\ & = \frac {\left (b^2-4 a c\right )^3}{576 c^4 d (b d+2 c d x)^{9/2}}-\frac {3 \left (b^2-4 a c\right )^2}{320 c^4 d^3 (b d+2 c d x)^{5/2}}+\frac {3 \left (b^2-4 a c\right )}{64 c^4 d^5 \sqrt {b d+2 c d x}}+\frac {(b d+2 c d x)^{3/2}}{192 c^4 d^7} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{11/2}} \, dx=\frac {5 b^6-60 a b^4 c+240 a^2 b^2 c^2-320 a^3 c^3-27 b^4 (b+2 c x)^2+216 a b^2 c (b+2 c x)^2-432 a^2 c^2 (b+2 c x)^2+135 b^2 (b+2 c x)^4-540 a c (b+2 c x)^4+15 (b+2 c x)^6}{2880 c^4 d (d (b+2 c x))^{9/2}} \]

[In]

Integrate[(a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^(11/2),x]

[Out]

(5*b^6 - 60*a*b^4*c + 240*a^2*b^2*c^2 - 320*a^3*c^3 - 27*b^4*(b + 2*c*x)^2 + 216*a*b^2*c*(b + 2*c*x)^2 - 432*a
^2*c^2*(b + 2*c*x)^2 + 135*b^2*(b + 2*c*x)^4 - 540*a*c*(b + 2*c*x)^4 + 15*(b + 2*c*x)^6)/(2880*c^4*d*(d*(b + 2
*c*x))^(9/2))

Maple [A] (verified)

Time = 2.39 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{3}-\frac {3 d^{2} \left (4 a c -b^{2}\right )}{\sqrt {2 c d x +b d}}-\frac {3 d^{4} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}{5 \left (2 c d x +b d \right )^{\frac {5}{2}}}-\frac {d^{6} \left (64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}{9 \left (2 c d x +b d \right )^{\frac {9}{2}}}}{64 c^{4} d^{7}}\)	\(132\)
default	\(\frac {\frac {\left (2 c d x +b d \right )^{\frac {3}{2}}}{3}-\frac {3 d^{2} \left (4 a c -b^{2}\right )}{\sqrt {2 c d x +b d}}-\frac {3 d^{4} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}{5 \left (2 c d x +b d \right )^{\frac {5}{2}}}-\frac {d^{6} \left (64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}{9 \left (2 c d x +b d \right )^{\frac {9}{2}}}}{64 c^{4} d^{7}}\)	\(132\)
pseudoelliptic	\(\frac {15 c^{6} x^{6}+\left (45 b \,x^{5}-135 a \,x^{4}\right ) c^{5}+\left (90 b^{2} x^{4}-270 a b \,x^{3}-27 a^{2} x^{2}\right ) c^{4}+\left (105 b^{3} x^{3}-189 a \,b^{2} x^{2}-27 a^{2} b x -5 a^{3}\right ) c^{3}+\left (63 b^{4} x^{2}-54 a \,b^{3} x -3 a^{2} b^{2}\right ) c^{2}-6 b^{4} \left (-3 b x +a \right ) c +2 b^{6}}{45 \sqrt {d \left (2 c x +b \right )}\, d^{5} \left (2 c x +b \right )^{4} c^{4}}\)	\(160\)
gosper	\(-\frac {\left (2 c x +b \right ) \left (-15 c^{6} x^{6}-45 b \,c^{5} x^{5}+135 a \,c^{5} x^{4}-90 b^{2} c^{4} x^{4}+270 a b \,c^{4} x^{3}-105 x^{3} b^{3} c^{3}+27 a^{2} c^{4} x^{2}+189 a \,b^{2} c^{3} x^{2}-63 x^{2} b^{4} c^{2}+27 a^{2} b \,c^{3} x +54 x a \,b^{3} c^{2}-18 x \,b^{5} c +5 c^{3} a^{3}+3 a^{2} b^{2} c^{2}+6 a \,b^{4} c -2 b^{6}\right )}{45 c^{4} \left (2 c d x +b d \right )^{\frac {11}{2}}}\)	\(174\)
trager	\(-\frac {\left (-15 c^{6} x^{6}-45 b \,c^{5} x^{5}+135 a \,c^{5} x^{4}-90 b^{2} c^{4} x^{4}+270 a b \,c^{4} x^{3}-105 x^{3} b^{3} c^{3}+27 a^{2} c^{4} x^{2}+189 a \,b^{2} c^{3} x^{2}-63 x^{2} b^{4} c^{2}+27 a^{2} b \,c^{3} x +54 x a \,b^{3} c^{2}-18 x \,b^{5} c +5 c^{3} a^{3}+3 a^{2} b^{2} c^{2}+6 a \,b^{4} c -2 b^{6}\right ) \sqrt {2 c d x +b d}}{45 d^{6} \left (2 c x +b \right )^{5} c^{4}}\)	\(179\)

[In]

int((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(11/2),x,method=_RETURNVERBOSE)

[Out]

1/64/c^4/d^7*(1/3*(2*c*d*x+b*d)^(3/2)-3*d^2*(4*a*c-b^2)/(2*c*d*x+b*d)^(1/2)-3/5*d^4*(16*a^2*c^2-8*a*b^2*c+b^4)
/(2*c*d*x+b*d)^(5/2)-1/9*d^6*(64*a^3*c^3-48*a^2*b^2*c^2+12*a*b^4*c-b^6)/(2*c*d*x+b*d)^(9/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (105) = 210\).

Time = 0.27 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.97 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{11/2}} \, dx=\frac {{\left (15 \, c^{6} x^{6} + 45 \, b c^{5} x^{5} + 2 \, b^{6} - 6 \, a b^{4} c - 3 \, a^{2} b^{2} c^{2} - 5 \, a^{3} c^{3} + 45 \, {\left (2 \, b^{2} c^{4} - 3 \, a c^{5}\right )} x^{4} + 15 \, {\left (7 \, b^{3} c^{3} - 18 \, a b c^{4}\right )} x^{3} + 9 \, {\left (7 \, b^{4} c^{2} - 21 \, a b^{2} c^{3} - 3 \, a^{2} c^{4}\right )} x^{2} + 9 \, {\left (2 \, b^{5} c - 6 \, a b^{3} c^{2} - 3 \, a^{2} b c^{3}\right )} x\right )} \sqrt {2 \, c d x + b d}}{45 \, {\left (32 \, c^{9} d^{6} x^{5} + 80 \, b c^{8} d^{6} x^{4} + 80 \, b^{2} c^{7} d^{6} x^{3} + 40 \, b^{3} c^{6} d^{6} x^{2} + 10 \, b^{4} c^{5} d^{6} x + b^{5} c^{4} d^{6}\right )}} \]

[In]

integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(11/2),x, algorithm="fricas")

[Out]

1/45*(15*c^6*x^6 + 45*b*c^5*x^5 + 2*b^6 - 6*a*b^4*c - 3*a^2*b^2*c^2 - 5*a^3*c^3 + 45*(2*b^2*c^4 - 3*a*c^5)*x^4
+ 15*(7*b^3*c^3 - 18*a*b*c^4)*x^3 + 9*(7*b^4*c^2 - 21*a*b^2*c^3 - 3*a^2*c^4)*x^2 + 9*(2*b^5*c - 6*a*b^3*c^2 -
3*a^2*b*c^3)*x)*sqrt(2*c*d*x + b*d)/(32*c^9*d^6*x^5 + 80*b*c^8*d^6*x^4 + 80*b^2*c^7*d^6*x^3 + 40*b^3*c^6*d^6*
x^2 + 10*b^4*c^5*d^6*x + b^5*c^4*d^6)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1731 vs. \(2 (119) = 238\).

Time = 1.16 (sec) , antiderivative size = 1731, normalized size of antiderivative = 14.31 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{11/2}} \, dx=\text {Too large to display} \]

[In]

integrate((c*x**2+b*x+a)**3/(2*c*d*x+b*d)**(11/2),x)

[Out]

Piecewise((-5*a**3*c**3*sqrt(b*d + 2*c*d*x)/(45*b**5*c**4*d**6 + 450*b**4*c**5*d**6*x + 1800*b**3*c**6*d**6*x*
*2 + 3600*b**2*c**7*d**6*x**3 + 3600*b*c**8*d**6*x**4 + 1440*c**9*d**6*x**5) - 3*a**2*b**2*c**2*sqrt(b*d + 2*c
*d*x)/(45*b**5*c**4*d**6 + 450*b**4*c**5*d**6*x + 1800*b**3*c**6*d**6*x**2 + 3600*b**2*c**7*d**6*x**3 + 3600*b
*c**8*d**6*x**4 + 1440*c**9*d**6*x**5) - 27*a**2*b*c**3*x*sqrt(b*d + 2*c*d*x)/(45*b**5*c**4*d**6 + 450*b**4*c*
*5*d**6*x + 1800*b**3*c**6*d**6*x**2 + 3600*b**2*c**7*d**6*x**3 + 3600*b*c**8*d**6*x**4 + 1440*c**9*d**6*x**5)
- 27*a**2*c**4*x**2*sqrt(b*d + 2*c*d*x)/(45*b**5*c**4*d**6 + 450*b**4*c**5*d**6*x + 1800*b**3*c**6*d**6*x**2
+ 3600*b**2*c**7*d**6*x**3 + 3600*b*c**8*d**6*x**4 + 1440*c**9*d**6*x**5) - 6*a*b**4*c*sqrt(b*d + 2*c*d*x)/(45
*b**5*c**4*d**6 + 450*b**4*c**5*d**6*x + 1800*b**3*c**6*d**6*x**2 + 3600*b**2*c**7*d**6*x**3 + 3600*b*c**8*d**
6*x**4 + 1440*c**9*d**6*x**5) - 54*a*b**3*c**2*x*sqrt(b*d + 2*c*d*x)/(45*b**5*c**4*d**6 + 450*b**4*c**5*d**6*x
+ 1800*b**3*c**6*d**6*x**2 + 3600*b**2*c**7*d**6*x**3 + 3600*b*c**8*d**6*x**4 + 1440*c**9*d**6*x**5) - 189*a*
b**2*c**3*x**2*sqrt(b*d + 2*c*d*x)/(45*b**5*c**4*d**6 + 450*b**4*c**5*d**6*x + 1800*b**3*c**6*d**6*x**2 + 3600
*b**2*c**7*d**6*x**3 + 3600*b*c**8*d**6*x**4 + 1440*c**9*d**6*x**5) - 270*a*b*c**4*x**3*sqrt(b*d + 2*c*d*x)/(4
5*b**5*c**4*d**6 + 450*b**4*c**5*d**6*x + 1800*b**3*c**6*d**6*x**2 + 3600*b**2*c**7*d**6*x**3 + 3600*b*c**8*d*
*6*x**4 + 1440*c**9*d**6*x**5) - 135*a*c**5*x**4*sqrt(b*d + 2*c*d*x)/(45*b**5*c**4*d**6 + 450*b**4*c**5*d**6*x
+ 1800*b**3*c**6*d**6*x**2 + 3600*b**2*c**7*d**6*x**3 + 3600*b*c**8*d**6*x**4 + 1440*c**9*d**6*x**5) + 2*b**6
*sqrt(b*d + 2*c*d*x)/(45*b**5*c**4*d**6 + 450*b**4*c**5*d**6*x + 1800*b**3*c**6*d**6*x**2 + 3600*b**2*c**7*d**
6*x**3 + 3600*b*c**8*d**6*x**4 + 1440*c**9*d**6*x**5) + 18*b**5*c*x*sqrt(b*d + 2*c*d*x)/(45*b**5*c**4*d**6 + 4
50*b**4*c**5*d**6*x + 1800*b**3*c**6*d**6*x**2 + 3600*b**2*c**7*d**6*x**3 + 3600*b*c**8*d**6*x**4 + 1440*c**9*
d**6*x**5) + 63*b**4*c**2*x**2*sqrt(b*d + 2*c*d*x)/(45*b**5*c**4*d**6 + 450*b**4*c**5*d**6*x + 1800*b**3*c**6*
d**6*x**2 + 3600*b**2*c**7*d**6*x**3 + 3600*b*c**8*d**6*x**4 + 1440*c**9*d**6*x**5) + 105*b**3*c**3*x**3*sqrt(
b*d + 2*c*d*x)/(45*b**5*c**4*d**6 + 450*b**4*c**5*d**6*x + 1800*b**3*c**6*d**6*x**2 + 3600*b**2*c**7*d**6*x**3
+ 3600*b*c**8*d**6*x**4 + 1440*c**9*d**6*x**5) + 90*b**2*c**4*x**4*sqrt(b*d + 2*c*d*x)/(45*b**5*c**4*d**6 + 4
50*b**4*c**5*d**6*x + 1800*b**3*c**6*d**6*x**2 + 3600*b**2*c**7*d**6*x**3 + 3600*b*c**8*d**6*x**4 + 1440*c**9*
d**6*x**5) + 45*b*c**5*x**5*sqrt(b*d + 2*c*d*x)/(45*b**5*c**4*d**6 + 450*b**4*c**5*d**6*x + 1800*b**3*c**6*d**
6*x**2 + 3600*b**2*c**7*d**6*x**3 + 3600*b*c**8*d**6*x**4 + 1440*c**9*d**6*x**5) + 15*c**6*x**6*sqrt(b*d + 2*c
*d*x)/(45*b**5*c**4*d**6 + 450*b**4*c**5*d**6*x + 1800*b**3*c**6*d**6*x**2 + 3600*b**2*c**7*d**6*x**3 + 3600*b
*c**8*d**6*x**4 + 1440*c**9*d**6*x**5), Ne(c, 0)), ((a**3*x + 3*a**2*b*x**2/2 + a*b**2*x**3 + b**3*x**4/4)/(b*
d)**(11/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{11/2}} \, dx=\frac {\frac {15 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}}{c^{3} d^{6}} + \frac {135 \, {\left (2 \, c d x + b d\right )}^{4} {\left (b^{2} - 4 \, a c\right )} - 27 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} {\left (2 \, c d x + b d\right )}^{2} d^{2} + 5 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{4}}{{\left (2 \, c d x + b d\right )}^{\frac {9}{2}} c^{3} d^{4}}}{2880 \, c d} \]

[In]

integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(11/2),x, algorithm="maxima")

[Out]

1/2880*(15*(2*c*d*x + b*d)^(3/2)/(c^3*d^6) + (135*(2*c*d*x + b*d)^4*(b^2 - 4*a*c) - 27*(b^4 - 8*a*b^2*c + 16*a
^2*c^2)*(2*c*d*x + b*d)^2*d^2 + 5*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^4)/((2*c*d*x + b*d)^(9/2)
*c^3*d^4))/(c*d)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.45 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{11/2}} \, dx=\frac {{\left (2 \, c d x + b d\right )}^{\frac {3}{2}}}{192 \, c^{4} d^{7}} + \frac {5 \, b^{6} d^{4} - 60 \, a b^{4} c d^{4} + 240 \, a^{2} b^{2} c^{2} d^{4} - 320 \, a^{3} c^{3} d^{4} - 27 \, {\left (2 \, c d x + b d\right )}^{2} b^{4} d^{2} + 216 \, {\left (2 \, c d x + b d\right )}^{2} a b^{2} c d^{2} - 432 \, {\left (2 \, c d x + b d\right )}^{2} a^{2} c^{2} d^{2} + 135 \, {\left (2 \, c d x + b d\right )}^{4} b^{2} - 540 \, {\left (2 \, c d x + b d\right )}^{4} a c}{2880 \, {\left (2 \, c d x + b d\right )}^{\frac {9}{2}} c^{4} d^{5}} \]

[In]

integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(11/2),x, algorithm="giac")

[Out]

1/192*(2*c*d*x + b*d)^(3/2)/(c^4*d^7) + 1/2880*(5*b^6*d^4 - 60*a*b^4*c*d^4 + 240*a^2*b^2*c^2*d^4 - 320*a^3*c^3
*d^4 - 27*(2*c*d*x + b*d)^2*b^4*d^2 + 216*(2*c*d*x + b*d)^2*a*b^2*c*d^2 - 432*(2*c*d*x + b*d)^2*a^2*c^2*d^2 +
135*(2*c*d*x + b*d)^4*b^2 - 540*(2*c*d*x + b*d)^4*a*c)/((2*c*d*x + b*d)^(9/2)*c^4*d^5)

Mupad [B] (verification not implemented)

Time = 9.39 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{11/2}} \, dx=\frac {{\left (b\,d+2\,c\,d\,x\right )}^{3/2}}{192\,c^4\,d^7}-\frac {{\left (b\,d+2\,c\,d\,x\right )}^2\,\left (\frac {48\,a^2\,c^2\,d^2}{5}-\frac {24\,a\,b^2\,c\,d^2}{5}+\frac {3\,b^4\,d^2}{5}\right )+{\left (b\,d+2\,c\,d\,x\right )}^4\,\left (12\,a\,c-3\,b^2\right )-\frac {b^6\,d^4}{9}+\frac {64\,a^3\,c^3\,d^4}{9}-\frac {16\,a^2\,b^2\,c^2\,d^4}{3}+\frac {4\,a\,b^4\,c\,d^4}{3}}{64\,c^4\,d^5\,{\left (b\,d+2\,c\,d\,x\right )}^{9/2}} \]

[In]

int((a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^(11/2),x)

[Out]

(b*d + 2*c*d*x)^(3/2)/(192*c^4*d^7) - ((b*d + 2*c*d*x)^2*((3*b^4*d^2)/5 + (48*a^2*c^2*d^2)/5 - (24*a*b^2*c*d^2
)/5) + (b*d + 2*c*d*x)^4*(12*a*c - 3*b^2) - (b^6*d^4)/9 + (64*a^3*c^3*d^4)/9 - (16*a^2*b^2*c^2*d^4)/3 + (4*a*b
^4*c*d^4)/3)/(64*c^4*d^5*(b*d + 2*c*d*x)^(9/2))