\(\int \frac {(a+b x+c x^2)^3}{(b d+2 c d x)^{5/2}} \, dx\) [1280]

Optimal result

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

[Out]

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

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

[In]

Int[(a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^(5/2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{5/2}} \, dx=\frac {15 b^6-180 a b^4 c+720 a^2 b^2 c^2-960 a^3 c^3+135 b^4 (b+2 c x)^2-1080 a b^2 c (b+2 c x)^2+2160 a^2 c^2 (b+2 c x)^2-27 b^2 (b+2 c x)^4+108 a c (b+2 c x)^4+5 (b+2 c x)^6}{2880 c^4 d (d (b+2 c x))^{3/2}} \]

[In]

Integrate[(a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^(5/2),x]

[Out]

(15*b^6 - 180*a*b^4*c + 720*a^2*b^2*c^2 - 960*a^3*c^3 + 135*b^4*(b + 2*c*x)^2 - 1080*a*b^2*c*(b + 2*c*x)^2 + 2
160*a^2*c^2*(b + 2*c*x)^2 - 27*b^2*(b + 2*c*x)^4 + 108*a*c*(b + 2*c*x)^4 + 5*(b + 2*c*x)^6)/(2880*c^4*d*(d*(b
+ 2*c*x))^(3/2))

Maple [A] (verified)

Time = 2.32 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.40

method	result	size
derivativedivides	\(\frac {48 a^{2} c^{2} d^{4} \sqrt {2 c d x +b d}-24 a \,b^{2} c \,d^{4} \sqrt {2 c d x +b d}+\frac {12 a c \,d^{2} \left (2 c d x +b d \right )^{\frac {5}{2}}}{5}+3 b^{4} d^{4} \sqrt {2 c d x +b d}-\frac {3 b^{2} d^{2} \left (2 c d x +b d \right )^{\frac {5}{2}}}{5}+\frac {\left (2 c d x +b d \right )^{\frac {9}{2}}}{9}-\frac {d^{6} \left (64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}{3 \left (2 c d x +b d \right )^{\frac {3}{2}}}}{64 c^{4} d^{7}}\)	\(170\)
default	\(\frac {48 a^{2} c^{2} d^{4} \sqrt {2 c d x +b d}-24 a \,b^{2} c \,d^{4} \sqrt {2 c d x +b d}+\frac {12 a c \,d^{2} \left (2 c d x +b d \right )^{\frac {5}{2}}}{5}+3 b^{4} d^{4} \sqrt {2 c d x +b d}-\frac {3 b^{2} d^{2} \left (2 c d x +b d \right )^{\frac {5}{2}}}{5}+\frac {\left (2 c d x +b d \right )^{\frac {9}{2}}}{9}-\frac {d^{6} \left (64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}{3 \left (2 c d x +b d \right )^{\frac {3}{2}}}}{64 c^{4} d^{7}}\)	\(170\)
gosper	\(-\frac {\left (2 c x +b \right ) \left (-5 c^{6} x^{6}-15 b \,c^{5} x^{5}-27 a \,c^{5} x^{4}-12 b^{2} c^{4} x^{4}-54 a b \,c^{4} x^{3}+x^{3} b^{3} c^{3}-135 a^{2} c^{4} x^{2}+27 a \,b^{2} c^{3} x^{2}-3 x^{2} b^{4} c^{2}-135 a^{2} b \,c^{3} x +54 x a \,b^{3} c^{2}-6 x \,b^{5} c +15 c^{3} a^{3}-45 a^{2} b^{2} c^{2}+18 a \,b^{4} c -2 b^{6}\right )}{45 c^{4} \left (2 c d x +b d \right )^{\frac {5}{2}}}\)	\(173\)
trager	\(-\frac {\left (-5 c^{6} x^{6}-15 b \,c^{5} x^{5}-27 a \,c^{5} x^{4}-12 b^{2} c^{4} x^{4}-54 a b \,c^{4} x^{3}+x^{3} b^{3} c^{3}-135 a^{2} c^{4} x^{2}+27 a \,b^{2} c^{3} x^{2}-3 x^{2} b^{4} c^{2}-135 a^{2} b \,c^{3} x +54 x a \,b^{3} c^{2}-6 x \,b^{5} c +15 c^{3} a^{3}-45 a^{2} b^{2} c^{2}+18 a \,b^{4} c -2 b^{6}\right ) \sqrt {2 c d x +b d}}{45 d^{3} c^{4} \left (2 c x +b \right )^{2}}\)	\(178\)
pseudoelliptic	\(\frac {5 c^{6} x^{6}+15 b \,c^{5} x^{5}+27 a \,c^{5} x^{4}+12 b^{2} c^{4} x^{4}+54 a b \,c^{4} x^{3}-x^{3} b^{3} c^{3}+135 a^{2} c^{4} x^{2}-27 a \,b^{2} c^{3} x^{2}+3 x^{2} b^{4} c^{2}+135 a^{2} b \,c^{3} x -54 x a \,b^{3} c^{2}+6 x \,b^{5} c -15 c^{3} a^{3}+45 a^{2} b^{2} c^{2}-18 a \,b^{4} c +2 b^{6}}{45 d^{2} \left (2 c x +b \right ) \sqrt {d \left (2 c x +b \right )}\, c^{4}}\)	\(178\)

[In]

int((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/64/c^4/d^7*(48*a^2*c^2*d^4*(2*c*d*x+b*d)^(1/2)-24*a*b^2*c*d^4*(2*c*d*x+b*d)^(1/2)+12/5*a*c*d^2*(2*c*d*x+b*d)
^(5/2)+3*b^4*d^4*(2*c*d*x+b*d)^(1/2)-3/5*b^2*d^2*(2*c*d*x+b*d)^(5/2)+1/9*(2*c*d*x+b*d)^(9/2)-1/3*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)^(3/2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.60 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{5/2}} \, dx=\frac {{\left (5 \, c^{6} x^{6} + 15 \, b c^{5} x^{5} + 2 \, b^{6} - 18 \, a b^{4} c + 45 \, a^{2} b^{2} c^{2} - 15 \, a^{3} c^{3} + 3 \, {\left (4 \, b^{2} c^{4} + 9 \, a c^{5}\right )} x^{4} - {\left (b^{3} c^{3} - 54 \, a b c^{4}\right )} x^{3} + 3 \, {\left (b^{4} c^{2} - 9 \, a b^{2} c^{3} + 45 \, a^{2} c^{4}\right )} x^{2} + 3 \, {\left (2 \, b^{5} c - 18 \, a b^{3} c^{2} + 45 \, a^{2} b c^{3}\right )} x\right )} \sqrt {2 \, c d x + b d}}{45 \, {\left (4 \, c^{6} d^{3} x^{2} + 4 \, b c^{5} d^{3} x + b^{2} c^{4} d^{3}\right )}} \]

[In]

integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(5/2),x, algorithm="fricas")

[Out]

1/45*(5*c^6*x^6 + 15*b*c^5*x^5 + 2*b^6 - 18*a*b^4*c + 45*a^2*b^2*c^2 - 15*a^3*c^3 + 3*(4*b^2*c^4 + 9*a*c^5)*x^
4 - (b^3*c^3 - 54*a*b*c^4)*x^3 + 3*(b^4*c^2 - 9*a*b^2*c^3 + 45*a^2*c^4)*x^2 + 3*(2*b^5*c - 18*a*b^3*c^2 + 45*a
^2*b*c^3)*x)*sqrt(2*c*d*x + b*d)/(4*c^6*d^3*x^2 + 4*b*c^5*d^3*x + b^2*c^4*d^3)

Sympy [A] (verification not implemented)

Time = 2.01 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.87 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{5/2}} \, dx=\begin {cases} \frac {- \frac {\left (4 a c - b^{2}\right )^{3}}{192 c^{3} \left (b d + 2 c d x\right )^{\frac {3}{2}}} + \frac {\sqrt {b d + 2 c d x} \left (48 a^{2} c^{2} - 24 a b^{2} c + 3 b^{4}\right )}{64 c^{3} d^{2}} + \frac {\left (12 a c - 3 b^{2}\right ) \left (b d + 2 c d x\right )^{\frac {5}{2}}}{320 c^{3} d^{4}} + \frac {\left (b d + 2 c d x\right )^{\frac {9}{2}}}{576 c^{3} d^{6}}}{c d} & \text {for}\: c d \neq 0 \\\frac {a^{3} x + \frac {3 a^{2} b x^{2}}{2} + \frac {b c^{2} x^{6}}{2} + \frac {c^{3} x^{7}}{7} + \frac {x^{5} \cdot \left (3 a c^{2} + 3 b^{2} c\right )}{5} + \frac {x^{4} \cdot \left (6 a b c + b^{3}\right )}{4} + \frac {x^{3} \cdot \left (3 a^{2} c + 3 a b^{2}\right )}{3}}{\left (b d\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate((c*x**2+b*x+a)**3/(2*c*d*x+b*d)**(5/2),x)

[Out]

Piecewise(((-(4*a*c - b**2)**3/(192*c**3*(b*d + 2*c*d*x)**(3/2)) + sqrt(b*d + 2*c*d*x)*(48*a**2*c**2 - 24*a*b*
*2*c + 3*b**4)/(64*c**3*d**2) + (12*a*c - 3*b**2)*(b*d + 2*c*d*x)**(5/2)/(320*c**3*d**4) + (b*d + 2*c*d*x)**(9
/2)/(576*c**3*d**6))/(c*d), Ne(c*d, 0)), ((a**3*x + 3*a**2*b*x**2/2 + b*c**2*x**6/2 + c**3*x**7/7 + x**5*(3*a*
c**2 + 3*b**2*c)/5 + x**4*(6*a*b*c + b**3)/4 + x**3*(3*a**2*c + 3*a*b**2)/3)/(b*d)**(5/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{5/2}} \, dx=\frac {\frac {15 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )}}{{\left (2 \, c d x + b d\right )}^{\frac {3}{2}} c^{3}} - \frac {27 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} {\left (b^{2} - 4 \, a c\right )} d^{2} - 135 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {2 \, c d x + b d} d^{4} - 5 \, {\left (2 \, c d x + b d\right )}^{\frac {9}{2}}}{c^{3} d^{6}}}{2880 \, c d} \]

[In]

integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(5/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.55 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{5/2}} \, dx=\frac {b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}{192 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} c^{4} d} + \frac {135 \, \sqrt {2 \, c d x + b d} b^{4} c^{32} d^{60} - 1080 \, \sqrt {2 \, c d x + b d} a b^{2} c^{33} d^{60} + 2160 \, \sqrt {2 \, c d x + b d} a^{2} c^{34} d^{60} - 27 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} b^{2} c^{32} d^{58} + 108 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} a c^{33} d^{58} + 5 \, {\left (2 \, c d x + b d\right )}^{\frac {9}{2}} c^{32} d^{56}}{2880 \, c^{36} d^{63}} \]

[In]

integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(5/2),x, algorithm="giac")

[Out]

1/192*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)/((2*c*d*x + b*d)^(3/2)*c^4*d) + 1/2880*(135*sqrt(2*c*d*
x + b*d)*b^4*c^32*d^60 - 1080*sqrt(2*c*d*x + b*d)*a*b^2*c^33*d^60 + 2160*sqrt(2*c*d*x + b*d)*a^2*c^34*d^60 - 2
7*(2*c*d*x + b*d)^(5/2)*b^2*c^32*d^58 + 108*(2*c*d*x + b*d)^(5/2)*a*c^33*d^58 + 5*(2*c*d*x + b*d)^(9/2)*c^32*d
^56)/(c^36*d^63)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{5/2}} \, dx=\frac {{\left (b\,d+2\,c\,d\,x\right )}^{9/2}}{576\,c^4\,d^7}+\frac {3\,{\left (b\,d+2\,c\,d\,x\right )}^{5/2}\,\left (4\,a\,c-b^2\right )}{320\,c^4\,d^5}+\frac {-\frac {64\,a^3\,c^3}{3}+16\,a^2\,b^2\,c^2-4\,a\,b^4\,c+\frac {b^6}{3}}{64\,c^4\,d\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}}+\frac {3\,\sqrt {b\,d+2\,c\,d\,x}\,{\left (4\,a\,c-b^2\right )}^2}{64\,c^4\,d^3} \]

[In]

int((a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^(5/2),x)

[Out]

(b*d + 2*c*d*x)^(9/2)/(576*c^4*d^7) + (3*(b*d + 2*c*d*x)^(5/2)*(4*a*c - b^2))/(320*c^4*d^5) + (b^6/3 - (64*a^3
*c^3)/3 + 16*a^2*b^2*c^2 - 4*a*b^4*c)/(64*c^4*d*(b*d + 2*c*d*x)^(3/2)) + (3*(b*d + 2*c*d*x)^(1/2)*(4*a*c - b^2
)^2)/(64*c^4*d^3)