\(\int \frac {(a+b x+c x^2)^3}{(b d+2 c d x)^{7/2}} \, dx\) [1281]

Optimal result

Integrand size = 26, antiderivative size = 121 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{7/2}} \, dx=\frac {\left (b^2-4 a c\right )^3}{320 c^4 d (b d+2 c d x)^{5/2}}-\frac {3 \left (b^2-4 a c\right )^2}{64 c^4 d^3 \sqrt {b d+2 c d x}}-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}{64 c^4 d^5}+\frac {(b d+2 c d x)^{7/2}}{448 c^4 d^7} \]

[Out]

1/320*(-4*a*c+b^2)^3/c^4/d/(2*c*d*x+b*d)^(5/2)-1/64*(-4*a*c+b^2)*(2*c*d*x+b*d)^(3/2)/c^4/d^5+1/448*(2*c*d*x+b*
d)^(7/2)/c^4/d^7-3/64*(-4*a*c+b^2)^2/c^4/d^3/(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} \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{7/2}} \, dx=-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}{64 c^4 d^5}-\frac {3 \left (b^2-4 a c\right )^2}{64 c^4 d^3 \sqrt {b d+2 c d x}}+\frac {\left (b^2-4 a c\right )^3}{320 c^4 d (b d+2 c d x)^{5/2}}+\frac {(b d+2 c d x)^{7/2}}{448 c^4 d^7} \]

[In]

Int[(a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^(7/2),x]

[Out]

(b^2 - 4*a*c)^3/(320*c^4*d*(b*d + 2*c*d*x)^(5/2)) - (3*(b^2 - 4*a*c)^2)/(64*c^4*d^3*Sqrt[b*d + 2*c*d*x]) - ((b
^2 - 4*a*c)*(b*d + 2*c*d*x)^(3/2))/(64*c^4*d^5) + (b*d + 2*c*d*x)^(7/2)/(448*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)^{7/2}}+\frac {3 \left (-b^2+4 a c\right )^2}{64 c^3 d^2 (b d+2 c d x)^{3/2}}+\frac {3 \left (-b^2+4 a c\right ) \sqrt {b d+2 c d x}}{64 c^3 d^4}+\frac {(b d+2 c d x)^{5/2}}{64 c^3 d^6}\right ) \, dx \\ & = \frac {\left (b^2-4 a c\right )^3}{320 c^4 d (b d+2 c d x)^{5/2}}-\frac {3 \left (b^2-4 a c\right )^2}{64 c^4 d^3 \sqrt {b d+2 c d x}}-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}}{64 c^4 d^5}+\frac {(b d+2 c d x)^{7/2}}{448 c^4 d^7} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{7/2}} \, dx=\frac {7 b^6-84 a b^4 c+336 a^2 b^2 c^2-448 a^3 c^3-105 b^4 (b+2 c x)^2+840 a b^2 c (b+2 c x)^2-1680 a^2 c^2 (b+2 c x)^2-35 b^2 (b+2 c x)^4+140 a c (b+2 c x)^4+5 (b+2 c x)^6}{2240 c^4 d (d (b+2 c x))^{5/2}} \]

[In]

Integrate[(a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^(7/2),x]

[Out]

(7*b^6 - 84*a*b^4*c + 336*a^2*b^2*c^2 - 448*a^3*c^3 - 105*b^4*(b + 2*c*x)^2 + 840*a*b^2*c*(b + 2*c*x)^2 - 1680
*a^2*c^2*(b + 2*c*x)^2 - 35*b^2*(b + 2*c*x)^4 + 140*a*c*(b + 2*c*x)^4 + 5*(b + 2*c*x)^6)/(2240*c^4*d*(d*(b + 2
*c*x))^(5/2))

Maple [A] (verified)

Time = 2.51 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.18

method	result	size
derivativedivides	\(\frac {4 a c \,d^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}-b^{2} d^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}+\frac {\left (2 c d x +b d \right )^{\frac {7}{2}}}{7}-\frac {3 d^{4} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}{\sqrt {2 c d x +b d}}-\frac {d^{6} \left (64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}{5 \left (2 c d x +b d \right )^{\frac {5}{2}}}}{64 c^{4} d^{7}}\)	\(143\)
default	\(\frac {4 a c \,d^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}-b^{2} d^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}+\frac {\left (2 c d x +b d \right )^{\frac {7}{2}}}{7}-\frac {3 d^{4} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}{\sqrt {2 c d x +b d}}-\frac {d^{6} \left (64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}{5 \left (2 c d x +b d \right )^{\frac {5}{2}}}}{64 c^{4} d^{7}}\)	\(143\)
pseudoelliptic	\(\frac {5 c^{6} x^{6}+\left (15 b \,x^{5}+35 a \,x^{4}\right ) c^{5}+\left (10 b^{2} x^{4}+70 a b \,x^{3}-105 a^{2} x^{2}\right ) c^{4}+\left (-5 b^{3} x^{3}+105 a \,b^{2} x^{2}-105 a^{2} b x -7 a^{3}\right ) c^{3}+\left (-15 b^{4} x^{2}+70 a \,b^{3} x -21 a^{2} b^{2}\right ) c^{2}+\left (-10 b^{5} x +14 a \,b^{4}\right ) c -2 b^{6}}{35 \sqrt {d \left (2 c x +b \right )}\, d^{3} \left (2 c x +b \right )^{2} c^{4}}\)	\(163\)
gosper	\(-\frac {\left (2 c x +b \right ) \left (-5 c^{6} x^{6}-15 b \,c^{5} x^{5}-35 a \,c^{5} x^{4}-10 b^{2} c^{4} x^{4}-70 a b \,c^{4} x^{3}+5 x^{3} b^{3} c^{3}+105 a^{2} c^{4} x^{2}-105 a \,b^{2} c^{3} x^{2}+15 x^{2} b^{4} c^{2}+105 a^{2} b \,c^{3} x -70 x a \,b^{3} c^{2}+10 x \,b^{5} c +7 c^{3} a^{3}+21 a^{2} b^{2} c^{2}-14 a \,b^{4} c +2 b^{6}\right )}{35 c^{4} \left (2 c d x +b d \right )^{\frac {7}{2}}}\)	\(174\)
trager	\(-\frac {\left (-5 c^{6} x^{6}-15 b \,c^{5} x^{5}-35 a \,c^{5} x^{4}-10 b^{2} c^{4} x^{4}-70 a b \,c^{4} x^{3}+5 x^{3} b^{3} c^{3}+105 a^{2} c^{4} x^{2}-105 a \,b^{2} c^{3} x^{2}+15 x^{2} b^{4} c^{2}+105 a^{2} b \,c^{3} x -70 x a \,b^{3} c^{2}+10 x \,b^{5} c +7 c^{3} a^{3}+21 a^{2} b^{2} c^{2}-14 a \,b^{4} c +2 b^{6}\right ) \sqrt {2 c d x +b d}}{35 d^{4} c^{4} \left (2 c x +b \right )^{3}}\)	\(179\)

[In]

int((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.72 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{7/2}} \, dx=\frac {{\left (5 \, c^{6} x^{6} + 15 \, b c^{5} x^{5} - 2 \, b^{6} + 14 \, a b^{4} c - 21 \, a^{2} b^{2} c^{2} - 7 \, a^{3} c^{3} + 5 \, {\left (2 \, b^{2} c^{4} + 7 \, a c^{5}\right )} x^{4} - 5 \, {\left (b^{3} c^{3} - 14 \, a b c^{4}\right )} x^{3} - 15 \, {\left (b^{4} c^{2} - 7 \, a b^{2} c^{3} + 7 \, a^{2} c^{4}\right )} x^{2} - 5 \, {\left (2 \, b^{5} c - 14 \, a b^{3} c^{2} + 21 \, a^{2} b c^{3}\right )} x\right )} \sqrt {2 \, c d x + b d}}{35 \, {\left (8 \, c^{7} d^{4} x^{3} + 12 \, b c^{6} d^{4} x^{2} + 6 \, b^{2} c^{5} d^{4} x + b^{3} c^{4} d^{4}\right )}} \]

[In]

integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(7/2),x, algorithm="fricas")

[Out]

1/35*(5*c^6*x^6 + 15*b*c^5*x^5 - 2*b^6 + 14*a*b^4*c - 21*a^2*b^2*c^2 - 7*a^3*c^3 + 5*(2*b^2*c^4 + 7*a*c^5)*x^4
- 5*(b^3*c^3 - 14*a*b*c^4)*x^3 - 15*(b^4*c^2 - 7*a*b^2*c^3 + 7*a^2*c^4)*x^2 - 5*(2*b^5*c - 14*a*b^3*c^2 + 21*
a^2*b*c^3)*x)*sqrt(2*c*d*x + b*d)/(8*c^7*d^4*x^3 + 12*b*c^6*d^4*x^2 + 6*b^2*c^5*d^4*x + b^3*c^4*d^4)

Sympy [A] (verification not implemented)

Time = 2.11 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.79 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{7/2}} \, dx=\begin {cases} \frac {- \frac {\left (4 a c - b^{2}\right )^{3}}{320 c^{3} \left (b d + 2 c d x\right )^{\frac {5}{2}}} - \frac {3 \left (4 a c - b^{2}\right )^{2}}{64 c^{3} d^{2} \sqrt {b d + 2 c d x}} + \frac {\left (12 a c - 3 b^{2}\right ) \left (b d + 2 c d x\right )^{\frac {3}{2}}}{192 c^{3} d^{4}} + \frac {\left (b d + 2 c d x\right )^{\frac {7}{2}}}{448 c^{3} d^{6}}}{c d} & \text {for}\: c d \neq 0 \\\frac {a^{3} x + \frac {3 a^{2} b x^{2}}{2} + \frac {b c^{2} x^{6}}{2} + \frac {c^{3} x^{7}}{7} + \frac {x^{5} \cdot \left (3 a c^{2} + 3 b^{2} c\right )}{5} + \frac {x^{4} \cdot \left (6 a b c + b^{3}\right )}{4} + \frac {x^{3} \cdot \left (3 a^{2} c + 3 a b^{2}\right )}{3}}{\left (b d\right )^{\frac {7}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate((c*x**2+b*x+a)**3/(2*c*d*x+b*d)**(7/2),x)

[Out]

Piecewise(((-(4*a*c - b**2)**3/(320*c**3*(b*d + 2*c*d*x)**(5/2)) - 3*(4*a*c - b**2)**2/(64*c**3*d**2*sqrt(b*d
+ 2*c*d*x)) + (12*a*c - 3*b**2)*(b*d + 2*c*d*x)**(3/2)/(192*c**3*d**4) + (b*d + 2*c*d*x)**(7/2)/(448*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)**(7/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.17 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{7/2}} \, dx=-\frac {\frac {7 \, {\left (15 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} {\left (2 \, c d x + b d\right )}^{2} - {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{2}\right )}}{{\left (2 \, c d x + b d\right )}^{\frac {5}{2}} c^{3} d^{2}} + \frac {5 \, {\left (7 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} {\left (b^{2} - 4 \, a c\right )} d^{2} - {\left (2 \, c d x + b d\right )}^{\frac {7}{2}}\right )}}{c^{3} d^{6}}}{2240 \, c d} \]

[In]

integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(7/2),x, algorithm="maxima")

[Out]

-1/2240*(7*(15*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*(2*c*d*x + b*d)^2 - (b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*
c^3)*d^2)/((2*c*d*x + b*d)^(5/2)*c^3*d^2) + 5*(7*(2*c*d*x + b*d)^(3/2)*(b^2 - 4*a*c)*d^2 - (2*c*d*x + b*d)^(7/
2))/(c^3*d^6))/(c*d)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.54 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{7/2}} \, dx=\frac {b^{6} d^{2} - 12 \, a b^{4} c d^{2} + 48 \, a^{2} b^{2} c^{2} d^{2} - 64 \, a^{3} c^{3} d^{2} - 15 \, {\left (2 \, c d x + b d\right )}^{2} b^{4} + 120 \, {\left (2 \, c d x + b d\right )}^{2} a b^{2} c - 240 \, {\left (2 \, c d x + b d\right )}^{2} a^{2} c^{2}}{320 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} c^{4} d^{3}} - \frac {7 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} c^{24} d^{44} - 28 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a c^{25} d^{44} - {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} c^{24} d^{42}}{448 \, c^{28} d^{49}} \]

[In]

integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^(7/2),x, algorithm="giac")

[Out]

1/320*(b^6*d^2 - 12*a*b^4*c*d^2 + 48*a^2*b^2*c^2*d^2 - 64*a^3*c^3*d^2 - 15*(2*c*d*x + b*d)^2*b^4 + 120*(2*c*d*
x + b*d)^2*a*b^2*c - 240*(2*c*d*x + b*d)^2*a^2*c^2)/((2*c*d*x + b*d)^(5/2)*c^4*d^3) - 1/448*(7*(2*c*d*x + b*d)
^(3/2)*b^2*c^24*d^44 - 28*(2*c*d*x + b*d)^(3/2)*a*c^25*d^44 - (2*c*d*x + b*d)^(7/2)*c^24*d^42)/(c^28*d^49)

Mupad [B] (verification not implemented)

Time = 9.22 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.40 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{7/2}} \, dx=-\frac {7\,a^3\,c^3+21\,a^2\,b^2\,c^2+105\,a^2\,b\,c^3\,x+105\,a^2\,c^4\,x^2-14\,a\,b^4\,c-70\,a\,b^3\,c^2\,x-105\,a\,b^2\,c^3\,x^2-70\,a\,b\,c^4\,x^3-35\,a\,c^5\,x^4+2\,b^6+10\,b^5\,c\,x+15\,b^4\,c^2\,x^2+5\,b^3\,c^3\,x^3-10\,b^2\,c^4\,x^4-15\,b\,c^5\,x^5-5\,c^6\,x^6}{35\,c^4\,d\,{\left (b\,d+2\,c\,d\,x\right )}^{5/2}} \]

[In]

int((a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^(7/2),x)

[Out]

-(2*b^6 + 7*a^3*c^3 - 5*c^6*x^6 - 35*a*c^5*x^4 - 15*b*c^5*x^5 + 21*a^2*b^2*c^2 + 105*a^2*c^4*x^2 + 15*b^4*c^2*
x^2 + 5*b^3*c^3*x^3 - 10*b^2*c^4*x^4 - 14*a*b^4*c + 10*b^5*c*x - 105*a*b^2*c^3*x^2 - 70*a*b^3*c^2*x + 105*a^2*
b*c^3*x - 70*a*b*c^4*x^3)/(35*c^4*d*(b*d + 2*c*d*x)^(5/2))