\(\int \frac {(a+b x+c x^2)^3}{(b d+2 c d x)^{3/2}} \, dx\) [1279]

Optimal result

Integrand size = 26, antiderivative size = 121 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{3/2}} \, dx=\frac {\left (b^2-4 a c\right )^3}{64 c^4 d \sqrt {b d+2 c d x}}+\frac {\left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}}{64 c^4 d^3}-\frac {3 \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}{448 c^4 d^5}+\frac {(b d+2 c d x)^{11/2}}{704 c^4 d^7} \]

[Out]

1/64*(-4*a*c+b^2)^2*(2*c*d*x+b*d)^(3/2)/c^4/d^3-3/448*(-4*a*c+b^2)*(2*c*d*x+b*d)^(7/2)/c^4/d^5+1/704*(2*c*d*x+
b*d)^(11/2)/c^4/d^7+1/64*(-4*a*c+b^2)^3/c^4/d/(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} \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{3/2}} \, dx=-\frac {3 \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}{448 c^4 d^5}+\frac {\left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}}{64 c^4 d^3}+\frac {\left (b^2-4 a c\right )^3}{64 c^4 d \sqrt {b d+2 c d x}}+\frac {(b d+2 c d x)^{11/2}}{704 c^4 d^7} \]

[In]

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

[Out]

(b^2 - 4*a*c)^3/(64*c^4*d*Sqrt[b*d + 2*c*d*x]) + ((b^2 - 4*a*c)^2*(b*d + 2*c*d*x)^(3/2))/(64*c^4*d^3) - (3*(b^
2 - 4*a*c)*(b*d + 2*c*d*x)^(7/2))/(448*c^4*d^5) + (b*d + 2*c*d*x)^(11/2)/(704*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)^{3/2}}+\frac {3 \left (-b^2+4 a c\right )^2 \sqrt {b d+2 c d x}}{64 c^3 d^2}+\frac {3 \left (-b^2+4 a c\right ) (b d+2 c d x)^{5/2}}{64 c^3 d^4}+\frac {(b d+2 c d x)^{9/2}}{64 c^3 d^6}\right ) \, dx \\ & = \frac {\left (b^2-4 a c\right )^3}{64 c^4 d \sqrt {b d+2 c d x}}+\frac {\left (b^2-4 a c\right )^2 (b d+2 c d x)^{3/2}}{64 c^4 d^3}-\frac {3 \left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}{448 c^4 d^5}+\frac {(b d+2 c d x)^{11/2}}{704 c^4 d^7} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{3/2}} \, dx=\frac {77 b^6-924 a b^4 c+3696 a^2 b^2 c^2-4928 a^3 c^3+77 b^4 (b+2 c x)^2-616 a b^2 c (b+2 c x)^2+1232 a^2 c^2 (b+2 c x)^2-33 b^2 (b+2 c x)^4+132 a c (b+2 c x)^4+7 (b+2 c x)^6}{4928 c^4 d \sqrt {d (b+2 c x)}} \]

[In]

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

[Out]

(77*b^6 - 924*a*b^4*c + 3696*a^2*b^2*c^2 - 4928*a^3*c^3 + 77*b^4*(b + 2*c*x)^2 - 616*a*b^2*c*(b + 2*c*x)^2 + 1
232*a^2*c^2*(b + 2*c*x)^2 - 33*b^2*(b + 2*c*x)^4 + 132*a*c*(b + 2*c*x)^4 + 7*(b + 2*c*x)^6)/(4928*c^4*d*Sqrt[d
*(b + 2*c*x)])

Maple [A] (verified)

Time = 2.30 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.40

method	result	size
derivativedivides	\(\frac {16 a^{2} c^{2} d^{4} \left (2 c d x +b d \right )^{\frac {3}{2}}-8 a \,b^{2} c \,d^{4} \left (2 c d x +b d \right )^{\frac {3}{2}}+\frac {12 a c \,d^{2} \left (2 c d x +b d \right )^{\frac {7}{2}}}{7}+b^{4} d^{4} \left (2 c d x +b d \right )^{\frac {3}{2}}-\frac {3 b^{2} d^{2} \left (2 c d x +b d \right )^{\frac {7}{2}}}{7}+\frac {\left (2 c d x +b d \right )^{\frac {11}{2}}}{11}-\frac {d^{6} \left (64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}{\sqrt {2 c d x +b d}}}{64 c^{4} d^{7}}\)	\(169\)
default	\(\frac {16 a^{2} c^{2} d^{4} \left (2 c d x +b d \right )^{\frac {3}{2}}-8 a \,b^{2} c \,d^{4} \left (2 c d x +b d \right )^{\frac {3}{2}}+\frac {12 a c \,d^{2} \left (2 c d x +b d \right )^{\frac {7}{2}}}{7}+b^{4} d^{4} \left (2 c d x +b d \right )^{\frac {3}{2}}-\frac {3 b^{2} d^{2} \left (2 c d x +b d \right )^{\frac {7}{2}}}{7}+\frac {\left (2 c d x +b d \right )^{\frac {11}{2}}}{11}-\frac {d^{6} \left (64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right )}{\sqrt {2 c d x +b d}}}{64 c^{4} d^{7}}\)	\(169\)
pseudoelliptic	\(\frac {7 c^{6} x^{6}+21 b \,c^{5} x^{5}+33 a \,c^{5} x^{4}+18 b^{2} c^{4} x^{4}+66 a b \,c^{4} x^{3}+x^{3} b^{3} c^{3}+77 a^{2} c^{4} x^{2}+11 a \,b^{2} c^{3} x^{2}-x^{2} b^{4} c^{2}+77 a^{2} b \,c^{3} x -22 x a \,b^{3} c^{2}+2 x \,b^{5} c -77 c^{3} a^{3}+77 a^{2} b^{2} c^{2}-22 a \,b^{4} c +2 b^{6}}{77 d \sqrt {d \left (2 c x +b \right )}\, c^{4}}\)	\(169\)
gosper	\(-\frac {\left (2 c x +b \right ) \left (-7 c^{6} x^{6}-21 b \,c^{5} x^{5}-33 a \,c^{5} x^{4}-18 b^{2} c^{4} x^{4}-66 a b \,c^{4} x^{3}-x^{3} b^{3} c^{3}-77 a^{2} c^{4} x^{2}-11 a \,b^{2} c^{3} x^{2}+x^{2} b^{4} c^{2}-77 a^{2} b \,c^{3} x +22 x a \,b^{3} c^{2}-2 x \,b^{5} c +77 c^{3} a^{3}-77 a^{2} b^{2} c^{2}+22 a \,b^{4} c -2 b^{6}\right )}{77 c^{4} \left (2 c d x +b d \right )^{\frac {3}{2}}}\)	\(173\)
trager	\(-\frac {\left (-7 c^{6} x^{6}-21 b \,c^{5} x^{5}-33 a \,c^{5} x^{4}-18 b^{2} c^{4} x^{4}-66 a b \,c^{4} x^{3}-x^{3} b^{3} c^{3}-77 a^{2} c^{4} x^{2}-11 a \,b^{2} c^{3} x^{2}+x^{2} b^{4} c^{2}-77 a^{2} b \,c^{3} x +22 x a \,b^{3} c^{2}-2 x \,b^{5} c +77 c^{3} a^{3}-77 a^{2} b^{2} c^{2}+22 a \,b^{4} c -2 b^{6}\right ) \sqrt {2 c d x +b d}}{77 d^{2} c^{4} \left (2 c x +b \right )}\)	\(178\)

[In]

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

[Out]

1/64/c^4/d^7*(16*a^2*c^2*d^4*(2*c*d*x+b*d)^(3/2)-8*a*b^2*c*d^4*(2*c*d*x+b*d)^(3/2)+12/7*a*c*d^2*(2*c*d*x+b*d)^
(7/2)+b^4*d^4*(2*c*d*x+b*d)^(3/2)-3/7*b^2*d^2*(2*c*d*x+b*d)^(7/2)+1/11*(2*c*d*x+b*d)^(11/2)-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)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{3/2}} \, dx=\frac {{\left (7 \, c^{6} x^{6} + 21 \, b c^{5} x^{5} + 2 \, b^{6} - 22 \, a b^{4} c + 77 \, a^{2} b^{2} c^{2} - 77 \, a^{3} c^{3} + 3 \, {\left (6 \, b^{2} c^{4} + 11 \, a c^{5}\right )} x^{4} + {\left (b^{3} c^{3} + 66 \, a b c^{4}\right )} x^{3} - {\left (b^{4} c^{2} - 11 \, a b^{2} c^{3} - 77 \, a^{2} c^{4}\right )} x^{2} + {\left (2 \, b^{5} c - 22 \, a b^{3} c^{2} + 77 \, a^{2} b c^{3}\right )} x\right )} \sqrt {2 \, c d x + b d}}{77 \, {\left (2 \, c^{5} d^{2} x + b c^{4} d^{2}\right )}} \]

[In]

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

[Out]

1/77*(7*c^6*x^6 + 21*b*c^5*x^5 + 2*b^6 - 22*a*b^4*c + 77*a^2*b^2*c^2 - 77*a^3*c^3 + 3*(6*b^2*c^4 + 11*a*c^5)*x
^4 + (b^3*c^3 + 66*a*b*c^4)*x^3 - (b^4*c^2 - 11*a*b^2*c^3 - 77*a^2*c^4)*x^2 + (2*b^5*c - 22*a*b^3*c^2 + 77*a^2
*b*c^3)*x)*sqrt(2*c*d*x + b*d)/(2*c^5*d^2*x + b*c^4*d^2)

Sympy [A] (verification not implemented)

Time = 1.98 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.87 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{3/2}} \, dx=\begin {cases} \frac {- \frac {\left (4 a c - b^{2}\right )^{3}}{64 c^{3} \sqrt {b d + 2 c d x}} + \frac {\left (b d + 2 c d x\right )^{\frac {3}{2}} \cdot \left (48 a^{2} c^{2} - 24 a b^{2} c + 3 b^{4}\right )}{192 c^{3} d^{2}} + \frac {\left (12 a c - 3 b^{2}\right ) \left (b d + 2 c d x\right )^{\frac {7}{2}}}{448 c^{3} d^{4}} + \frac {\left (b d + 2 c d x\right )^{\frac {11}{2}}}{704 c^{3} d^{6}}}{c d} & \text {for}\: c d \neq 0 \\\frac {a^{3} x + \frac {3 a^{2} b x^{2}}{2} + \frac {b c^{2} x^{6}}{2} + \frac {c^{3} x^{7}}{7} + \frac {x^{5} \cdot \left (3 a c^{2} + 3 b^{2} c\right )}{5} + \frac {x^{4} \cdot \left (6 a b c + b^{3}\right )}{4} + \frac {x^{3} \cdot \left (3 a^{2} c + 3 a b^{2}\right )}{3}}{\left (b d\right )^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise(((-(4*a*c - b**2)**3/(64*c**3*sqrt(b*d + 2*c*d*x)) + (b*d + 2*c*d*x)**(3/2)*(48*a**2*c**2 - 24*a*b**
2*c + 3*b**4)/(192*c**3*d**2) + (12*a*c - 3*b**2)*(b*d + 2*c*d*x)**(7/2)/(448*c**3*d**4) + (b*d + 2*c*d*x)**(1
1/2)/(704*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)**(3/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{3/2}} \, dx=\frac {\frac {77 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )}}{\sqrt {2 \, c d x + b d} c^{3}} - \frac {33 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} {\left (b^{2} - 4 \, a c\right )} d^{2} - 77 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} d^{4} - 7 \, {\left (2 \, c d x + b d\right )}^{\frac {11}{2}}}{c^{3} d^{6}}}{4928 \, c d} \]

[In]

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

[Out]

1/4928*(77*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)/(sqrt(2*c*d*x + b*d)*c^3) - (33*(2*c*d*x + b*d)^(7
/2)*(b^2 - 4*a*c)*d^2 - 77*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*(2*c*d*x + b*d)^(3/2)*d^4 - 7*(2*c*d*x + b*d)^(11/2)
)/(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 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{3/2}} \, dx=\frac {b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}{64 \, \sqrt {2 \, c d x + b d} c^{4} d} + \frac {77 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{4} c^{40} d^{74} - 616 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a b^{2} c^{41} d^{74} + 1232 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a^{2} c^{42} d^{74} - 33 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} b^{2} c^{40} d^{72} + 132 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} a c^{41} d^{72} + 7 \, {\left (2 \, c d x + b d\right )}^{\frac {11}{2}} c^{40} d^{70}}{4928 \, c^{44} d^{77}} \]

[In]

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

[Out]

1/64*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)/(sqrt(2*c*d*x + b*d)*c^4*d) + 1/4928*(77*(2*c*d*x + b*d)
^(3/2)*b^4*c^40*d^74 - 616*(2*c*d*x + b*d)^(3/2)*a*b^2*c^41*d^74 + 1232*(2*c*d*x + b*d)^(3/2)*a^2*c^42*d^74 -
33*(2*c*d*x + b*d)^(7/2)*b^2*c^40*d^72 + 132*(2*c*d*x + b*d)^(7/2)*a*c^41*d^72 + 7*(2*c*d*x + b*d)^(11/2)*c^40
*d^70)/(c^44*d^77)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^{3/2}} \, dx=\frac {{\left (b\,d+2\,c\,d\,x\right )}^{11/2}}{704\,c^4\,d^7}+\frac {3\,{\left (b\,d+2\,c\,d\,x\right )}^{7/2}\,\left (4\,a\,c-b^2\right )}{448\,c^4\,d^5}+\frac {{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,{\left (4\,a\,c-b^2\right )}^2}{64\,c^4\,d^3}+\frac {-64\,a^3\,c^3+48\,a^2\,b^2\,c^2-12\,a\,b^4\,c+b^6}{64\,c^4\,d\,\sqrt {b\,d+2\,c\,d\,x}} \]

[In]

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

[Out]

(b*d + 2*c*d*x)^(11/2)/(704*c^4*d^7) + (3*(b*d + 2*c*d*x)^(7/2)*(4*a*c - b^2))/(448*c^4*d^5) + ((b*d + 2*c*d*x
)^(3/2)*(4*a*c - b^2)^2)/(64*c^4*d^3) + (b^6 - 64*a^3*c^3 + 48*a^2*b^2*c^2 - 12*a*b^4*c)/(64*c^4*d*(b*d + 2*c*
d*x)^(1/2))