\(\int (b d+2 c d x)^{3/2} (a+b x+c x^2) \, dx\) [1262]

Optimal result

Integrand size = 24, antiderivative size = 55 \[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right ) \, dx=-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}{20 c^2 d}+\frac {(b d+2 c d x)^{9/2}}{36 c^2 d^3} \]

[Out]

-1/20*(-4*a*c+b^2)*(2*c*d*x+b*d)^(5/2)/c^2/d+1/36*(2*c*d*x+b*d)^(9/2)/c^2/d^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {697} \[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right ) \, dx=\frac {(b d+2 c d x)^{9/2}}{36 c^2 d^3}-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}{20 c^2 d} \]

[In]

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

[Out]

-1/20*((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(5/2))/(c^2*d) + (b*d + 2*c*d*x)^(9/2)/(36*c^2*d^3)

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 ) (b d+2 c d x)^{3/2}}{4 c}+\frac {(b d+2 c d x)^{7/2}}{4 c d^2}\right ) \, dx \\ & = -\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}{20 c^2 d}+\frac {(b d+2 c d x)^{9/2}}{36 c^2 d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.82 \[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right ) \, dx=\frac {(b+2 c x) (d (b+2 c x))^{3/2} \left (-9 b^2+36 a c+5 (b+2 c x)^2\right )}{180 c^2} \]

[In]

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

[Out]

((b + 2*c*x)*(d*(b + 2*c*x))^(3/2)*(-9*b^2 + 36*a*c + 5*(b + 2*c*x)^2))/(180*c^2)

Maple [A] (verified)

Time = 2.40 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84

method	result	size
gosper	\(\frac {\left (2 c x +b \right ) \left (5 c^{2} x^{2}+5 b c x +9 a c -b^{2}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}{45 c^{2}}\)	\(46\)
pseudoelliptic	\(\frac {d \left (2 c x +b \right )^{2} \sqrt {d \left (2 c x +b \right )}\, \left (5 c^{2} x^{2}+5 b c x +9 a c -b^{2}\right )}{45 c^{2}}\)	\(48\)
derivativedivides	\(\frac {\frac {\left (2 c d x +b d \right )^{\frac {9}{2}}}{9}+\frac {\left (4 a c \,d^{2}-b^{2} d^{2}\right ) \left (2 c d x +b d \right )^{\frac {5}{2}}}{5}}{4 c^{2} d^{3}}\)	\(52\)
default	\(\frac {\frac {\left (2 c d x +b d \right )^{\frac {9}{2}}}{9}+\frac {\left (4 a c \,d^{2}-b^{2} d^{2}\right ) \left (2 c d x +b d \right )^{\frac {5}{2}}}{5}}{4 c^{2} d^{3}}\)	\(52\)
trager	\(\frac {d \left (20 c^{4} x^{4}+40 b \,c^{3} x^{3}+36 x^{2} c^{3} a +21 b^{2} c^{2} x^{2}+36 a b \,c^{2} x +b^{3} c x +9 a \,b^{2} c -b^{4}\right ) \sqrt {2 c d x +b d}}{45 c^{2}}\)	\(82\)

[In]

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

[Out]

1/45*(2*c*x+b)*(5*c^2*x^2+5*b*c*x+9*a*c-b^2)*(2*c*d*x+b*d)^(3/2)/c^2

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.58 \[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right ) \, dx=\frac {{\left (20 \, c^{4} d x^{4} + 40 \, b c^{3} d x^{3} + 3 \, {\left (7 \, b^{2} c^{2} + 12 \, a c^{3}\right )} d x^{2} + {\left (b^{3} c + 36 \, a b c^{2}\right )} d x - {\left (b^{4} - 9 \, a b^{2} c\right )} d\right )} \sqrt {2 \, c d x + b d}}{45 \, c^{2}} \]

[In]

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

[Out]

1/45*(20*c^4*d*x^4 + 40*b*c^3*d*x^3 + 3*(7*b^2*c^2 + 12*a*c^3)*d*x^2 + (b^3*c + 36*a*b*c^2)*d*x - (b^4 - 9*a*b
^2*c)*d)*sqrt(2*c*d*x + b*d)/c^2

Sympy [A] (verification not implemented)

Time = 0.75 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.33 \[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right ) \, dx=\begin {cases} \frac {\frac {\left (4 a c - b^{2}\right ) \left (b d + 2 c d x\right )^{\frac {5}{2}}}{20 c} + \frac {\left (b d + 2 c d x\right )^{\frac {9}{2}}}{36 c d^{2}}}{c d} & \text {for}\: c d \neq 0 \\\left (b d\right )^{\frac {3}{2}} \left (a x + \frac {b x^{2}}{2} + \frac {c x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((((4*a*c - b**2)*(b*d + 2*c*d*x)**(5/2)/(20*c) + (b*d + 2*c*d*x)**(9/2)/(36*c*d**2))/(c*d), Ne(c*d,
0)), ((b*d)**(3/2)*(a*x + b*x**2/2 + c*x**3/3), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84 \[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right ) \, dx=-\frac {9 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} {\left (b^{2} - 4 \, a c\right )} d^{2} - 5 \, {\left (2 \, c d x + b d\right )}^{\frac {9}{2}}}{180 \, c^{2} d^{3}} \]

[In]

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

[Out]

-1/180*(9*(2*c*d*x + b*d)^(5/2)*(b^2 - 4*a*c)*d^2 - 5*(2*c*d*x + b*d)^(9/2))/(c^2*d^3)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 376 vs. \(2 (47) = 94\).

Time = 0.27 (sec) , antiderivative size = 376, normalized size of antiderivative = 6.84 \[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right ) \, dx=\frac {1260 \, \sqrt {2 \, c d x + b d} a b^{2} d - 840 \, {\left (3 \, \sqrt {2 \, c d x + b d} b d - {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}\right )} a b - \frac {210 \, {\left (3 \, \sqrt {2 \, c d x + b d} b d - {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}\right )} b^{3}}{c} + \frac {84 \, {\left (15 \, \sqrt {2 \, c d x + b d} b^{2} d^{2} - 10 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b d + 3 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}\right )} a}{d} + \frac {105 \, {\left (15 \, \sqrt {2 \, c d x + b d} b^{2} d^{2} - 10 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b d + 3 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}\right )} b^{2}}{c d} - \frac {36 \, {\left (35 \, \sqrt {2 \, c d x + b d} b^{3} d^{3} - 35 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} d^{2} + 21 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} b d - 5 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}}\right )} b}{c d^{2}} + \frac {315 \, \sqrt {2 \, c d x + b d} b^{4} d^{4} - 420 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{3} d^{3} + 378 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} b^{2} d^{2} - 180 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} b d + 35 \, {\left (2 \, c d x + b d\right )}^{\frac {9}{2}}}{c d^{3}}}{1260 \, c} \]

[In]

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

[Out]

1/1260*(1260*sqrt(2*c*d*x + b*d)*a*b^2*d - 840*(3*sqrt(2*c*d*x + b*d)*b*d - (2*c*d*x + b*d)^(3/2))*a*b - 210*(
3*sqrt(2*c*d*x + b*d)*b*d - (2*c*d*x + b*d)^(3/2))*b^3/c + 84*(15*sqrt(2*c*d*x + b*d)*b^2*d^2 - 10*(2*c*d*x +
b*d)^(3/2)*b*d + 3*(2*c*d*x + b*d)^(5/2))*a/d + 105*(15*sqrt(2*c*d*x + b*d)*b^2*d^2 - 10*(2*c*d*x + b*d)^(3/2)
*b*d + 3*(2*c*d*x + b*d)^(5/2))*b^2/(c*d) - 36*(35*sqrt(2*c*d*x + b*d)*b^3*d^3 - 35*(2*c*d*x + b*d)^(3/2)*b^2*
d^2 + 21*(2*c*d*x + b*d)^(5/2)*b*d - 5*(2*c*d*x + b*d)^(7/2))*b/(c*d^2) + (315*sqrt(2*c*d*x + b*d)*b^4*d^4 - 4
20*(2*c*d*x + b*d)^(3/2)*b^3*d^3 + 378*(2*c*d*x + b*d)^(5/2)*b^2*d^2 - 180*(2*c*d*x + b*d)^(7/2)*b*d + 35*(2*c
*d*x + b*d)^(9/2))/(c*d^3))/c

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71 \[ \int (b d+2 c d x)^{3/2} \left (a+b x+c x^2\right ) \, dx=\frac {{\left (b\,d+2\,c\,d\,x\right )}^{5/2}\,\left (36\,a\,c+5\,{\left (b+2\,c\,x\right )}^2-9\,b^2\right )}{180\,c^2\,d} \]

[In]

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

[Out]

((b*d + 2*c*d*x)^(5/2)*(36*a*c + 5*(b + 2*c*x)^2 - 9*b^2))/(180*c^2*d)