\(\int (b d+2 c d x)^{5/2} (a+b x+c x^2) \, dx\) [1261]

Optimal result

Integrand size = 24, antiderivative size = 55 \[ \int (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right ) \, dx=-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}{28 c^2 d}+\frac {(b d+2 c d x)^{11/2}}{44 c^2 d^3} \]

[Out]

-1/28*(-4*a*c+b^2)*(2*c*d*x+b*d)^(7/2)/c^2/d+1/44*(2*c*d*x+b*d)^(11/2)/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} \[ \int (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right ) \, dx=\frac {(b d+2 c d x)^{11/2}}{44 c^2 d^3}-\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}{28 c^2 d} \]

[In]

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

[Out]

-1/28*((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(7/2))/(c^2*d) + (b*d + 2*c*d*x)^(11/2)/(44*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)^{5/2}}{4 c}+\frac {(b d+2 c d x)^{9/2}}{4 c d^2}\right ) \, dx \\ & = -\frac {\left (b^2-4 a c\right ) (b d+2 c d x)^{7/2}}{28 c^2 d}+\frac {(b d+2 c d x)^{11/2}}{44 c^2 d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.82 \[ \int (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right ) \, dx=\frac {(b+2 c x) (d (b+2 c x))^{5/2} \left (-11 b^2+44 a c+7 (b+2 c x)^2\right )}{308 c^2} \]

[In]

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

[Out]

((b + 2*c*x)*(d*(b + 2*c*x))^(5/2)*(-11*b^2 + 44*a*c + 7*(b + 2*c*x)^2))/(308*c^2)

Maple [A] (verified)

Time = 2.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84

method	result	size
gosper	\(\frac {\left (2 c x +b \right ) \left (7 c^{2} x^{2}+7 b c x +11 a c -b^{2}\right ) \left (2 c d x +b d \right )^{\frac {5}{2}}}{77 c^{2}}\)	\(46\)
pseudoelliptic	\(\frac {d^{2} \left (2 c x +b \right )^{3} \sqrt {d \left (2 c x +b \right )}\, \left (7 c^{2} x^{2}+7 b c x +11 a c -b^{2}\right )}{77 c^{2}}\)	\(50\)
derivativedivides	\(\frac {\frac {\left (2 c d x +b d \right )^{\frac {11}{2}}}{11}+\frac {\left (4 a c \,d^{2}-b^{2} d^{2}\right ) \left (2 c d x +b d \right )^{\frac {7}{2}}}{7}}{4 c^{2} d^{3}}\)	\(52\)
default	\(\frac {\frac {\left (2 c d x +b d \right )^{\frac {11}{2}}}{11}+\frac {\left (4 a c \,d^{2}-b^{2} d^{2}\right ) \left (2 c d x +b d \right )^{\frac {7}{2}}}{7}}{4 c^{2} d^{3}}\)	\(52\)
trager	\(\frac {d^{2} \left (56 c^{5} x^{5}+140 b \,x^{4} c^{4}+88 a \,c^{4} x^{3}+118 b^{2} c^{3} x^{3}+132 a b \,c^{3} x^{2}+37 x^{2} b^{3} c^{2}+66 a \,c^{2} b^{2} x +c x \,b^{4}+11 a \,b^{3} c -b^{5}\right ) \sqrt {2 c d x +b d}}{77 c^{2}}\)	\(107\)

[In]

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

[Out]

1/77*(2*c*x+b)*(7*c^2*x^2+7*b*c*x+11*a*c-b^2)*(2*c*d*x+b*d)^(5/2)/c^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 122 vs. \(2 (47) = 94\).

Time = 0.25 (sec) , antiderivative size = 122, normalized size of antiderivative = 2.22 \[ \int (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right ) \, dx=\frac {{\left (56 \, c^{5} d^{2} x^{5} + 140 \, b c^{4} d^{2} x^{4} + 2 \, {\left (59 \, b^{2} c^{3} + 44 \, a c^{4}\right )} d^{2} x^{3} + {\left (37 \, b^{3} c^{2} + 132 \, a b c^{3}\right )} d^{2} x^{2} + {\left (b^{4} c + 66 \, a b^{2} c^{2}\right )} d^{2} x - {\left (b^{5} - 11 \, a b^{3} c\right )} d^{2}\right )} \sqrt {2 \, c d x + b d}}{77 \, c^{2}} \]

[In]

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

[Out]

1/77*(56*c^5*d^2*x^5 + 140*b*c^4*d^2*x^4 + 2*(59*b^2*c^3 + 44*a*c^4)*d^2*x^3 + (37*b^3*c^2 + 132*a*b*c^3)*d^2*
x^2 + (b^4*c + 66*a*b^2*c^2)*d^2*x - (b^5 - 11*a*b^3*c)*d^2)*sqrt(2*c*d*x + b*d)/c^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 289 vs. \(2 (49) = 98\).

Time = 0.41 (sec) , antiderivative size = 289, normalized size of antiderivative = 5.25 \[ \int (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right ) \, dx=\begin {cases} \frac {a b^{3} d^{2} \sqrt {b d + 2 c d x}}{7 c} + \frac {6 a b^{2} d^{2} x \sqrt {b d + 2 c d x}}{7} + \frac {12 a b c d^{2} x^{2} \sqrt {b d + 2 c d x}}{7} + \frac {8 a c^{2} d^{2} x^{3} \sqrt {b d + 2 c d x}}{7} - \frac {b^{5} d^{2} \sqrt {b d + 2 c d x}}{77 c^{2}} + \frac {b^{4} d^{2} x \sqrt {b d + 2 c d x}}{77 c} + \frac {37 b^{3} d^{2} x^{2} \sqrt {b d + 2 c d x}}{77} + \frac {118 b^{2} c d^{2} x^{3} \sqrt {b d + 2 c d x}}{77} + \frac {20 b c^{2} d^{2} x^{4} \sqrt {b d + 2 c d x}}{11} + \frac {8 c^{3} d^{2} x^{5} \sqrt {b d + 2 c d x}}{11} & \text {for}\: c \neq 0 \\\left (b d\right )^{\frac {5}{2}} \left (a x + \frac {b x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a*b**3*d**2*sqrt(b*d + 2*c*d*x)/(7*c) + 6*a*b**2*d**2*x*sqrt(b*d + 2*c*d*x)/7 + 12*a*b*c*d**2*x**2*
sqrt(b*d + 2*c*d*x)/7 + 8*a*c**2*d**2*x**3*sqrt(b*d + 2*c*d*x)/7 - b**5*d**2*sqrt(b*d + 2*c*d*x)/(77*c**2) + b
**4*d**2*x*sqrt(b*d + 2*c*d*x)/(77*c) + 37*b**3*d**2*x**2*sqrt(b*d + 2*c*d*x)/77 + 118*b**2*c*d**2*x**3*sqrt(b
*d + 2*c*d*x)/77 + 20*b*c**2*d**2*x**4*sqrt(b*d + 2*c*d*x)/11 + 8*c**3*d**2*x**5*sqrt(b*d + 2*c*d*x)/11, Ne(c,
0)), ((b*d)**(5/2)*(a*x + b*x**2/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84 \[ \int (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right ) \, dx=-\frac {11 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} {\left (b^{2} - 4 \, a c\right )} d^{2} - 7 \, {\left (2 \, c d x + b d\right )}^{\frac {11}{2}}}{308 \, c^{2} d^{3}} \]

[In]

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

[Out]

-1/308*(11*(2*c*d*x + b*d)^(7/2)*(b^2 - 4*a*c)*d^2 - 7*(2*c*d*x + b*d)^(11/2))/(c^2*d^3)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 567 vs. \(2 (47) = 94\).

Time = 0.27 (sec) , antiderivative size = 567, normalized size of antiderivative = 10.31 \[ \int (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right ) \, dx=\frac {13860 \, \sqrt {2 \, c d x + b d} a b^{3} d^{2} - 13860 \, {\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^{2} d - \frac {2310 \, {\left (3 \, \sqrt {2 \, c d x + b d} b d - {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}\right )} b^{4} d}{c} + 2772 \, {\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 b + \frac {1617 \, {\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^{3}}{c} - \frac {396 \, {\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 )} a}{d} - \frac {891 \, {\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^{2}}{c d} + \frac {55 \, {\left (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}}\right )} b}{c d^{2}} - \frac {5 \, {\left (693 \, \sqrt {2 \, c d x + b d} b^{5} d^{5} - 1155 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{4} d^{4} + 1386 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} b^{3} d^{3} - 990 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} b^{2} d^{2} + 385 \, {\left (2 \, c d x + b d\right )}^{\frac {9}{2}} b d - 63 \, {\left (2 \, c d x + b d\right )}^{\frac {11}{2}}\right )}}{c d^{3}}}{13860 \, c} \]

[In]

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

[Out]

1/13860*(13860*sqrt(2*c*d*x + b*d)*a*b^3*d^2 - 13860*(3*sqrt(2*c*d*x + b*d)*b*d - (2*c*d*x + b*d)^(3/2))*a*b^2
*d - 2310*(3*sqrt(2*c*d*x + b*d)*b*d - (2*c*d*x + b*d)^(3/2))*b^4*d/c + 2772*(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*b + 1617*(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^3/c - 396*(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))*a/d - 891*(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^2/(c*d)
+ 55*(315*sqrt(2*c*d*x + b*d)*b^4*d^4 - 420*(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))*b/(c*d^2) - 5*(693*sqrt(2*c*d*x + b*d)*b^5*d^5 - 1
155*(2*c*d*x + b*d)^(3/2)*b^4*d^4 + 1386*(2*c*d*x + b*d)^(5/2)*b^3*d^3 - 990*(2*c*d*x + b*d)^(7/2)*b^2*d^2 + 3
85*(2*c*d*x + b*d)^(9/2)*b*d - 63*(2*c*d*x + b*d)^(11/2))/(c*d^3))/c

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71 \[ \int (b d+2 c d x)^{5/2} \left (a+b x+c x^2\right ) \, dx=\frac {{\left (b\,d+2\,c\,d\,x\right )}^{7/2}\,\left (44\,a\,c+7\,{\left (b+2\,c\,x\right )}^2-11\,b^2\right )}{308\,c^2\,d} \]

[In]

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

[Out]

((b*d + 2*c*d*x)^(7/2)*(44*a*c + 7*(b + 2*c*x)^2 - 11*b^2))/(308*c^2*d)