\(\int \sqrt {b d+2 c d x} (a+b x+c x^2)^3 \, dx\) [1277]

Optimal result

Integrand size = 26, antiderivative size = 121 \[ \int \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^3 \, dx=-\frac {\left (b^2-4 a c\right )^3 (b d+2 c d x)^{3/2}}{192 c^4 d}+\frac {3 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{7/2}}{448 c^4 d^3}-\frac {3 \left (b^2-4 a c\right ) (b d+2 c d x)^{11/2}}{704 c^4 d^5}+\frac {(b d+2 c d x)^{15/2}}{960 c^4 d^7} \]

[Out]

-1/192*(-4*a*c+b^2)^3*(2*c*d*x+b*d)^(3/2)/c^4/d+3/448*(-4*a*c+b^2)^2*(2*c*d*x+b*d)^(7/2)/c^4/d^3-3/704*(-4*a*c
+b^2)*(2*c*d*x+b*d)^(11/2)/c^4/d^5+1/960*(2*c*d*x+b*d)^(15/2)/c^4/d^7

Rubi [A] (verified)

Time = 0.04 (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} \[ \int \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^3 \, dx=-\frac {3 \left (b^2-4 a c\right ) (b d+2 c d x)^{11/2}}{704 c^4 d^5}+\frac {3 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{7/2}}{448 c^4 d^3}-\frac {\left (b^2-4 a c\right )^3 (b d+2 c d x)^{3/2}}{192 c^4 d}+\frac {(b d+2 c d x)^{15/2}}{960 c^4 d^7} \]

[In]

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

[Out]

-1/192*((b^2 - 4*a*c)^3*(b*d + 2*c*d*x)^(3/2))/(c^4*d) + (3*(b^2 - 4*a*c)^2*(b*d + 2*c*d*x)^(7/2))/(448*c^4*d^
3) - (3*(b^2 - 4*a*c)*(b*d + 2*c*d*x)^(11/2))/(704*c^4*d^5) + (b*d + 2*c*d*x)^(15/2)/(960*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 \sqrt {b d+2 c d x}}{64 c^3}+\frac {3 \left (-b^2+4 a c\right )^2 (b d+2 c d x)^{5/2}}{64 c^3 d^2}+\frac {3 \left (-b^2+4 a c\right ) (b d+2 c d x)^{9/2}}{64 c^3 d^4}+\frac {(b d+2 c d x)^{13/2}}{64 c^3 d^6}\right ) \, dx \\ & = -\frac {\left (b^2-4 a c\right )^3 (b d+2 c d x)^{3/2}}{192 c^4 d}+\frac {3 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{7/2}}{448 c^4 d^3}-\frac {3 \left (b^2-4 a c\right ) (b d+2 c d x)^{11/2}}{704 c^4 d^5}+\frac {(b d+2 c d x)^{15/2}}{960 c^4 d^7} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.10 \[ \int \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^3 \, dx=\frac {(d (b+2 c x))^{3/2} \left (-385 b^6+4620 a b^4 c-18480 a^2 b^2 c^2+24640 a^3 c^3+495 b^4 (b+2 c x)^2-3960 a b^2 c (b+2 c x)^2+7920 a^2 c^2 (b+2 c x)^2-315 b^2 (b+2 c x)^4+1260 a c (b+2 c x)^4+77 (b+2 c x)^6\right )}{73920 c^4 d} \]

[In]

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

[Out]

((d*(b + 2*c*x))^(3/2)*(-385*b^6 + 4620*a*b^4*c - 18480*a^2*b^2*c^2 + 24640*a^3*c^3 + 495*b^4*(b + 2*c*x)^2 -
3960*a*b^2*c*(b + 2*c*x)^2 + 7920*a^2*c^2*(b + 2*c*x)^2 - 315*b^2*(b + 2*c*x)^4 + 1260*a*c*(b + 2*c*x)^4 + 77*
(b + 2*c*x)^6))/(73920*c^4*d)

Maple [A] (verified)

Time = 2.66 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.22

method	result	size
derivativedivides	\(\frac {\frac {\left (2 c d x +b d \right )^{\frac {15}{2}}}{15}+\frac {\left (12 a c \,d^{2}-3 b^{2} d^{2}\right ) \left (2 c d x +b d \right )^{\frac {11}{2}}}{11}+\frac {\left (\left (4 a c \,d^{2}-b^{2} d^{2}\right ) \left (8 a c \,d^{2}-2 b^{2} d^{2}\right )+\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{2}\right ) \left (2 c d x +b d \right )^{\frac {7}{2}}}{7}+\frac {\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{3} \left (2 c d x +b d \right )^{\frac {3}{2}}}{3}}{64 d^{7} c^{4}}\)	\(148\)
default	\(\frac {\frac {\left (2 c d x +b d \right )^{\frac {15}{2}}}{15}+\frac {\left (12 a c \,d^{2}-3 b^{2} d^{2}\right ) \left (2 c d x +b d \right )^{\frac {11}{2}}}{11}+\frac {\left (\left (4 a c \,d^{2}-b^{2} d^{2}\right ) \left (8 a c \,d^{2}-2 b^{2} d^{2}\right )+\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{2}\right ) \left (2 c d x +b d \right )^{\frac {7}{2}}}{7}+\frac {\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{3} \left (2 c d x +b d \right )^{\frac {3}{2}}}{3}}{64 d^{7} c^{4}}\)	\(148\)
pseudoelliptic	\(\frac {\left (2 c x +b \right ) \sqrt {d \left (2 c x +b \right )}\, \left (77 c^{6} x^{6}+231 b \,c^{5} x^{5}+315 a \,c^{5} x^{4}+210 b^{2} c^{4} x^{4}+630 a b \,c^{4} x^{3}+35 x^{3} b^{3} c^{3}+495 a^{2} c^{4} x^{2}+225 a \,b^{2} c^{3} x^{2}-15 x^{2} b^{4} c^{2}+495 a^{2} b \,c^{3} x -90 x a \,b^{3} c^{2}+6 x \,b^{5} c +385 c^{3} a^{3}-165 a^{2} b^{2} c^{2}+30 a \,b^{4} c -2 b^{6}\right )}{1155 c^{4}}\)	\(173\)
gosper	\(\frac {\left (2 c x +b \right ) \left (77 c^{6} x^{6}+231 b \,c^{5} x^{5}+315 a \,c^{5} x^{4}+210 b^{2} c^{4} x^{4}+630 a b \,c^{4} x^{3}+35 x^{3} b^{3} c^{3}+495 a^{2} c^{4} x^{2}+225 a \,b^{2} c^{3} x^{2}-15 x^{2} b^{4} c^{2}+495 a^{2} b \,c^{3} x -90 x a \,b^{3} c^{2}+6 x \,b^{5} c +385 c^{3} a^{3}-165 a^{2} b^{2} c^{2}+30 a \,b^{4} c -2 b^{6}\right ) \sqrt {2 c d x +b d}}{1155 c^{4}}\)	\(174\)
trager	\(\frac {\left (154 c^{7} x^{7}+539 b \,c^{6} x^{6}+630 a \,c^{6} x^{5}+651 b^{2} c^{5} x^{5}+1575 a b \,c^{5} x^{4}+280 b^{3} c^{4} x^{4}+990 a^{2} c^{5} x^{3}+1080 a \,b^{2} c^{4} x^{3}+5 b^{4} c^{3} x^{3}+1485 a^{2} b \,c^{4} x^{2}+45 a \,b^{3} c^{3} x^{2}-3 b^{5} c^{2} x^{2}+770 a^{3} c^{4} x +165 a^{2} b^{2} c^{3} x -30 c^{2} a \,b^{4} x +2 b^{6} c x +385 a^{3} c^{3} b -165 a^{2} c^{2} b^{3}+30 a \,b^{5} c -2 b^{7}\right ) \sqrt {2 c d x +b d}}{1155 c^{4}}\)	\(215\)

[In]

int((2*c*d*x+b*d)^(1/2)*(c*x^2+b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

1/64/d^7/c^4*(1/15*(2*c*d*x+b*d)^(15/2)+1/11*(12*a*c*d^2-3*b^2*d^2)*(2*c*d*x+b*d)^(11/2)+1/7*((4*a*c*d^2-b^2*d
^2)*(8*a*c*d^2-2*b^2*d^2)+(4*a*c*d^2-b^2*d^2)^2)*(2*c*d*x+b*d)^(7/2)+1/3*(4*a*c*d^2-b^2*d^2)^3*(2*c*d*x+b*d)^(
3/2))

Fricas [A] (verification not implemented)

none

Time = 0.63 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.69 \[ \int \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^3 \, dx=\frac {{\left (154 \, c^{7} x^{7} + 539 \, b c^{6} x^{6} - 2 \, b^{7} + 30 \, a b^{5} c - 165 \, a^{2} b^{3} c^{2} + 385 \, a^{3} b c^{3} + 21 \, {\left (31 \, b^{2} c^{5} + 30 \, a c^{6}\right )} x^{5} + 35 \, {\left (8 \, b^{3} c^{4} + 45 \, a b c^{5}\right )} x^{4} + 5 \, {\left (b^{4} c^{3} + 216 \, a b^{2} c^{4} + 198 \, a^{2} c^{5}\right )} x^{3} - 3 \, {\left (b^{5} c^{2} - 15 \, a b^{3} c^{3} - 495 \, a^{2} b c^{4}\right )} x^{2} + {\left (2 \, b^{6} c - 30 \, a b^{4} c^{2} + 165 \, a^{2} b^{2} c^{3} + 770 \, a^{3} c^{4}\right )} x\right )} \sqrt {2 \, c d x + b d}}{1155 \, c^{4}} \]

[In]

integrate((2*c*d*x+b*d)^(1/2)*(c*x^2+b*x+a)^3,x, algorithm="fricas")

[Out]

1/1155*(154*c^7*x^7 + 539*b*c^6*x^6 - 2*b^7 + 30*a*b^5*c - 165*a^2*b^3*c^2 + 385*a^3*b*c^3 + 21*(31*b^2*c^5 +
30*a*c^6)*x^5 + 35*(8*b^3*c^4 + 45*a*b*c^5)*x^4 + 5*(b^4*c^3 + 216*a*b^2*c^4 + 198*a^2*c^5)*x^3 - 3*(b^5*c^2 -
15*a*b^3*c^3 - 495*a^2*b*c^4)*x^2 + (2*b^6*c - 30*a*b^4*c^2 + 165*a^2*b^2*c^3 + 770*a^3*c^4)*x)*sqrt(2*c*d*x
+ b*d)/c^4

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (119) = 238\).

Time = 1.07 (sec) , antiderivative size = 248, normalized size of antiderivative = 2.05 \[ \int \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^3 \, dx=\begin {cases} \frac {\frac {\left (b d + 2 c d x\right )^{\frac {3}{2}} \cdot \left (64 a^{3} c^{3} - 48 a^{2} b^{2} c^{2} + 12 a b^{4} c - b^{6}\right )}{192 c^{3}} + \frac {\left (b d + 2 c d x\right )^{\frac {7}{2}} \cdot \left (48 a^{2} c^{2} - 24 a b^{2} c + 3 b^{4}\right )}{448 c^{3} d^{2}} + \frac {\left (12 a c - 3 b^{2}\right ) \left (b d + 2 c d x\right )^{\frac {11}{2}}}{704 c^{3} d^{4}} + \frac {\left (b d + 2 c d x\right )^{\frac {15}{2}}}{960 c^{3} d^{6}}}{c d} & \text {for}\: c d \neq 0 \\\sqrt {b d} \left (a^{3} x + \frac {3 a^{2} b x^{2}}{2} + \frac {b c^{2} x^{6}}{2} + \frac {c^{3} x^{7}}{7} + \frac {x^{5} \cdot \left (3 a c^{2} + 3 b^{2} c\right )}{5} + \frac {x^{4} \cdot \left (6 a b c + b^{3}\right )}{4} + \frac {x^{3} \cdot \left (3 a^{2} c + 3 a b^{2}\right )}{3}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate((2*c*d*x+b*d)**(1/2)*(c*x**2+b*x+a)**3,x)

[Out]

Piecewise((((b*d + 2*c*d*x)**(3/2)*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)/(192*c**3) + (b*d +
2*c*d*x)**(7/2)*(48*a**2*c**2 - 24*a*b**2*c + 3*b**4)/(448*c**3*d**2) + (12*a*c - 3*b**2)*(b*d + 2*c*d*x)**(1
1/2)/(704*c**3*d**4) + (b*d + 2*c*d*x)**(15/2)/(960*c**3*d**6))/(c*d), Ne(c*d, 0)), (sqrt(b*d)*(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), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.05 \[ \int \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^3 \, dx=-\frac {315 \, {\left (2 \, c d x + b d\right )}^{\frac {11}{2}} {\left (b^{2} - 4 \, a c\right )} d^{2} - 495 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} d^{4} + 385 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} d^{6} - 77 \, {\left (2 \, c d x + b d\right )}^{\frac {15}{2}}}{73920 \, c^{4} d^{7}} \]

[In]

integrate((2*c*d*x+b*d)^(1/2)*(c*x^2+b*x+a)^3,x, algorithm="maxima")

[Out]

-1/73920*(315*(2*c*d*x + b*d)^(11/2)*(b^2 - 4*a*c)*d^2 - 495*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*(2*c*d*x + b*d)^(7
/2)*d^4 + 385*(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)*d^6 - 77*(2*c*d*x + b*d)^
(15/2))/(c^4*d^7)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1165 vs. \(2 (105) = 210\).

Time = 0.28 (sec) , antiderivative size = 1165, normalized size of antiderivative = 9.63 \[ \int \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^3 \, dx=\text {Too large to display} \]

[In]

integrate((2*c*d*x+b*d)^(1/2)*(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

1/2882880*(2882880*sqrt(2*c*d*x + b*d)*a^3*b - 960960*(3*sqrt(2*c*d*x + b*d)*b*d - (2*c*d*x + b*d)^(3/2))*a^3/
d - 1441440*(3*sqrt(2*c*d*x + b*d)*b*d - (2*c*d*x + b*d)^(3/2))*a^2*b^2/(c*d) + 144144*(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^3/(c^2*d^2) + 432432*(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^2*b/(c*d^2) - 10296*(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
^4/(c^3*d^3) - 123552*(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*b^2/(c^2*d^3) - 61776*(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^2/(c*d^3) + 2860*(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^3/(c^3*d^4) + 8580*(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))*a*b/(c^2*d^4) - 1170*(693*sqrt(2*c*d*x + b*d)*b^5*d^5 - 1155*(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 + 385*(2*c*d*x + b*d)^(9/2)*b*d - 63*(2*c*d*x + b*
d)^(11/2))*b^2/(c^3*d^5) - 780*(693*sqrt(2*c*d*x + b*d)*b^5*d^5 - 1155*(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 + 385*(2*c*d*x + b*d)^(9/2)*b*d - 63*(2*c*d*x
+ b*d)^(11/2))*a/(c^2*d^5) + 105*(3003*sqrt(2*c*d*x + b*d)*b^6*d^6 - 6006*(2*c*d*x + b*d)^(3/2)*b^5*d^5 + 9009
*(2*c*d*x + b*d)^(5/2)*b^4*d^4 - 8580*(2*c*d*x + b*d)^(7/2)*b^3*d^3 + 5005*(2*c*d*x + b*d)^(9/2)*b^2*d^2 - 163
8*(2*c*d*x + b*d)^(11/2)*b*d + 231*(2*c*d*x + b*d)^(13/2))*b/(c^3*d^6) - 7*(6435*sqrt(2*c*d*x + b*d)*b^7*d^7 -
15015*(2*c*d*x + b*d)^(3/2)*b^6*d^6 + 27027*(2*c*d*x + b*d)^(5/2)*b^5*d^5 - 32175*(2*c*d*x + b*d)^(7/2)*b^4*d
^4 + 25025*(2*c*d*x + b*d)^(9/2)*b^3*d^3 - 12285*(2*c*d*x + b*d)^(11/2)*b^2*d^2 + 3465*(2*c*d*x + b*d)^(13/2)*
b*d - 429*(2*c*d*x + b*d)^(15/2))/(c^3*d^7))/c

Mupad [B] (verification not implemented)

Time = 9.18 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.92 \[ \int \sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^3 \, dx=\frac {{\left (b\,d+2\,c\,d\,x\right )}^{15/2}}{960\,c^4\,d^7}+\frac {3\,{\left (b\,d+2\,c\,d\,x\right )}^{11/2}\,\left (4\,a\,c-b^2\right )}{704\,c^4\,d^5}+\frac {{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,{\left (4\,a\,c-b^2\right )}^3}{192\,c^4\,d}+\frac {3\,{\left (b\,d+2\,c\,d\,x\right )}^{7/2}\,{\left (4\,a\,c-b^2\right )}^2}{448\,c^4\,d^3} \]

[In]

int((b*d + 2*c*d*x)^(1/2)*(a + b*x + c*x^2)^3,x)

[Out]

(b*d + 2*c*d*x)^(15/2)/(960*c^4*d^7) + (3*(b*d + 2*c*d*x)^(11/2)*(4*a*c - b^2))/(704*c^4*d^5) + ((b*d + 2*c*d*
x)^(3/2)*(4*a*c - b^2)^3)/(192*c^4*d) + (3*(b*d + 2*c*d*x)^(7/2)*(4*a*c - b^2)^2)/(448*c^4*d^3)