\(\int (b d+2 c d x)^3 (a+b x+c x^2)^{3/2} \, dx\) [1204]

Optimal result

Integrand size = 26, antiderivative size = 59 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {4}{35} \left (b^2-4 a c\right ) d^3 \left (a+b x+c x^2\right )^{5/2}+\frac {2}{7} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{5/2} \]

[Out]

4/35*(-4*a*c+b^2)*d^3*(c*x^2+b*x+a)^(5/2)+2/7*d^3*(2*c*x+b)^2*(c*x^2+b*x+a)^(5/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {706, 643} \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {4}{35} d^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{5/2}+\frac {2}{7} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{5/2} \]

[In]

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

[Out]

(4*(b^2 - 4*a*c)*d^3*(a + b*x + c*x^2)^(5/2))/35 + (2*d^3*(b + 2*c*x)^2*(a + b*x + c*x^2)^(5/2))/7

Rule 643

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*((a + b*x + c*x^2)^(p +
1)/(b*(p + 1))), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 706

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2*d*(d + e*x)^(m - 1
)*((a + b*x + c*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] + Dist[d^2*(m - 1)*((b^2 - 4*a*c)/(b^2*(m + 2*p + 1))), In
t[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[
2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

Rubi steps \begin{align*} \text {integral}& = \frac {2}{7} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{5/2}+\frac {1}{7} \left (2 \left (b^2-4 a c\right ) d^2\right ) \int (b d+2 c d x) \left (a+b x+c x^2\right )^{3/2} \, dx \\ & = \frac {4}{35} \left (b^2-4 a c\right ) d^3 \left (a+b x+c x^2\right )^{5/2}+\frac {2}{7} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{5/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.84 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.75 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^{3/2} \, dx=-\frac {2}{35} d^3 (a+x (b+c x))^{5/2} \left (-7 b^2-20 b c x+4 c \left (2 a-5 c x^2\right )\right ) \]

[In]

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

[Out]

(-2*d^3*(a + x*(b + c*x))^(5/2)*(-7*b^2 - 20*b*c*x + 4*c*(2*a - 5*c*x^2)))/35

Maple [A] (verified)

Time = 2.32 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.69

method	result	size
gosper	\(-\frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} \left (-20 c^{2} x^{2}-20 b c x +8 a c -7 b^{2}\right ) d^{3}}{35}\)	\(41\)
pseudoelliptic	\(-\frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} \left (-20 c^{2} x^{2}-20 b c x +8 a c -7 b^{2}\right ) d^{3}}{35}\)	\(41\)
trager	\(d^{3} \left (\frac {8}{7} c^{4} x^{6}+\frac {24}{7} b \,c^{3} x^{5}+\frac {64}{35} c^{3} a \,x^{4}+\frac {134}{35} b^{2} c^{2} x^{4}+\frac {128}{35} a b \,c^{2} x^{3}+\frac {68}{35} b^{3} c \,x^{3}+\frac {8}{35} a^{2} c^{2} x^{2}+\frac {92}{35} a \,b^{2} c \,x^{2}+\frac {2}{5} b^{4} x^{2}+\frac {8}{35} a^{2} b c x +\frac {4}{5} a \,b^{3} x -\frac {16}{35} c \,a^{3}+\frac {2}{5} a^{2} b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\)	\(132\)
risch	\(-\frac {2 d^{3} \left (-20 c^{4} x^{6}-60 b \,c^{3} x^{5}-32 c^{3} a \,x^{4}-67 b^{2} c^{2} x^{4}-64 a b \,c^{2} x^{3}-34 b^{3} c \,x^{3}-4 a^{2} c^{2} x^{2}-46 a \,b^{2} c \,x^{2}-7 b^{4} x^{2}-4 a^{2} b c x -14 a \,b^{3} x +8 c \,a^{3}-7 a^{2} b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{35}\)	\(133\)
default	\(\text {Expression too large to display}\)	\(939\)

[In]

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

[Out]

-2/35*(c*x^2+b*x+a)^(5/2)*(-20*c^2*x^2-20*b*c*x+8*a*c-7*b^2)*d^3

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (51) = 102\).

Time = 0.66 (sec) , antiderivative size = 149, normalized size of antiderivative = 2.53 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {2}{35} \, {\left (20 \, c^{4} d^{3} x^{6} + 60 \, b c^{3} d^{3} x^{5} + {\left (67 \, b^{2} c^{2} + 32 \, a c^{3}\right )} d^{3} x^{4} + 2 \, {\left (17 \, b^{3} c + 32 \, a b c^{2}\right )} d^{3} x^{3} + {\left (7 \, b^{4} + 46 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} d^{3} x^{2} + 2 \, {\left (7 \, a b^{3} + 2 \, a^{2} b c\right )} d^{3} x + {\left (7 \, a^{2} b^{2} - 8 \, a^{3} c\right )} d^{3}\right )} \sqrt {c x^{2} + b x + a} \]

[In]

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

[Out]

2/35*(20*c^4*d^3*x^6 + 60*b*c^3*d^3*x^5 + (67*b^2*c^2 + 32*a*c^3)*d^3*x^4 + 2*(17*b^3*c + 32*a*b*c^2)*d^3*x^3
+ (7*b^4 + 46*a*b^2*c + 4*a^2*c^2)*d^3*x^2 + 2*(7*a*b^3 + 2*a^2*b*c)*d^3*x + (7*a^2*b^2 - 8*a^3*c)*d^3)*sqrt(c
*x^2 + b*x + a)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 371 vs. \(2 (58) = 116\).

Time = 0.23 (sec) , antiderivative size = 371, normalized size of antiderivative = 6.29 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^{3/2} \, dx=- \frac {16 a^{3} c d^{3} \sqrt {a + b x + c x^{2}}}{35} + \frac {2 a^{2} b^{2} d^{3} \sqrt {a + b x + c x^{2}}}{5} + \frac {8 a^{2} b c d^{3} x \sqrt {a + b x + c x^{2}}}{35} + \frac {8 a^{2} c^{2} d^{3} x^{2} \sqrt {a + b x + c x^{2}}}{35} + \frac {4 a b^{3} d^{3} x \sqrt {a + b x + c x^{2}}}{5} + \frac {92 a b^{2} c d^{3} x^{2} \sqrt {a + b x + c x^{2}}}{35} + \frac {128 a b c^{2} d^{3} x^{3} \sqrt {a + b x + c x^{2}}}{35} + \frac {64 a c^{3} d^{3} x^{4} \sqrt {a + b x + c x^{2}}}{35} + \frac {2 b^{4} d^{3} x^{2} \sqrt {a + b x + c x^{2}}}{5} + \frac {68 b^{3} c d^{3} x^{3} \sqrt {a + b x + c x^{2}}}{35} + \frac {134 b^{2} c^{2} d^{3} x^{4} \sqrt {a + b x + c x^{2}}}{35} + \frac {24 b c^{3} d^{3} x^{5} \sqrt {a + b x + c x^{2}}}{7} + \frac {8 c^{4} d^{3} x^{6} \sqrt {a + b x + c x^{2}}}{7} \]

[In]

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

[Out]

-16*a**3*c*d**3*sqrt(a + b*x + c*x**2)/35 + 2*a**2*b**2*d**3*sqrt(a + b*x + c*x**2)/5 + 8*a**2*b*c*d**3*x*sqrt
(a + b*x + c*x**2)/35 + 8*a**2*c**2*d**3*x**2*sqrt(a + b*x + c*x**2)/35 + 4*a*b**3*d**3*x*sqrt(a + b*x + c*x**
2)/5 + 92*a*b**2*c*d**3*x**2*sqrt(a + b*x + c*x**2)/35 + 128*a*b*c**2*d**3*x**3*sqrt(a + b*x + c*x**2)/35 + 64
*a*c**3*d**3*x**4*sqrt(a + b*x + c*x**2)/35 + 2*b**4*d**3*x**2*sqrt(a + b*x + c*x**2)/5 + 68*b**3*c*d**3*x**3*
sqrt(a + b*x + c*x**2)/35 + 134*b**2*c**2*d**3*x**4*sqrt(a + b*x + c*x**2)/35 + 24*b*c**3*d**3*x**5*sqrt(a + b
*x + c*x**2)/7 + 8*c**4*d**3*x**6*sqrt(a + b*x + c*x**2)/7

Maxima [F(-2)]

Exception generated. \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^{3/2} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.98 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {2}{5} \, {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} b^{2} d^{3} + \frac {8}{7} \, {\left (c x^{2} + b x + a\right )}^{\frac {7}{2}} c d^{3} - \frac {8}{5} \, {\left (c x^{2} + b x + a\right )}^{\frac {5}{2}} a c d^{3} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.06 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.98 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {8\,c\,d^3\,{\left (c\,x^2+b\,x+a\right )}^{7/2}}{7}+\frac {2\,b^2\,d^3\,{\left (c\,x^2+b\,x+a\right )}^{5/2}}{5}-\frac {8\,a\,c\,d^3\,{\left (c\,x^2+b\,x+a\right )}^{5/2}}{5} \]

[In]

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

[Out]

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