\(\int (b d+2 c d x)^2 (a+b x+c x^2)^{3/2} \, dx\) [1205]

Optimal result

Integrand size = 26, antiderivative size = 165 \[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {\left (b^2-4 a c\right )^2 d^2 (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^2}-\frac {\left (b^2-4 a c\right ) d^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{64 c^2}+\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{12 c}+\frac {\left (b^2-4 a c\right )^3 d^2 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{5/2}} \]

[Out]

1/12*d^2*(2*c*x+b)^3*(c*x^2+b*x+a)^(3/2)/c+1/256*(-4*a*c+b^2)^3*d^2*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a
)^(1/2))/c^(5/2)+1/128*(-4*a*c+b^2)^2*d^2*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c^2-1/64*(-4*a*c+b^2)*d^2*(2*c*x+b)^3*
(c*x^2+b*x+a)^(1/2)/c^2

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {699, 706, 635, 212} \[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {d^2 \left (b^2-4 a c\right )^3 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{5/2}}+\frac {d^2 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^2}-\frac {d^2 \left (b^2-4 a c\right ) (b+2 c x)^3 \sqrt {a+b x+c x^2}}{64 c^2}+\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{12 c} \]

[In]

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

[Out]

((b^2 - 4*a*c)^2*d^2*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(128*c^2) - ((b^2 - 4*a*c)*d^2*(b + 2*c*x)^3*Sqrt[a +
b*x + c*x^2])/(64*c^2) + (d^2*(b + 2*c*x)^3*(a + b*x + c*x^2)^(3/2))/(12*c) + ((b^2 - 4*a*c)^3*d^2*ArcTanh[(b
+ 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(256*c^(5/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 699

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

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 {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{12 c}-\frac {\left (b^2-4 a c\right ) \int (b d+2 c d x)^2 \sqrt {a+b x+c x^2} \, dx}{8 c} \\ & = -\frac {\left (b^2-4 a c\right ) d^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{64 c^2}+\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{12 c}+\frac {\left (b^2-4 a c\right )^2 \int \frac {(b d+2 c d x)^2}{\sqrt {a+b x+c x^2}} \, dx}{128 c^2} \\ & = \frac {\left (b^2-4 a c\right )^2 d^2 (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^2}-\frac {\left (b^2-4 a c\right ) d^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{64 c^2}+\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{12 c}+\frac {\left (\left (b^2-4 a c\right )^3 d^2\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{256 c^2} \\ & = \frac {\left (b^2-4 a c\right )^2 d^2 (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^2}-\frac {\left (b^2-4 a c\right ) d^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{64 c^2}+\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{12 c}+\frac {\left (\left (b^2-4 a c\right )^3 d^2\right ) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{128 c^2} \\ & = \frac {\left (b^2-4 a c\right )^2 d^2 (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^2}-\frac {\left (b^2-4 a c\right ) d^2 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{64 c^2}+\frac {d^2 (b+2 c x)^3 \left (a+b x+c x^2\right )^{3/2}}{12 c}+\frac {\left (b^2-4 a c\right )^3 d^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.97 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.93 \[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {d^2 \left (\sqrt {c} (b+2 c x) \sqrt {a+x (b+c x)} \left (-3 b^4+8 b^3 c x+32 b c^2 x \left (7 a+8 c x^2\right )+8 b^2 c \left (4 a+17 c x^2\right )+16 c^2 \left (3 a^2+14 a c x^2+8 c^2 x^4\right )\right )+3 \left (b^2-4 a c\right )^3 \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )\right )}{384 c^{5/2}} \]

[In]

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

[Out]

(d^2*(Sqrt[c]*(b + 2*c*x)*Sqrt[a + x*(b + c*x)]*(-3*b^4 + 8*b^3*c*x + 32*b*c^2*x*(7*a + 8*c*x^2) + 8*b^2*c*(4*
a + 17*c*x^2) + 16*c^2*(3*a^2 + 14*a*c*x^2 + 8*c^2*x^4)) + 3*(b^2 - 4*a*c)^3*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a] + S
qrt[a + x*(b + c*x)])]))/(384*c^(5/2))

Maple [A] (verified)

Time = 2.67 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.17

method	result	size
risch	\(\frac {\left (256 c^{5} x^{5}+640 b \,x^{4} c^{4}+448 a \,c^{4} x^{3}+528 b^{2} c^{3} x^{3}+672 a b \,c^{3} x^{2}+152 x^{2} b^{3} c^{2}+96 a^{2} c^{3} x +288 a \,c^{2} b^{2} x +2 c x \,b^{4}+48 a^{2} b \,c^{2}+32 a \,b^{3} c -3 b^{5}\right ) \sqrt {c \,x^{2}+b x +a}\, d^{2}}{384 c^{2}}-\frac {\left (64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) d^{2}}{256 c^{\frac {5}{2}}}\)	\(193\)
default	\(d^{2} \left (b^{2} \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )+4 c^{2} \left (\frac {x \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}{6 c}-\frac {7 b \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\right )}{12 c}-\frac {a \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{6 c}\right )+4 b c \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}{5 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 c}\right )\right )\)	\(510\)

[In]

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

[Out]

1/384/c^2*(256*c^5*x^5+640*b*c^4*x^4+448*a*c^4*x^3+528*b^2*c^3*x^3+672*a*b*c^3*x^2+152*b^3*c^2*x^2+96*a^2*c^3*
x+288*a*b^2*c^2*x+2*b^4*c*x+48*a^2*b*c^2+32*a*b^3*c-3*b^5)*(c*x^2+b*x+a)^(1/2)*d^2-1/256*(64*a^3*c^3-48*a^2*b^
2*c^2+12*a*b^4*c-b^6)/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d^2

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 473, normalized size of antiderivative = 2.87 \[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right )^{3/2} \, dx=\left [-\frac {3 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt {c} d^{2} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (256 \, c^{6} d^{2} x^{5} + 640 \, b c^{5} d^{2} x^{4} + 16 \, {\left (33 \, b^{2} c^{4} + 28 \, a c^{5}\right )} d^{2} x^{3} + 8 \, {\left (19 \, b^{3} c^{3} + 84 \, a b c^{4}\right )} d^{2} x^{2} + 2 \, {\left (b^{4} c^{2} + 144 \, a b^{2} c^{3} + 48 \, a^{2} c^{4}\right )} d^{2} x - {\left (3 \, b^{5} c - 32 \, a b^{3} c^{2} - 48 \, a^{2} b c^{3}\right )} d^{2}\right )} \sqrt {c x^{2} + b x + a}}{1536 \, c^{3}}, -\frac {3 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt {-c} d^{2} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (256 \, c^{6} d^{2} x^{5} + 640 \, b c^{5} d^{2} x^{4} + 16 \, {\left (33 \, b^{2} c^{4} + 28 \, a c^{5}\right )} d^{2} x^{3} + 8 \, {\left (19 \, b^{3} c^{3} + 84 \, a b c^{4}\right )} d^{2} x^{2} + 2 \, {\left (b^{4} c^{2} + 144 \, a b^{2} c^{3} + 48 \, a^{2} c^{4}\right )} d^{2} x - {\left (3 \, b^{5} c - 32 \, a b^{3} c^{2} - 48 \, a^{2} b c^{3}\right )} d^{2}\right )} \sqrt {c x^{2} + b x + a}}{768 \, c^{3}}\right ] \]

[In]

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

[Out]

[-1/1536*(3*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*sqrt(c)*d^2*log(-8*c^2*x^2 - 8*b*c*x - b^2 + 4*sq
rt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 4*(256*c^6*d^2*x^5 + 640*b*c^5*d^2*x^4 + 16*(33*b^2*c^4 + 2
8*a*c^5)*d^2*x^3 + 8*(19*b^3*c^3 + 84*a*b*c^4)*d^2*x^2 + 2*(b^4*c^2 + 144*a*b^2*c^3 + 48*a^2*c^4)*d^2*x - (3*b
^5*c - 32*a*b^3*c^2 - 48*a^2*b*c^3)*d^2)*sqrt(c*x^2 + b*x + a))/c^3, -1/768*(3*(b^6 - 12*a*b^4*c + 48*a^2*b^2*
c^2 - 64*a^3*c^3)*sqrt(-c)*d^2*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c))
- 2*(256*c^6*d^2*x^5 + 640*b*c^5*d^2*x^4 + 16*(33*b^2*c^4 + 28*a*c^5)*d^2*x^3 + 8*(19*b^3*c^3 + 84*a*b*c^4)*d^
2*x^2 + 2*(b^4*c^2 + 144*a*b^2*c^3 + 48*a^2*c^4)*d^2*x - (3*b^5*c - 32*a*b^3*c^2 - 48*a^2*b*c^3)*d^2)*sqrt(c*x
^2 + b*x + a))/c^3]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1042 vs. \(2 (153) = 306\).

Time = 0.74 (sec) , antiderivative size = 1042, normalized size of antiderivative = 6.32 \[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right )^{3/2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((sqrt(a + b*x + c*x**2)*(5*b*c**2*d**2*x**4/3 + 2*c**3*d**2*x**5/3 + x**3*(14*a*c**3*d**2/3 + 11*b**
2*c**2*d**2/2)/(4*c) + x**2*(28*a*b*c**2*d**2/3 + 6*b**3*c*d**2 - 7*b*(14*a*c**3*d**2/3 + 11*b**2*c**2*d**2/2)
/(8*c))/(3*c) + x*(4*a**2*c**2*d**2 + 10*a*b**2*c*d**2 - 3*a*(14*a*c**3*d**2/3 + 11*b**2*c**2*d**2/2)/(4*c) +
b**4*d**2 - 5*b*(28*a*b*c**2*d**2/3 + 6*b**3*c*d**2 - 7*b*(14*a*c**3*d**2/3 + 11*b**2*c**2*d**2/2)/(8*c))/(6*c
))/(2*c) + (4*a**2*b*c*d**2 + 2*a*b**3*d**2 - 2*a*(28*a*b*c**2*d**2/3 + 6*b**3*c*d**2 - 7*b*(14*a*c**3*d**2/3
+ 11*b**2*c**2*d**2/2)/(8*c))/(3*c) - 3*b*(4*a**2*c**2*d**2 + 10*a*b**2*c*d**2 - 3*a*(14*a*c**3*d**2/3 + 11*b*
*2*c**2*d**2/2)/(4*c) + b**4*d**2 - 5*b*(28*a*b*c**2*d**2/3 + 6*b**3*c*d**2 - 7*b*(14*a*c**3*d**2/3 + 11*b**2*
c**2*d**2/2)/(8*c))/(6*c))/(4*c))/c) + (a**2*b**2*d**2 - a*(4*a**2*c**2*d**2 + 10*a*b**2*c*d**2 - 3*a*(14*a*c*
*3*d**2/3 + 11*b**2*c**2*d**2/2)/(4*c) + b**4*d**2 - 5*b*(28*a*b*c**2*d**2/3 + 6*b**3*c*d**2 - 7*b*(14*a*c**3*
d**2/3 + 11*b**2*c**2*d**2/2)/(8*c))/(6*c))/(2*c) - b*(4*a**2*b*c*d**2 + 2*a*b**3*d**2 - 2*a*(28*a*b*c**2*d**2
/3 + 6*b**3*c*d**2 - 7*b*(14*a*c**3*d**2/3 + 11*b**2*c**2*d**2/2)/(8*c))/(3*c) - 3*b*(4*a**2*c**2*d**2 + 10*a*
b**2*c*d**2 - 3*a*(14*a*c**3*d**2/3 + 11*b**2*c**2*d**2/2)/(4*c) + b**4*d**2 - 5*b*(28*a*b*c**2*d**2/3 + 6*b**
3*c*d**2 - 7*b*(14*a*c**3*d**2/3 + 11*b**2*c**2*d**2/2)/(8*c))/(6*c))/(4*c))/(2*c))*Piecewise((log(b + 2*sqrt(
c)*sqrt(a + b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(a - b**2/(4*c), 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/
(2*c) + x)**2), True)), Ne(c, 0)), (2*(4*c**2*d**2*(a + b*x)**(9/2)/(9*b**2) + (a + b*x)**(7/2)*(-8*a*c**2*d**
2 + 4*b**2*c*d**2)/(7*b**2) + (a + b*x)**(5/2)*(4*a**2*c**2*d**2 - 4*a*b**2*c*d**2 + b**4*d**2)/(5*b**2))/b, N
e(b, 0)), (4*a**(3/2)*c**2*d**2*x**3/3, True))

Maxima [F(-2)]

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

[In]

integrate((2*c*d*x+b*d)^2*(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 = 0.37 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.56 \[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right )^{3/2} \, dx=\frac {1}{384} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (2 \, c^{3} d^{2} x + 5 \, b c^{2} d^{2}\right )} x + \frac {33 \, b^{2} c^{6} d^{2} + 28 \, a c^{7} d^{2}}{c^{5}}\right )} x + \frac {19 \, b^{3} c^{5} d^{2} + 84 \, a b c^{6} d^{2}}{c^{5}}\right )} x + \frac {b^{4} c^{4} d^{2} + 144 \, a b^{2} c^{5} d^{2} + 48 \, a^{2} c^{6} d^{2}}{c^{5}}\right )} x - \frac {3 \, b^{5} c^{3} d^{2} - 32 \, a b^{3} c^{4} d^{2} - 48 \, a^{2} b c^{5} d^{2}}{c^{5}}\right )} - \frac {{\left (b^{6} d^{2} - 12 \, a b^{4} c d^{2} + 48 \, a^{2} b^{2} c^{2} d^{2} - 64 \, a^{3} c^{3} d^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{256 \, c^{\frac {5}{2}}} \]

[In]

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

[Out]

1/384*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*(2*c^3*d^2*x + 5*b*c^2*d^2)*x + (33*b^2*c^6*d^2 + 28*a*c^7*d^2)/c^5)*x
+ (19*b^3*c^5*d^2 + 84*a*b*c^6*d^2)/c^5)*x + (b^4*c^4*d^2 + 144*a*b^2*c^5*d^2 + 48*a^2*c^6*d^2)/c^5)*x - (3*b
^5*c^3*d^2 - 32*a*b^3*c^4*d^2 - 48*a^2*b*c^5*d^2)/c^5) - 1/256*(b^6*d^2 - 12*a*b^4*c*d^2 + 48*a^2*b^2*c^2*d^2
- 64*a^3*c^3*d^2)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(5/2)

Mupad [F(-1)]

Timed out. \[ \int (b d+2 c d x)^2 \left (a+b x+c x^2\right )^{3/2} \, dx=\int {\left (b\,d+2\,c\,d\,x\right )}^2\,{\left (c\,x^2+b\,x+a\right )}^{3/2} \,d x \]

[In]

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

[Out]

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