\(\int \frac {(b d+2 c d x)^3}{\sqrt {a+b x+c x^2}} \, dx\) [1235]

Optimal result

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

[Out]

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

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} \[ \int \frac {(b d+2 c d x)^3}{\sqrt {a+b x+c x^2}} \, dx=\frac {4}{3} d^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}+\frac {2}{3} d^3 (b+2 c x)^2 \sqrt {a+b x+c x^2} \]

[In]

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

[Out]

(4*(b^2 - 4*a*c)*d^3*Sqrt[a + b*x + c*x^2])/3 + (2*d^3*(b + 2*c*x)^2*Sqrt[a + b*x + c*x^2])/3

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}{3} d^3 (b+2 c x)^2 \sqrt {a+b x+c x^2}+\frac {1}{3} \left (2 \left (b^2-4 a c\right ) d^2\right ) \int \frac {b d+2 c d x}{\sqrt {a+b x+c x^2}} \, dx \\ & = \frac {4}{3} \left (b^2-4 a c\right ) d^3 \sqrt {a+b x+c x^2}+\frac {2}{3} d^3 (b+2 c x)^2 \sqrt {a+b x+c x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.73 \[ \int \frac {(b d+2 c d x)^3}{\sqrt {a+b x+c x^2}} \, dx=\frac {2}{3} d^3 \sqrt {a+x (b+c x)} \left (3 b^2+4 b c x+4 c \left (-2 a+c x^2\right )\right ) \]

[In]

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

[Out]

(2*d^3*Sqrt[a + x*(b + c*x)]*(3*b^2 + 4*b*c*x + 4*c*(-2*a + c*x^2)))/3

Maple [A] (verified)

Time = 2.31 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.68

method	result	size
trager	\(d^{3} \left (\frac {8}{3} c^{2} x^{2}+\frac {8}{3} b c x -\frac {16}{3} a c +2 b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\)	\(40\)
gosper	\(-\frac {2 d^{3} \left (-4 c^{2} x^{2}-4 b c x +8 a c -3 b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{3}\)	\(41\)
risch	\(-\frac {2 d^{3} \left (-4 c^{2} x^{2}-4 b c x +8 a c -3 b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{3}\)	\(41\)
pseudoelliptic	\(-\frac {2 d^{3} \left (-4 c^{2} x^{2}-4 b c x +8 a c -3 b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{3}\)	\(41\)
default	\(d^{3} \left (\frac {b^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+8 c^{3} \left (\frac {x^{2} \sqrt {c \,x^{2}+b x +a}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{6 c}-\frac {2 a \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{3 c}\right )+6 b^{2} c \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )+12 b \,c^{2} \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )\right )\)	\(392\)

[In]

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

[Out]

d^3*(8/3*c^2*x^2+8/3*b*c*x-16/3*a*c+2*b^2)*(c*x^2+b*x+a)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.81 \[ \int \frac {(b d+2 c d x)^3}{\sqrt {a+b x+c x^2}} \, dx=\frac {2}{3} \, {\left (4 \, c^{2} d^{3} x^{2} + 4 \, b c d^{3} x + {\left (3 \, b^{2} - 8 \, a c\right )} d^{3}\right )} \sqrt {c x^{2} + b x + a} \]

[In]

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

[Out]

2/3*(4*c^2*d^3*x^2 + 4*b*c*d^3*x + (3*b^2 - 8*a*c)*d^3)*sqrt(c*x^2 + b*x + a)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.64 \[ \int \frac {(b d+2 c d x)^3}{\sqrt {a+b x+c x^2}} \, dx=- \frac {16 a c d^{3} \sqrt {a + b x + c x^{2}}}{3} + 2 b^{2} d^{3} \sqrt {a + b x + c x^{2}} + \frac {8 b c d^{3} x \sqrt {a + b x + c x^{2}}}{3} + \frac {8 c^{2} d^{3} x^{2} \sqrt {a + b x + c x^{2}}}{3} \]

[In]

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

[Out]

-16*a*c*d**3*sqrt(a + b*x + c*x**2)/3 + 2*b**2*d**3*sqrt(a + b*x + c*x**2) + 8*b*c*d**3*x*sqrt(a + b*x + c*x**
2)/3 + 8*c**2*d**3*x**2*sqrt(a + b*x + c*x**2)/3

Maxima [F(-2)]

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

[In]

integrate((2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(1/2),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 = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.98 \[ \int \frac {(b d+2 c d x)^3}{\sqrt {a+b x+c x^2}} \, dx=2 \, \sqrt {c x^{2} + b x + a} b^{2} d^{3} + \frac {8}{3} \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} c d^{3} - 8 \, \sqrt {c x^{2} + b x + a} a c d^{3} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.60 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.81 \[ \int \frac {(b d+2 c d x)^3}{\sqrt {a+b x+c x^2}} \, dx=\left (\frac {8\,c^2\,d^3\,x^2}{3}-\frac {2\,d^3\,\left (8\,a\,c-3\,b^2\right )}{3}+\frac {8\,b\,c\,d^3\,x}{3}\right )\,\sqrt {c\,x^2+b\,x+a} \]

[In]

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

[Out]

((8*c^2*d^3*x^2)/3 - (2*d^3*(8*a*c - 3*b^2))/3 + (8*b*c*d^3*x)/3)*(a + b*x + c*x^2)^(1/2)