\(\int \frac {(b d+2 c d x)^3}{(a+b x+c x^2)^{5/2}} \, dx\) [1253]

Optimal result

Integrand size = 26, antiderivative size = 52 \[ \int \frac {(b d+2 c d x)^3}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 d^3 (b+2 c x)^2}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {16 c d^3}{3 \sqrt {a+b x+c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 52, 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 = {700, 643} \[ \int \frac {(b d+2 c d x)^3}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {16 c d^3}{3 \sqrt {a+b x+c x^2}}-\frac {2 d^3 (b+2 c x)^2}{3 \left (a+b x+c x^2\right )^{3/2}} \]

[In]

Int[(b*d + 2*c*d*x)^3/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*d^3*(b + 2*c*x)^2)/(3*(a + b*x + c*x^2)^(3/2)) - (16*c*d^3)/(3*Sqrt[a + b*x + c*x^2])

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 700

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 d^3 (b+2 c x)^2}{3 \left (a+b x+c x^2\right )^{3/2}}+\frac {1}{3} \left (8 c d^2\right ) \int \frac {b d+2 c d x}{\left (a+b x+c x^2\right )^{3/2}} \, dx \\ & = -\frac {2 d^3 (b+2 c x)^2}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac {16 c d^3}{3 \sqrt {a+b x+c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.81 \[ \int \frac {(b d+2 c d x)^3}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 d^3 \left (b^2+12 b c x+4 c \left (2 a+3 c x^2\right )\right )}{3 (a+x (b+c x))^{3/2}} \]

[In]

Integrate[(b*d + 2*c*d*x)^3/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*d^3*(b^2 + 12*b*c*x + 4*c*(2*a + 3*c*x^2)))/(3*(a + x*(b + c*x))^(3/2))

Maple [A] (verified)

Time = 2.47 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.75

method	result	size
gosper	\(-\frac {2 d^{3} \left (12 c^{2} x^{2}+12 b c x +8 a c +b^{2}\right )}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}\)	\(39\)
trager	\(-\frac {2 d^{3} \left (12 c^{2} x^{2}+12 b c x +8 a c +b^{2}\right )}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}\)	\(39\)
pseudoelliptic	\(-\frac {2 d^{3} \left (12 c^{2} x^{2}+12 b c x +8 a c +b^{2}\right )}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}\)	\(39\)
default	\(d^{3} \left (b^{3} \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )+8 c^{3} \left (-\frac {x^{2}}{c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {b \left (-\frac {x}{2 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {1}{3 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )}{4 c}+\frac {a \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )}{2 c}+\frac {2 a \left (-\frac {1}{3 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )}{c}\right )+6 b^{2} c \left (-\frac {1}{3 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )+12 b \,c^{2} \left (-\frac {x}{2 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {1}{3 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}-\frac {b \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )}{4 c}+\frac {a \left (\frac {\frac {4 c x}{3}+\frac {2 b}{3}}{\left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 \left (4 a c -b^{2}\right )^{2} \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}\right )\right )\)	\(680\)

[In]

int((2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/3*d^3*(12*c^2*x^2+12*b*c*x+8*a*c+b^2)/(c*x^2+b*x+a)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.60 \[ \int \frac {(b d+2 c d x)^3}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (12 \, c^{2} d^{3} x^{2} + 12 \, b c d^{3} x + {\left (b^{2} + 8 \, a c\right )} d^{3}\right )} \sqrt {c x^{2} + b x + a}}{3 \, {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}} \]

[In]

integrate((2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

-2/3*(12*c^2*d^3*x^2 + 12*b*c*d^3*x + (b^2 + 8*a*c)*d^3)*sqrt(c*x^2 + b*x + a)/(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x
+ (b^2 + 2*a*c)*x^2 + a^2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (51) = 102\).

Time = 0.61 (sec) , antiderivative size = 320, normalized size of antiderivative = 6.15 \[ \int \frac {(b d+2 c d x)^3}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\begin {cases} - \frac {16 a c d^{3}}{3 a \sqrt {a + b x + c x^{2}} + 3 b x \sqrt {a + b x + c x^{2}} + 3 c x^{2} \sqrt {a + b x + c x^{2}}} - \frac {2 b^{2} d^{3}}{3 a \sqrt {a + b x + c x^{2}} + 3 b x \sqrt {a + b x + c x^{2}} + 3 c x^{2} \sqrt {a + b x + c x^{2}}} - \frac {24 b c d^{3} x}{3 a \sqrt {a + b x + c x^{2}} + 3 b x \sqrt {a + b x + c x^{2}} + 3 c x^{2} \sqrt {a + b x + c x^{2}}} - \frac {24 c^{2} d^{3} x^{2}}{3 a \sqrt {a + b x + c x^{2}} + 3 b x \sqrt {a + b x + c x^{2}} + 3 c x^{2} \sqrt {a + b x + c x^{2}}} & \text {for}\: a \neq - x \left (b + c x\right ) \\\tilde {\infty } b^{3} d^{3} x + \tilde {\infty } b^{2} c d^{3} x^{2} + \tilde {\infty } b c^{2} d^{3} x^{3} + \tilde {\infty } c^{3} d^{3} x^{4} & \text {otherwise} \end {cases} \]

[In]

integrate((2*c*d*x+b*d)**3/(c*x**2+b*x+a)**(5/2),x)

[Out]

Piecewise((-16*a*c*d**3/(3*a*sqrt(a + b*x + c*x**2) + 3*b*x*sqrt(a + b*x + c*x**2) + 3*c*x**2*sqrt(a + b*x + c
*x**2)) - 2*b**2*d**3/(3*a*sqrt(a + b*x + c*x**2) + 3*b*x*sqrt(a + b*x + c*x**2) + 3*c*x**2*sqrt(a + b*x + c*x
**2)) - 24*b*c*d**3*x/(3*a*sqrt(a + b*x + c*x**2) + 3*b*x*sqrt(a + b*x + c*x**2) + 3*c*x**2*sqrt(a + b*x + c*x
**2)) - 24*c**2*d**3*x**2/(3*a*sqrt(a + b*x + c*x**2) + 3*b*x*sqrt(a + b*x + c*x**2) + 3*c*x**2*sqrt(a + b*x +
c*x**2)), Ne(a, -x*(b + c*x))), (zoo*b**3*d**3*x + zoo*b**2*c*d**3*x**2 + zoo*b*c**2*d**3*x**3 + zoo*c**3*d**
3*x**4, True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {(b d+2 c d x)^3}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(5/2),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 = 0.29 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.29 \[ \int \frac {(b d+2 c d x)^3}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (b^{2} d^{7} - 4 \, a c d^{7} + 12 \, {\left (a d^{2} + {\left (c d x^{2} + b d x\right )} d\right )} c d^{5}\right )}}{3 \, {\left (a d^{2} + {\left (c d x^{2} + b d x\right )} d\right )}^{\frac {3}{2}} {\left | d \right |}} \]

[In]

integrate((2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

-2/3*(b^2*d^7 - 4*a*c*d^7 + 12*(a*d^2 + (c*d*x^2 + b*d*x)*d)*c*d^5)/((a*d^2 + (c*d*x^2 + b*d*x)*d)^(3/2)*abs(d
))

Mupad [B] (verification not implemented)

Time = 9.63 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.19 \[ \int \frac {(b d+2 c d x)^3}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2\,b^2\,d^3+24\,c\,d^3\,\left (c\,x^2+b\,x+a\right )-8\,a\,c\,d^3}{\sqrt {c\,x^2+b\,x+a}\,\left (3\,c\,x^2+3\,b\,x+3\,a\right )} \]

[In]

int((b*d + 2*c*d*x)^3/(a + b*x + c*x^2)^(5/2),x)

[Out]

-(2*b^2*d^3 + 24*c*d^3*(a + b*x + c*x^2) - 8*a*c*d^3)/((a + b*x + c*x^2)^(1/2)*(3*a + 3*b*x + 3*c*x^2))