\(\int \frac {(b d+2 c d x)^5}{(a+b x+c x^2)^3} \, dx\) [1182]

Optimal result

Integrand size = 24, antiderivative size = 73 \[ \int \frac {(b d+2 c d x)^5}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {d^5 (b+2 c x)^4}{2 \left (a+b x+c x^2\right )^2}-\frac {4 c d^5 (b+2 c x)^2}{a+b x+c x^2}+16 c^2 d^5 \log \left (a+b x+c x^2\right ) \]

[Out]

-1/2*d^5*(2*c*x+b)^4/(c*x^2+b*x+a)^2-4*c*d^5*(2*c*x+b)^2/(c*x^2+b*x+a)+16*c^2*d^5*ln(c*x^2+b*x+a)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {700, 642} \[ \int \frac {(b d+2 c d x)^5}{\left (a+b x+c x^2\right )^3} \, dx=16 c^2 d^5 \log \left (a+b x+c x^2\right )-\frac {4 c d^5 (b+2 c x)^2}{a+b x+c x^2}-\frac {d^5 (b+2 c x)^4}{2 \left (a+b x+c x^2\right )^2} \]

[In]

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

[Out]

-1/2*(d^5*(b + 2*c*x)^4)/(a + b*x + c*x^2)^2 - (4*c*d^5*(b + 2*c*x)^2)/(a + b*x + c*x^2) + 16*c^2*d^5*Log[a +
b*x + c*x^2]

Rule 642

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

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 {d^5 (b+2 c x)^4}{2 \left (a+b x+c x^2\right )^2}+\left (4 c d^2\right ) \int \frac {(b d+2 c d x)^3}{\left (a+b x+c x^2\right )^2} \, dx \\ & = -\frac {d^5 (b+2 c x)^4}{2 \left (a+b x+c x^2\right )^2}-\frac {4 c d^5 (b+2 c x)^2}{a+b x+c x^2}+\left (16 c^2 d^4\right ) \int \frac {b d+2 c d x}{a+b x+c x^2} \, dx \\ & = -\frac {d^5 (b+2 c x)^4}{2 \left (a+b x+c x^2\right )^2}-\frac {4 c d^5 (b+2 c x)^2}{a+b x+c x^2}+16 c^2 d^5 \log \left (a+b x+c x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.89 \[ \int \frac {(b d+2 c d x)^5}{\left (a+b x+c x^2\right )^3} \, dx=d^5 \left (-\frac {\left (b^2-4 a c\right ) \left (b^2+16 b c x+4 c \left (3 a+4 c x^2\right )\right )}{2 (a+x (b+c x))^2}+16 c^2 \log (a+x (b+c x))\right ) \]

[In]

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

[Out]

d^5*(-1/2*((b^2 - 4*a*c)*(b^2 + 16*b*c*x + 4*c*(3*a + 4*c*x^2)))/(a + x*(b + c*x))^2 + 16*c^2*Log[a + x*(b + c
*x)])

Maple [A] (verified)

Time = 2.52 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.22

method	result	size
default	\(d^{5} \left (\frac {8 c^{2} \left (4 a c -b^{2}\right ) x^{2}+8 b c \left (4 a c -b^{2}\right ) x +24 a^{2} c^{2}-4 a \,b^{2} c -\frac {b^{4}}{2}}{\left (c \,x^{2}+b x +a \right )^{2}}+16 c^{2} \ln \left (c \,x^{2}+b x +a \right )\right )\)	\(89\)
risch	\(\frac {8 d^{5} c^{2} \left (4 a c -b^{2}\right ) x^{2}+8 b c \,d^{5} \left (4 a c -b^{2}\right ) x +\frac {d^{5} \left (48 a^{2} c^{2}-8 a \,b^{2} c -b^{4}\right )}{2}}{\left (c \,x^{2}+b x +a \right )^{2}}+16 c^{2} d^{5} \ln \left (c \,x^{2}+b x +a \right )\)	\(100\)
norman	\(\frac {\frac {\left (32 a \,c^{5} d^{5}-8 c^{4} b^{2} d^{5}\right ) x^{2}}{c^{2}}+\frac {48 a^{2} d^{5} c^{4}-8 a \,b^{2} c^{3} d^{5}-b^{4} c^{2} d^{5}}{2 c^{2}}+\frac {2 b \left (16 a \,c^{4} d^{5}-4 b^{2} c^{3} d^{5}\right ) x}{c^{2}}}{\left (c \,x^{2}+b x +a \right )^{2}}+16 c^{2} d^{5} \ln \left (c \,x^{2}+b x +a \right )\)	\(131\)
parallelrisch	\(\frac {32 \ln \left (c \,x^{2}+b x +a \right ) x^{4} c^{6} d^{5}+64 \ln \left (c \,x^{2}+b x +a \right ) x^{3} b \,c^{5} d^{5}+64 \ln \left (c \,x^{2}+b x +a \right ) x^{2} a \,c^{5} d^{5}+32 \ln \left (c \,x^{2}+b x +a \right ) x^{2} b^{2} c^{4} d^{5}+64 \ln \left (c \,x^{2}+b x +a \right ) x a b \,c^{4} d^{5}+64 x^{2} a \,c^{5} d^{5}-16 x^{2} b^{2} c^{4} d^{5}+32 \ln \left (c \,x^{2}+b x +a \right ) a^{2} c^{4} d^{5}+64 x a b \,c^{4} d^{5}-16 x \,b^{3} c^{3} d^{5}+48 a^{2} d^{5} c^{4}-8 a \,b^{2} c^{3} d^{5}-b^{4} c^{2} d^{5}}{2 c^{2} \left (c \,x^{2}+b x +a \right )^{2}}\)	\(239\)

[In]

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

[Out]

d^5*((8*c^2*(4*a*c-b^2)*x^2+8*b*c*(4*a*c-b^2)*x+24*a^2*c^2-4*a*b^2*c-1/2*b^4)/(c*x^2+b*x+a)^2+16*c^2*ln(c*x^2+
b*x+a))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (71) = 142\).

Time = 0.27 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.49 \[ \int \frac {(b d+2 c d x)^5}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {16 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{5} x^{2} + 16 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} d^{5} x + {\left (b^{4} + 8 \, a b^{2} c - 48 \, a^{2} c^{2}\right )} d^{5} - 32 \, {\left (c^{4} d^{5} x^{4} + 2 \, b c^{3} d^{5} x^{3} + 2 \, a b c^{2} d^{5} x + a^{2} c^{2} d^{5} + {\left (b^{2} c^{2} + 2 \, a c^{3}\right )} d^{5} x^{2}\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}} \]

[In]

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

[Out]

-1/2*(16*(b^2*c^2 - 4*a*c^3)*d^5*x^2 + 16*(b^3*c - 4*a*b*c^2)*d^5*x + (b^4 + 8*a*b^2*c - 48*a^2*c^2)*d^5 - 32*
(c^4*d^5*x^4 + 2*b*c^3*d^5*x^3 + 2*a*b*c^2*d^5*x + a^2*c^2*d^5 + (b^2*c^2 + 2*a*c^3)*d^5*x^2)*log(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. 141 vs. \(2 (70) = 140\).

Time = 2.12 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.93 \[ \int \frac {(b d+2 c d x)^5}{\left (a+b x+c x^2\right )^3} \, dx=16 c^{2} d^{5} \log {\left (a + b x + c x^{2} \right )} + \frac {48 a^{2} c^{2} d^{5} - 8 a b^{2} c d^{5} - b^{4} d^{5} + x^{2} \cdot \left (64 a c^{3} d^{5} - 16 b^{2} c^{2} d^{5}\right ) + x \left (64 a b c^{2} d^{5} - 16 b^{3} c d^{5}\right )}{2 a^{2} + 4 a b x + 4 b c x^{3} + 2 c^{2} x^{4} + x^{2} \cdot \left (4 a c + 2 b^{2}\right )} \]

[In]

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

[Out]

16*c**2*d**5*log(a + b*x + c*x**2) + (48*a**2*c**2*d**5 - 8*a*b**2*c*d**5 - b**4*d**5 + x**2*(64*a*c**3*d**5 -
16*b**2*c**2*d**5) + x*(64*a*b*c**2*d**5 - 16*b**3*c*d**5))/(2*a**2 + 4*a*b*x + 4*b*c*x**3 + 2*c**2*x**4 + x*
*2*(4*a*c + 2*b**2))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.70 \[ \int \frac {(b d+2 c d x)^5}{\left (a+b x+c x^2\right )^3} \, dx=16 \, c^{2} d^{5} \log \left (c x^{2} + b x + a\right ) - \frac {16 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{5} x^{2} + 16 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} d^{5} x + {\left (b^{4} + 8 \, a b^{2} c - 48 \, a^{2} c^{2}\right )} d^{5}}{2 \, {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}} \]

[In]

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

[Out]

16*c^2*d^5*log(c*x^2 + b*x + a) - 1/2*(16*(b^2*c^2 - 4*a*c^3)*d^5*x^2 + 16*(b^3*c - 4*a*b*c^2)*d^5*x + (b^4 +
8*a*b^2*c - 48*a^2*c^2)*d^5)/(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.51 \[ \int \frac {(b d+2 c d x)^5}{\left (a+b x+c x^2\right )^3} \, dx=16 \, c^{2} d^{5} \log \left (c x^{2} + b x + a\right ) - \frac {b^{4} d^{5} + 8 \, a b^{2} c d^{5} - 48 \, a^{2} c^{2} d^{5} + 16 \, {\left (b^{2} c^{2} d^{5} - 4 \, a c^{3} d^{5}\right )} x^{2} + 16 \, {\left (b^{3} c d^{5} - 4 \, a b c^{2} d^{5}\right )} x}{2 \, {\left (c x^{2} + b x + a\right )}^{2}} \]

[In]

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

[Out]

16*c^2*d^5*log(c*x^2 + b*x + a) - 1/2*(b^4*d^5 + 8*a*b^2*c*d^5 - 48*a^2*c^2*d^5 + 16*(b^2*c^2*d^5 - 4*a*c^3*d^
5)*x^2 + 16*(b^3*c*d^5 - 4*a*b*c^2*d^5)*x)/(c*x^2 + b*x + a)^2

Mupad [B] (verification not implemented)

Time = 9.39 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.86 \[ \int \frac {(b d+2 c d x)^5}{\left (a+b x+c x^2\right )^3} \, dx=\frac {x^2\,\left (32\,a\,c^3\,d^5-8\,b^2\,c^2\,d^5\right )-\frac {b^4\,d^5}{2}+8\,b\,x\,\left (4\,a\,c^2\,d^5-b^2\,c\,d^5\right )+24\,a^2\,c^2\,d^5-4\,a\,b^2\,c\,d^5}{x^2\,\left (b^2+2\,a\,c\right )+a^2+c^2\,x^4+2\,a\,b\,x+2\,b\,c\,x^3}+16\,c^2\,d^5\,\ln \left (c\,x^2+b\,x+a\right ) \]

[In]

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

[Out]

(x^2*(32*a*c^3*d^5 - 8*b^2*c^2*d^5) - (b^4*d^5)/2 + 8*b*x*(4*a*c^2*d^5 - b^2*c*d^5) + 24*a^2*c^2*d^5 - 4*a*b^2
*c*d^5)/(x^2*(2*a*c + b^2) + a^2 + c^2*x^4 + 2*a*b*x + 2*b*c*x^3) + 16*c^2*d^5*log(a + b*x + c*x^2)