\(\int \frac {(b d+2 c d x)^5}{a+b x+c x^2} \, dx\) [1157]

Optimal result

Integrand size = 24, antiderivative size = 61 \[ \int \frac {(b d+2 c d x)^5}{a+b x+c x^2} \, dx=\left (b^2-4 a c\right ) d^5 (b+2 c x)^2+\frac {1}{2} d^5 (b+2 c x)^4+\left (b^2-4 a c\right )^2 d^5 \log \left (a+b x+c x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 61, 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 = {706, 642} \[ \int \frac {(b d+2 c d x)^5}{a+b x+c x^2} \, dx=d^5 \left (b^2-4 a c\right )^2 \log \left (a+b x+c x^2\right )+d^5 \left (b^2-4 a c\right ) (b+2 c x)^2+\frac {1}{2} d^5 (b+2 c x)^4 \]

[In]

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

[Out]

(b^2 - 4*a*c)*d^5*(b + 2*c*x)^2 + (d^5*(b + 2*c*x)^4)/2 + (b^2 - 4*a*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 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 {1}{2} d^5 (b+2 c x)^4+\left (\left (b^2-4 a c\right ) d^2\right ) \int \frac {(b d+2 c d x)^3}{a+b x+c x^2} \, dx \\ & = \left (b^2-4 a c\right ) d^5 (b+2 c x)^2+\frac {1}{2} d^5 (b+2 c x)^4+\left (\left (b^2-4 a c\right )^2 d^4\right ) \int \frac {b d+2 c d x}{a+b x+c x^2} \, dx \\ & = \left (b^2-4 a c\right ) d^5 (b+2 c x)^2+\frac {1}{2} d^5 (b+2 c x)^4+\left (b^2-4 a c\right )^2 d^5 \log \left (a+b x+c x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89 \[ \int \frac {(b d+2 c d x)^5}{a+b x+c x^2} \, dx=d^5 \left (8 c x (b+c x) \left (b^2+b c x+c \left (-2 a+c x^2\right )\right )+\left (b^2-4 a c\right )^2 \log (a+x (b+c x))\right ) \]

[In]

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

[Out]

d^5*(8*c*x*(b + c*x)*(b^2 + b*c*x + c*(-2*a + c*x^2)) + (b^2 - 4*a*c)^2*Log[a + x*(b + c*x)])

Maple [A] (verified)

Time = 2.43 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.46

method	result	size
default	\(d^{5} \left (8 c^{4} x^{4}+16 b \,c^{3} x^{3}-16 x^{2} a \,c^{3}+16 b^{2} c^{2} x^{2}-16 a b \,c^{2} x +8 b^{3} c x +\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \ln \left (c \,x^{2}+b x +a \right )\right )\)	\(89\)
norman	\(\left (-16 d^{5} c^{3} a +16 b^{2} c^{2} d^{5}\right ) x^{2}+8 c^{4} d^{5} x^{4}+16 b \,c^{3} d^{5} x^{3}-8 b \,d^{5} c \left (2 a c -b^{2}\right ) x +\left (16 d^{5} a^{2} c^{2}-8 a \,d^{5} c \,b^{2}+d^{5} b^{4}\right ) \ln \left (c \,x^{2}+b x +a \right )\)	\(109\)
parallelrisch	\(8 c^{4} d^{5} x^{4}+16 b \,c^{3} d^{5} x^{3}-16 a \,c^{3} d^{5} x^{2}+16 b^{2} c^{2} d^{5} x^{2}+16 \ln \left (c \,x^{2}+b x +a \right ) a^{2} c^{2} d^{5}-8 \ln \left (c \,x^{2}+b x +a \right ) a \,b^{2} c \,d^{5}+\ln \left (c \,x^{2}+b x +a \right ) b^{4} d^{5}-16 a b \,c^{2} d^{5} x +8 b^{3} c \,d^{5} x\)	\(133\)
risch	\(8 c^{4} d^{5} x^{4}+16 b \,c^{3} d^{5} x^{3}-16 a \,c^{3} d^{5} x^{2}+16 b^{2} c^{2} d^{5} x^{2}-16 a b \,c^{2} d^{5} x +8 b^{3} c \,d^{5} x +8 d^{5} a^{2} c^{2}-8 a \,d^{5} c \,b^{2}+2 d^{5} b^{4}+16 \ln \left (c \,x^{2}+b x +a \right ) a^{2} c^{2} d^{5}-8 \ln \left (c \,x^{2}+b x +a \right ) a \,b^{2} c \,d^{5}+\ln \left (c \,x^{2}+b x +a \right ) b^{4} d^{5}\)	\(162\)

[In]

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

[Out]

d^5*(8*c^4*x^4+16*b*c^3*x^3-16*x^2*a*c^3+16*b^2*c^2*x^2-16*a*b*c^2*x+8*b^3*c*x+(16*a^2*c^2-8*a*b^2*c+b^4)*ln(c
*x^2+b*x+a))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.62 \[ \int \frac {(b d+2 c d x)^5}{a+b x+c x^2} \, dx=8 \, c^{4} d^{5} x^{4} + 16 \, b c^{3} d^{5} x^{3} + 16 \, {\left (b^{2} c^{2} - a c^{3}\right )} d^{5} x^{2} + 8 \, {\left (b^{3} c - 2 \, a b c^{2}\right )} d^{5} x + {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} d^{5} \log \left (c x^{2} + b x + a\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.62 \[ \int \frac {(b d+2 c d x)^5}{a+b x+c x^2} \, dx=16 b c^{3} d^{5} x^{3} + 8 c^{4} d^{5} x^{4} + d^{5} \left (4 a c - b^{2}\right )^{2} \log {\left (a + b x + c x^{2} \right )} + x^{2} \left (- 16 a c^{3} d^{5} + 16 b^{2} c^{2} d^{5}\right ) + x \left (- 16 a b c^{2} d^{5} + 8 b^{3} c d^{5}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.62 \[ \int \frac {(b d+2 c d x)^5}{a+b x+c x^2} \, dx=8 \, c^{4} d^{5} x^{4} + 16 \, b c^{3} d^{5} x^{3} + 16 \, {\left (b^{2} c^{2} - a c^{3}\right )} d^{5} x^{2} + 8 \, {\left (b^{3} c - 2 \, a b c^{2}\right )} d^{5} x + {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} d^{5} \log \left (c x^{2} + b x + a\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.93 \[ \int \frac {(b d+2 c d x)^5}{a+b x+c x^2} \, dx={\left (b^{4} d^{5} - 8 \, a b^{2} c d^{5} + 16 \, a^{2} c^{2} d^{5}\right )} \log \left (c x^{2} + b x + a\right ) + \frac {8 \, {\left (c^{8} d^{5} x^{4} + 2 \, b c^{7} d^{5} x^{3} + 2 \, b^{2} c^{6} d^{5} x^{2} - 2 \, a c^{7} d^{5} x^{2} + b^{3} c^{5} d^{5} x - 2 \, a b c^{6} d^{5} x\right )}}{c^{4}} \]

[In]

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

[Out]

(b^4*d^5 - 8*a*b^2*c*d^5 + 16*a^2*c^2*d^5)*log(c*x^2 + b*x + a) + 8*(c^8*d^5*x^4 + 2*b*c^7*d^5*x^3 + 2*b^2*c^6
*d^5*x^2 - 2*a*c^7*d^5*x^2 + b^3*c^5*d^5*x - 2*a*b*c^6*d^5*x)/c^4

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.28 \[ \int \frac {(b d+2 c d x)^5}{a+b x+c x^2} \, dx=\ln \left (c\,x^2+b\,x+a\right )\,\left (16\,a^2\,c^2\,d^5-8\,a\,b^2\,c\,d^5+b^4\,d^5\right )-x^2\,\left (16\,a\,c^3\,d^5-16\,b^2\,c^2\,d^5\right )+x\,\left (40\,b^3\,c\,d^5+\frac {b\,\left (32\,a\,c^3\,d^5-32\,b^2\,c^2\,d^5\right )}{c}-48\,a\,b\,c^2\,d^5\right )+8\,c^4\,d^5\,x^4+16\,b\,c^3\,d^5\,x^3 \]

[In]

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

[Out]

log(a + b*x + c*x^2)*(b^4*d^5 + 16*a^2*c^2*d^5 - 8*a*b^2*c*d^5) - x^2*(16*a*c^3*d^5 - 16*b^2*c^2*d^5) + x*(40*
b^3*c*d^5 + (b*(32*a*c^3*d^5 - 32*b^2*c^2*d^5))/c - 48*a*b*c^2*d^5) + 8*c^4*d^5*x^4 + 16*b*c^3*d^5*x^3