\(\int \frac {(a+b x)^5}{a c+(b c+a d) x+b d x^2} \, dx\) [1800]

Optimal result

Integrand size = 29, antiderivative size = 98 \[ \int \frac {(a+b x)^5}{a c+(b c+a d) x+b d x^2} \, dx=-\frac {b (b c-a d)^3 x}{d^4}+\frac {(b c-a d)^2 (a+b x)^2}{2 d^3}-\frac {(b c-a d) (a+b x)^3}{3 d^2}+\frac {(a+b x)^4}{4 d}+\frac {(b c-a d)^4 \log (c+d x)}{d^5} \]

[Out]

-b*(-a*d+b*c)^3*x/d^4+1/2*(-a*d+b*c)^2*(b*x+a)^2/d^3-1/3*(-a*d+b*c)*(b*x+a)^3/d^2+1/4*(b*x+a)^4/d+(-a*d+b*c)^4
*ln(d*x+c)/d^5

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {640, 45} \[ \int \frac {(a+b x)^5}{a c+(b c+a d) x+b d x^2} \, dx=\frac {(b c-a d)^4 \log (c+d x)}{d^5}-\frac {b x (b c-a d)^3}{d^4}+\frac {(a+b x)^2 (b c-a d)^2}{2 d^3}-\frac {(a+b x)^3 (b c-a d)}{3 d^2}+\frac {(a+b x)^4}{4 d} \]

[In]

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

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 640

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

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.17 \[ \int \frac {(a+b x)^5}{a c+(b c+a d) x+b d x^2} \, dx=\frac {b d x \left (48 a^3 d^3+36 a^2 b d^2 (-2 c+d x)+8 a b^2 d \left (6 c^2-3 c d x+2 d^2 x^2\right )+b^3 \left (-12 c^3+6 c^2 d x-4 c d^2 x^2+3 d^3 x^3\right )\right )+12 (b c-a d)^4 \log (c+d x)}{12 d^5} \]

[In]

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

[Out]

(b*d*x*(48*a^3*d^3 + 36*a^2*b*d^2*(-2*c + d*x) + 8*a*b^2*d*(6*c^2 - 3*c*d*x + 2*d^2*x^2) + b^3*(-12*c^3 + 6*c^
2*d*x - 4*c*d^2*x^2 + 3*d^3*x^3)) + 12*(b*c - a*d)^4*Log[c + d*x])/(12*d^5)

Maple [A] (verified)

Time = 2.48 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.71

method	result	size
norman	\(\frac {b \left (4 a^{3} d^{3}-6 a^{2} b c \,d^{2}+4 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) x}{d^{4}}+\frac {b^{4} x^{4}}{4 d}+\frac {b^{2} \left (6 a^{2} d^{2}-4 a b c d +b^{2} c^{2}\right ) x^{2}}{2 d^{3}}+\frac {b^{3} \left (4 a d -b c \right ) x^{3}}{3 d^{2}}+\frac {\left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right ) \ln \left (d x +c \right )}{d^{5}}\)	\(168\)
default	\(\frac {b \left (\frac {d^{3} x^{4} b^{3}}{4}+\frac {\left (\left (2 a d -b c \right ) b^{2} d^{2}+2 a \,b^{2} d^{3}\right ) x^{3}}{3}+\frac {\left (2 \left (2 a d -b c \right ) a b \,d^{2}+b d \left (2 a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )\right ) x^{2}}{2}+\left (2 a d -b c \right ) \left (2 a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) x \right )}{d^{4}}+\frac {\left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right ) \ln \left (d x +c \right )}{d^{5}}\)	\(189\)
risch	\(\frac {b^{4} x^{4}}{4 d}+\frac {4 b^{3} x^{3} a}{3 d}-\frac {b^{4} x^{3} c}{3 d^{2}}+\frac {3 b^{2} x^{2} a^{2}}{d}-\frac {2 b^{3} x^{2} a c}{d^{2}}+\frac {b^{4} x^{2} c^{2}}{2 d^{3}}+\frac {4 b \,a^{3} x}{d}-\frac {6 b^{2} a^{2} c x}{d^{2}}+\frac {4 b^{3} a \,c^{2} x}{d^{3}}-\frac {b^{4} c^{3} x}{d^{4}}+\frac {\ln \left (d x +c \right ) a^{4}}{d}-\frac {4 \ln \left (d x +c \right ) a^{3} b c}{d^{2}}+\frac {6 \ln \left (d x +c \right ) a^{2} b^{2} c^{2}}{d^{3}}-\frac {4 \ln \left (d x +c \right ) a \,b^{3} c^{3}}{d^{4}}+\frac {\ln \left (d x +c \right ) b^{4} c^{4}}{d^{5}}\)	\(209\)
parallelrisch	\(\frac {3 b^{4} x^{4} d^{4}+16 x^{3} a \,b^{3} d^{4}-4 x^{3} b^{4} c \,d^{3}+36 x^{2} a^{2} b^{2} d^{4}-24 x^{2} a \,b^{3} c \,d^{3}+6 x^{2} b^{4} c^{2} d^{2}+12 \ln \left (d x +c \right ) a^{4} d^{4}-48 \ln \left (d x +c \right ) a^{3} b c \,d^{3}+72 \ln \left (d x +c \right ) a^{2} b^{2} c^{2} d^{2}-48 \ln \left (d x +c \right ) a \,b^{3} c^{3} d +12 \ln \left (d x +c \right ) b^{4} c^{4}+48 a^{3} b \,d^{4} x -72 a^{2} b^{2} c \,d^{3} x +48 a \,b^{3} c^{2} d^{2} x -12 b^{4} c^{3} d x}{12 d^{5}}\)	\(209\)

[In]

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

[Out]

b*(4*a^3*d^3-6*a^2*b*c*d^2+4*a*b^2*c^2*d-b^3*c^3)/d^4*x+1/4*b^4/d*x^4+1/2*b^2/d^3*(6*a^2*d^2-4*a*b*c*d+b^2*c^2
)*x^2+1/3*b^3/d^2*(4*a*d-b*c)*x^3+(a^4*d^4-4*a^3*b*c*d^3+6*a^2*b^2*c^2*d^2-4*a*b^3*c^3*d+b^4*c^4)/d^5*ln(d*x+c
)

Fricas [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.83 \[ \int \frac {(a+b x)^5}{a c+(b c+a d) x+b d x^2} \, dx=\frac {3 \, b^{4} d^{4} x^{4} - 4 \, {\left (b^{4} c d^{3} - 4 \, a b^{3} d^{4}\right )} x^{3} + 6 \, {\left (b^{4} c^{2} d^{2} - 4 \, a b^{3} c d^{3} + 6 \, a^{2} b^{2} d^{4}\right )} x^{2} - 12 \, {\left (b^{4} c^{3} d - 4 \, a b^{3} c^{2} d^{2} + 6 \, a^{2} b^{2} c d^{3} - 4 \, a^{3} b d^{4}\right )} x + 12 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \log \left (d x + c\right )}{12 \, d^{5}} \]

[In]

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

[Out]

1/12*(3*b^4*d^4*x^4 - 4*(b^4*c*d^3 - 4*a*b^3*d^4)*x^3 + 6*(b^4*c^2*d^2 - 4*a*b^3*c*d^3 + 6*a^2*b^2*d^4)*x^2 -
12*(b^4*c^3*d - 4*a*b^3*c^2*d^2 + 6*a^2*b^2*c*d^3 - 4*a^3*b*d^4)*x + 12*(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c
^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*log(d*x + c))/d^5

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.39 \[ \int \frac {(a+b x)^5}{a c+(b c+a d) x+b d x^2} \, dx=\frac {b^{4} x^{4}}{4 d} + x^{3} \cdot \left (\frac {4 a b^{3}}{3 d} - \frac {b^{4} c}{3 d^{2}}\right ) + x^{2} \cdot \left (\frac {3 a^{2} b^{2}}{d} - \frac {2 a b^{3} c}{d^{2}} + \frac {b^{4} c^{2}}{2 d^{3}}\right ) + x \left (\frac {4 a^{3} b}{d} - \frac {6 a^{2} b^{2} c}{d^{2}} + \frac {4 a b^{3} c^{2}}{d^{3}} - \frac {b^{4} c^{3}}{d^{4}}\right ) + \frac {\left (a d - b c\right )^{4} \log {\left (c + d x \right )}}{d^{5}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.81 \[ \int \frac {(a+b x)^5}{a c+(b c+a d) x+b d x^2} \, dx=\frac {3 \, b^{4} d^{3} x^{4} - 4 \, {\left (b^{4} c d^{2} - 4 \, a b^{3} d^{3}\right )} x^{3} + 6 \, {\left (b^{4} c^{2} d - 4 \, a b^{3} c d^{2} + 6 \, a^{2} b^{2} d^{3}\right )} x^{2} - 12 \, {\left (b^{4} c^{3} - 4 \, a b^{3} c^{2} d + 6 \, a^{2} b^{2} c d^{2} - 4 \, a^{3} b d^{3}\right )} x}{12 \, d^{4}} + \frac {{\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \log \left (d x + c\right )}{d^{5}} \]

[In]

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

[Out]

1/12*(3*b^4*d^3*x^4 - 4*(b^4*c*d^2 - 4*a*b^3*d^3)*x^3 + 6*(b^4*c^2*d - 4*a*b^3*c*d^2 + 6*a^2*b^2*d^3)*x^2 - 12
*(b^4*c^3 - 4*a*b^3*c^2*d + 6*a^2*b^2*c*d^2 - 4*a^3*b*d^3)*x)/d^4 + (b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d
^2 - 4*a^3*b*c*d^3 + a^4*d^4)*log(d*x + c)/d^5

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.88 \[ \int \frac {(a+b x)^5}{a c+(b c+a d) x+b d x^2} \, dx=\frac {3 \, b^{4} d^{3} x^{4} - 4 \, b^{4} c d^{2} x^{3} + 16 \, a b^{3} d^{3} x^{3} + 6 \, b^{4} c^{2} d x^{2} - 24 \, a b^{3} c d^{2} x^{2} + 36 \, a^{2} b^{2} d^{3} x^{2} - 12 \, b^{4} c^{3} x + 48 \, a b^{3} c^{2} d x - 72 \, a^{2} b^{2} c d^{2} x + 48 \, a^{3} b d^{3} x}{12 \, d^{4}} + \frac {{\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \log \left ({\left | d x + c \right |}\right )}{d^{5}} \]

[In]

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

[Out]

1/12*(3*b^4*d^3*x^4 - 4*b^4*c*d^2*x^3 + 16*a*b^3*d^3*x^3 + 6*b^4*c^2*d*x^2 - 24*a*b^3*c*d^2*x^2 + 36*a^2*b^2*d
^3*x^2 - 12*b^4*c^3*x + 48*a*b^3*c^2*d*x - 72*a^2*b^2*c*d^2*x + 48*a^3*b*d^3*x)/d^4 + (b^4*c^4 - 4*a*b^3*c^3*d
+ 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*log(abs(d*x + c))/d^5

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.93 \[ \int \frac {(a+b x)^5}{a c+(b c+a d) x+b d x^2} \, dx=x^3\,\left (\frac {4\,a\,b^3}{3\,d}-\frac {b^4\,c}{3\,d^2}\right )+x\,\left (\frac {4\,a^3\,b}{d}+\frac {c\,\left (\frac {c\,\left (\frac {4\,a\,b^3}{d}-\frac {b^4\,c}{d^2}\right )}{d}-\frac {6\,a^2\,b^2}{d}\right )}{d}\right )-x^2\,\left (\frac {c\,\left (\frac {4\,a\,b^3}{d}-\frac {b^4\,c}{d^2}\right )}{2\,d}-\frac {3\,a^2\,b^2}{d}\right )+\frac {\ln \left (c+d\,x\right )\,\left (a^4\,d^4-4\,a^3\,b\,c\,d^3+6\,a^2\,b^2\,c^2\,d^2-4\,a\,b^3\,c^3\,d+b^4\,c^4\right )}{d^5}+\frac {b^4\,x^4}{4\,d} \]

[In]

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

[Out]

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