\(\int \frac {(a+b x)^4}{a c+(b c+a d) x+b d x^2} \, dx\) [1801]

Optimal result

Integrand size = 29, antiderivative size = 74 \[ \int \frac {(a+b x)^4}{a c+(b c+a d) x+b d x^2} \, dx=\frac {b (b c-a d)^2 x}{d^3}-\frac {(b c-a d) (a+b x)^2}{2 d^2}+\frac {(a+b x)^3}{3 d}-\frac {(b c-a d)^3 \log (c+d x)}{d^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 74, 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} \[ \int \frac {(a+b x)^4}{a c+(b c+a d) x+b d x^2} \, dx=-\frac {(b c-a d)^3 \log (c+d x)}{d^4}+\frac {b x (b c-a d)^2}{d^3}-\frac {(a+b x)^2 (b c-a d)}{2 d^2}+\frac {(a+b x)^3}{3 d} \]

[In]

Int[(a + b*x)^4/(a*c + (b*c + a*d)*x + b*d*x^2),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^4}{a c+(b c+a d) x+b d x^2} \, dx=\frac {b d x \left (18 a^2 d^2+9 a b d (-2 c+d x)+b^2 \left (6 c^2-3 c d x+2 d^2 x^2\right )\right )-6 (b c-a d)^3 \log (c+d x)}{6 d^4} \]

[In]

Integrate[(a + b*x)^4/(a*c + (b*c + a*d)*x + b*d*x^2),x]

[Out]

(b*d*x*(18*a^2*d^2 + 9*a*b*d*(-2*c + d*x) + b^2*(6*c^2 - 3*c*d*x + 2*d^2*x^2)) - 6*(b*c - a*d)^3*Log[c + d*x])
/(6*d^4)

Maple [A] (verified)

Time = 2.56 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.45

method	result	size
norman	\(\frac {b \left (3 a^{2} d^{2}-3 a b c d +b^{2} c^{2}\right ) x}{d^{3}}+\frac {b^{3} x^{3}}{3 d}+\frac {b^{2} \left (3 a d -b c \right ) x^{2}}{2 d^{2}}+\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (d x +c \right )}{d^{4}}\)	\(107\)
default	\(\frac {b \left (\frac {1}{3} d^{2} x^{3} b^{2}+\frac {3}{2} x^{2} a b \,d^{2}-\frac {1}{2} x^{2} b^{2} c d +3 a^{2} d^{2} x -3 a b c d x +b^{2} c^{2} x \right )}{d^{3}}+\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (d x +c \right )}{d^{4}}\)	\(109\)
risch	\(\frac {b^{3} x^{3}}{3 d}+\frac {3 b^{2} x^{2} a}{2 d}-\frac {b^{3} x^{2} c}{2 d^{2}}+\frac {3 b \,a^{2} x}{d}-\frac {3 b^{2} a c x}{d^{2}}+\frac {b^{3} c^{2} x}{d^{3}}+\frac {\ln \left (d x +c \right ) a^{3}}{d}-\frac {3 \ln \left (d x +c \right ) a^{2} b c}{d^{2}}+\frac {3 \ln \left (d x +c \right ) a \,b^{2} c^{2}}{d^{3}}-\frac {\ln \left (d x +c \right ) b^{3} c^{3}}{d^{4}}\)	\(133\)
parallelrisch	\(\frac {2 d^{3} x^{3} b^{3}+9 x^{2} a \,b^{2} d^{3}-3 x^{2} b^{3} c \,d^{2}+6 \ln \left (d x +c \right ) a^{3} d^{3}-18 \ln \left (d x +c \right ) a^{2} b c \,d^{2}+18 \ln \left (d x +c \right ) a \,b^{2} c^{2} d -6 \ln \left (d x +c \right ) b^{3} c^{3}+18 x \,a^{2} b \,d^{3}-18 x a \,b^{2} c \,d^{2}+6 x \,b^{3} c^{2} d}{6 d^{4}}\)	\(133\)

[In]

int((b*x+a)^4/(b*d*x^2+(a*d+b*c)*x+a*c),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.55 \[ \int \frac {(a+b x)^4}{a c+(b c+a d) x+b d x^2} \, dx=\frac {2 \, b^{3} d^{3} x^{3} - 3 \, {\left (b^{3} c d^{2} - 3 \, a b^{2} d^{3}\right )} x^{2} + 6 \, {\left (b^{3} c^{2} d - 3 \, a b^{2} c d^{2} + 3 \, a^{2} b d^{3}\right )} x - 6 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (d x + c\right )}{6 \, d^{4}} \]

[In]

integrate((b*x+a)^4/(a*c+(a*d+b*c)*x+b*d*x^2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b x)^4}{a c+(b c+a d) x+b d x^2} \, dx=\frac {b^{3} x^{3}}{3 d} + x^{2} \cdot \left (\frac {3 a b^{2}}{2 d} - \frac {b^{3} c}{2 d^{2}}\right ) + x \left (\frac {3 a^{2} b}{d} - \frac {3 a b^{2} c}{d^{2}} + \frac {b^{3} c^{2}}{d^{3}}\right ) + \frac {\left (a d - b c\right )^{3} \log {\left (c + d x \right )}}{d^{4}} \]

[In]

integrate((b*x+a)**4/(a*c+(a*d+b*c)*x+b*d*x**2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.54 \[ \int \frac {(a+b x)^4}{a c+(b c+a d) x+b d x^2} \, dx=\frac {2 \, b^{3} d^{2} x^{3} - 3 \, {\left (b^{3} c d - 3 \, a b^{2} d^{2}\right )} x^{2} + 6 \, {\left (b^{3} c^{2} - 3 \, a b^{2} c d + 3 \, a^{2} b d^{2}\right )} x}{6 \, d^{3}} - \frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (d x + c\right )}{d^{4}} \]

[In]

integrate((b*x+a)^4/(a*c+(a*d+b*c)*x+b*d*x^2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.57 \[ \int \frac {(a+b x)^4}{a c+(b c+a d) x+b d x^2} \, dx=\frac {2 \, b^{3} d^{2} x^{3} - 3 \, b^{3} c d x^{2} + 9 \, a b^{2} d^{2} x^{2} + 6 \, b^{3} c^{2} x - 18 \, a b^{2} c d x + 18 \, a^{2} b d^{2} x}{6 \, d^{3}} - \frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left ({\left | d x + c \right |}\right )}{d^{4}} \]

[In]

integrate((b*x+a)^4/(a*c+(a*d+b*c)*x+b*d*x^2),x, algorithm="giac")

[Out]

1/6*(2*b^3*d^2*x^3 - 3*b^3*c*d*x^2 + 9*a*b^2*d^2*x^2 + 6*b^3*c^2*x - 18*a*b^2*c*d*x + 18*a^2*b*d^2*x)/d^3 - (b
^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(abs(d*x + c))/d^4

Mupad [B] (verification not implemented)

Time = 9.88 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.59 \[ \int \frac {(a+b x)^4}{a c+(b c+a d) x+b d x^2} \, dx=x^2\,\left (\frac {3\,a\,b^2}{2\,d}-\frac {b^3\,c}{2\,d^2}\right )+x\,\left (\frac {3\,a^2\,b}{d}-\frac {c\,\left (\frac {3\,a\,b^2}{d}-\frac {b^3\,c}{d^2}\right )}{d}\right )+\frac {\ln \left (c+d\,x\right )\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )}{d^4}+\frac {b^3\,x^3}{3\,d} \]

[In]

int((a + b*x)^4/(a*c + x*(a*d + b*c) + b*d*x^2),x)

[Out]

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