\(\int \frac {(a+b x^3) (c+d x+e x^2+f x^3+g x^4+h x^5)}{x^4} \, dx\) [381]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 36, antiderivative size = 86 \[ \int \frac {\left (a+b x^3\right ) \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^4} \, dx=-\frac {a c}{3 x^3}-\frac {a d}{2 x^2}-\frac {a e}{x}+(b d+a g) x+\frac {1}{2} (b e+a h) x^2+\frac {1}{3} b f x^3+\frac {1}{4} b g x^4+\frac {1}{5} b h x^5+(b c+a f) \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {1834} \[ \int \frac {\left (a+b x^3\right ) \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^4} \, dx=\log (x) (a f+b c)+x (a g+b d)+\frac {1}{2} x^2 (a h+b e)-\frac {a c}{3 x^3}-\frac {a d}{2 x^2}-\frac {a e}{x}+\frac {1}{3} b f x^3+\frac {1}{4} b g x^4+\frac {1}{5} b h x^5 \]

[In]

Int[((a + b*x^3)*(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5))/x^4,x]

[Out]

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

Rule 1834

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a +
 b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = \int \left (b d \left (1+\frac {a g}{b d}\right )+\frac {a c}{x^4}+\frac {a d}{x^3}+\frac {a e}{x^2}+\frac {b c+a f}{x}+(b e+a h) x+b f x^2+b g x^3+b h x^4\right ) \, dx \\ & = -\frac {a c}{3 x^3}-\frac {a d}{2 x^2}-\frac {a e}{x}+(b d+a g) x+\frac {1}{2} (b e+a h) x^2+\frac {1}{3} b f x^3+\frac {1}{4} b g x^4+\frac {1}{5} b h x^5+(b c+a f) \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a+b x^3\right ) \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^4} \, dx=-\frac {a \left (2 c+3 x \left (d+2 e x-x^3 (2 g+h x)\right )\right )}{6 x^3}+\frac {1}{60} b x \left (60 d+x \left (30 e+x \left (20 f+15 g x+12 h x^2\right )\right )\right )+(b c+a f) \log (x) \]

[In]

Integrate[((a + b*x^3)*(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5))/x^4,x]

[Out]

-1/6*(a*(2*c + 3*x*(d + 2*e*x - x^3*(2*g + h*x))))/x^3 + (b*x*(60*d + x*(30*e + x*(20*f + 15*g*x + 12*h*x^2)))
)/60 + (b*c + a*f)*Log[x]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.88

method result size
default \(\frac {b h \,x^{5}}{5}+\frac {b g \,x^{4}}{4}+\frac {f \,x^{3} b}{3}+\frac {a h \,x^{2}}{2}+\frac {b e \,x^{2}}{2}+a g x +b d x +\left (a f +b c \right ) \ln \left (x \right )-\frac {a c}{3 x^{3}}-\frac {a e}{x}-\frac {a d}{2 x^{2}}\) \(76\)
risch \(\frac {b h \,x^{5}}{5}+\frac {b g \,x^{4}}{4}+\frac {f \,x^{3} b}{3}+\frac {a h \,x^{2}}{2}+\frac {b e \,x^{2}}{2}+a g x +b d x +\frac {-a e \,x^{2}-\frac {1}{2} a d x -\frac {1}{3} a c}{x^{3}}+\ln \left (x \right ) a f +\ln \left (x \right ) b c\) \(76\)
norman \(\frac {\left (\frac {a h}{2}+\frac {b e}{2}\right ) x^{5}+\left (a g +b d \right ) x^{4}-\frac {a c}{3}-\frac {a d x}{2}-a e \,x^{2}+\frac {b f \,x^{6}}{3}+\frac {b g \,x^{7}}{4}+\frac {b h \,x^{8}}{5}}{x^{3}}+\left (a f +b c \right ) \ln \left (x \right )\) \(78\)
parallelrisch \(\frac {12 b h \,x^{8}+15 b g \,x^{7}+20 b f \,x^{6}+30 a h \,x^{5}+30 b e \,x^{5}+60 \ln \left (x \right ) x^{3} a f +60 \ln \left (x \right ) x^{3} b c +60 a g \,x^{4}+60 b d \,x^{4}-60 a e \,x^{2}-30 a d x -20 a c}{60 x^{3}}\) \(90\)

[In]

int((b*x^3+a)*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+b x^3\right ) \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^4} \, dx=\frac {12 \, b h x^{8} + 15 \, b g x^{7} + 20 \, b f x^{6} + 30 \, {\left (b e + a h\right )} x^{5} + 60 \, {\left (b d + a g\right )} x^{4} + 60 \, {\left (b c + a f\right )} x^{3} \log \left (x\right ) - 60 \, a e x^{2} - 30 \, a d x - 20 \, a c}{60 \, x^{3}} \]

[In]

integrate((b*x^3+a)*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^4,x, algorithm="fricas")

[Out]

1/60*(12*b*h*x^8 + 15*b*g*x^7 + 20*b*f*x^6 + 30*(b*e + a*h)*x^5 + 60*(b*d + a*g)*x^4 + 60*(b*c + a*f)*x^3*log(
x) - 60*a*e*x^2 - 30*a*d*x - 20*a*c)/x^3

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+b x^3\right ) \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^4} \, dx=\frac {b f x^{3}}{3} + \frac {b g x^{4}}{4} + \frac {b h x^{5}}{5} + x^{2} \left (\frac {a h}{2} + \frac {b e}{2}\right ) + x \left (a g + b d\right ) + \left (a f + b c\right ) \log {\left (x \right )} + \frac {- 2 a c - 3 a d x - 6 a e x^{2}}{6 x^{3}} \]

[In]

integrate((b*x**3+a)*(h*x**5+g*x**4+f*x**3+e*x**2+d*x+c)/x**4,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^3\right ) \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^4} \, dx=\frac {1}{5} \, b h x^{5} + \frac {1}{4} \, b g x^{4} + \frac {1}{3} \, b f x^{3} + \frac {1}{2} \, {\left (b e + a h\right )} x^{2} + {\left (b d + a g\right )} x + {\left (b c + a f\right )} \log \left (x\right ) - \frac {6 \, a e x^{2} + 3 \, a d x + 2 \, a c}{6 \, x^{3}} \]

[In]

integrate((b*x^3+a)*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^4,x, algorithm="maxima")

[Out]

1/5*b*h*x^5 + 1/4*b*g*x^4 + 1/3*b*f*x^3 + 1/2*(b*e + a*h)*x^2 + (b*d + a*g)*x + (b*c + a*f)*log(x) - 1/6*(6*a*
e*x^2 + 3*a*d*x + 2*a*c)/x^3

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b x^3\right ) \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^4} \, dx=\frac {1}{5} \, b h x^{5} + \frac {1}{4} \, b g x^{4} + \frac {1}{3} \, b f x^{3} + \frac {1}{2} \, b e x^{2} + \frac {1}{2} \, a h x^{2} + b d x + a g x + {\left (b c + a f\right )} \log \left ({\left | x \right |}\right ) - \frac {6 \, a e x^{2} + 3 \, a d x + 2 \, a c}{6 \, x^{3}} \]

[In]

integrate((b*x^3+a)*(h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/x^4,x, algorithm="giac")

[Out]

1/5*b*h*x^5 + 1/4*b*g*x^4 + 1/3*b*f*x^3 + 1/2*b*e*x^2 + 1/2*a*h*x^2 + b*d*x + a*g*x + (b*c + a*f)*log(abs(x))
- 1/6*(6*a*e*x^2 + 3*a*d*x + 2*a*c)/x^3

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^3\right ) \left (c+d x+e x^2+f x^3+g x^4+h x^5\right )}{x^4} \, dx=x\,\left (b\,d+a\,g\right )-\frac {a\,e\,x^2+\frac {a\,d\,x}{2}+\frac {a\,c}{3}}{x^3}+x^2\,\left (\frac {b\,e}{2}+\frac {a\,h}{2}\right )+\ln \left (x\right )\,\left (b\,c+a\,f\right )+\frac {b\,h\,x^5}{5}+\frac {b\,f\,x^3}{3}+\frac {b\,g\,x^4}{4} \]

[In]

int(((a + b*x^3)*(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5))/x^4,x)

[Out]

x*(b*d + a*g) - ((a*c)/3 + (a*d*x)/2 + a*e*x^2)/x^3 + x^2*((b*e)/2 + (a*h)/2) + log(x)*(b*c + a*f) + (b*h*x^5)
/5 + (b*f*x^3)/3 + (b*g*x^4)/4