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

Optimal. Leaf size=86 \[ \log (x) (a f+b c)+x (a g+b d)+\frac {1}{2} x^2 (a h+b e)-\frac {a c}{3 x^3}-\frac {a d}{2 x^2}-\frac {a e}{x}+\frac {1}{3} b f x^3+\frac {1}{4} b g x^4+\frac {1}{5} b h x^5 \]

[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)

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Rubi [A]  time = 0.07, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {1820} \[ \log (x) (a f+b c)+x (a g+b d)+\frac {1}{2} x^2 (a h+b e)-\frac {a c}{3 x^3}-\frac {a d}{2 x^2}-\frac {a e}{x}+\frac {1}{3} b f x^3+\frac {1}{4} b g x^4+\frac {1}{5} b h x^5 \]

Antiderivative was successfully verified.

[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]

-(a*c)/(3*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 1820

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*} \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 &=\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*}

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Mathematica [A]  time = 0.07, size = 76, normalized size = 0.88 \[ \log (x) (a f+b c)-\frac {a \left (2 c+3 x \left (d+2 e x-\left (x^3 (2 g+h x)\right )\right )\right )}{6 x^3}+\frac {1}{60} b x \left (60 d+x \left (30 e+x \left (20 f+15 g x+12 h x^2\right )\right )\right ) \]

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.94, size = 81, normalized size = 0.94 \[ \frac {12 \, b h x^{8} + 15 \, b g x^{7} + 20 \, b f x^{6} + 30 \, {\left (b e + a h\right )} x^{5} + 60 \, {\left (b d + a g\right )} x^{4} + 60 \, {\left (b c + a f\right )} x^{3} \log \relax (x) - 60 \, a e x^{2} - 30 \, a d x - 20 \, a c}{60 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.17, size = 79, normalized size = 0.92 \[ \frac {1}{5} \, b h x^{5} + \frac {1}{4} \, b g x^{4} + \frac {1}{3} \, b f x^{3} + \frac {1}{2} \, a h x^{2} + \frac {1}{2} \, b x^{2} e + b d x + a g x + {\left (b c + a f\right )} \log \left ({\left | x \right |}\right ) - \frac {6 \, a x^{2} e + 3 \, a d x + 2 \, a c}{6 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

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maple [A]  time = 0.05, size = 76, normalized size = 0.88 \[ \frac {b h \,x^{5}}{5}+\frac {b g \,x^{4}}{4}+\frac {b f \,x^{3}}{3}+\frac {a h \,x^{2}}{2}+\frac {b e \,x^{2}}{2}+a f \ln \relax (x )+a g x +b c \ln \relax (x )+b d x -\frac {a e}{x}-\frac {a d}{2 x^{2}}-\frac {a c}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.36, size = 75, normalized size = 0.87 \[ \frac {1}{5} \, b h x^{5} + \frac {1}{4} \, b g x^{4} + \frac {1}{3} \, b f x^{3} + \frac {1}{2} \, {\left (b e + a h\right )} x^{2} + {\left (b d + a g\right )} x + {\left (b c + a f\right )} \log \relax (x) - \frac {6 \, a e x^{2} + 3 \, a d x + 2 \, a c}{6 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 0.04, size = 75, normalized size = 0.87 \[ x\,\left (b\,d+a\,g\right )-\frac {a\,e\,x^2+\frac {a\,d\,x}{2}+\frac {a\,c}{3}}{x^3}+x^2\,\left (\frac {b\,e}{2}+\frac {a\,h}{2}\right )+\ln \relax (x)\,\left (b\,c+a\,f\right )+\frac {b\,h\,x^5}{5}+\frac {b\,f\,x^3}{3}+\frac {b\,g\,x^4}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.67, size = 83, normalized size = 0.97 \[ \frac {b f x^{3}}{3} + \frac {b g x^{4}}{4} + \frac {b h x^{5}}{5} + x^{2} \left (\frac {a h}{2} + \frac {b e}{2}\right ) + x \left (a g + b d\right ) + \left (a f + b c\right ) \log {\relax (x )} + \frac {- 2 a c - 3 a d x - 6 a e x^{2}}{6 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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