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

Optimal. Leaf size=258 \[ -\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (\sqrt [3]{b} (b d-a g)-\sqrt [3]{a} (b e-a h)\right )}{6 a^{2/3} b^{5/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (\sqrt [3]{b} (b d-a g)-\sqrt [3]{a} (b e-a h)\right )}{3 a^{2/3} b^{5/3}}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (a^{4/3} (-h)+\sqrt [3]{a} b e-a \sqrt [3]{b} g+b^{4/3} d\right )}{\sqrt {3} a^{2/3} b^{5/3}}-\frac {(b c-a f) \log \left (a+b x^3\right )}{3 a b}+\frac {c \log (x)}{a}+\frac {g x}{b}+\frac {h x^2}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.47, antiderivative size = 256, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 9, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {1834, 1871, 1860, 31, 634, 617, 204, 628, 260} \[ -\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (-\frac {\sqrt [3]{a} (b e-a h)}{\sqrt [3]{b}}-a g+b d\right )}{6 a^{2/3} b^{4/3}}+\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (\sqrt [3]{b} (b d-a g)-\sqrt [3]{a} (b e-a h)\right )}{3 a^{2/3} b^{5/3}}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (a^{4/3} (-h)+\sqrt [3]{a} b e-a \sqrt [3]{b} g+b^{4/3} d\right )}{\sqrt {3} a^{2/3} b^{5/3}}-\frac {(b c-a f) \log \left (a+b x^3\right )}{3 a b}+\frac {c \log (x)}{a}+\frac {g x}{b}+\frac {h x^2}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(g*x)/b + (h*x^2)/(2*b) - ((b^(4/3)*d + a^(1/3)*b*e - a*b^(1/3)*g - a^(4/3)*h)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/
(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(2/3)*b^(5/3)) + (c*Log[x])/a + ((b^(1/3)*(b*d - a*g) - a^(1/3)*(b*e - a*h))*Lo
g[a^(1/3) + b^(1/3)*x])/(3*a^(2/3)*b^(5/3)) - ((b*d - a*g - (a^(1/3)*(b*e - a*h))/b^(1/3))*Log[a^(2/3) - a^(1/
3)*b^(1/3)*x + b^(2/3)*x^2])/(6*a^(2/3)*b^(4/3)) - ((b*c - a*f)*Log[a + b*x^3])/(3*a*b)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

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 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1834

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

Rule 1860

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[a/b, 3]], s = Denominator[R
t[a/b, 3]]}, -Dist[(r*(B*r - A*s))/(3*a*s), Int[1/(r + s*x), x], x] + Dist[r/(3*a*s), Int[(r*(B*r + 2*A*s) + s
*(B*r - A*s)*x)/(r^2 - r*s*x + s^2*x^2), x], x]] /; FreeQ[{a, b, A, B}, x] && NeQ[a*B^3 - b*A^3, 0] && PosQ[a/
b]

Rule 1871

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,
2]}, Int[(A + B*x)/(a + b*x^3), x] + Dist[C, Int[x^2/(a + b*x^3), x], x] /; EqQ[a*B^3 - b*A^3, 0] ||  !Ration
alQ[a/b]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2]

Rubi steps

\begin {align*} \int \frac {c+d x+e x^2+f x^3+g x^4+h x^5}{x \left (a+b x^3\right )} \, dx &=\int \left (\frac {g}{b}+\frac {c}{a x}+\frac {h x}{b}+\frac {a (b d-a g)+a (b e-a h) x-b (b c-a f) x^2}{a b \left (a+b x^3\right )}\right ) \, dx\\ &=\frac {g x}{b}+\frac {h x^2}{2 b}+\frac {c \log (x)}{a}+\frac {\int \frac {a (b d-a g)+a (b e-a h) x-b (b c-a f) x^2}{a+b x^3} \, dx}{a b}\\ &=\frac {g x}{b}+\frac {h x^2}{2 b}+\frac {c \log (x)}{a}+\frac {\int \frac {a (b d-a g)+a (b e-a h) x}{a+b x^3} \, dx}{a b}-\frac {(b c-a f) \int \frac {x^2}{a+b x^3} \, dx}{a}\\ &=\frac {g x}{b}+\frac {h x^2}{2 b}+\frac {c \log (x)}{a}-\frac {(b c-a f) \log \left (a+b x^3\right )}{3 a b}+\frac {\int \frac {\sqrt [3]{a} \left (2 a \sqrt [3]{b} (b d-a g)+a^{4/3} (b e-a h)\right )+\sqrt [3]{b} \left (-a \sqrt [3]{b} (b d-a g)+a^{4/3} (b e-a h)\right ) x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{3 a^{5/3} b^{4/3}}+\frac {\left (b d-a g-\frac {\sqrt [3]{a} (b e-a h)}{\sqrt [3]{b}}\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 a^{2/3} b}\\ &=\frac {g x}{b}+\frac {h x^2}{2 b}+\frac {c \log (x)}{a}+\frac {\left (b d-a g-\frac {\sqrt [3]{a} (b e-a h)}{\sqrt [3]{b}}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} b^{4/3}}-\frac {(b c-a f) \log \left (a+b x^3\right )}{3 a b}+\frac {\left (b^{4/3} d+\sqrt [3]{a} b e-a \sqrt [3]{b} g-a^{4/3} h\right ) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{2 \sqrt [3]{a} b^{4/3}}-\frac {\left (b d-a g-\frac {\sqrt [3]{a} (b e-a h)}{\sqrt [3]{b}}\right ) \int \frac {-\sqrt [3]{a} \sqrt [3]{b}+2 b^{2/3} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{6 a^{2/3} b^{4/3}}\\ &=\frac {g x}{b}+\frac {h x^2}{2 b}+\frac {c \log (x)}{a}+\frac {\left (b d-a g-\frac {\sqrt [3]{a} (b e-a h)}{\sqrt [3]{b}}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} b^{4/3}}-\frac {\left (b d-a g-\frac {\sqrt [3]{a} (b e-a h)}{\sqrt [3]{b}}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{2/3} b^{4/3}}-\frac {(b c-a f) \log \left (a+b x^3\right )}{3 a b}+\frac {\left (b^{4/3} d+\sqrt [3]{a} b e-a \sqrt [3]{b} g-a^{4/3} h\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{a^{2/3} b^{5/3}}\\ &=\frac {g x}{b}+\frac {h x^2}{2 b}-\frac {\left (b^{4/3} d+\sqrt [3]{a} b e-a \sqrt [3]{b} g-a^{4/3} h\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{2/3} b^{5/3}}+\frac {c \log (x)}{a}+\frac {\left (b d-a g-\frac {\sqrt [3]{a} (b e-a h)}{\sqrt [3]{b}}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{2/3} b^{4/3}}-\frac {\left (b d-a g-\frac {\sqrt [3]{a} (b e-a h)}{\sqrt [3]{b}}\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{2/3} b^{4/3}}-\frac {(b c-a f) \log \left (a+b x^3\right )}{3 a b}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 258, normalized size = 1.00 \[ \frac {-\sqrt [3]{a} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right ) \left (a^{4/3} h-\sqrt [3]{a} b e-a \sqrt [3]{b} g+b^{4/3} d\right )+2 \sqrt [3]{a} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \left (a^{4/3} h-\sqrt [3]{a} b e-a \sqrt [3]{b} g+b^{4/3} d\right )+2 \sqrt {3} \sqrt [3]{a} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right ) \left (a^{4/3} h-\sqrt [3]{a} b e+a \sqrt [3]{b} g-b^{4/3} d\right )-2 b^{2/3} (b c-a f) \log \left (a+b x^3\right )+6 a b^{2/3} g x+3 a b^{2/3} h x^2+6 b^{5/3} c \log (x)}{6 a b^{5/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a*b^(2/3)*g*x + 3*a*b^(2/3)*h*x^2 + 2*Sqrt[3]*a^(1/3)*(-(b^(4/3)*d) - a^(1/3)*b*e + a*b^(1/3)*g + a^(4/3)*h
)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] + 6*b^(5/3)*c*Log[x] + 2*a^(1/3)*(b^(4/3)*d - a^(1/3)*b*e - a*b^
(1/3)*g + a^(4/3)*h)*Log[a^(1/3) + b^(1/3)*x] - a^(1/3)*(b^(4/3)*d - a^(1/3)*b*e - a*b^(1/3)*g + a^(4/3)*h)*Lo
g[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2] - 2*b^(2/3)*(b*c - a*f)*Log[a + b*x^3])/(6*a*b^(5/3))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [A]  time = 0.19, size = 281, normalized size = 1.09 \[ \frac {c \log \left ({\left | x \right |}\right )}{a} - \frac {\sqrt {3} {\left (b^{2} d - a b g + \left (-a b^{2}\right )^{\frac {1}{3}} a h - \left (-a b^{2}\right )^{\frac {1}{3}} b e\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, \left (-a b^{2}\right )^{\frac {2}{3}} b} - \frac {{\left (b^{2} d - a b g - \left (-a b^{2}\right )^{\frac {1}{3}} a h + \left (-a b^{2}\right )^{\frac {1}{3}} b e\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, \left (-a b^{2}\right )^{\frac {2}{3}} b} - \frac {{\left (b c - a f\right )} \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a b} + \frac {b h x^{2} + 2 \, b g x}{2 \, b^{2}} + \frac {{\left (a^{3} b^{2} h \left (-\frac {a}{b}\right )^{\frac {1}{3}} - a^{2} b^{3} \left (-\frac {a}{b}\right )^{\frac {1}{3}} e - a^{2} b^{3} d + a^{3} b^{2} g\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, a^{3} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 426, normalized size = 1.65 \[ \frac {h \,x^{2}}{2 b}-\frac {\sqrt {3}\, a g \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{2}}-\frac {a g \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{2}}+\frac {a g \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{2}}-\frac {\sqrt {3}\, a h \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{2}}+\frac {a h \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{2}}-\frac {a h \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{2}}+\frac {c \ln \relax (x )}{a}-\frac {c \ln \left (b \,x^{3}+a \right )}{3 a}+\frac {\sqrt {3}\, d \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} b}+\frac {d \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} b}-\frac {d \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {2}{3}} b}+\frac {\sqrt {3}\, e \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} b}-\frac {e \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} b}+\frac {e \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {1}{3}} b}+\frac {f \ln \left (b \,x^{3}+a \right )}{3 b}+\frac {g x}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.02, size = 290, normalized size = 1.12 \[ \frac {c \log \relax (x)}{a} + \frac {h x^{2} + 2 \, g x}{2 \, b} + \frac {\sqrt {3} {\left (a b e \left (\frac {a}{b}\right )^{\frac {2}{3}} - a^{2} h \left (\frac {a}{b}\right )^{\frac {2}{3}} + a b d \left (\frac {a}{b}\right )^{\frac {1}{3}} - a^{2} g \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a^{2} b} - \frac {{\left (2 \, b^{2} c \left (\frac {a}{b}\right )^{\frac {2}{3}} - 2 \, a b f \left (\frac {a}{b}\right )^{\frac {2}{3}} - a b e \left (\frac {a}{b}\right )^{\frac {1}{3}} + a^{2} h \left (\frac {a}{b}\right )^{\frac {1}{3}} + a b d - a^{2} g\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (b^{2} c \left (\frac {a}{b}\right )^{\frac {2}{3}} - a b f \left (\frac {a}{b}\right )^{\frac {2}{3}} + a b e \left (\frac {a}{b}\right )^{\frac {1}{3}} - a^{2} h \left (\frac {a}{b}\right )^{\frac {1}{3}} - a b d + a^{2} g\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, a b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.10, size = 1731, normalized size = 6.71 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

symsum(log(b^2*c*d^2 - root(27*a^3*b^5*z^3 - 27*a^3*b^4*f*z^2 + 27*a^2*b^5*c*z^2 + 9*a^4*b^2*g*h*z - 9*a^3*b^3
*e*g*z - 9*a^3*b^3*d*h*z - 18*a^2*b^4*c*f*z + 9*a^2*b^4*d*e*z + 9*a*b^5*c^2*z + 9*a^3*b^3*f^2*z - 3*a^4*b*f*g*
h + 3*a*b^4*c*d*e + 3*a^3*b^2*e*f*g + 3*a^3*b^2*d*f*h + 3*a^3*b^2*c*g*h - 3*a^2*b^3*d*e*f - 3*a^2*b^3*c*e*g -
3*a^2*b^3*c*d*h + 3*a^4*b*e*h^2 - 3*a*b^4*c^2*f - 3*a^3*b^2*e^2*h - 3*a^3*b^2*d*g^2 + 3*a^2*b^3*d^2*g + 3*a^2*
b^3*c*f^2 + a^2*b^3*e^3 + a^4*b*g^3 + b^5*c^3 - a^3*b^2*f^3 - a*b^4*d^3 - a^5*h^3, z, k)*(a^3*g^2 - root(27*a^
3*b^5*z^3 - 27*a^3*b^4*f*z^2 + 27*a^2*b^5*c*z^2 + 9*a^4*b^2*g*h*z - 9*a^3*b^3*e*g*z - 9*a^3*b^3*d*h*z - 18*a^2
*b^4*c*f*z + 9*a^2*b^4*d*e*z + 9*a*b^5*c^2*z + 9*a^3*b^3*f^2*z - 3*a^4*b*f*g*h + 3*a*b^4*c*d*e + 3*a^3*b^2*e*f
*g + 3*a^3*b^2*d*f*h + 3*a^3*b^2*c*g*h - 3*a^2*b^3*d*e*f - 3*a^2*b^3*c*e*g - 3*a^2*b^3*c*d*h + 3*a^4*b*e*h^2 -
3*a*b^4*c^2*f - 3*a^3*b^2*e^2*h - 3*a^3*b^2*d*g^2 + 3*a^2*b^3*d^2*g + 3*a^2*b^3*c*f^2 + a^2*b^3*e^3 + a^4*b*g
^3 + b^5*c^3 - a^3*b^2*f^3 - a*b^4*d^3 - a^5*h^3, z, k)*((x*(33*a^2*b^4*f - 24*a*b^5*c))/b^2 + 3*a^2*b^2*e - 3
*a^3*b*h - 36*root(27*a^3*b^5*z^3 - 27*a^3*b^4*f*z^2 + 27*a^2*b^5*c*z^2 + 9*a^4*b^2*g*h*z - 9*a^3*b^3*e*g*z -
9*a^3*b^3*d*h*z - 18*a^2*b^4*c*f*z + 9*a^2*b^4*d*e*z + 9*a*b^5*c^2*z + 9*a^3*b^3*f^2*z - 3*a^4*b*f*g*h + 3*a*b
^4*c*d*e + 3*a^3*b^2*e*f*g + 3*a^3*b^2*d*f*h + 3*a^3*b^2*c*g*h - 3*a^2*b^3*d*e*f - 3*a^2*b^3*c*e*g - 3*a^2*b^3
*c*d*h + 3*a^4*b*e*h^2 - 3*a*b^4*c^2*f - 3*a^3*b^2*e^2*h - 3*a^3*b^2*d*g^2 + 3*a^2*b^3*d^2*g + 3*a^2*b^3*c*f^2
+ a^2*b^3*e^3 + a^4*b*g^3 + b^5*c^3 - a^3*b^2*f^3 - a*b^4*d^3 - a^5*h^3, z, k)*a^2*b^3*x) + (x*(4*b^5*c^2 + 1
0*a^2*b^3*f^2 - 14*a*b^4*c*f + 10*a*b^4*d*e - 10*a^2*b^3*d*h - 10*a^2*b^3*e*g + 10*a^3*b^2*g*h))/b^2 + a*b^2*d
^2 - a^3*f*h + 2*a*b^2*c*e - 2*a^2*b*c*h - 2*a^2*b*d*g + a^2*b*e*f) - b^2*c^2*e + a^2*c*g^2 + (x*(b^4*d^3 + a^
4*h^3 - a*b^3*e^3 - a^3*b*g^3 + b^4*c^2*f + a^2*b^2*f^3 + 3*a^2*b^2*d*g^2 + 3*a^2*b^2*e^2*h - 2*b^4*c*d*e - 2*
a*b^3*c*f^2 - 3*a*b^3*d^2*g - 3*a^3*b*e*h^2 - 2*a^2*b^2*c*g*h - 3*a^2*b^2*d*f*h - 3*a^2*b^2*e*f*g + 2*a*b^3*c*
d*h + 2*a*b^3*c*e*g + 3*a*b^3*d*e*f + 3*a^3*b*f*g*h))/b^2 + a*b*c^2*h - a^2*c*f*h - 2*a*b*c*d*g + a*b*c*e*f)*r
oot(27*a^3*b^5*z^3 - 27*a^3*b^4*f*z^2 + 27*a^2*b^5*c*z^2 + 9*a^4*b^2*g*h*z - 9*a^3*b^3*e*g*z - 9*a^3*b^3*d*h*z
- 18*a^2*b^4*c*f*z + 9*a^2*b^4*d*e*z + 9*a*b^5*c^2*z + 9*a^3*b^3*f^2*z - 3*a^4*b*f*g*h + 3*a*b^4*c*d*e + 3*a^
3*b^2*e*f*g + 3*a^3*b^2*d*f*h + 3*a^3*b^2*c*g*h - 3*a^2*b^3*d*e*f - 3*a^2*b^3*c*e*g - 3*a^2*b^3*c*d*h + 3*a^4*
b*e*h^2 - 3*a*b^4*c^2*f - 3*a^3*b^2*e^2*h - 3*a^3*b^2*d*g^2 + 3*a^2*b^3*d^2*g + 3*a^2*b^3*c*f^2 + a^2*b^3*e^3
+ a^4*b*g^3 + b^5*c^3 - a^3*b^2*f^3 - a*b^4*d^3 - a^5*h^3, z, k), k, 1, 3) + (h*x^2)/(2*b) + (c*log(x))/a + (g
*x)/b

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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