3.1.59 \(\int \frac {1}{(a+\frac {c}{x^2}+\frac {b}{x}) x^3 (d+e x)} \, dx\)

Optimal. Leaf size=158 \[ \frac {\left (a b d+2 a c e+b^2 (-e)\right ) \tanh ^{-1}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}-\frac {e^2 \log (d+e x)}{d \left (a d^2-b d e+c e^2\right )}-\frac {(a d-b e) \log \left (a x^2+b x+c\right )}{2 c \left (a d^2-e (b d-c e)\right )}+\frac {\log (x)}{c d} \]

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Rubi [A]  time = 0.27, antiderivative size = 159, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1569, 893, 634, 618, 206, 628} \begin {gather*} \frac {\left (a b d+2 a c e+b^2 (-e)\right ) \tanh ^{-1}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}-\frac {e^2 \log (d+e x)}{d \left (a d^2-e (b d-c e)\right )}-\frac {(a d-b e) \log \left (a x^2+b x+c\right )}{2 c \left (a d^2-e (b d-c e)\right )}+\frac {\log (x)}{c d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + c/x^2 + b/x)*x^3*(d + e*x)),x]

[Out]

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

Rule 206

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; 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 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1569

Int[(x_)^(m_.)*((a_.) + (b_.)*(x_)^(mn_.) + (c_.)*(x_)^(mn2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_.))^(q_.), x_Symbo
l] :> Int[x^(m - 2*n*p)*(d + e*x^n)^q*(c + b*x^n + a*x^(2*n))^p, x] /; FreeQ[{a, b, c, d, e, m, n, q}, x] && E
qQ[mn, -n] && EqQ[mn2, 2*mn] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^3 (d+e x)} \, dx &=\int \frac {1}{x (d+e x) \left (c+b x+a x^2\right )} \, dx\\ &=\int \left (\frac {1}{c d x}+\frac {e^3}{d \left (-a d^2+e (b d-c e)\right ) (d+e x)}+\frac {b^2 e-a (b d+c e)-a (a d-b e) x}{c \left (a d^2-e (b d-c e)\right ) \left (c+b x+a x^2\right )}\right ) \, dx\\ &=\frac {\log (x)}{c d}-\frac {e^2 \log (d+e x)}{d \left (a d^2-e (b d-c e)\right )}+\frac {\int \frac {b^2 e-a (b d+c e)-a (a d-b e) x}{c+b x+a x^2} \, dx}{c \left (a d^2-b d e+c e^2\right )}\\ &=\frac {\log (x)}{c d}-\frac {e^2 \log (d+e x)}{d \left (a d^2-e (b d-c e)\right )}+\frac {\left (-a b d+b^2 e-2 a c e\right ) \int \frac {1}{c+b x+a x^2} \, dx}{2 c \left (a d^2-b d e+c e^2\right )}-\frac {(a d-b e) \int \frac {b+2 a x}{c+b x+a x^2} \, dx}{2 c \left (a d^2-e (b d-c e)\right )}\\ &=\frac {\log (x)}{c d}-\frac {e^2 \log (d+e x)}{d \left (a d^2-e (b d-c e)\right )}-\frac {(a d-b e) \log \left (c+b x+a x^2\right )}{2 c \left (a d^2-e (b d-c e)\right )}-\frac {\left (-a b d+b^2 e-2 a c e\right ) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 a x\right )}{c \left (a d^2-b d e+c e^2\right )}\\ &=\frac {\left (a b d-b^2 e+2 a c e\right ) \tanh ^{-1}\left (\frac {b+2 a x}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c} \left (a d^2-b d e+c e^2\right )}+\frac {\log (x)}{c d}-\frac {e^2 \log (d+e x)}{d \left (a d^2-e (b d-c e)\right )}-\frac {(a d-b e) \log \left (c+b x+a x^2\right )}{2 c \left (a d^2-e (b d-c e)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 152, normalized size = 0.96 \begin {gather*} -\frac {\sqrt {4 a c-b^2} \left (-2 \log (x) \left (a d^2+e (c e-b d)\right )+d (a d-b e) \log (x (a x+b)+c)+2 c e^2 \log (d+e x)\right )+2 d \left (a b d+2 a c e+b^2 (-e)\right ) \tan ^{-1}\left (\frac {2 a x+b}{\sqrt {4 a c-b^2}}\right )}{2 c d \sqrt {4 a c-b^2} \left (a d^2+e (c e-b d)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + c/x^2 + b/x)*x^3*(d + e*x)),x]

[Out]

-1/2*(2*d*(a*b*d - b^2*e + 2*a*c*e)*ArcTan[(b + 2*a*x)/Sqrt[-b^2 + 4*a*c]] + Sqrt[-b^2 + 4*a*c]*(-2*(a*d^2 + e
*(-(b*d) + c*e))*Log[x] + 2*c*e^2*Log[d + e*x] + d*(a*d - b*e)*Log[c + x*(b + a*x)]))/(c*Sqrt[-b^2 + 4*a*c]*d*
(a*d^2 + e*(-(b*d) + c*e)))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a+\frac {c}{x^2}+\frac {b}{x}\right ) x^3 (d+e x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/((a + c/x^2 + b/x)*x^3*(d + e*x)),x]

[Out]

IntegrateAlgebraic[1/((a + c/x^2 + b/x)*x^3*(d + e*x)), x]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+c/x^2+b/x)/x^3/(e*x+d),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.35, size = 164, normalized size = 1.04 \begin {gather*} -\frac {{\left (a d - b e\right )} \log \left (a x^{2} + b x + c\right )}{2 \, {\left (a c d^{2} - b c d e + c^{2} e^{2}\right )}} - \frac {e^{3} \log \left ({\left | x e + d \right |}\right )}{a d^{3} e - b d^{2} e^{2} + c d e^{3}} - \frac {{\left (a b d - b^{2} e + 2 \, a c e\right )} \arctan \left (\frac {2 \, a x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a c d^{2} - b c d e + c^{2} e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {\log \left ({\left | x \right |}\right )}{c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+c/x^2+b/x)/x^3/(e*x+d),x, algorithm="giac")

[Out]

-1/2*(a*d - b*e)*log(a*x^2 + b*x + c)/(a*c*d^2 - b*c*d*e + c^2*e^2) - e^3*log(abs(x*e + d))/(a*d^3*e - b*d^2*e
^2 + c*d*e^3) - (a*b*d - b^2*e + 2*a*c*e)*arctan((2*a*x + b)/sqrt(-b^2 + 4*a*c))/((a*c*d^2 - b*c*d*e + c^2*e^2
)*sqrt(-b^2 + 4*a*c)) + log(abs(x))/(c*d)

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maple [A]  time = 0.01, size = 285, normalized size = 1.80 \begin {gather*} -\frac {a b d \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right ) \sqrt {4 a c -b^{2}}\, c}-\frac {2 a e \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right ) \sqrt {4 a c -b^{2}}}+\frac {b^{2} e \arctan \left (\frac {2 a x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right ) \sqrt {4 a c -b^{2}}\, c}-\frac {a d \ln \left (a \,x^{2}+b x +c \right )}{2 \left (a \,d^{2}-d e b +c \,e^{2}\right ) c}+\frac {b e \ln \left (a \,x^{2}+b x +c \right )}{2 \left (a \,d^{2}-d e b +c \,e^{2}\right ) c}-\frac {e^{2} \ln \left (e x +d \right )}{\left (a \,d^{2}-d e b +c \,e^{2}\right ) d}+\frac {\ln \relax (x )}{c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+c/x^2+b/x)/x^3/(e*x+d),x)

[Out]

-1/2/(a*d^2-b*d*e+c*e^2)/c*a*ln(a*x^2+b*x+c)*d+1/2/(a*d^2-b*d*e+c*e^2)/c*ln(a*x^2+b*x+c)*b*e-1/(a*d^2-b*d*e+c*
e^2)/c/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*a*b*d-2/(a*d^2-b*d*e+c*e^2)/(4*a*c-b^2)^(1/2)*arc
tan((2*a*x+b)/(4*a*c-b^2)^(1/2))*a*e+1/(a*d^2-b*d*e+c*e^2)/c/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1
/2))*b^2*e+ln(x)/c/d-e^2*ln(e*x+d)/d/(a*d^2-b*d*e+c*e^2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+c/x^2+b/x)/x^3/(e*x+d),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 5.40, size = 2399, normalized size = 15.18

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(d + e*x)*(a + b/x + c/x^2)),x)

[Out]

(log(b^3*c^3*e^5 - 6*a^4*c^2*d^5 + 2*a^3*b^2*c*d^5 + 8*a^2*c^4*d*e^4 - b^4*c^2*d*e^4 - 2*b^5*c*d^2*e^3 + 2*a^3
*b^3*d^5*x + 8*a^2*c^4*e^5*x + b^4*c^2*e^5*x - 2*b^6*d^2*e^3*x + b^2*c^3*e^5*(b^2 - 4*a*c)^(1/2) + 18*a^3*c^3*
d^3*e^2 - 4*a*b*c^4*e^5 - 4*a*c^4*e^5*(b^2 - 4*a*c)^(1/2) - 5*a^2*c^3*d^2*e^3*(b^2 - 4*a*c)^(1/2) - 7*a^4*b*c*
d^5*x - b^5*c*d*e^4*x - 27*a^2*b^2*c^2*d^3*e^2 + 2*a^3*b*c*d^5*(b^2 - 4*a*c)^(1/2) - 3*a^4*c*d^5*x*(b^2 - 4*a*
c)^(1/2) + 2*a*b^2*c^3*d*e^4 + 6*a*b^4*c*d^3*e^2 - 6*a^2*b^3*c*d^4*e + 21*a^3*b*c^2*d^4*e - 6*a*b^2*c^3*e^5*x
+ 6*a*b^5*d^3*e^2*x - 6*a^2*b^4*d^4*e*x - 14*a^4*c^2*d^4*e*x + 7*a^3*c^2*d^4*e*(b^2 - 4*a*c)^(1/2) - b^3*c^2*d
*e^4*(b^2 - 4*a*c)^(1/2) - 2*b^4*c*d^2*e^3*(b^2 - 4*a*c)^(1/2) + 2*a^3*b^2*d^5*x*(b^2 - 4*a*c)^(1/2) + b^3*c^2
*e^5*x*(b^2 - 4*a*c)^(1/2) - 2*b^5*d^2*e^3*x*(b^2 - 4*a*c)^(1/2) + 13*a*b^3*c^2*d^2*e^3 - 21*a^2*b*c^3*d^2*e^3
+ 10*a^3*c^3*d^2*e^3*x + 6*a*b^3*c*d^3*e^2*(b^2 - 4*a*c)^(1/2) - 6*a^2*b^2*c*d^4*e*(b^2 - 4*a*c)^(1/2) + 6*a*
b^4*d^3*e^2*x*(b^2 - 4*a*c)^(1/2) - 6*a^2*b^3*d^4*e*x*(b^2 - 4*a*c)^(1/2) + 4*a^2*c^3*d*e^4*x*(b^2 - 4*a*c)^(1
/2) - 32*a^2*b^3*c*d^3*e^2*x + 35*a^3*b*c^2*d^3*e^2*x + 7*a*b^2*c^2*d^2*e^3*(b^2 - 4*a*c)^(1/2) - 13*a^2*b*c^2
*d^3*e^2*(b^2 - 4*a*c)^(1/2) + 9*a^3*c^2*d^3*e^2*x*(b^2 - 4*a*c)^(1/2) - 27*a^2*b^2*c^2*d^2*e^3*x + 4*a*b*c^3*
d*e^4*(b^2 - 4*a*c)^(1/2) - 4*a*b*c^3*e^5*x*(b^2 - 4*a*c)^(1/2) - b^4*c*d*e^4*x*(b^2 - 4*a*c)^(1/2) + 5*a*b^3*
c^2*d*e^4*x + 14*a*b^4*c*d^2*e^3*x - 4*a^2*b*c^3*d*e^4*x + 26*a^3*b^2*c*d^4*e*x + 14*a^3*b*c*d^4*e*x*(b^2 - 4*
a*c)^(1/2) + 3*a*b^2*c^2*d*e^4*x*(b^2 - 4*a*c)^(1/2) + 10*a*b^3*c*d^2*e^3*x*(b^2 - 4*a*c)^(1/2) - 13*a^2*b*c^2
*d^2*e^3*x*(b^2 - 4*a*c)^(1/2) - 20*a^2*b^2*c*d^3*e^2*x*(b^2 - 4*a*c)^(1/2))*(d*((a*b^2)/2 - 2*a^2*c + (a*b*(b
^2 - 4*a*c)^(1/2))/2) - (b^3*e)/2 - (b^2*e*(b^2 - 4*a*c)^(1/2))/2 + a*c*e*(b^2 - 4*a*c)^(1/2) + 2*a*b*c*e))/(4
*a*c^3*e^2 + 4*a^2*c^2*d^2 - b^2*c^2*e^2 + b^3*c*d*e - a*b^2*c*d^2 - 4*a*b*c^2*d*e) - (log(6*a^4*c^2*d^5 - b^3
*c^3*e^5 - 2*a^3*b^2*c*d^5 - 8*a^2*c^4*d*e^4 + b^4*c^2*d*e^4 + 2*b^5*c*d^2*e^3 - 2*a^3*b^3*d^5*x - 8*a^2*c^4*e
^5*x - b^4*c^2*e^5*x + 2*b^6*d^2*e^3*x + b^2*c^3*e^5*(b^2 - 4*a*c)^(1/2) - 18*a^3*c^3*d^3*e^2 + 4*a*b*c^4*e^5
- 4*a*c^4*e^5*(b^2 - 4*a*c)^(1/2) - 5*a^2*c^3*d^2*e^3*(b^2 - 4*a*c)^(1/2) + 7*a^4*b*c*d^5*x + b^5*c*d*e^4*x +
27*a^2*b^2*c^2*d^3*e^2 + 2*a^3*b*c*d^5*(b^2 - 4*a*c)^(1/2) - 3*a^4*c*d^5*x*(b^2 - 4*a*c)^(1/2) - 2*a*b^2*c^3*d
*e^4 - 6*a*b^4*c*d^3*e^2 + 6*a^2*b^3*c*d^4*e - 21*a^3*b*c^2*d^4*e + 6*a*b^2*c^3*e^5*x - 6*a*b^5*d^3*e^2*x + 6*
a^2*b^4*d^4*e*x + 14*a^4*c^2*d^4*e*x + 7*a^3*c^2*d^4*e*(b^2 - 4*a*c)^(1/2) - b^3*c^2*d*e^4*(b^2 - 4*a*c)^(1/2)
- 2*b^4*c*d^2*e^3*(b^2 - 4*a*c)^(1/2) + 2*a^3*b^2*d^5*x*(b^2 - 4*a*c)^(1/2) + b^3*c^2*e^5*x*(b^2 - 4*a*c)^(1/
2) - 2*b^5*d^2*e^3*x*(b^2 - 4*a*c)^(1/2) - 13*a*b^3*c^2*d^2*e^3 + 21*a^2*b*c^3*d^2*e^3 - 10*a^3*c^3*d^2*e^3*x
+ 6*a*b^3*c*d^3*e^2*(b^2 - 4*a*c)^(1/2) - 6*a^2*b^2*c*d^4*e*(b^2 - 4*a*c)^(1/2) + 6*a*b^4*d^3*e^2*x*(b^2 - 4*a
*c)^(1/2) - 6*a^2*b^3*d^4*e*x*(b^2 - 4*a*c)^(1/2) + 4*a^2*c^3*d*e^4*x*(b^2 - 4*a*c)^(1/2) + 32*a^2*b^3*c*d^3*e
^2*x - 35*a^3*b*c^2*d^3*e^2*x + 7*a*b^2*c^2*d^2*e^3*(b^2 - 4*a*c)^(1/2) - 13*a^2*b*c^2*d^3*e^2*(b^2 - 4*a*c)^(
1/2) + 9*a^3*c^2*d^3*e^2*x*(b^2 - 4*a*c)^(1/2) + 27*a^2*b^2*c^2*d^2*e^3*x + 4*a*b*c^3*d*e^4*(b^2 - 4*a*c)^(1/2
) - 4*a*b*c^3*e^5*x*(b^2 - 4*a*c)^(1/2) - b^4*c*d*e^4*x*(b^2 - 4*a*c)^(1/2) - 5*a*b^3*c^2*d*e^4*x - 14*a*b^4*c
*d^2*e^3*x + 4*a^2*b*c^3*d*e^4*x - 26*a^3*b^2*c*d^4*e*x + 14*a^3*b*c*d^4*e*x*(b^2 - 4*a*c)^(1/2) + 3*a*b^2*c^2
*d*e^4*x*(b^2 - 4*a*c)^(1/2) + 10*a*b^3*c*d^2*e^3*x*(b^2 - 4*a*c)^(1/2) - 13*a^2*b*c^2*d^2*e^3*x*(b^2 - 4*a*c)
^(1/2) - 20*a^2*b^2*c*d^3*e^2*x*(b^2 - 4*a*c)^(1/2))*((b^3*e)/2 + d*(2*a^2*c - (a*b^2)/2 + (a*b*(b^2 - 4*a*c)^
(1/2))/2) - (b^2*e*(b^2 - 4*a*c)^(1/2))/2 + a*c*e*(b^2 - 4*a*c)^(1/2) - 2*a*b*c*e))/(4*a*c^3*e^2 + 4*a^2*c^2*d
^2 - b^2*c^2*e^2 + b^3*c*d*e - a*b^2*c*d^2 - 4*a*b*c^2*d*e) - (e^2*log(d + e*x))/(a*d^3 - b*d^2*e + c*d*e^2) +
log(x)/(c*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+c/x**2+b/x)/x**3/(e*x+d),x)

[Out]

Timed out

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