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

Optimal. Leaf size=124 \[ \frac {(b d-2 c e) \tanh ^{-1}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac {d \log \left (a x^2+b x+c\right )}{2 \left (a d^2-e (b d-c e)\right )}-\frac {d \log (d+e x)}{a d^2-e (b d-c e)} \]

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Rubi [A]  time = 0.14, antiderivative size = 124, normalized size of antiderivative = 1.00, 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, 800, 634, 618, 206, 628} \begin {gather*} \frac {(b d-2 c e) \tanh ^{-1}\left (\frac {2 a x+b}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac {d \log \left (a x^2+b x+c\right )}{2 \left (a d^2-e (b d-c e)\right )}-\frac {d \log (d+e x)}{a d^2-e (b d-c e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*d - 2*c*e)*ArcTanh[(b + 2*a*x)/Sqrt[b^2 - 4*a*c]])/(Sqrt[b^2 - 4*a*c]*(a*d^2 - e*(b*d - c*e))) - (d*Log[d
+ e*x])/(a*d^2 - e*(b*d - c*e)) + (d*Log[c + b*x + a*x^2])/(2*(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 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

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

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Mathematica [A]  time = 0.07, size = 107, normalized size = 0.86 \begin {gather*} \frac {d \sqrt {4 a c-b^2} (2 \log (d+e x)-\log (x (a x+b)+c))+2 (b d-2 c e) \tan ^{-1}\left (\frac {2 a x+b}{\sqrt {4 a c-b^2}}\right )}{2 \sqrt {4 a c-b^2} \left (e (b d-c e)-a d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*d - 2*c*e)*ArcTan[(b + 2*a*x)/Sqrt[-b^2 + 4*a*c]] + Sqrt[-b^2 + 4*a*c]*d*(2*Log[d + e*x] - Log[c + x*(b
+ a*x)]))/(2*Sqrt[-b^2 + 4*a*c]*(-(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 (d+e x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 2.17, size = 305, normalized size = 2.46 \begin {gather*} \left [\frac {{\left (b^{2} - 4 \, a c\right )} d \log \left (a x^{2} + b x + c\right ) - 2 \, {\left (b^{2} - 4 \, a c\right )} d \log \left (e x + d\right ) - \sqrt {b^{2} - 4 \, a c} {\left (b d - 2 \, c e\right )} \log \left (\frac {2 \, a^{2} x^{2} + 2 \, a b x + b^{2} - 2 \, a c - \sqrt {b^{2} - 4 \, a c} {\left (2 \, a x + b\right )}}{a x^{2} + b x + c}\right )}{2 \, {\left ({\left (a b^{2} - 4 \, a^{2} c\right )} d^{2} - {\left (b^{3} - 4 \, a b c\right )} d e + {\left (b^{2} c - 4 \, a c^{2}\right )} e^{2}\right )}}, \frac {{\left (b^{2} - 4 \, a c\right )} d \log \left (a x^{2} + b x + c\right ) - 2 \, {\left (b^{2} - 4 \, a c\right )} d \log \left (e x + d\right ) + 2 \, \sqrt {-b^{2} + 4 \, a c} {\left (b d - 2 \, c e\right )} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, a x + b\right )}}{b^{2} - 4 \, a c}\right )}{2 \, {\left ({\left (a b^{2} - 4 \, a^{2} c\right )} d^{2} - {\left (b^{3} - 4 \, a b c\right )} d e + {\left (b^{2} c - 4 \, a c^{2}\right )} e^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.39, size = 127, normalized size = 1.02 \begin {gather*} -\frac {d e \log \left ({\left | x e + d \right |}\right )}{a d^{2} e - b d e^{2} + c e^{3}} + \frac {d \log \left (a x^{2} + b x + c\right )}{2 \, {\left (a d^{2} - b d e + c e^{2}\right )}} - \frac {{\left (b d - 2 \, c e\right )} \arctan \left (\frac {2 \, a x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a d^{2} - b d e + c e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 169, normalized size = 1.36 \begin {gather*} -\frac {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}}}+\frac {2 c 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 {d \ln \left (e x +d \right )}{a \,d^{2}-d e b +c \,e^{2}}+\frac {d \ln \left (a \,x^{2}+b x +c \right )}{2 a \,d^{2}-2 d e b +2 c \,e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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/(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 = 3.41, size = 801, normalized size = 6.46 \begin {gather*} \frac {\ln \left (a\,e\,x-\frac {\left (d\,\left (\frac {b\,\sqrt {b^2-4\,a\,c}}{2}-2\,a\,c+\frac {b^2}{2}\right )-c\,e\,\sqrt {b^2-4\,a\,c}\right )\,\left (x\,\left (d\,a^2\,e+b\,a\,e^2\right )+\frac {\left (d\,\left (\frac {b\,\sqrt {b^2-4\,a\,c}}{2}-2\,a\,c+\frac {b^2}{2}\right )-c\,e\,\sqrt {b^2-4\,a\,c}\right )\,\left (x\,\left (2\,a^3\,d^2\,e-2\,a^2\,b\,d\,e^2-6\,c\,a^2\,e^3+2\,a\,b^2\,e^3\right )+a\,b\,c\,e^3+a\,b^2\,d\,e^2+a^2\,b\,d^2\,e-8\,a^2\,c\,d\,e^2\right )}{-4\,a^2\,c\,d^2+a\,b^2\,d^2+4\,a\,b\,c\,d\,e-4\,a\,c^2\,e^2-b^3\,d\,e+b^2\,c\,e^2}+a\,c\,e^2+a\,b\,d\,e\right )}{-4\,a^2\,c\,d^2+a\,b^2\,d^2+4\,a\,b\,c\,d\,e-4\,a\,c^2\,e^2-b^3\,d\,e+b^2\,c\,e^2}\right )\,\left (d\,\left (\frac {b\,\sqrt {b^2-4\,a\,c}}{2}-2\,a\,c+\frac {b^2}{2}\right )-c\,e\,\sqrt {b^2-4\,a\,c}\right )}{-4\,a^2\,c\,d^2+a\,b^2\,d^2+4\,a\,b\,c\,d\,e-4\,a\,c^2\,e^2-b^3\,d\,e+b^2\,c\,e^2}-\frac {\ln \left (\frac {\left (d\,\left (2\,a\,c+\frac {b\,\sqrt {b^2-4\,a\,c}}{2}-\frac {b^2}{2}\right )-c\,e\,\sqrt {b^2-4\,a\,c}\right )\,\left (x\,\left (d\,a^2\,e+b\,a\,e^2\right )-\frac {\left (d\,\left (2\,a\,c+\frac {b\,\sqrt {b^2-4\,a\,c}}{2}-\frac {b^2}{2}\right )-c\,e\,\sqrt {b^2-4\,a\,c}\right )\,\left (x\,\left (2\,a^3\,d^2\,e-2\,a^2\,b\,d\,e^2-6\,c\,a^2\,e^3+2\,a\,b^2\,e^3\right )+a\,b\,c\,e^3+a\,b^2\,d\,e^2+a^2\,b\,d^2\,e-8\,a^2\,c\,d\,e^2\right )}{-4\,a^2\,c\,d^2+a\,b^2\,d^2+4\,a\,b\,c\,d\,e-4\,a\,c^2\,e^2-b^3\,d\,e+b^2\,c\,e^2}+a\,c\,e^2+a\,b\,d\,e\right )}{-4\,a^2\,c\,d^2+a\,b^2\,d^2+4\,a\,b\,c\,d\,e-4\,a\,c^2\,e^2-b^3\,d\,e+b^2\,c\,e^2}+a\,e\,x\right )\,\left (d\,\left (2\,a\,c+\frac {b\,\sqrt {b^2-4\,a\,c}}{2}-\frac {b^2}{2}\right )-c\,e\,\sqrt {b^2-4\,a\,c}\right )}{-4\,a^2\,c\,d^2+a\,b^2\,d^2+4\,a\,b\,c\,d\,e-4\,a\,c^2\,e^2-b^3\,d\,e+b^2\,c\,e^2}-\frac {d\,\ln \left (d+e\,x\right )}{a\,d^2-b\,d\,e+c\,e^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(a*e*x - ((d*((b*(b^2 - 4*a*c)^(1/2))/2 - 2*a*c + b^2/2) - c*e*(b^2 - 4*a*c)^(1/2))*(x*(a*b*e^2 + a^2*d*e)
+ ((d*((b*(b^2 - 4*a*c)^(1/2))/2 - 2*a*c + b^2/2) - c*e*(b^2 - 4*a*c)^(1/2))*(x*(2*a*b^2*e^3 - 6*a^2*c*e^3 +
2*a^3*d^2*e - 2*a^2*b*d*e^2) + a*b*c*e^3 + a*b^2*d*e^2 + a^2*b*d^2*e - 8*a^2*c*d*e^2))/(a*b^2*d^2 - 4*a^2*c*d^
2 - 4*a*c^2*e^2 + b^2*c*e^2 - b^3*d*e + 4*a*b*c*d*e) + a*c*e^2 + a*b*d*e))/(a*b^2*d^2 - 4*a^2*c*d^2 - 4*a*c^2*
e^2 + b^2*c*e^2 - b^3*d*e + 4*a*b*c*d*e))*(d*((b*(b^2 - 4*a*c)^(1/2))/2 - 2*a*c + b^2/2) - c*e*(b^2 - 4*a*c)^(
1/2)))/(a*b^2*d^2 - 4*a^2*c*d^2 - 4*a*c^2*e^2 + b^2*c*e^2 - b^3*d*e + 4*a*b*c*d*e) - (log(((d*(2*a*c + (b*(b^2
- 4*a*c)^(1/2))/2 - b^2/2) - c*e*(b^2 - 4*a*c)^(1/2))*(x*(a*b*e^2 + a^2*d*e) - ((d*(2*a*c + (b*(b^2 - 4*a*c)^
(1/2))/2 - b^2/2) - c*e*(b^2 - 4*a*c)^(1/2))*(x*(2*a*b^2*e^3 - 6*a^2*c*e^3 + 2*a^3*d^2*e - 2*a^2*b*d*e^2) + a*
b*c*e^3 + a*b^2*d*e^2 + a^2*b*d^2*e - 8*a^2*c*d*e^2))/(a*b^2*d^2 - 4*a^2*c*d^2 - 4*a*c^2*e^2 + b^2*c*e^2 - b^3
*d*e + 4*a*b*c*d*e) + a*c*e^2 + a*b*d*e))/(a*b^2*d^2 - 4*a^2*c*d^2 - 4*a*c^2*e^2 + b^2*c*e^2 - b^3*d*e + 4*a*b
*c*d*e) + a*e*x)*(d*(2*a*c + (b*(b^2 - 4*a*c)^(1/2))/2 - b^2/2) - c*e*(b^2 - 4*a*c)^(1/2)))/(a*b^2*d^2 - 4*a^2
*c*d^2 - 4*a*c^2*e^2 + b^2*c*e^2 - b^3*d*e + 4*a*b*c*d*e) - (d*log(d + e*x))/(a*d^2 + c*e^2 - b*d*e)

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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/(e*x+d),x)

[Out]

Timed out

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