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

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

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

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Rubi [A]  time = 0.270758, antiderivative size = 159, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {1569, 893, 634, 618, 206, 628} \[ \frac{\left (a b d+2 a c e+b^2 (-e)\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{c \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )}-\frac{e^2 \log (d+e x)}{d \left (a d^2-e (b d-c e)\right )}-\frac{(a d-b e) \log \left (a x^2+b x+c\right )}{2 c \left (a d^2-e (b d-c e)\right )}+\frac{\log (x)}{c d} \]

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 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]

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 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 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 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 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]

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*}

Mathematica [A]  time = 0.188878, size = 152, normalized size = 0.96 \[ -\frac{\sqrt{4 a c-b^2} \left (-2 \log (x) \left (a d^2+e (c e-b d)\right )+d (a d-b e) \log (x (a x+b)+c)+2 c e^2 \log (d+e x)\right )+2 d \left (a b d+2 a c e+b^2 (-e)\right ) \tan ^{-1}\left (\frac{2 a x+b}{\sqrt{4 a c-b^2}}\right )}{2 c d \sqrt{4 a c-b^2} \left (a d^2+e (c e-b d)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 285, normalized size = 1.8 \begin{align*} -{\frac{{e}^{2}\ln \left ( ex+d \right ) }{d \left ( a{d}^{2}-bde+{e}^{2}c \right ) }}-{\frac{a\ln \left ( a{x}^{2}+bx+c \right ) d}{ \left ( 2\,a{d}^{2}-2\,bde+2\,{e}^{2}c \right ) c}}+{\frac{\ln \left ( a{x}^{2}+bx+c \right ) be}{ \left ( 2\,a{d}^{2}-2\,bde+2\,{e}^{2}c \right ) c}}-{\frac{abd}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) c}\arctan \left ({(2\,ax+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}-2\,{\frac{ae}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) \sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,ax+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+{\frac{{b}^{2}e}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) c}\arctan \left ({(2\,ax+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+{\frac{\ln \left ( x \right ) }{cd}} \end{align*}

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]

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

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

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [A]  time = 1.1042, size = 221, normalized size = 1.4 \begin{align*} -\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{align*}

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)