3.68 \(\int \frac{1}{(a+\frac{c}{x^2}+\frac{b}{x}) x^4 (d+e x)} \, dx\)

Optimal. Leaf size=193 \[ \frac{\left (2 a^2 c d-a b (b d+3 c e)+b^3 e\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{c^2 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac{\left (a b d+a c e+b^2 (-e)\right ) \log \left (a x^2+b x+c\right )}{2 c^2 \left (a d^2-e (b d-c e)\right )}+\frac{e^3 \log (d+e x)}{d^2 \left (a d^2-e (b d-c e)\right )}-\frac{\log (x) (b d+c e)}{c^2 d^2}-\frac{1}{c d x} \]

[Out]

-(1/(c*d*x)) + ((2*a^2*c*d + b^3*e - a*b*(b*d + 3*c*e))*ArcTanh[(b + 2*a*x)/Sqrt[b^2 - 4*a*c]])/(c^2*Sqrt[b^2
- 4*a*c]*(a*d^2 - e*(b*d - c*e))) - ((b*d + c*e)*Log[x])/(c^2*d^2) + (e^3*Log[d + e*x])/(d^2*(a*d^2 - e*(b*d -
c*e))) + ((a*b*d - b^2*e + a*c*e)*Log[c + b*x + a*x^2])/(2*c^2*(a*d^2 - e*(b*d - c*e)))

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Rubi [A]  time = 0.343094, antiderivative size = 193, normalized size of antiderivative = 1., 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} \[ \frac{\left (2 a^2 c d-a b (b d+3 c e)+b^3 e\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{c^2 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac{\left (a b d+a c e+b^2 (-e)\right ) \log \left (a x^2+b x+c\right )}{2 c^2 \left (a d^2-e (b d-c e)\right )}+\frac{e^3 \log (d+e x)}{d^2 \left (a d^2-e (b d-c e)\right )}-\frac{\log (x) (b d+c e)}{c^2 d^2}-\frac{1}{c d x} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + c/x^2 + b/x)*x^4*(d + e*x)),x]

[Out]

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

Mathematica [A]  time = 0.177771, size = 194, normalized size = 1.01 \[ \frac{\left (2 a^2 c d-a b (b d+3 c e)+b^3 e\right ) \tan ^{-1}\left (\frac{2 a x+b}{\sqrt{4 a c-b^2}}\right )}{c^2 \sqrt{4 a c-b^2} \left (e (b d-c e)-a d^2\right )}+\frac{\left (a b d+a c e+b^2 (-e)\right ) \log (x (a x+b)+c)}{2 c^2 \left (a d^2+e (c e-b d)\right )}+\frac{e^3 \log (d+e x)}{a d^4+d^2 e (c e-b d)}-\frac{\log (x) (b d+c e)}{c^2 d^2}-\frac{1}{c d x} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + c/x^2 + b/x)*x^4*(d + e*x)),x]

[Out]

-(1/(c*d*x)) + ((2*a^2*c*d + b^3*e - a*b*(b*d + 3*c*e))*ArcTan[(b + 2*a*x)/Sqrt[-b^2 + 4*a*c]])/(c^2*Sqrt[-b^2
+ 4*a*c]*(-(a*d^2) + e*(b*d - c*e))) - ((b*d + c*e)*Log[x])/(c^2*d^2) + (e^3*Log[d + e*x])/(a*d^4 + d^2*e*(-(
b*d) + c*e)) + ((a*b*d - b^2*e + a*c*e)*Log[c + x*(b + a*x)])/(2*c^2*(a*d^2 + e*(-(b*d) + c*e)))

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Maple [B]  time = 0.011, size = 412, normalized size = 2.1 \begin{align*}{\frac{{e}^{3}\ln \left ( ex+d \right ) }{{d}^{2} \left ( a{d}^{2}-bde+{e}^{2}c \right ) }}+{\frac{a\ln \left ( a{x}^{2}+bx+c \right ) bd}{ \left ( 2\,a{d}^{2}-2\,bde+2\,{e}^{2}c \right ){c}^{2}}}+{\frac{a\ln \left ( a{x}^{2}+bx+c \right ) e}{ \left ( 2\,a{d}^{2}-2\,bde+2\,{e}^{2}c \right ) c}}-{\frac{\ln \left ( a{x}^{2}+bx+c \right ){b}^{2}e}{ \left ( 2\,a{d}^{2}-2\,bde+2\,{e}^{2}c \right ){c}^{2}}}-2\,{\frac{{a}^{2}d}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) c\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,ax+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+{\frac{a{b}^{2}d}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ){c}^{2}}\arctan \left ({(2\,ax+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+3\,{\frac{abe}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) c\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,ax+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-{\frac{{b}^{3}e}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ){c}^{2}}\arctan \left ({(2\,ax+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}-{\frac{1}{cdx}}-{\frac{b\ln \left ( x \right ) }{{c}^{2}d}}-{\frac{\ln \left ( x \right ) e}{c{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+c/x^2+b/x)/x^4/(e*x+d),x)

[Out]

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

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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^4/(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^4/(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**4/(e*x+d),x)

[Out]

Timed out

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Giac [A]  time = 1.10823, size = 284, normalized size = 1.47 \begin{align*} \frac{{\left (a b d - b^{2} e + a c e\right )} \log \left (a x^{2} + b x + c\right )}{2 \,{\left (a c^{2} d^{2} - b c^{2} d e + c^{3} e^{2}\right )}} + \frac{e^{4} \log \left ({\left | x e + d \right |}\right )}{a d^{4} e - b d^{3} e^{2} + c d^{2} e^{3}} + \frac{{\left (a b^{2} d - 2 \, a^{2} c d - b^{3} e + 3 \, a b c e\right )} \arctan \left (\frac{2 \, a x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (a c^{2} d^{2} - b c^{2} d e + c^{3} e^{2}\right )} \sqrt{-b^{2} + 4 \, a c}} - \frac{{\left (b d + c e\right )} \log \left ({\left | x \right |}\right )}{c^{2} d^{2}} - \frac{1}{c d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+c/x^2+b/x)/x^4/(e*x+d),x, algorithm="giac")

[Out]

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