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

Optimal. Leaf size=149 \[ -\frac{\left (-2 a c d+b^2 d-b c e\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{a \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac{d^2 \log (d+e x)}{e \left (a d^2-b d e+c e^2\right )}-\frac{(b d-c e) \log \left (a x^2+b x+c\right )}{2 a \left (a d^2-e (b d-c e)\right )} \]

[Out]

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

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Rubi [A]  time = 0.210117, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {1445, 1628, 634, 618, 206, 628} \[ -\frac{\left (-2 a c d+b^2 d-b c e\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{a \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac{d^2 \log (d+e x)}{e \left (a d^2-b d e+c e^2\right )}-\frac{(b d-c e) \log \left (a x^2+b x+c\right )}{2 a \left (a d^2-e (b d-c e)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1445

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

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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

Mathematica [A]  time = 0.127595, size = 132, normalized size = 0.89 \[ -\frac{\sqrt{4 a c-b^2} \left (e (b d-c e) \log (x (a x+b)+c)-2 a d^2 \log (d+e x)\right )+2 e \left (2 a c d+b^2 (-d)+b c e\right ) \tan ^{-1}\left (\frac{2 a x+b}{\sqrt{4 a c-b^2}}\right )}{2 a e \sqrt{4 a c-b^2} \left (a d^2+e (c e-b d)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 275, normalized size = 1.9 \begin{align*}{\frac{{d}^{2}\ln \left ( ex+d \right ) }{e \left ( a{d}^{2}-bde+{e}^{2}c \right ) }}-{\frac{\ln \left ( a{x}^{2}+bx+c \right ) bd}{ \left ( 2\,a{d}^{2}-2\,bde+2\,{e}^{2}c \right ) a}}+{\frac{\ln \left ( a{x}^{2}+bx+c \right ) ce}{ \left ( 2\,a{d}^{2}-2\,bde+2\,{e}^{2}c \right ) a}}-2\,{\frac{cd}{ \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}d}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) a}\arctan \left ({(2\,ax+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}-{\frac{bce}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) a}\arctan \left ({(2\,ax+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 9.36673, size = 872, normalized size = 5.85 \begin{align*} \left [\frac{2 \,{\left (a b^{2} - 4 \, a^{2} c\right )} d^{2} \log \left (e x + d\right ) +{\left (b c e^{2} -{\left (b^{2} - 2 \, a c\right )} d e\right )} \sqrt{b^{2} - 4 \, a c} \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 ) -{\left ({\left (b^{3} - 4 \, a b c\right )} d e -{\left (b^{2} c - 4 \, a c^{2}\right )} e^{2}\right )} \log \left (a x^{2} + b x + c\right )}{2 \,{\left ({\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d^{2} e -{\left (a b^{3} - 4 \, a^{2} b c\right )} d e^{2} +{\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} e^{3}\right )}}, \frac{2 \,{\left (a b^{2} - 4 \, a^{2} c\right )} d^{2} \log \left (e x + d\right ) + 2 \,{\left (b c e^{2} -{\left (b^{2} - 2 \, a c\right )} d e\right )} \sqrt{-b^{2} + 4 \, a c} \arctan \left (-\frac{\sqrt{-b^{2} + 4 \, a c}{\left (2 \, a x + b\right )}}{b^{2} - 4 \, a c}\right ) -{\left ({\left (b^{3} - 4 \, a b c\right )} d e -{\left (b^{2} c - 4 \, a c^{2}\right )} e^{2}\right )} \log \left (a x^{2} + b x + c\right )}{2 \,{\left ({\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d^{2} e -{\left (a b^{3} - 4 \, a^{2} b c\right )} d e^{2} +{\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} e^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.11749, size = 201, normalized size = 1.35 \begin{align*} \frac{d^{2} \log \left ({\left | x e + d \right |}\right )}{a d^{2} e - b d e^{2} + c e^{3}} - \frac{{\left (b d - c e\right )} \log \left (a x^{2} + b x + c\right )}{2 \,{\left (a^{2} d^{2} - a b d e + a c e^{2}\right )}} + \frac{{\left (b^{2} d - 2 \, a c d - b c e\right )} \arctan \left (\frac{2 \, a x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (a^{2} d^{2} - a b d e + a c e^{2}\right )} \sqrt{-b^{2} + 4 \, a c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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