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

Optimal. Leaf size=176 \[ \frac{\left (-3 a b c d+2 a c^2 e-b^2 c e+b^3 d\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{a^2 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac{\left (-a c d+b^2 d-b c e\right ) \log \left (a x^2+b x+c\right )}{2 a^2 \left (a d^2-e (b d-c e)\right )}-\frac{d^3 \log (d+e x)}{e^2 \left (a d^2-e (b d-c e)\right )}+\frac{x}{a e} \]

[Out]

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

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Rubi [A]  time = 0.285269, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {1569, 1628, 634, 618, 206, 628} \[ \frac{\left (-3 a b c d+2 a c^2 e-b^2 c e+b^3 d\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{a^2 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )}+\frac{\left (-a c d+b^2 d-b c e\right ) \log \left (a x^2+b x+c\right )}{2 a^2 \left (a d^2-e (b d-c e)\right )}-\frac{d^3 \log (d+e x)}{e^2 \left (a d^2-e (b d-c e)\right )}+\frac{x}{a e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.187473, size = 178, normalized size = 1.01 \[ \frac{\left (-3 a b c d+2 a c^2 e-b^2 c e+b^3 d\right ) \tan ^{-1}\left (\frac{2 a x+b}{\sqrt{4 a c-b^2}}\right )}{a^2 \sqrt{4 a c-b^2} \left (-a d^2+b d e-c e^2\right )}+\frac{\left (-a c d+b^2 d-b c e\right ) \log \left (a x^2+b x+c\right )}{2 a^2 \left (a d^2-b d e+c e^2\right )}-\frac{d^3 \log (d+e x)}{e^2 \left (a d^2-b d e+c e^2\right )}+\frac{x}{a e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.007, size = 388, normalized size = 2.2 \begin{align*}{\frac{x}{ae}}-{\frac{{d}^{3}\ln \left ( ex+d \right ) }{{e}^{2} \left ( a{d}^{2}-bde+{e}^{2}c \right ) }}-{\frac{\ln \left ( a{x}^{2}+bx+c \right ) cd}{ \left ( 2\,a{d}^{2}-2\,bde+2\,{e}^{2}c \right ) a}}+{\frac{\ln \left ( a{x}^{2}+bx+c \right ){b}^{2}d}{ \left ( 2\,a{d}^{2}-2\,bde+2\,{e}^{2}c \right ){a}^{2}}}-{\frac{\ln \left ( a{x}^{2}+bx+c \right ) bce}{ \left ( 2\,a{d}^{2}-2\,bde+2\,{e}^{2}c \right ){a}^{2}}}+3\,{\frac{bcd}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) a\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,ax+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-2\,{\frac{{c}^{2}e}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ) a\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,ax+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-{\frac{{b}^{3}d}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ){a}^{2}}\arctan \left ({(2\,ax+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+{\frac{{b}^{2}ce}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ){a}^{2}}\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(x/(a+c/x^2+b/x)/(e*x+d),x)

[Out]

x/a/e-1/e^2*d^3/(a*d^2-b*d*e+c*e^2)*ln(e*x+d)-1/2/(a*d^2-b*d*e+c*e^2)/a*ln(a*x^2+b*x+c)*c*d+1/2/(a*d^2-b*d*e+c
*e^2)/a^2*ln(a*x^2+b*x+c)*b^2*d-1/2/(a*d^2-b*d*e+c*e^2)/a^2*ln(a*x^2+b*x+c)*b*c*e+3/(a*d^2-b*d*e+c*e^2)/a/(4*a
*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*b*c*d-2/(a*d^2-b*d*e+c*e^2)/a/(4*a*c-b^2)^(1/2)*arctan((2*a*
x+b)/(4*a*c-b^2)^(1/2))*c^2*e-1/(a*d^2-b*d*e+c*e^2)/a^2/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*
b^3*d+1/(a*d^2-b*d*e+c*e^2)/a^2/(4*a*c-b^2)^(1/2)*arctan((2*a*x+b)/(4*a*c-b^2)^(1/2))*b^2*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(x/(a+c/x^2+b/x)/(e*x+d),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 30.6945, size = 1237, normalized size = 7.03 \begin{align*} \left [-\frac{2 \,{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d^{3} \log \left (e x + d\right ) -{\left ({\left (b^{3} - 3 \, a b c\right )} d e^{2} -{\left (b^{2} c - 2 \, a c^{2}\right )} e^{3}\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 ) - 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 )} x -{\left ({\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} d e^{2} -{\left (b^{3} c - 4 \, a b c^{2}\right )} e^{3}\right )} \log \left (a x^{2} + b x + c\right )}{2 \,{\left ({\left (a^{3} b^{2} - 4 \, a^{4} c\right )} d^{2} e^{2} -{\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} d e^{3} +{\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} e^{4}\right )}}, -\frac{2 \,{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} d^{3} \log \left (e x + d\right ) - 2 \,{\left ({\left (b^{3} - 3 \, a b c\right )} d e^{2} -{\left (b^{2} c - 2 \, a c^{2}\right )} e^{3}\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 ) - 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 )} x -{\left ({\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} d e^{2} -{\left (b^{3} c - 4 \, a b c^{2}\right )} e^{3}\right )} \log \left (a x^{2} + b x + c\right )}{2 \,{\left ({\left (a^{3} b^{2} - 4 \, a^{4} c\right )} d^{2} e^{2} -{\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} d e^{3} +{\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} e^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*(a^2*b^2 - 4*a^3*c)*d^3*log(e*x + d) - ((b^3 - 3*a*b*c)*d*e^2 - (b^2*c - 2*a*c^2)*e^3)*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)) - 2*((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)*x - ((b^4 - 5*a*b^2*c + 4*a^2*c^2)*d*
e^2 - (b^3*c - 4*a*b*c^2)*e^3)*log(a*x^2 + b*x + c))/((a^3*b^2 - 4*a^4*c)*d^2*e^2 - (a^2*b^3 - 4*a^3*b*c)*d*e^
3 + (a^2*b^2*c - 4*a^3*c^2)*e^4), -1/2*(2*(a^2*b^2 - 4*a^3*c)*d^3*log(e*x + d) - 2*((b^3 - 3*a*b*c)*d*e^2 - (b
^2*c - 2*a*c^2)*e^3)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*a*x + b)/(b^2 - 4*a*c)) - 2*((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)*x - ((b^4 - 5*a*b^2*c + 4*a^2*c^2)*d*e
^2 - (b^3*c - 4*a*b*c^2)*e^3)*log(a*x^2 + b*x + c))/((a^3*b^2 - 4*a^4*c)*d^2*e^2 - (a^2*b^3 - 4*a^3*b*c)*d*e^3
+ (a^2*b^2*c - 4*a^3*c^2)*e^4)]

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

[Out]

Timed out

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Giac [A]  time = 1.09625, size = 250, normalized size = 1.42 \begin{align*} -\frac{d^{3} \log \left ({\left | x e + d \right |}\right )}{a d^{2} e^{2} - b d e^{3} + c e^{4}} + \frac{x e^{\left (-1\right )}}{a} + \frac{{\left (b^{2} d - a c d - b c e\right )} \log \left (a x^{2} + b x + c\right )}{2 \,{\left (a^{3} d^{2} - a^{2} b d e + a^{2} c e^{2}\right )}} - \frac{{\left (b^{3} d - 3 \, a b c d - b^{2} c e + 2 \, a c^{2} e\right )} \arctan \left (\frac{2 \, a x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (a^{3} d^{2} - a^{2} b d e + a^{2} c e^{2}\right )} \sqrt{-b^{2} + 4 \, a c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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