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

Optimal. Leaf size=218 \[ -\frac{\left (-2 a b c d+a c^2 e-b^2 c e+b^3 d\right ) \log \left (a x^2+b x+c\right )}{2 a^3 \left (a d^2-e (b d-c e)\right )}-\frac{\left (2 a^2 c^2 d-4 a b^2 c d+3 a b c^2 e-b^3 c e+b^4 d\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{a^3 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )}-\frac{x (a d+b e)}{a^2 e^2}+\frac{d^4 \log (d+e x)}{e^3 \left (a d^2-e (b d-c e)\right )}+\frac{x^2}{2 a e} \]

[Out]

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

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Rubi [A]  time = 0.395218, antiderivative size = 218, 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, 1628, 634, 618, 206, 628} \[ -\frac{\left (-2 a b c d+a c^2 e-b^2 c e+b^3 d\right ) \log \left (a x^2+b x+c\right )}{2 a^3 \left (a d^2-e (b d-c e)\right )}-\frac{\left (2 a^2 c^2 d-4 a b^2 c d+3 a b c^2 e-b^3 c e+b^4 d\right ) \tanh ^{-1}\left (\frac{2 a x+b}{\sqrt{b^2-4 a c}}\right )}{a^3 \sqrt{b^2-4 a c} \left (a d^2-e (b d-c e)\right )}-\frac{x (a d+b e)}{a^2 e^2}+\frac{d^4 \log (d+e x)}{e^3 \left (a d^2-e (b d-c e)\right )}+\frac{x^2}{2 a e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.18102, size = 218, normalized size = 1. \[ \frac{\left (2 a b c d-a c^2 e+b^2 c e+b^3 (-d)\right ) \log (x (a x+b)+c)}{2 a^3 \left (a d^2+e (c e-b d)\right )}+\frac{\left (2 a^2 c^2 d-4 a b^2 c d+3 a b c^2 e-b^3 c e+b^4 d\right ) \tan ^{-1}\left (\frac{2 a x+b}{\sqrt{4 a c-b^2}}\right )}{a^3 \sqrt{4 a c-b^2} \left (a d^2+e (c e-b d)\right )}-\frac{x (a d+b e)}{a^2 e^2}+\frac{d^4 \log (d+e x)}{e^3 \left (a d^2+e (c e-b d)\right )}+\frac{x^2}{2 a e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.006, size = 512, normalized size = 2.4 \begin{align*}{\frac{{x}^{2}}{2\,ae}}-{\frac{dx}{a{e}^{2}}}-{\frac{bx}{{a}^{2}e}}+{\frac{{d}^{4}\ln \left ( ex+d \right ) }{{e}^{3} \left ( a{d}^{2}-bde+{e}^{2}c \right ) }}+{\frac{\ln \left ( a{x}^{2}+bx+c \right ) bcd}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ){a}^{2}}}-{\frac{\ln \left ( a{x}^{2}+bx+c \right ){c}^{2}e}{ \left ( 2\,a{d}^{2}-2\,bde+2\,{e}^{2}c \right ){a}^{2}}}-{\frac{\ln \left ( a{x}^{2}+bx+c \right ){b}^{3}d}{ \left ( 2\,a{d}^{2}-2\,bde+2\,{e}^{2}c \right ){a}^{3}}}+{\frac{\ln \left ( a{x}^{2}+bx+c \right ){b}^{2}ce}{ \left ( 2\,a{d}^{2}-2\,bde+2\,{e}^{2}c \right ){a}^{3}}}+2\,{\frac{d{c}^{2}}{ \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 ) }-4\,{\frac{{b}^{2}cd}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ){a}^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,ax+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+3\,{\frac{b{c}^{2}e}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ){a}^{2}\sqrt{4\,ac-{b}^{2}}}\arctan \left ({\frac{2\,ax+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+{\frac{{b}^{4}d}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ){a}^{3}}\arctan \left ({(2\,ax+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}-{\frac{{b}^{3}ce}{ \left ( a{d}^{2}-bde+{e}^{2}c \right ){a}^{3}}\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^2/(a+c/x^2+b/x)/(e*x+d),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 84.0832, size = 1623, normalized size = 7.44 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(a^3*b^2 - 4*a^4*c)*d^4*log(e*x + d) + ((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)*x^2 + ((b^4 - 4*a*b^2*c + 2*a^2*c^2)*d*e^3 - (b^3*c - 3*a*b*c^2)*e^4)*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^3*b^2
- 4*a^4*c)*d^3*e - (a*b^4 - 5*a^2*b^2*c + 4*a^3*c^2)*d*e^3 + (a*b^3*c - 4*a^2*b*c^2)*e^4)*x - ((b^5 - 6*a*b^3*
c + 8*a^2*b*c^2)*d*e^3 - (b^4*c - 5*a*b^2*c^2 + 4*a^2*c^3)*e^4)*log(a*x^2 + b*x + c))/((a^4*b^2 - 4*a^5*c)*d^2
*e^3 - (a^3*b^3 - 4*a^4*b*c)*d*e^4 + (a^3*b^2*c - 4*a^4*c^2)*e^5), 1/2*(2*(a^3*b^2 - 4*a^4*c)*d^4*log(e*x + d)
+ ((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)*x^2 - 2*((b^4 - 4
*a*b^2*c + 2*a^2*c^2)*d*e^3 - (b^3*c - 3*a*b*c^2)*e^4)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*a*x +
b)/(b^2 - 4*a*c)) - 2*((a^3*b^2 - 4*a^4*c)*d^3*e - (a*b^4 - 5*a^2*b^2*c + 4*a^3*c^2)*d*e^3 + (a*b^3*c - 4*a^2*
b*c^2)*e^4)*x - ((b^5 - 6*a*b^3*c + 8*a^2*b*c^2)*d*e^3 - (b^4*c - 5*a*b^2*c^2 + 4*a^2*c^3)*e^4)*log(a*x^2 + b*
x + c))/((a^4*b^2 - 4*a^5*c)*d^2*e^3 - (a^3*b^3 - 4*a^4*b*c)*d*e^4 + (a^3*b^2*c - 4*a^4*c^2)*e^5)]

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

[Out]

Timed out

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Giac [A]  time = 1.11492, size = 302, normalized size = 1.39 \begin{align*} \frac{d^{4} \log \left ({\left | x e + d \right |}\right )}{a d^{2} e^{3} - b d e^{4} + c e^{5}} - \frac{{\left (b^{3} d - 2 \, a b c d - b^{2} c e + a c^{2} e\right )} \log \left (a x^{2} + b x + c\right )}{2 \,{\left (a^{4} d^{2} - a^{3} b d e + a^{3} c e^{2}\right )}} + \frac{{\left (b^{4} d - 4 \, a b^{2} c d + 2 \, a^{2} c^{2} d - b^{3} c e + 3 \, a b c^{2} e\right )} \arctan \left (\frac{2 \, a x + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (a^{4} d^{2} - a^{3} b d e + a^{3} c e^{2}\right )} \sqrt{-b^{2} + 4 \, a c}} + \frac{{\left (a x^{2} e - 2 \, a d x - 2 \, b x e\right )} e^{\left (-2\right )}}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^4*log(abs(x*e + d))/(a*d^2*e^3 - b*d*e^4 + c*e^5) - 1/2*(b^3*d - 2*a*b*c*d - b^2*c*e + a*c^2*e)*log(a*x^2 +
b*x + c)/(a^4*d^2 - a^3*b*d*e + a^3*c*e^2) + (b^4*d - 4*a*b^2*c*d + 2*a^2*c^2*d - b^3*c*e + 3*a*b*c^2*e)*arcta
n((2*a*x + b)/sqrt(-b^2 + 4*a*c))/((a^4*d^2 - a^3*b*d*e + a^3*c*e^2)*sqrt(-b^2 + 4*a*c)) + 1/2*(a*x^2*e - 2*a*
d*x - 2*b*x*e)*e^(-2)/a^2