3.3.84 \(\int \frac {a+b x^2+c x^4}{x^2 (d+e x^2)^2} \, dx\) [284]

Optimal. Leaf size=89 \[ -\frac {a}{d^2 x}-\frac {\left (c d^2-b d e+a e^2\right ) x}{2 d^2 e \left (d+e x^2\right )}+\frac {\left (c d^2+e (b d-3 a e)\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} e^{3/2}} \]

[Out]

-a/d^2/x-1/2*(a*e^2-b*d*e+c*d^2)*x/d^2/e/(e*x^2+d)+1/2*(c*d^2+e*(-3*a*e+b*d))*arctan(x*e^(1/2)/d^(1/2))/d^(5/2
)/e^(3/2)

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Rubi [A]
time = 0.08, antiderivative size = 86, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {1273, 464, 211} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (e (b d-3 a e)+c d^2\right )}{2 d^{5/2} e^{3/2}}-\frac {x \left (\frac {c}{e}-\frac {b d-a e}{d^2}\right )}{2 \left (d+e x^2\right )}-\frac {a}{d^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(a/(d^2*x)) - ((c/e - (b*d - a*e)/d^2)*x)/(2*(d + e*x^2)) + ((c*d^2 + e*(b*d - 3*a*e))*ArcTan[(Sqrt[e]*x)/Sqr
t[d]])/(2*d^(5/2)*e^(3/2))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 1273

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[(-d)^(m
/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2*(-d)^(-m/2 + 1)*e^(2*
p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x],
x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rubi steps

\begin {align*} \int \frac {a+b x^2+c x^4}{x^2 \left (d+e x^2\right )^2} \, dx &=-\frac {\left (\frac {c}{e}-\frac {b d-a e}{d^2}\right ) x}{2 \left (d+e x^2\right )}-\frac {\int \frac {-2 a d e^2-e \left (c d^2+e (b d-a e)\right ) x^2}{x^2 \left (d+e x^2\right )} \, dx}{2 d^2 e^2}\\ &=-\frac {a}{d^2 x}-\frac {\left (\frac {c}{e}-\frac {b d-a e}{d^2}\right ) x}{2 \left (d+e x^2\right )}+\frac {1}{2} \left (\frac {c}{e}+\frac {b d-3 a e}{d^2}\right ) \int \frac {1}{d+e x^2} \, dx\\ &=-\frac {a}{d^2 x}-\frac {\left (\frac {c}{e}-\frac {b d-a e}{d^2}\right ) x}{2 \left (d+e x^2\right )}+\frac {\left (c d^2+e (b d-3 a e)\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} e^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 89, normalized size = 1.00 \begin {gather*} -\frac {a}{d^2 x}-\frac {\left (c d^2-b d e+a e^2\right ) x}{2 d^2 e \left (d+e x^2\right )}+\frac {\left (c d^2+b d e-3 a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{5/2} e^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(a/(d^2*x)) - ((c*d^2 - b*d*e + a*e^2)*x)/(2*d^2*e*(d + e*x^2)) + ((c*d^2 + b*d*e - 3*a*e^2)*ArcTan[(Sqrt[e]*
x)/Sqrt[d]])/(2*d^(5/2)*e^(3/2))

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Maple [A]
time = 0.14, size = 85, normalized size = 0.96

method	result	size
default	\(-\frac {\frac {\left (a \,e^{2}-d e b +c \,d^{2}\right ) x}{2 e \left (e \,x^{2}+d \right )}+\frac {\left (3 a \,e^{2}-d e b -c \,d^{2}\right ) \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 e \sqrt {d e}}}{d^{2}}-\frac {a}{d^{2} x}\)	\(85\)
risch	\(\frac {-\frac {\left (3 a \,e^{2}-d e b +c \,d^{2}\right ) x^{2}}{2 d^{2} e}-\frac {a}{d}}{x \left (e \,x^{2}+d \right )}-\frac {3 e \ln \left (-\sqrt {-d e}\, x -d \right ) a}{4 \sqrt {-d e}\, d^{2}}+\frac {\ln \left (-\sqrt {-d e}\, x -d \right ) b}{4 \sqrt {-d e}\, d}+\frac {\ln \left (-\sqrt {-d e}\, x -d \right ) c}{4 \sqrt {-d e}\, e}+\frac {3 e \ln \left (-\sqrt {-d e}\, x +d \right ) a}{4 \sqrt {-d e}\, d^{2}}-\frac {\ln \left (-\sqrt {-d e}\, x +d \right ) b}{4 \sqrt {-d e}\, d}-\frac {\ln \left (-\sqrt {-d e}\, x +d \right ) c}{4 \sqrt {-d e}\, e}\)	\(202\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 81, normalized size = 0.91 \begin {gather*} \frac {{\left (c d^{2} + b d e - 3 \, a e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {3}{2}\right )}}{2 \, d^{\frac {5}{2}}} - \frac {{\left (c d^{2} - b d e + 3 \, a e^{2}\right )} x^{2} + 2 \, a d e}{2 \, {\left (d^{2} x^{3} e^{2} + d^{3} x e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(c*d^2 + b*d*e - 3*a*e^2)*arctan(x*e^(1/2)/sqrt(d))*e^(-3/2)/d^(5/2) - 1/2*((c*d^2 - b*d*e + 3*a*e^2)*x^2
+ 2*a*d*e)/(d^2*x^3*e^2 + d^3*x*e)

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Fricas [A]
time = 0.38, size = 274, normalized size = 3.08 \begin {gather*} \left [-\frac {2 \, c d^{3} x^{2} e + 6 \, a d x^{2} e^{3} - {\left (c d^{3} x - 3 \, a x^{3} e^{3} + {\left (b d x^{3} - 3 \, a d x\right )} e^{2} + {\left (c d^{2} x^{3} + b d^{2} x\right )} e\right )} \sqrt {-d e} \log \left (\frac {x^{2} e + 2 \, \sqrt {-d e} x - d}{x^{2} e + d}\right ) - 2 \, {\left (b d^{2} x^{2} - 2 \, a d^{2}\right )} e^{2}}{4 \, {\left (d^{3} x^{3} e^{3} + d^{4} x e^{2}\right )}}, -\frac {c d^{3} x^{2} e + 3 \, a d x^{2} e^{3} - {\left (c d^{3} x - 3 \, a x^{3} e^{3} + {\left (b d x^{3} - 3 \, a d x\right )} e^{2} + {\left (c d^{2} x^{3} + b d^{2} x\right )} e\right )} \sqrt {d} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\frac {1}{2}} - {\left (b d^{2} x^{2} - 2 \, a d^{2}\right )} e^{2}}{2 \, {\left (d^{3} x^{3} e^{3} + d^{4} x e^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.61, size = 155, normalized size = 1.74 \begin {gather*} \frac {\sqrt {- \frac {1}{d^{5} e^{3}}} \cdot \left (3 a e^{2} - b d e - c d^{2}\right ) \log {\left (- d^{3} e \sqrt {- \frac {1}{d^{5} e^{3}}} + x \right )}}{4} - \frac {\sqrt {- \frac {1}{d^{5} e^{3}}} \cdot \left (3 a e^{2} - b d e - c d^{2}\right ) \log {\left (d^{3} e \sqrt {- \frac {1}{d^{5} e^{3}}} + x \right )}}{4} + \frac {- 2 a d e + x^{2} \left (- 3 a e^{2} + b d e - c d^{2}\right )}{2 d^{3} e x + 2 d^{2} e^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2+a)/x**2/(e*x**2+d)**2,x)

[Out]

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

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Giac [A]
time = 4.41, size = 83, normalized size = 0.93 \begin {gather*} \frac {{\left (c d^{2} + b d e - 3 \, a e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {3}{2}\right )}}{2 \, d^{\frac {5}{2}}} - \frac {{\left (c d^{2} x^{2} - b d x^{2} e + 3 \, a x^{2} e^{2} + 2 \, a d e\right )} e^{\left (-1\right )}}{2 \, {\left (x^{3} e + d x\right )} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(c*d^2 + b*d*e - 3*a*e^2)*arctan(x*e^(1/2)/sqrt(d))*e^(-3/2)/d^(5/2) - 1/2*(c*d^2*x^2 - b*d*x^2*e + 3*a*x^
2*e^2 + 2*a*d*e)*e^(-1)/((x^3*e + d*x)*d^2)

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Mupad [B]
time = 0.37, size = 81, normalized size = 0.91 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (c\,d^2+b\,d\,e-3\,a\,e^2\right )}{2\,d^{5/2}\,e^{3/2}}-\frac {\frac {a}{d}+\frac {x^2\,\left (c\,d^2-b\,d\,e+3\,a\,e^2\right )}{2\,d^2\,e}}{e\,x^3+d\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((e^(1/2)*x)/d^(1/2))*(c*d^2 - 3*a*e^2 + b*d*e))/(2*d^(5/2)*e^(3/2)) - (a/d + (x^2*(3*a*e^2 + c*d^2 - b*d
*e))/(2*d^2*e))/(d*x + e*x^3)

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