3.3.83 \(\int \frac {a+b x^2+c x^4}{(d+e x^2)^2} \, dx\) [283]

Optimal. Leaf size=83 \[ \frac {c x}{e^2}+\frac {\left (a+\frac {d (c d-b e)}{e^2}\right ) x}{2 d \left (d+e x^2\right )}-\frac {\left (3 c d^2-e (b d+a e)\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{3/2} e^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 87, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1171, 396, 211} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (3 c d^2-e (a e+b d)\right )}{2 d^{3/2} e^{5/2}}+\frac {x \left (a e^2-b d e+c d^2\right )}{2 d e^2 \left (d+e x^2\right )}+\frac {c x}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 1171

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1
)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 88, normalized size = 1.06 \begin {gather*} \frac {c x}{e^2}+\frac {\left (c d^2-b d e+a e^2\right ) x}{2 d e^2 \left (d+e x^2\right )}-\frac {\left (3 c d^2-b d e-a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{3/2} e^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.13, size = 79, normalized size = 0.95

method	result	size
default	\(\frac {c x}{e^{2}}+\frac {\frac {\left (a \,e^{2}-d e b +c \,d^{2}\right ) x}{2 d \left (e \,x^{2}+d \right )}+\frac {\left (a \,e^{2}+d e b -3 c \,d^{2}\right ) \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 d \sqrt {d e}}}{e^{2}}\)	\(79\)
risch	\(\frac {c x}{e^{2}}+\frac {\left (a \,e^{2}-d e b +c \,d^{2}\right ) x}{2 d \,e^{2} \left (e \,x^{2}+d \right )}-\frac {\ln \left (e x +\sqrt {-d e}\right ) a}{4 \sqrt {-d e}\, d}-\frac {\ln \left (e x +\sqrt {-d e}\right ) b}{4 e \sqrt {-d e}}+\frac {3 d \ln \left (e x +\sqrt {-d e}\right ) c}{4 e^{2} \sqrt {-d e}}+\frac {\ln \left (-e x +\sqrt {-d e}\right ) a}{4 \sqrt {-d e}\, d}+\frac {\ln \left (-e x +\sqrt {-d e}\right ) b}{4 e \sqrt {-d e}}-\frac {3 d \ln \left (-e x +\sqrt {-d e}\right ) c}{4 e^{2} \sqrt {-d e}}\)	\(185\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 74, normalized size = 0.89 \begin {gather*} c x e^{\left (-2\right )} - \frac {{\left (3 \, c d^{2} - b d e - a e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {5}{2}\right )}}{2 \, d^{\frac {3}{2}}} + \frac {{\left (c d^{2} - b d e + a e^{2}\right )} x}{2 \, {\left (d x^{2} e^{3} + d^{2} e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 266, normalized size = 3.20 \begin {gather*} \left [\frac {6 \, c d^{3} x e + 2 \, a d x e^{3} + {\left (3 \, c d^{3} - a x^{2} e^{3} - {\left (b d x^{2} + a d\right )} e^{2} + {\left (3 \, c d^{2} x^{2} - b d^{2}\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 (2 \, c d^{2} x^{3} - b d^{2} x\right )} e^{2}}{4 \, {\left (d^{2} x^{2} e^{4} + d^{3} e^{3}\right )}}, \frac {3 \, c d^{3} x e + a d x e^{3} - {\left (3 \, c d^{3} - a x^{2} e^{3} - {\left (b d x^{2} + a d\right )} e^{2} + {\left (3 \, c d^{2} x^{2} - b d^{2}\right )} e\right )} \sqrt {d} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\frac {1}{2}} + {\left (2 \, c d^{2} x^{3} - b d^{2} x\right )} e^{2}}{2 \, {\left (d^{2} x^{2} e^{4} + d^{3} e^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (75) = 150\).
time = 0.43, size = 153, normalized size = 1.84 \begin {gather*} \frac {c x}{e^{2}} + \frac {x \left (a e^{2} - b d e + c d^{2}\right )}{2 d^{2} e^{2} + 2 d e^{3} x^{2}} - \frac {\sqrt {- \frac {1}{d^{3} e^{5}}} \left (a e^{2} + b d e - 3 c d^{2}\right ) \log {\left (- d^{2} e^{2} \sqrt {- \frac {1}{d^{3} e^{5}}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{d^{3} e^{5}}} \left (a e^{2} + b d e - 3 c d^{2}\right ) \log {\left (d^{2} e^{2} \sqrt {- \frac {1}{d^{3} e^{5}}} + x \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 4.11, size = 75, normalized size = 0.90 \begin {gather*} c x e^{\left (-2\right )} - \frac {{\left (3 \, c d^{2} - b d e - a e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {5}{2}\right )}}{2 \, d^{\frac {3}{2}}} + \frac {{\left (c d^{2} x - b d x e + a x e^{2}\right )} e^{\left (-2\right )}}{2 \, {\left (x^{2} e + d\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.36, size = 77, normalized size = 0.93 \begin {gather*} \frac {c\,x}{e^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (-3\,c\,d^2+b\,d\,e+a\,e^2\right )}{2\,d^{3/2}\,e^{5/2}}+\frac {x\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}{2\,d\,\left (e^3\,x^2+d\,e^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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