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

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

[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 = 0.08, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1814, 388, 205} \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (3 c d^2-e (a e+b d)\right )}{2 d^{3/2} e^{5/2}}+\frac {x \left (a+\frac {d (c d-b e)}{e^2}\right )}{2 d \left (d+e x^2\right )}+\frac {c x}{e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

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

Rule 388

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 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {a+x^2 \left (b+c x^2\right )}{\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 = 0.02, size = 88, normalized size = 1.06 \[ \frac {x \left (a e^2-b d e+c d^2\right )}{2 d e^2 \left (d+e x^2\right )}-\frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (-a e^2-b d e+3 c d^2\right )}{2 d^{3/2} e^{5/2}}+\frac {c x}{e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + x^2*(b + c*x^2))/(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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fricas [A]  time = 0.61, size = 268, normalized size = 3.23 \[ \left [\frac {4 \, c d^{2} e^{2} x^{3} + {\left (3 \, c d^{3} - b d^{2} e - a d e^{2} + {\left (3 \, c d^{2} e - b d e^{2} - a e^{3}\right )} x^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) + 2 \, {\left (3 \, c d^{3} e - b d^{2} e^{2} + a d e^{3}\right )} x}{4 \, {\left (d^{2} e^{4} x^{2} + d^{3} e^{3}\right )}}, \frac {2 \, c d^{2} e^{2} x^{3} - {\left (3 \, c d^{3} - b d^{2} e - a d e^{2} + {\left (3 \, c d^{2} e - b d e^{2} - a e^{3}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) + {\left (3 \, c d^{3} e - b d^{2} e^{2} + a d e^{3}\right )} x}{2 \, {\left (d^{2} e^{4} x^{2} + d^{3} e^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 75, normalized size = 0.90 \[ c x e^{\left (-2\right )} - \frac {{\left (3 \, c d^{2} - b d e - a e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {5}{2}\right )}}{2 \, d^{\frac {3}{2}}} + \frac {{\left (c d^{2} x - b d x e + a x e^{2}\right )} e^{\left (-2\right )}}{2 \, {\left (x^{2} e + d\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+x^2*(c*x^2+b))/(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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maple [A]  time = 0.01, size = 118, normalized size = 1.42 \[ \frac {a x}{2 \left (e \,x^{2}+d \right ) d}+\frac {a \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, d}-\frac {b x}{2 \left (e \,x^{2}+d \right ) e}+\frac {b \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, e}+\frac {c d x}{2 \left (e \,x^{2}+d \right ) e^{2}}-\frac {3 c d \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, e^{2}}+\frac {c x}{e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.37, size = 84, normalized size = 1.01 \[ \frac {{\left (c d^{2} - b d e + a e^{2}\right )} x}{2 \, {\left (d e^{3} x^{2} + d^{2} e^{2}\right )}} + \frac {c x}{e^{2}} - \frac {{\left (3 \, c d^{2} - b d e - a e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \, \sqrt {d e} d e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.11, size = 77, normalized size = 0.93 \[ \frac {c\,x}{e^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (-3\,c\,d^2+b\,d\,e+a\,e^2\right )}{2\,d^{3/2}\,e^{5/2}}+\frac {x\,\left (c\,d^2-b\,d\,e+a\,e^2\right )}{2\,d\,\left (e^3\,x^2+d\,e^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + x^2*(b + c*x^2))/(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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sympy [B]  time = 0.86, size = 153, normalized size = 1.84 \[ \frac {c x}{e^{2}} + \frac {x \left (a e^{2} - b d e + c d^{2}\right )}{2 d^{2} e^{2} + 2 d e^{3} x^{2}} - \frac {\sqrt {- \frac {1}{d^{3} e^{5}}} \left (a e^{2} + b d e - 3 c d^{2}\right ) \log {\left (- d^{2} e^{2} \sqrt {- \frac {1}{d^{3} e^{5}}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{d^{3} e^{5}}} \left (a e^{2} + b d e - 3 c d^{2}\right ) \log {\left (d^{2} e^{2} \sqrt {- \frac {1}{d^{3} e^{5}}} + x \right )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+x**2*(c*x**2+b))/(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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