3.251 \(\int \frac{a+b x^2+c x^4}{(d+e x^2)^4} \, dx\)

Optimal. Leaf size=150 \[ \frac{x \left (e (5 a e+b d)+c d^2\right )}{16 d^3 e^2 \left (d+e x^2\right )}-\frac{x \left (7 c d^2-e (5 a e+b d)\right )}{24 d^2 e^2 \left (d+e x^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (e (5 a e+b d)+c d^2\right )}{16 d^{7/2} e^{5/2}}+\frac{x \left (a+\frac{d (c d-b e)}{e^2}\right )}{6 d \left (d+e x^2\right )^3} \]

[Out]

((a + (d*(c*d - b*e))/e^2)*x)/(6*d*(d + e*x^2)^3) - ((7*c*d^2 - e*(b*d + 5*a*e))*x)/(24*d^2*e^2*(d + e*x^2)^2)
+ ((c*d^2 + e*(b*d + 5*a*e))*x)/(16*d^3*e^2*(d + e*x^2)) + ((c*d^2 + e*(b*d + 5*a*e))*ArcTan[(Sqrt[e]*x)/Sqrt
[d]])/(16*d^(7/2)*e^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.205394, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {1157, 385, 199, 205} \[ \frac{x \left (e (5 a e+b d)+c d^2\right )}{16 d^3 e^2 \left (d+e x^2\right )}-\frac{x \left (7 c d^2-e (5 a e+b d)\right )}{24 d^2 e^2 \left (d+e x^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (e (5 a e+b d)+c d^2\right )}{16 d^{7/2} e^{5/2}}+\frac{x \left (a+\frac{d (c d-b e)}{e^2}\right )}{6 d \left (d+e x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + (d*(c*d - b*e))/e^2)*x)/(6*d*(d + e*x^2)^3) - ((7*c*d^2 - e*(b*d + 5*a*e))*x)/(24*d^2*e^2*(d + e*x^2)^2)
+ ((c*d^2 + e*(b*d + 5*a*e))*x)/(16*d^3*e^2*(d + e*x^2)) + ((c*d^2 + e*(b*d + 5*a*e))*ArcTan[(Sqrt[e]*x)/Sqrt
[d]])/(16*d^(7/2)*e^(5/2))

Rule 1157

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] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 385

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

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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]

Rubi steps

\begin{align*} \int \frac{a+b x^2+c x^4}{\left (d+e x^2\right )^4} \, dx &=\frac{\left (a+\frac{d (c d-b e)}{e^2}\right ) x}{6 d \left (d+e x^2\right )^3}-\frac{\int \frac{-5 a+\frac{d (c d-b e)}{e^2}-\frac{6 c d x^2}{e}}{\left (d+e x^2\right )^3} \, dx}{6 d}\\ &=\frac{\left (a+\frac{d (c d-b e)}{e^2}\right ) x}{6 d \left (d+e x^2\right )^3}-\frac{\left (7 c d^2-e (b d+5 a e)\right ) x}{24 d^2 e^2 \left (d+e x^2\right )^2}+\frac{\left (c d^2+e (b d+5 a e)\right ) \int \frac{1}{\left (d+e x^2\right )^2} \, dx}{8 d^2 e^2}\\ &=\frac{\left (a+\frac{d (c d-b e)}{e^2}\right ) x}{6 d \left (d+e x^2\right )^3}-\frac{\left (7 c d^2-e (b d+5 a e)\right ) x}{24 d^2 e^2 \left (d+e x^2\right )^2}+\frac{\left (c d^2+e (b d+5 a e)\right ) x}{16 d^3 e^2 \left (d+e x^2\right )}+\frac{\left (c d^2+e (b d+5 a e)\right ) \int \frac{1}{d+e x^2} \, dx}{16 d^3 e^2}\\ &=\frac{\left (a+\frac{d (c d-b e)}{e^2}\right ) x}{6 d \left (d+e x^2\right )^3}-\frac{\left (7 c d^2-e (b d+5 a e)\right ) x}{24 d^2 e^2 \left (d+e x^2\right )^2}+\frac{\left (c d^2+e (b d+5 a e)\right ) x}{16 d^3 e^2 \left (d+e x^2\right )}+\frac{\left (c d^2+e (b d+5 a e)\right ) \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{16 d^{7/2} e^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.13153, size = 142, normalized size = 0.95 \[ \frac{x \left (e \left (a e \left (33 d^2+40 d e x^2+15 e^2 x^4\right )+b d \left (-3 d^2+8 d e x^2+3 e^2 x^4\right )\right )+c d^2 \left (-3 d^2-8 d e x^2+3 e^2 x^4\right )\right )}{48 d^3 e^2 \left (d+e x^2\right )^3}+\frac{\tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (e (5 a e+b d)+c d^2\right )}{16 d^{7/2} e^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(c*d^2*(-3*d^2 - 8*d*e*x^2 + 3*e^2*x^4) + e*(b*d*(-3*d^2 + 8*d*e*x^2 + 3*e^2*x^4) + a*e*(33*d^2 + 40*d*e*x^
2 + 15*e^2*x^4))))/(48*d^3*e^2*(d + e*x^2)^3) + ((c*d^2 + e*(b*d + 5*a*e))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(16*d^
(7/2)*e^(5/2))

________________________________________________________________________________________

Maple [A]  time = 0.01, size = 158, normalized size = 1.1 \begin{align*}{\frac{1}{ \left ( e{x}^{2}+d \right ) ^{3}} \left ({\frac{ \left ( 5\,a{e}^{2}+deb+c{d}^{2} \right ){x}^{5}}{16\,{d}^{3}}}+{\frac{ \left ( 5\,a{e}^{2}+deb-c{d}^{2} \right ){x}^{3}}{6\,{d}^{2}e}}+{\frac{ \left ( 11\,a{e}^{2}-deb-c{d}^{2} \right ) x}{16\,d{e}^{2}}} \right ) }+{\frac{5\,a}{16\,{d}^{3}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}+{\frac{b}{16\,{d}^{2}e}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}}+{\frac{c}{16\,d{e}^{2}}\arctan \left ({ex{\frac{1}{\sqrt{de}}}} \right ){\frac{1}{\sqrt{de}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/16*(5*a*e^2+b*d*e+c*d^2)/d^3*x^5+1/6*(5*a*e^2+b*d*e-c*d^2)/d^2/e*x^3+1/16*(11*a*e^2-b*d*e-c*d^2)/d/e^2*x)/(
e*x^2+d)^3+5/16/d^3/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*a+1/16/d^2/e/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*b+1/1
6/d/e^2/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*c

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.67425, size = 1091, normalized size = 7.27 \begin{align*} \left [\frac{6 \,{\left (c d^{3} e^{3} + b d^{2} e^{4} + 5 \, a d e^{5}\right )} x^{5} - 16 \,{\left (c d^{4} e^{2} - b d^{3} e^{3} - 5 \, a d^{2} e^{4}\right )} x^{3} - 3 \,{\left ({\left (c d^{2} e^{3} + b d e^{4} + 5 \, a e^{5}\right )} x^{6} + c d^{5} + b d^{4} e + 5 \, a d^{3} e^{2} + 3 \,{\left (c d^{3} e^{2} + b d^{2} e^{3} + 5 \, a d e^{4}\right )} x^{4} + 3 \,{\left (c d^{4} e + b d^{3} e^{2} + 5 \, a d^{2} 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 ) - 6 \,{\left (c d^{5} e + b d^{4} e^{2} - 11 \, a d^{3} e^{3}\right )} x}{96 \,{\left (d^{4} e^{6} x^{6} + 3 \, d^{5} e^{5} x^{4} + 3 \, d^{6} e^{4} x^{2} + d^{7} e^{3}\right )}}, \frac{3 \,{\left (c d^{3} e^{3} + b d^{2} e^{4} + 5 \, a d e^{5}\right )} x^{5} - 8 \,{\left (c d^{4} e^{2} - b d^{3} e^{3} - 5 \, a d^{2} e^{4}\right )} x^{3} + 3 \,{\left ({\left (c d^{2} e^{3} + b d e^{4} + 5 \, a e^{5}\right )} x^{6} + c d^{5} + b d^{4} e + 5 \, a d^{3} e^{2} + 3 \,{\left (c d^{3} e^{2} + b d^{2} e^{3} + 5 \, a d e^{4}\right )} x^{4} + 3 \,{\left (c d^{4} e + b d^{3} e^{2} + 5 \, a d^{2} e^{3}\right )} x^{2}\right )} \sqrt{d e} \arctan \left (\frac{\sqrt{d e} x}{d}\right ) - 3 \,{\left (c d^{5} e + b d^{4} e^{2} - 11 \, a d^{3} e^{3}\right )} x}{48 \,{\left (d^{4} e^{6} x^{6} + 3 \, d^{5} e^{5} x^{4} + 3 \, d^{6} e^{4} x^{2} + d^{7} e^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(6*(c*d^3*e^3 + b*d^2*e^4 + 5*a*d*e^5)*x^5 - 16*(c*d^4*e^2 - b*d^3*e^3 - 5*a*d^2*e^4)*x^3 - 3*((c*d^2*e^
3 + b*d*e^4 + 5*a*e^5)*x^6 + c*d^5 + b*d^4*e + 5*a*d^3*e^2 + 3*(c*d^3*e^2 + b*d^2*e^3 + 5*a*d*e^4)*x^4 + 3*(c*
d^4*e + b*d^3*e^2 + 5*a*d^2*e^3)*x^2)*sqrt(-d*e)*log((e*x^2 - 2*sqrt(-d*e)*x - d)/(e*x^2 + d)) - 6*(c*d^5*e +
b*d^4*e^2 - 11*a*d^3*e^3)*x)/(d^4*e^6*x^6 + 3*d^5*e^5*x^4 + 3*d^6*e^4*x^2 + d^7*e^3), 1/48*(3*(c*d^3*e^3 + b*d
^2*e^4 + 5*a*d*e^5)*x^5 - 8*(c*d^4*e^2 - b*d^3*e^3 - 5*a*d^2*e^4)*x^3 + 3*((c*d^2*e^3 + b*d*e^4 + 5*a*e^5)*x^6
+ c*d^5 + b*d^4*e + 5*a*d^3*e^2 + 3*(c*d^3*e^2 + b*d^2*e^3 + 5*a*d*e^4)*x^4 + 3*(c*d^4*e + b*d^3*e^2 + 5*a*d^
2*e^3)*x^2)*sqrt(d*e)*arctan(sqrt(d*e)*x/d) - 3*(c*d^5*e + b*d^4*e^2 - 11*a*d^3*e^3)*x)/(d^4*e^6*x^6 + 3*d^5*e
^5*x^4 + 3*d^6*e^4*x^2 + d^7*e^3)]

________________________________________________________________________________________

Sympy [A]  time = 3.32007, size = 241, normalized size = 1.61 \begin{align*} - \frac{\sqrt{- \frac{1}{d^{7} e^{5}}} \left (5 a e^{2} + b d e + c d^{2}\right ) \log{\left (- d^{4} e^{2} \sqrt{- \frac{1}{d^{7} e^{5}}} + x \right )}}{32} + \frac{\sqrt{- \frac{1}{d^{7} e^{5}}} \left (5 a e^{2} + b d e + c d^{2}\right ) \log{\left (d^{4} e^{2} \sqrt{- \frac{1}{d^{7} e^{5}}} + x \right )}}{32} + \frac{x^{5} \left (15 a e^{4} + 3 b d e^{3} + 3 c d^{2} e^{2}\right ) + x^{3} \left (40 a d e^{3} + 8 b d^{2} e^{2} - 8 c d^{3} e\right ) + x \left (33 a d^{2} e^{2} - 3 b d^{3} e - 3 c d^{4}\right )}{48 d^{6} e^{2} + 144 d^{5} e^{3} x^{2} + 144 d^{4} e^{4} x^{4} + 48 d^{3} e^{5} x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-1/(d**7*e**5))*(5*a*e**2 + b*d*e + c*d**2)*log(-d**4*e**2*sqrt(-1/(d**7*e**5)) + x)/32 + sqrt(-1/(d**7*
e**5))*(5*a*e**2 + b*d*e + c*d**2)*log(d**4*e**2*sqrt(-1/(d**7*e**5)) + x)/32 + (x**5*(15*a*e**4 + 3*b*d*e**3
+ 3*c*d**2*e**2) + x**3*(40*a*d*e**3 + 8*b*d**2*e**2 - 8*c*d**3*e) + x*(33*a*d**2*e**2 - 3*b*d**3*e - 3*c*d**4
))/(48*d**6*e**2 + 144*d**5*e**3*x**2 + 144*d**4*e**4*x**4 + 48*d**3*e**5*x**6)

________________________________________________________________________________________

Giac [A]  time = 1.17237, size = 181, normalized size = 1.21 \begin{align*} \frac{{\left (c d^{2} + b d e + 5 \, a e^{2}\right )} \arctan \left (\frac{x e^{\frac{1}{2}}}{\sqrt{d}}\right ) e^{\left (-\frac{5}{2}\right )}}{16 \, d^{\frac{7}{2}}} + \frac{{\left (3 \, c d^{2} x^{5} e^{2} + 3 \, b d x^{5} e^{3} - 8 \, c d^{3} x^{3} e + 15 \, a x^{5} e^{4} + 8 \, b d^{2} x^{3} e^{2} - 3 \, c d^{4} x + 40 \, a d x^{3} e^{3} - 3 \, b d^{3} x e + 33 \, a d^{2} x e^{2}\right )} e^{\left (-2\right )}}{48 \,{\left (x^{2} e + d\right )}^{3} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(c*d^2 + b*d*e + 5*a*e^2)*arctan(x*e^(1/2)/sqrt(d))*e^(-5/2)/d^(7/2) + 1/48*(3*c*d^2*x^5*e^2 + 3*b*d*x^5*
e^3 - 8*c*d^3*x^3*e + 15*a*x^5*e^4 + 8*b*d^2*x^3*e^2 - 3*c*d^4*x + 40*a*d*x^3*e^3 - 3*b*d^3*x*e + 33*a*d^2*x*e
^2)*e^(-2)/((x^2*e + d)^3*d^3)