3.3.91 \(\int \frac {a+b x^2+c x^4}{(d+e x^2)^3} \, dx\) [291]

Optimal. Leaf size=115 \[ \frac {\left (a+\frac {d (c d-b e)}{e^2}\right ) x}{4 d \left (d+e x^2\right )^2}-\frac {\left (5 c d^2-e (b d+3 a e)\right ) x}{8 d^2 e^2 \left (d+e x^2\right )}+\frac {\left (3 c d^2+e (b d+3 a e)\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{5/2}} \]

[Out]

1/4*(a+d*(-b*e+c*d)/e^2)*x/d/(e*x^2+d)^2-1/8*(5*c*d^2-e*(3*a*e+b*d))*x/d^2/e^2/(e*x^2+d)+1/8*(3*c*d^2+e*(3*a*e
+b*d))*arctan(x*e^(1/2)/d^(1/2))/d^(5/2)/e^(5/2)

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Rubi [A]
time = 0.08, antiderivative size = 119, normalized size of antiderivative = 1.03, 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, 393, 211} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (e (3 a e+b d)+3 c d^2\right )}{8 d^{5/2} e^{5/2}}-\frac {x \left (5 c d^2-e (3 a e+b d)\right )}{8 d^2 e^2 \left (d+e x^2\right )}+\frac {x \left (a e^2-b d e+c d^2\right )}{4 d e^2 \left (d+e x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)/(d + e*x^2)^3,x]

[Out]

((c*d^2 - b*d*e + a*e^2)*x)/(4*d*e^2*(d + e*x^2)^2) - ((5*c*d^2 - e*(b*d + 3*a*e))*x)/(8*d^2*e^2*(d + e*x^2))
+ ((3*c*d^2 + e*(b*d + 3*a*e))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^(5/2)*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 393

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

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Mathematica [A]
time = 0.06, size = 110, normalized size = 0.96 \begin {gather*} \frac {x \left (-c d^2 \left (3 d+5 e x^2\right )+e \left (b d \left (-d+e x^2\right )+a e \left (5 d+3 e x^2\right )\right )\right )}{8 d^2 e^2 \left (d+e x^2\right )^2}+\frac {\left (3 c d^2+e (b d+3 a e)\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2 + c*x^4)/(d + e*x^2)^3,x]

[Out]

(x*(-(c*d^2*(3*d + 5*e*x^2)) + e*(b*d*(-d + e*x^2) + a*e*(5*d + 3*e*x^2))))/(8*d^2*e^2*(d + e*x^2)^2) + ((3*c*
d^2 + e*(b*d + 3*a*e))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^(5/2)*e^(5/2))

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Maple [A]
time = 0.13, size = 107, normalized size = 0.93

method	result	size
default	\(\frac {\frac {\left (3 a \,e^{2}+d e b -5 c \,d^{2}\right ) x^{3}}{8 d^{2} e}+\frac {\left (5 a \,e^{2}-d e b -3 c \,d^{2}\right ) x}{8 d \,e^{2}}}{\left (e \,x^{2}+d \right )^{2}}+\frac {\left (3 a \,e^{2}+d e b +3 c \,d^{2}\right ) \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 d^{2} e^{2} \sqrt {d e}}\)	\(107\)
risch	\(\frac {\frac {\left (3 a \,e^{2}+d e b -5 c \,d^{2}\right ) x^{3}}{8 d^{2} e}+\frac {\left (5 a \,e^{2}-d e b -3 c \,d^{2}\right ) x}{8 d \,e^{2}}}{\left (e \,x^{2}+d \right )^{2}}-\frac {3 \ln \left (e x +\sqrt {-d e}\right ) a}{16 \sqrt {-d e}\, d^{2}}-\frac {\ln \left (e x +\sqrt {-d e}\right ) b}{16 \sqrt {-d e}\, e d}-\frac {3 \ln \left (e x +\sqrt {-d e}\right ) c}{16 \sqrt {-d e}\, e^{2}}+\frac {3 \ln \left (-e x +\sqrt {-d e}\right ) a}{16 \sqrt {-d e}\, d^{2}}+\frac {\ln \left (-e x +\sqrt {-d e}\right ) b}{16 \sqrt {-d e}\, e d}+\frac {3 \ln \left (-e x +\sqrt {-d e}\right ) c}{16 \sqrt {-d e}\, e^{2}}\)	\(215\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)/(e*x^2+d)^3,x,method=_RETURNVERBOSE)

[Out]

(1/8*(3*a*e^2+b*d*e-5*c*d^2)/d^2/e*x^3+1/8*(5*a*e^2-b*d*e-3*c*d^2)/d/e^2*x)/(e*x^2+d)^2+1/8*(3*a*e^2+b*d*e+3*c
*d^2)/d^2/e^2/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))

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Maxima [A]
time = 0.51, size = 110, normalized size = 0.96 \begin {gather*} \frac {{\left (3 \, 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 {5}{2}\right )}}{8 \, d^{\frac {5}{2}}} - \frac {{\left (5 \, c d^{2} e - b d e^{2} - 3 \, a e^{3}\right )} x^{3} + {\left (3 \, c d^{3} + b d^{2} e - 5 \, a d e^{2}\right )} x}{8 \, {\left (d^{2} x^{4} e^{4} + 2 \, d^{3} x^{2} e^{3} + d^{4} 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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 389, normalized size = 3.38 \begin {gather*} \left [-\frac {6 \, c d^{4} x e - 6 \, a d x^{3} e^{4} + {\left (3 \, a x^{4} e^{4} + 3 \, c d^{4} + {\left (b d x^{4} + 6 \, a d x^{2}\right )} e^{3} + {\left (3 \, c d^{2} x^{4} + 2 \, b d^{2} x^{2} + 3 \, a d^{2}\right )} e^{2} + {\left (6 \, c d^{3} x^{2} + b d^{3}\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^{3} + 5 \, a d^{2} x\right )} e^{3} + 2 \, {\left (5 \, c d^{3} x^{3} + b d^{3} x\right )} e^{2}}{16 \, {\left (d^{3} x^{4} e^{5} + 2 \, d^{4} x^{2} e^{4} + d^{5} e^{3}\right )}}, -\frac {3 \, c d^{4} x e - 3 \, a d x^{3} e^{4} - {\left (3 \, a x^{4} e^{4} + 3 \, c d^{4} + {\left (b d x^{4} + 6 \, a d x^{2}\right )} e^{3} + {\left (3 \, c d^{2} x^{4} + 2 \, b d^{2} x^{2} + 3 \, a d^{2}\right )} e^{2} + {\left (6 \, c d^{3} x^{2} + b d^{3}\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^{3} + 5 \, a d^{2} x\right )} e^{3} + {\left (5 \, c d^{3} x^{3} + b d^{3} x\right )} e^{2}}{8 \, {\left (d^{3} x^{4} e^{5} + 2 \, d^{4} x^{2} e^{4} + d^{5} 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)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.78, size = 196, normalized size = 1.70 \begin {gather*} - \frac {\sqrt {- \frac {1}{d^{5} e^{5}}} \cdot \left (3 a e^{2} + b d e + 3 c d^{2}\right ) \log {\left (- d^{3} e^{2} \sqrt {- \frac {1}{d^{5} e^{5}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{d^{5} e^{5}}} \cdot \left (3 a e^{2} + b d e + 3 c d^{2}\right ) \log {\left (d^{3} e^{2} \sqrt {- \frac {1}{d^{5} e^{5}}} + x \right )}}{16} + \frac {x^{3} \cdot \left (3 a e^{3} + b d e^{2} - 5 c d^{2} e\right ) + x \left (5 a d e^{2} - b d^{2} e - 3 c d^{3}\right )}{8 d^{4} e^{2} + 16 d^{3} e^{3} x^{2} + 8 d^{2} e^{4} x^{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)**3,x)

[Out]

-sqrt(-1/(d**5*e**5))*(3*a*e**2 + b*d*e + 3*c*d**2)*log(-d**3*e**2*sqrt(-1/(d**5*e**5)) + x)/16 + sqrt(-1/(d**
5*e**5))*(3*a*e**2 + b*d*e + 3*c*d**2)*log(d**3*e**2*sqrt(-1/(d**5*e**5)) + x)/16 + (x**3*(3*a*e**3 + b*d*e**2
- 5*c*d**2*e) + x*(5*a*d*e**2 - b*d**2*e - 3*c*d**3))/(8*d**4*e**2 + 16*d**3*e**3*x**2 + 8*d**2*e**4*x**4)

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Giac [A]
time = 3.23, size = 101, normalized size = 0.88 \begin {gather*} \frac {{\left (3 \, 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 {5}{2}\right )}}{8 \, d^{\frac {5}{2}}} - \frac {{\left (5 \, c d^{2} x^{3} e - b d x^{3} e^{2} + 3 \, c d^{3} x - 3 \, a x^{3} e^{3} + b d^{2} x e - 5 \, a d x e^{2}\right )} e^{\left (-2\right )}}{8 \, {\left (x^{2} e + d\right )}^{2} d^{2}} \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)^3,x, algorithm="giac")

[Out]

1/8*(3*c*d^2 + b*d*e + 3*a*e^2)*arctan(x*e^(1/2)/sqrt(d))*e^(-5/2)/d^(5/2) - 1/8*(5*c*d^2*x^3*e - b*d*x^3*e^2
+ 3*c*d^3*x - 3*a*x^3*e^3 + b*d^2*x*e - 5*a*d*x*e^2)*e^(-2)/((x^2*e + d)^2*d^2)

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Mupad [B]
time = 0.38, size = 112, normalized size = 0.97 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (3\,c\,d^2+b\,d\,e+3\,a\,e^2\right )}{8\,d^{5/2}\,e^{5/2}}-\frac {\frac {x\,\left (3\,c\,d^2+b\,d\,e-5\,a\,e^2\right )}{8\,d\,e^2}-\frac {x^3\,\left (-5\,c\,d^2+b\,d\,e+3\,a\,e^2\right )}{8\,d^2\,e}}{d^2+2\,d\,e\,x^2+e^2\,x^4} \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)^3,x)

[Out]

(atan((e^(1/2)*x)/d^(1/2))*(3*a*e^2 + 3*c*d^2 + b*d*e))/(8*d^(5/2)*e^(5/2)) - ((x*(3*c*d^2 - 5*a*e^2 + b*d*e))
/(8*d*e^2) - (x^3*(3*a*e^2 - 5*c*d^2 + b*d*e))/(8*d^2*e))/(d^2 + e^2*x^4 + 2*d*e*x^2)

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