3.3.93 \(\int \frac {a+b x^2+c x^4}{x^4 (d+e x^2)^3} \, dx\) [293]

Optimal. Leaf size=142 \[ -\frac {a}{3 d^3 x^3}-\frac {b d-3 a e}{d^4 x}+\frac {\left (c d^2-b d e+a e^2\right ) x}{4 d^3 \left (d+e x^2\right )^2}+\frac {\left (3 c d^2-e (7 b d-11 a e)\right ) x}{8 d^4 \left (d+e x^2\right )}+\frac {\left (3 c d^2-15 b d e+35 a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{9/2} \sqrt {e}} \]

[Out]

-1/3*a/d^3/x^3+(3*a*e-b*d)/d^4/x+1/4*(a*e^2-b*d*e+c*d^2)*x/d^3/(e*x^2+d)^2+1/8*(3*c*d^2-e*(-11*a*e+7*b*d))*x/d
^4/(e*x^2+d)+1/8*(35*a*e^2-15*b*d*e+3*c*d^2)*arctan(x*e^(1/2)/d^(1/2))/d^(9/2)/e^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {1273, 1275, 211} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (35 a e^2-15 b d e+3 c d^2\right )}{8 d^{9/2} \sqrt {e}}+\frac {x \left (3 c d^2-e (7 b d-11 a e)\right )}{8 d^4 \left (d+e x^2\right )}+\frac {x \left (a e^2-b d e+c d^2\right )}{4 d^3 \left (d+e x^2\right )^2}-\frac {b d-3 a e}{d^4 x}-\frac {a}{3 d^3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*a/(d^3*x^3) - (b*d - 3*a*e)/(d^4*x) + ((c*d^2 - b*d*e + a*e^2)*x)/(4*d^3*(d + e*x^2)^2) + ((3*c*d^2 - e*(
7*b*d - 11*a*e))*x)/(8*d^4*(d + e*x^2)) + ((3*c*d^2 - 15*b*d*e + 35*a*e^2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^(
9/2)*Sqrt[e])

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 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]

Rule 1275

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rubi steps

\begin {align*} \int \frac {a+b x^2+c x^4}{x^4 \left (d+e x^2\right )^3} \, dx &=\frac {\left (c d^2-b d e+a e^2\right ) x}{4 d^3 \left (d+e x^2\right )^2}+\frac {\int \frac {4 a d^2 e^2+4 d e^2 (b d-a e) x^2+3 e^2 \left (c d^2-b d e+a e^2\right ) x^4}{x^4 \left (d+e x^2\right )^2} \, dx}{4 d^3 e^2}\\ &=\frac {\left (c d^2-b d e+a e^2\right ) x}{4 d^3 \left (d+e x^2\right )^2}+\frac {\left (3 c d^2-e (7 b d-11 a e)\right ) x}{8 d^4 \left (d+e x^2\right )}+\frac {\int \frac {8 a d^4 e^4+8 d^3 e^4 (b d-2 a e) x^2+d^2 e^4 \left (3 c d^2-e (7 b d-11 a e)\right ) x^4}{x^4 \left (d+e x^2\right )} \, dx}{8 d^6 e^4}\\ &=\frac {\left (c d^2-b d e+a e^2\right ) x}{4 d^3 \left (d+e x^2\right )^2}+\frac {\left (3 c d^2-e (7 b d-11 a e)\right ) x}{8 d^4 \left (d+e x^2\right )}+\frac {\int \left (\frac {8 a d^3 e^4}{x^4}+\frac {8 d^2 e^4 (b d-3 a e)}{x^2}+\frac {d^2 e^4 \left (3 c d^2-15 b d e+35 a e^2\right )}{d+e x^2}\right ) \, dx}{8 d^6 e^4}\\ &=-\frac {a}{3 d^3 x^3}-\frac {b d-3 a e}{d^4 x}+\frac {\left (c d^2-b d e+a e^2\right ) x}{4 d^3 \left (d+e x^2\right )^2}+\frac {\left (3 c d^2-e (7 b d-11 a e)\right ) x}{8 d^4 \left (d+e x^2\right )}+\frac {\left (3 c d^2-15 b d e+35 a e^2\right ) \int \frac {1}{d+e x^2} \, dx}{8 d^4}\\ &=-\frac {a}{3 d^3 x^3}-\frac {b d-3 a e}{d^4 x}+\frac {\left (c d^2-b d e+a e^2\right ) x}{4 d^3 \left (d+e x^2\right )^2}+\frac {\left (3 c d^2-e (7 b d-11 a e)\right ) x}{8 d^4 \left (d+e x^2\right )}+\frac {\left (3 c d^2-15 b d e+35 a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{9/2} \sqrt {e}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 141, normalized size = 0.99 \begin {gather*} -\frac {a}{3 d^3 x^3}+\frac {-b d+3 a e}{d^4 x}+\frac {\left (c d^2-b d e+a e^2\right ) x}{4 d^3 \left (d+e x^2\right )^2}+\frac {\left (3 c d^2-7 b d e+11 a e^2\right ) x}{8 d^4 \left (d+e x^2\right )}+\frac {\left (3 c d^2-15 b d e+35 a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{9/2} \sqrt {e}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*a/(d^3*x^3) + (-(b*d) + 3*a*e)/(d^4*x) + ((c*d^2 - b*d*e + a*e^2)*x)/(4*d^3*(d + e*x^2)^2) + ((3*c*d^2 -
7*b*d*e + 11*a*e^2)*x)/(8*d^4*(d + e*x^2)) + ((3*c*d^2 - 15*b*d*e + 35*a*e^2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*
d^(9/2)*Sqrt[e])

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Maple [A]
time = 0.14, size = 124, normalized size = 0.87

method	result	size
default	\(\frac {\frac {\left (\frac {11}{8} a \,e^{3}-\frac {7}{8} d \,e^{2} b +\frac {3}{8} c \,d^{2} e \right ) x^{3}+\frac {d \left (13 a \,e^{2}-9 d e b +5 c \,d^{2}\right ) x}{8}}{\left (e \,x^{2}+d \right )^{2}}+\frac {\left (35 a \,e^{2}-15 d e b +3 c \,d^{2}\right ) \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \sqrt {d e}}}{d^{4}}-\frac {a}{3 d^{3} x^{3}}-\frac {-3 a e +b d}{d^{4} x}\)	\(124\)
risch	\(\frac {\frac {e \left (35 a \,e^{2}-15 d e b +3 c \,d^{2}\right ) x^{6}}{8 d^{4}}+\frac {5 \left (35 a \,e^{2}-15 d e b +3 c \,d^{2}\right ) x^{4}}{24 d^{3}}+\frac {\left (7 a e -3 b d \right ) x^{2}}{3 d^{2}}-\frac {a}{3 d}}{x^{3} \left (e \,x^{2}+d \right )^{2}}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (d^{9} e \,\textit {\_Z}^{2}+1225 a^{2} e^{4}-1050 a b d \,e^{3}+210 a c \,d^{2} e^{2}+225 b^{2} d^{2} e^{2}-90 b c \,d^{3} e +9 c^{2} d^{4}\right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} d^{9} e +2450 a^{2} e^{4}-2100 a b d \,e^{3}+420 a c \,d^{2} e^{2}+450 b^{2} d^{2} e^{2}-180 b c \,d^{3} e +18 c^{2} d^{4}\right ) x +\left (-35 a \,d^{5} e^{2}+15 b \,d^{6} e -3 c \,d^{7}\right ) \textit {\_R} \right )\right )}{16}\)	\(254\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d^4*(((11/8*a*e^3-7/8*d*e^2*b+3/8*c*d^2*e)*x^3+1/8*d*(13*a*e^2-9*b*d*e+5*c*d^2)*x)/(e*x^2+d)^2+1/8*(35*a*e^2
-15*b*d*e+3*c*d^2)/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2)))-1/3*a/d^3/x^3-(-3*a*e+b*d)/d^4/x

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Maxima [A]
time = 0.50, size = 143, normalized size = 1.01 \begin {gather*} \frac {3 \, {\left (3 \, c d^{2} e - 15 \, b d e^{2} + 35 \, a e^{3}\right )} x^{6} + 5 \, {\left (3 \, c d^{3} - 15 \, b d^{2} e + 35 \, a d e^{2}\right )} x^{4} - 8 \, a d^{3} - 8 \, {\left (3 \, b d^{3} - 7 \, a d^{2} e\right )} x^{2}}{24 \, {\left (d^{4} x^{7} e^{2} + 2 \, d^{5} x^{5} e + d^{6} x^{3}\right )}} + \frac {{\left (3 \, c d^{2} - 15 \, b d e + 35 \, a e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )}}{8 \, d^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/x^4/(e*x^2+d)^3,x, algorithm="maxima")

[Out]

1/24*(3*(3*c*d^2*e - 15*b*d*e^2 + 35*a*e^3)*x^6 + 5*(3*c*d^3 - 15*b*d^2*e + 35*a*d*e^2)*x^4 - 8*a*d^3 - 8*(3*b
*d^3 - 7*a*d^2*e)*x^2)/(d^4*x^7*e^2 + 2*d^5*x^5*e + d^6*x^3) + 1/8*(3*c*d^2 - 15*b*d*e + 35*a*e^2)*arctan(x*e^
(1/2)/sqrt(d))*e^(-1/2)/d^(9/2)

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Fricas [A]
time = 0.37, size = 491, normalized size = 3.46 \begin {gather*} \left [\frac {210 \, a d x^{6} e^{4} - 3 \, {\left (35 \, a x^{7} e^{4} + 3 \, c d^{4} x^{3} - 5 \, {\left (3 \, b d x^{7} - 14 \, a d x^{5}\right )} e^{3} + {\left (3 \, c d^{2} x^{7} - 30 \, b d^{2} x^{5} + 35 \, a d^{2} x^{3}\right )} e^{2} + 3 \, {\left (2 \, c d^{3} x^{5} - 5 \, b d^{3} x^{3}\right )} e\right )} \sqrt {-d e} \log \left (\frac {x^{2} e - 2 \, \sqrt {-d e} x - d}{x^{2} e + d}\right ) - 10 \, {\left (9 \, b d^{2} x^{6} - 35 \, a d^{2} x^{4}\right )} e^{3} + 2 \, {\left (9 \, c d^{3} x^{6} - 75 \, b d^{3} x^{4} + 56 \, a d^{3} x^{2}\right )} e^{2} + 2 \, {\left (15 \, c d^{4} x^{4} - 24 \, b d^{4} x^{2} - 8 \, a d^{4}\right )} e}{48 \, {\left (d^{5} x^{7} e^{3} + 2 \, d^{6} x^{5} e^{2} + d^{7} x^{3} e\right )}}, \frac {105 \, a d x^{6} e^{4} + 3 \, {\left (35 \, a x^{7} e^{4} + 3 \, c d^{4} x^{3} - 5 \, {\left (3 \, b d x^{7} - 14 \, a d x^{5}\right )} e^{3} + {\left (3 \, c d^{2} x^{7} - 30 \, b d^{2} x^{5} + 35 \, a d^{2} x^{3}\right )} e^{2} + 3 \, {\left (2 \, c d^{3} x^{5} - 5 \, b d^{3} x^{3}\right )} e\right )} \sqrt {d} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\frac {1}{2}} - 5 \, {\left (9 \, b d^{2} x^{6} - 35 \, a d^{2} x^{4}\right )} e^{3} + {\left (9 \, c d^{3} x^{6} - 75 \, b d^{3} x^{4} + 56 \, a d^{3} x^{2}\right )} e^{2} + {\left (15 \, c d^{4} x^{4} - 24 \, b d^{4} x^{2} - 8 \, a d^{4}\right )} e}{24 \, {\left (d^{5} x^{7} e^{3} + 2 \, d^{6} x^{5} e^{2} + d^{7} x^{3} e\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/x^4/(e*x^2+d)^3,x, algorithm="fricas")

[Out]

[1/48*(210*a*d*x^6*e^4 - 3*(35*a*x^7*e^4 + 3*c*d^4*x^3 - 5*(3*b*d*x^7 - 14*a*d*x^5)*e^3 + (3*c*d^2*x^7 - 30*b*
d^2*x^5 + 35*a*d^2*x^3)*e^2 + 3*(2*c*d^3*x^5 - 5*b*d^3*x^3)*e)*sqrt(-d*e)*log((x^2*e - 2*sqrt(-d*e)*x - d)/(x^
2*e + d)) - 10*(9*b*d^2*x^6 - 35*a*d^2*x^4)*e^3 + 2*(9*c*d^3*x^6 - 75*b*d^3*x^4 + 56*a*d^3*x^2)*e^2 + 2*(15*c*
d^4*x^4 - 24*b*d^4*x^2 - 8*a*d^4)*e)/(d^5*x^7*e^3 + 2*d^6*x^5*e^2 + d^7*x^3*e), 1/24*(105*a*d*x^6*e^4 + 3*(35*
a*x^7*e^4 + 3*c*d^4*x^3 - 5*(3*b*d*x^7 - 14*a*d*x^5)*e^3 + (3*c*d^2*x^7 - 30*b*d^2*x^5 + 35*a*d^2*x^3)*e^2 + 3
*(2*c*d^3*x^5 - 5*b*d^3*x^3)*e)*sqrt(d)*arctan(x*e^(1/2)/sqrt(d))*e^(1/2) - 5*(9*b*d^2*x^6 - 35*a*d^2*x^4)*e^3
+ (9*c*d^3*x^6 - 75*b*d^3*x^4 + 56*a*d^3*x^2)*e^2 + (15*c*d^4*x^4 - 24*b*d^4*x^2 - 8*a*d^4)*e)/(d^5*x^7*e^3 +
2*d^6*x^5*e^2 + d^7*x^3*e)]

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Sympy [A]
time = 1.42, size = 214, normalized size = 1.51 \begin {gather*} - \frac {\sqrt {- \frac {1}{d^{9} e}} \left (35 a e^{2} - 15 b d e + 3 c d^{2}\right ) \log {\left (- d^{5} \sqrt {- \frac {1}{d^{9} e}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{d^{9} e}} \left (35 a e^{2} - 15 b d e + 3 c d^{2}\right ) \log {\left (d^{5} \sqrt {- \frac {1}{d^{9} e}} + x \right )}}{16} + \frac {- 8 a d^{3} + x^{6} \cdot \left (105 a e^{3} - 45 b d e^{2} + 9 c d^{2} e\right ) + x^{4} \cdot \left (175 a d e^{2} - 75 b d^{2} e + 15 c d^{3}\right ) + x^{2} \cdot \left (56 a d^{2} e - 24 b d^{3}\right )}{24 d^{6} x^{3} + 48 d^{5} e x^{5} + 24 d^{4} e^{2} x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2+a)/x**4/(e*x**2+d)**3,x)

[Out]

-sqrt(-1/(d**9*e))*(35*a*e**2 - 15*b*d*e + 3*c*d**2)*log(-d**5*sqrt(-1/(d**9*e)) + x)/16 + sqrt(-1/(d**9*e))*(
35*a*e**2 - 15*b*d*e + 3*c*d**2)*log(d**5*sqrt(-1/(d**9*e)) + x)/16 + (-8*a*d**3 + x**6*(105*a*e**3 - 45*b*d*e
**2 + 9*c*d**2*e) + x**4*(175*a*d*e**2 - 75*b*d**2*e + 15*c*d**3) + x**2*(56*a*d**2*e - 24*b*d**3))/(24*d**6*x
**3 + 48*d**5*e*x**5 + 24*d**4*e**2*x**7)

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Giac [A]
time = 4.46, size = 128, normalized size = 0.90 \begin {gather*} \frac {{\left (3 \, c d^{2} - 15 \, b d e + 35 \, a e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )}}{8 \, d^{\frac {9}{2}}} + \frac {3 \, c d^{2} x^{3} e - 7 \, b d x^{3} e^{2} + 5 \, c d^{3} x + 11 \, a x^{3} e^{3} - 9 \, b d^{2} x e + 13 \, a d x e^{2}}{8 \, {\left (x^{2} e + d\right )}^{2} d^{4}} - \frac {3 \, b d x^{2} - 9 \, a x^{2} e + a d}{3 \, d^{4} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)/x^4/(e*x^2+d)^3,x, algorithm="giac")

[Out]

1/8*(3*c*d^2 - 15*b*d*e + 35*a*e^2)*arctan(x*e^(1/2)/sqrt(d))*e^(-1/2)/d^(9/2) + 1/8*(3*c*d^2*x^3*e - 7*b*d*x^
3*e^2 + 5*c*d^3*x + 11*a*x^3*e^3 - 9*b*d^2*x*e + 13*a*d*x*e^2)/((x^2*e + d)^2*d^4) - 1/3*(3*b*d*x^2 - 9*a*x^2*
e + a*d)/(d^4*x^3)

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Mupad [B]
time = 0.40, size = 138, normalized size = 0.97 \begin {gather*} \frac {\frac {x^2\,\left (7\,a\,e-3\,b\,d\right )}{3\,d^2}-\frac {a}{3\,d}+\frac {5\,x^4\,\left (3\,c\,d^2-15\,b\,d\,e+35\,a\,e^2\right )}{24\,d^3}+\frac {e\,x^6\,\left (3\,c\,d^2-15\,b\,d\,e+35\,a\,e^2\right )}{8\,d^4}}{d^2\,x^3+2\,d\,e\,x^5+e^2\,x^7}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (3\,c\,d^2-15\,b\,d\,e+35\,a\,e^2\right )}{8\,d^{9/2}\,\sqrt {e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x^2*(7*a*e - 3*b*d))/(3*d^2) - a/(3*d) + (5*x^4*(35*a*e^2 + 3*c*d^2 - 15*b*d*e))/(24*d^3) + (e*x^6*(35*a*e^2
+ 3*c*d^2 - 15*b*d*e))/(8*d^4))/(d^2*x^3 + e^2*x^7 + 2*d*e*x^5) + (atan((e^(1/2)*x)/d^(1/2))*(35*a*e^2 + 3*c*
d^2 - 15*b*d*e))/(8*d^(9/2)*e^(1/2))

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