3.250 \(\int \frac{1}{x^3 (d+e x^2) (a+c x^4)^2} \, dx\)

Optimal. Leaf size=236 \[ -\frac{c^{3/2} d \left (2 a e^2+c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{2 a^{5/2} \left (a e^2+c d^2\right )^2}-\frac{c^{3/2} d \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{4 a^{5/2} \left (a e^2+c d^2\right )}-\frac{c \left (a e+c d x^2\right )}{4 a^2 \left (a+c x^4\right ) \left (a e^2+c d^2\right )}+\frac{c e \left (2 a e^2+c d^2\right ) \log \left (a+c x^4\right )}{4 a^2 \left (a e^2+c d^2\right )^2}-\frac{e \log (x)}{a^2 d^2}-\frac{1}{2 a^2 d x^2}+\frac{e^5 \log \left (d+e x^2\right )}{2 d^2 \left (a e^2+c d^2\right )^2} \]

[Out]

-1/(2*a^2*d*x^2) - (c*(a*e + c*d*x^2))/(4*a^2*(c*d^2 + a*e^2)*(a + c*x^4)) - (c^(3/2)*d*ArcTan[(Sqrt[c]*x^2)/S
qrt[a]])/(4*a^(5/2)*(c*d^2 + a*e^2)) - (c^(3/2)*d*(c*d^2 + 2*a*e^2)*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(2*a^(5/2)*
(c*d^2 + a*e^2)^2) - (e*Log[x])/(a^2*d^2) + (e^5*Log[d + e*x^2])/(2*d^2*(c*d^2 + a*e^2)^2) + (c*e*(c*d^2 + 2*a
*e^2)*Log[a + c*x^4])/(4*a^2*(c*d^2 + a*e^2)^2)

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Rubi [A]  time = 0.261094, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {1252, 894, 639, 205, 635, 260} \[ -\frac{c^{3/2} d \left (2 a e^2+c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{2 a^{5/2} \left (a e^2+c d^2\right )^2}-\frac{c^{3/2} d \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{4 a^{5/2} \left (a e^2+c d^2\right )}-\frac{c \left (a e+c d x^2\right )}{4 a^2 \left (a+c x^4\right ) \left (a e^2+c d^2\right )}+\frac{c e \left (2 a e^2+c d^2\right ) \log \left (a+c x^4\right )}{4 a^2 \left (a e^2+c d^2\right )^2}-\frac{e \log (x)}{a^2 d^2}-\frac{1}{2 a^2 d x^2}+\frac{e^5 \log \left (d+e x^2\right )}{2 d^2 \left (a e^2+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(d + e*x^2)*(a + c*x^4)^2),x]

[Out]

-1/(2*a^2*d*x^2) - (c*(a*e + c*d*x^2))/(4*a^2*(c*d^2 + a*e^2)*(a + c*x^4)) - (c^(3/2)*d*ArcTan[(Sqrt[c]*x^2)/S
qrt[a]])/(4*a^(5/2)*(c*d^2 + a*e^2)) - (c^(3/2)*d*(c*d^2 + 2*a*e^2)*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(2*a^(5/2)*
(c*d^2 + a*e^2)^2) - (e*Log[x])/(a^2*d^2) + (e^5*Log[d + e*x^2])/(2*d^2*(c*d^2 + a*e^2)^2) + (c*e*(c*d^2 + 2*a
*e^2)*Log[a + c*x^4])/(4*a^2*(c*d^2 + a*e^2)^2)

Rule 1252

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

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 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [A]  time = 0.450885, size = 248, normalized size = 1.05 \[ \frac{1}{4} \left (\frac{c^{3/2} d \left (5 a e^2+3 c d^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{a^{5/2} \left (a e^2+c d^2\right )^2}+\frac{c^{3/2} d \left (5 a e^2+3 c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{a^{5/2} \left (a e^2+c d^2\right )^2}-\frac{c \left (a e+c d x^2\right )}{a^2 \left (a+c x^4\right ) \left (a e^2+c d^2\right )}+\frac{c \left (2 a e^3+c d^2 e\right ) \log \left (a+c x^4\right )}{a^2 \left (a e^2+c d^2\right )^2}-\frac{4 e \log (x)}{a^2 d^2}-\frac{2}{a^2 d x^2}+\frac{2 e^5 \log \left (d+e x^2\right )}{\left (a d e^2+c d^3\right )^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(d + e*x^2)*(a + c*x^4)^2),x]

[Out]

(-2/(a^2*d*x^2) - (c*(a*e + c*d*x^2))/(a^2*(c*d^2 + a*e^2)*(a + c*x^4)) + (c^(3/2)*d*(3*c*d^2 + 5*a*e^2)*ArcTa
n[1 - (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(a^(5/2)*(c*d^2 + a*e^2)^2) + (c^(3/2)*d*(3*c*d^2 + 5*a*e^2)*ArcTan[1 + (S
qrt[2]*c^(1/4)*x)/a^(1/4)])/(a^(5/2)*(c*d^2 + a*e^2)^2) - (4*e*Log[x])/(a^2*d^2) + (2*e^5*Log[d + e*x^2])/(c*d
^3 + a*d*e^2)^2 + (c*(c*d^2*e + 2*a*e^3)*Log[a + c*x^4])/(a^2*(c*d^2 + a*e^2)^2))/4

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Maple [A]  time = 0.025, size = 332, normalized size = 1.4 \begin{align*} -{\frac{{c}^{2}{x}^{2}{e}^{2}d}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}a \left ( c{x}^{4}+a \right ) }}-{\frac{{c}^{3}{x}^{2}{d}^{3}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}{a}^{2} \left ( c{x}^{4}+a \right ) }}-{\frac{{e}^{3}c}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( c{x}^{4}+a \right ) }}-{\frac{e{d}^{2}{c}^{2}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}a \left ( c{x}^{4}+a \right ) }}+{\frac{c\ln \left ( c{x}^{4}+a \right ){e}^{3}}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}a}}+{\frac{{c}^{2}\ln \left ( c{x}^{4}+a \right ) e{d}^{2}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}{a}^{2}}}-{\frac{5\,{e}^{2}{c}^{2}d}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}a}\arctan \left ({c{x}^{2}{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}-{\frac{3\,{c}^{3}{d}^{3}}{4\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}{a}^{2}}\arctan \left ({c{x}^{2}{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}-{\frac{1}{2\,d{a}^{2}{x}^{2}}}-{\frac{\ln \left ( x \right ) e}{{d}^{2}{a}^{2}}}+{\frac{{e}^{5}\ln \left ( e{x}^{2}+d \right ) }{2\,{d}^{2} \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(e*x^2+d)/(c*x^4+a)^2,x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(e*x^2+d)/(c*x^4+a)^2,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(e*x**2+d)/(c*x**4+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.10956, size = 464, normalized size = 1.97 \begin{align*} \frac{{\left (c^{2} d^{2} e + 2 \, a c e^{3}\right )} \log \left (c x^{4} + a\right )}{4 \,{\left (a^{2} c^{2} d^{4} + 2 \, a^{3} c d^{2} e^{2} + a^{4} e^{4}\right )}} + \frac{e^{6} \log \left ({\left | x^{2} e + d \right |}\right )}{2 \,{\left (c^{2} d^{6} e + 2 \, a c d^{4} e^{3} + a^{2} d^{2} e^{5}\right )}} - \frac{{\left (3 \, c^{3} d^{3} + 5 \, a c^{2} d e^{2}\right )} \arctan \left (\frac{c x^{2}}{\sqrt{a c}}\right )}{4 \,{\left (a^{2} c^{2} d^{4} + 2 \, a^{3} c d^{2} e^{2} + a^{4} e^{4}\right )} \sqrt{a c}} - \frac{9 \, c^{3} d^{5} x^{4} + 15 \, a c^{2} d^{3} x^{4} e^{2} - 2 \, a^{2} c x^{6} e^{5} + 3 \, a c^{2} d^{4} x^{2} e + 6 \, a^{2} c d x^{4} e^{4} + 6 \, a c^{2} d^{5} + 3 \, a^{2} c d^{2} x^{2} e^{3} + 12 \, a^{2} c d^{3} e^{2} - 2 \, a^{3} x^{2} e^{5} + 6 \, a^{3} d e^{4}}{12 \,{\left (a^{2} c^{2} d^{6} + 2 \, a^{3} c d^{4} e^{2} + a^{4} d^{2} e^{4}\right )}{\left (c x^{6} + a x^{2}\right )}} - \frac{e \log \left (x^{2}\right )}{2 \, a^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(e*x^2+d)/(c*x^4+a)^2,x, algorithm="giac")

[Out]

1/4*(c^2*d^2*e + 2*a*c*e^3)*log(c*x^4 + a)/(a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4) + 1/2*e^6*log(abs(x^2*e +
d))/(c^2*d^6*e + 2*a*c*d^4*e^3 + a^2*d^2*e^5) - 1/4*(3*c^3*d^3 + 5*a*c^2*d*e^2)*arctan(c*x^2/sqrt(a*c))/((a^2
*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4)*sqrt(a*c)) - 1/12*(9*c^3*d^5*x^4 + 15*a*c^2*d^3*x^4*e^2 - 2*a^2*c*x^6*e^
5 + 3*a*c^2*d^4*x^2*e + 6*a^2*c*d*x^4*e^4 + 6*a*c^2*d^5 + 3*a^2*c*d^2*x^2*e^3 + 12*a^2*c*d^3*e^2 - 2*a^3*x^2*e
^5 + 6*a^3*d*e^4)/((a^2*c^2*d^6 + 2*a^3*c*d^4*e^2 + a^4*d^2*e^4)*(c*x^6 + a*x^2)) - 1/2*e*log(x^2)/(a^2*d^2)