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

Optimal. Leaf size=129 \[ -\frac {c^{3/2} d \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 a^{3/2} \left (a e^2+c d^2\right )}+\frac {c e \log \left (a+c x^4\right )}{4 a \left (a e^2+c d^2\right )}+\frac {e^3 \log \left (d+e x^2\right )}{2 d^2 \left (a e^2+c d^2\right )}-\frac {e \log (x)}{a d^2}-\frac {1}{2 a d x^2} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1252, 894, 635, 205, 260} \[ -\frac {c^{3/2} d \tan ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 a^{3/2} \left (a e^2+c d^2\right )}+\frac {e^3 \log \left (d+e x^2\right )}{2 d^2 \left (a e^2+c d^2\right )}+\frac {c e \log \left (a+c x^4\right )}{4 a \left (a e^2+c d^2\right )}-\frac {e \log (x)}{a d^2}-\frac {1}{2 a d x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 169, normalized size = 1.31 \[ \frac {2 c^{3/2} d^3 x^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )+2 c^{3/2} d^3 x^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )+\sqrt {a} \left (-4 e x^2 \log (x) \left (a e^2+c d^2\right )+c d^2 e x^2 \log \left (a+c x^4\right )+2 a e^3 x^2 \log \left (d+e x^2\right )-2 a d e^2-2 c d^3\right )}{4 a^{3/2} d^2 x^2 \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 68.84, size = 265, normalized size = 2.05 \[ \left [\frac {c d^{3} x^{2} \sqrt {-\frac {c}{a}} \log \left (\frac {c x^{4} - 2 \, a x^{2} \sqrt {-\frac {c}{a}} - a}{c x^{4} + a}\right ) + c d^{2} e x^{2} \log \left (c x^{4} + a\right ) + 2 \, a e^{3} x^{2} \log \left (e x^{2} + d\right ) - 2 \, c d^{3} - 2 \, a d e^{2} - 4 \, {\left (c d^{2} e + a e^{3}\right )} x^{2} \log \relax (x)}{4 \, {\left (a c d^{4} + a^{2} d^{2} e^{2}\right )} x^{2}}, \frac {2 \, c d^{3} x^{2} \sqrt {\frac {c}{a}} \arctan \left (\frac {a \sqrt {\frac {c}{a}}}{c x^{2}}\right ) + c d^{2} e x^{2} \log \left (c x^{4} + a\right ) + 2 \, a e^{3} x^{2} \log \left (e x^{2} + d\right ) - 2 \, c d^{3} - 2 \, a d e^{2} - 4 \, {\left (c d^{2} e + a e^{3}\right )} x^{2} \log \relax (x)}{4 \, {\left (a c d^{4} + a^{2} d^{2} e^{2}\right )} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.30, size = 132, normalized size = 1.02 \[ -\frac {c^{2} d \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{2 \, {\left (a c d^{2} + a^{2} e^{2}\right )} \sqrt {a c}} + \frac {c e \log \left (c x^{4} + a\right )}{4 \, {\left (a c d^{2} + a^{2} e^{2}\right )}} + \frac {e^{4} \log \left ({\left | x^{2} e + d \right |}\right )}{2 \, {\left (c d^{4} e + a d^{2} e^{3}\right )}} - \frac {e \log \left (x^{2}\right )}{2 \, a d^{2}} + \frac {x^{2} e - d}{2 \, a d^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 119, normalized size = 0.92 \[ -\frac {c^{2} d \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{2 \left (a \,e^{2}+c \,d^{2}\right ) \sqrt {a c}\, a}+\frac {c e \ln \left (c \,x^{4}+a \right )}{4 \left (a \,e^{2}+c \,d^{2}\right ) a}+\frac {e^{3} \ln \left (e \,x^{2}+d \right )}{2 \left (a \,e^{2}+c \,d^{2}\right ) d^{2}}-\frac {e \ln \relax (x )}{a \,d^{2}}-\frac {1}{2 a d \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.00, size = 120, normalized size = 0.93 \[ \frac {e^{3} \log \left (e x^{2} + d\right )}{2 \, {\left (c d^{4} + a d^{2} e^{2}\right )}} - \frac {c^{2} d \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{2 \, {\left (a c d^{2} + a^{2} e^{2}\right )} \sqrt {a c}} + \frac {c e \log \left (c x^{4} + a\right )}{4 \, {\left (a c d^{2} + a^{2} e^{2}\right )}} - \frac {e \log \left (x^{2}\right )}{2 \, a d^{2}} - \frac {1}{2 \, a d x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.38, size = 820, normalized size = 6.36 \[ \frac {\ln \left (a^6\,c^{12}\,d^{16}\,x^2+64\,a^{14}\,c^4\,e^{16}\,x^2+a^2\,c^7\,d^{16}\,{\left (-a^3\,c^3\right )}^{3/2}-64\,a^{13}\,c^2\,e^{16}\,\sqrt {-a^3\,c^3}+63\,a^3\,d^8\,e^8\,{\left (-a^3\,c^3\right )}^{5/2}+224\,a^9\,d^2\,e^{14}\,{\left (-a^3\,c^3\right )}^{3/2}-28\,c^3\,d^{14}\,e^2\,{\left (-a^3\,c^3\right )}^{5/2}+28\,a^7\,c^{11}\,d^{14}\,e^2\,x^2+114\,a^8\,c^{10}\,d^{12}\,e^4\,x^2+108\,a^9\,c^9\,d^{10}\,e^6\,x^2-63\,a^{10}\,c^8\,d^8\,e^8\,x^2-32\,a^{11}\,c^7\,d^6\,e^{10}\,x^2+212\,a^{12}\,c^6\,d^4\,e^{12}\,x^2+224\,a^{13}\,c^5\,d^2\,e^{14}\,x^2-114\,a\,c^2\,d^{12}\,e^4\,{\left (-a^3\,c^3\right )}^{5/2}-108\,a^2\,c\,d^{10}\,e^6\,{\left (-a^3\,c^3\right )}^{5/2}+212\,a^8\,c\,d^4\,e^{12}\,{\left (-a^3\,c^3\right )}^{3/2}-32\,a^7\,c^2\,d^6\,e^{10}\,{\left (-a^3\,c^3\right )}^{3/2}\right )\,\left (d\,\sqrt {-a^3\,c^3}+a^2\,c\,e\right )}{4\,a^4\,e^2+4\,c\,a^3\,d^2}-\frac {\ln \left (a^6\,c^{12}\,d^{16}\,x^2+64\,a^{14}\,c^4\,e^{16}\,x^2-a^2\,c^7\,d^{16}\,{\left (-a^3\,c^3\right )}^{3/2}+64\,a^{13}\,c^2\,e^{16}\,\sqrt {-a^3\,c^3}-63\,a^3\,d^8\,e^8\,{\left (-a^3\,c^3\right )}^{5/2}-224\,a^9\,d^2\,e^{14}\,{\left (-a^3\,c^3\right )}^{3/2}+28\,c^3\,d^{14}\,e^2\,{\left (-a^3\,c^3\right )}^{5/2}+28\,a^7\,c^{11}\,d^{14}\,e^2\,x^2+114\,a^8\,c^{10}\,d^{12}\,e^4\,x^2+108\,a^9\,c^9\,d^{10}\,e^6\,x^2-63\,a^{10}\,c^8\,d^8\,e^8\,x^2-32\,a^{11}\,c^7\,d^6\,e^{10}\,x^2+212\,a^{12}\,c^6\,d^4\,e^{12}\,x^2+224\,a^{13}\,c^5\,d^2\,e^{14}\,x^2+114\,a\,c^2\,d^{12}\,e^4\,{\left (-a^3\,c^3\right )}^{5/2}+108\,a^2\,c\,d^{10}\,e^6\,{\left (-a^3\,c^3\right )}^{5/2}-212\,a^8\,c\,d^4\,e^{12}\,{\left (-a^3\,c^3\right )}^{3/2}+32\,a^7\,c^2\,d^6\,e^{10}\,{\left (-a^3\,c^3\right )}^{3/2}\right )\,\left (d\,\sqrt {-a^3\,c^3}-a^2\,c\,e\right )}{4\,\left (a^4\,e^2+c\,a^3\,d^2\right )}+\frac {e^3\,\ln \left (e\,x^2+d\right )}{2\,c\,d^4+2\,a\,d^2\,e^2}-\frac {1}{2\,a\,d\,x^2}-\frac {e\,\ln \relax (x)}{a\,d^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(a^6*c^12*d^16*x^2 + 64*a^14*c^4*e^16*x^2 + a^2*c^7*d^16*(-a^3*c^3)^(3/2) - 64*a^13*c^2*e^16*(-a^3*c^3)^(1
/2) + 63*a^3*d^8*e^8*(-a^3*c^3)^(5/2) + 224*a^9*d^2*e^14*(-a^3*c^3)^(3/2) - 28*c^3*d^14*e^2*(-a^3*c^3)^(5/2) +
28*a^7*c^11*d^14*e^2*x^2 + 114*a^8*c^10*d^12*e^4*x^2 + 108*a^9*c^9*d^10*e^6*x^2 - 63*a^10*c^8*d^8*e^8*x^2 - 3
2*a^11*c^7*d^6*e^10*x^2 + 212*a^12*c^6*d^4*e^12*x^2 + 224*a^13*c^5*d^2*e^14*x^2 - 114*a*c^2*d^12*e^4*(-a^3*c^3
)^(5/2) - 108*a^2*c*d^10*e^6*(-a^3*c^3)^(5/2) + 212*a^8*c*d^4*e^12*(-a^3*c^3)^(3/2) - 32*a^7*c^2*d^6*e^10*(-a^
3*c^3)^(3/2))*(d*(-a^3*c^3)^(1/2) + a^2*c*e))/(4*a^4*e^2 + 4*a^3*c*d^2) - (log(a^6*c^12*d^16*x^2 + 64*a^14*c^4
*e^16*x^2 - a^2*c^7*d^16*(-a^3*c^3)^(3/2) + 64*a^13*c^2*e^16*(-a^3*c^3)^(1/2) - 63*a^3*d^8*e^8*(-a^3*c^3)^(5/2
) - 224*a^9*d^2*e^14*(-a^3*c^3)^(3/2) + 28*c^3*d^14*e^2*(-a^3*c^3)^(5/2) + 28*a^7*c^11*d^14*e^2*x^2 + 114*a^8*
c^10*d^12*e^4*x^2 + 108*a^9*c^9*d^10*e^6*x^2 - 63*a^10*c^8*d^8*e^8*x^2 - 32*a^11*c^7*d^6*e^10*x^2 + 212*a^12*c
^6*d^4*e^12*x^2 + 224*a^13*c^5*d^2*e^14*x^2 + 114*a*c^2*d^12*e^4*(-a^3*c^3)^(5/2) + 108*a^2*c*d^10*e^6*(-a^3*c
^3)^(5/2) - 212*a^8*c*d^4*e^12*(-a^3*c^3)^(3/2) + 32*a^7*c^2*d^6*e^10*(-a^3*c^3)^(3/2))*(d*(-a^3*c^3)^(1/2) -
a^2*c*e))/(4*(a^4*e^2 + a^3*c*d^2)) + (e^3*log(d + e*x^2))/(2*c*d^4 + 2*a*d^2*e^2) - 1/(2*a*d*x^2) - (e*log(x)
)/(a*d^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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