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

Optimal. Leaf size=241 \[ -\frac{3 d \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{16 \sqrt{2} a^{7/4} \sqrt [4]{c}}+\frac{3 d \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{16 \sqrt{2} a^{7/4} \sqrt [4]{c}}-\frac{3 d \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt{2} a^{7/4} \sqrt [4]{c}}+\frac{3 d \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt{2} a^{7/4} \sqrt [4]{c}}+\frac{e \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{4 a^{3/2} \sqrt{c}}+\frac{x (d+e x)}{4 a \left (a+c x^4\right )} \]

[Out]

(x*(d + e*x))/(4*a*(a + c*x^4)) + (e*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(4*a^(3/2)*Sqrt[c]) - (3*d*ArcTan[1 - (Sqr
t[2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(7/4)*c^(1/4)) + (3*d*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[
2]*a^(7/4)*c^(1/4)) - (3*d*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(16*Sqrt[2]*a^(7/4)*c^(1/4)
) + (3*d*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(16*Sqrt[2]*a^(7/4)*c^(1/4))

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Rubi [A]  time = 0.189006, antiderivative size = 241, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {1855, 1876, 211, 1165, 628, 1162, 617, 204, 275, 205} \[ -\frac{3 d \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{16 \sqrt{2} a^{7/4} \sqrt [4]{c}}+\frac{3 d \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{16 \sqrt{2} a^{7/4} \sqrt [4]{c}}-\frac{3 d \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt{2} a^{7/4} \sqrt [4]{c}}+\frac{3 d \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt{2} a^{7/4} \sqrt [4]{c}}+\frac{e \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{4 a^{3/2} \sqrt{c}}+\frac{x (d+e x)}{4 a \left (a+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)/(a + c*x^4)^2,x]

[Out]

(x*(d + e*x))/(4*a*(a + c*x^4)) + (e*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(4*a^(3/2)*Sqrt[c]) - (3*d*ArcTan[1 - (Sqr
t[2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(7/4)*c^(1/4)) + (3*d*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(8*Sqrt[
2]*a^(7/4)*c^(1/4)) - (3*d*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(16*Sqrt[2]*a^(7/4)*c^(1/4)
) + (3*d*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/(16*Sqrt[2]*a^(7/4)*c^(1/4))

Rule 1855

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(x*Pq*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Di
st[1/(a*n*(p + 1)), Int[ExpandToSum[n*(p + 1)*Pq + D[x*Pq, x], x]*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b},
x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[Expon[Pq, x], n - 1]

Rule 1876

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[(x^ii*(Coeff[Pq, x, ii] + Coeff[Pq, x, n/2 + ii
]*x^(n/2)))/(a + b*x^n), {ii, 0, n/2 - 1}]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ
[n/2, 0] && Expon[Pq, x] < n

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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{d+e x}{\left (a+c x^4\right )^2} \, dx &=\frac{x (d+e x)}{4 a \left (a+c x^4\right )}-\frac{\int \frac{-3 d-2 e x}{a+c x^4} \, dx}{4 a}\\ &=\frac{x (d+e x)}{4 a \left (a+c x^4\right )}-\frac{\int \left (-\frac{3 d}{a+c x^4}-\frac{2 e x}{a+c x^4}\right ) \, dx}{4 a}\\ &=\frac{x (d+e x)}{4 a \left (a+c x^4\right )}+\frac{(3 d) \int \frac{1}{a+c x^4} \, dx}{4 a}+\frac{e \int \frac{x}{a+c x^4} \, dx}{2 a}\\ &=\frac{x (d+e x)}{4 a \left (a+c x^4\right )}+\frac{(3 d) \int \frac{\sqrt{a}-\sqrt{c} x^2}{a+c x^4} \, dx}{8 a^{3/2}}+\frac{(3 d) \int \frac{\sqrt{a}+\sqrt{c} x^2}{a+c x^4} \, dx}{8 a^{3/2}}+\frac{e \operatorname{Subst}\left (\int \frac{1}{a+c x^2} \, dx,x,x^2\right )}{4 a}\\ &=\frac{x (d+e x)}{4 a \left (a+c x^4\right )}+\frac{e \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{4 a^{3/2} \sqrt{c}}+\frac{(3 d) \int \frac{1}{\frac{\sqrt{a}}{\sqrt{c}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx}{16 a^{3/2} \sqrt{c}}+\frac{(3 d) \int \frac{1}{\frac{\sqrt{a}}{\sqrt{c}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{c}}+x^2} \, dx}{16 a^{3/2} \sqrt{c}}-\frac{(3 d) \int \frac{\frac{\sqrt{2} \sqrt [4]{a}}{\sqrt [4]{c}}+2 x}{-\frac{\sqrt{a}}{\sqrt{c}}-\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{c}}-x^2} \, dx}{16 \sqrt{2} a^{7/4} \sqrt [4]{c}}-\frac{(3 d) \int \frac{\frac{\sqrt{2} \sqrt [4]{a}}{\sqrt [4]{c}}-2 x}{-\frac{\sqrt{a}}{\sqrt{c}}+\frac{\sqrt{2} \sqrt [4]{a} x}{\sqrt [4]{c}}-x^2} \, dx}{16 \sqrt{2} a^{7/4} \sqrt [4]{c}}\\ &=\frac{x (d+e x)}{4 a \left (a+c x^4\right )}+\frac{e \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{4 a^{3/2} \sqrt{c}}-\frac{3 d \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{c} x^2\right )}{16 \sqrt{2} a^{7/4} \sqrt [4]{c}}+\frac{3 d \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{c} x^2\right )}{16 \sqrt{2} a^{7/4} \sqrt [4]{c}}+\frac{(3 d) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt{2} a^{7/4} \sqrt [4]{c}}-\frac{(3 d) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt{2} a^{7/4} \sqrt [4]{c}}\\ &=\frac{x (d+e x)}{4 a \left (a+c x^4\right )}+\frac{e \tan ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{a}}\right )}{4 a^{3/2} \sqrt{c}}-\frac{3 d \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt{2} a^{7/4} \sqrt [4]{c}}+\frac{3 d \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt{2} a^{7/4} \sqrt [4]{c}}-\frac{3 d \log \left (\sqrt{a}-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{c} x^2\right )}{16 \sqrt{2} a^{7/4} \sqrt [4]{c}}+\frac{3 d \log \left (\sqrt{a}+\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{c} x^2\right )}{16 \sqrt{2} a^{7/4} \sqrt [4]{c}}\\ \end{align*}

Mathematica [A]  time = 0.199239, size = 224, normalized size = 0.93 \[ \frac{\frac{8 a^{3/4} x (d+e x)}{a+c x^4}-\frac{2 \left (4 \sqrt [4]{a} e+3 \sqrt{2} \sqrt [4]{c} d\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{\sqrt{c}}+\frac{2 \left (3 \sqrt{2} \sqrt [4]{c} d-4 \sqrt [4]{a} e\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{\sqrt{c}}-\frac{3 \sqrt{2} d \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{\sqrt [4]{c}}+\frac{3 \sqrt{2} d \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt{a}+\sqrt{c} x^2\right )}{\sqrt [4]{c}}}{32 a^{7/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)/(a + c*x^4)^2,x]

[Out]

((8*a^(3/4)*x*(d + e*x))/(a + c*x^4) - (2*(3*Sqrt[2]*c^(1/4)*d + 4*a^(1/4)*e)*ArcTan[1 - (Sqrt[2]*c^(1/4)*x)/a
^(1/4)])/Sqrt[c] + (2*(3*Sqrt[2]*c^(1/4)*d - 4*a^(1/4)*e)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/Sqrt[c] - (
3*Sqrt[2]*d*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/c^(1/4) + (3*Sqrt[2]*d*Log[Sqrt[a] + Sqrt[
2]*a^(1/4)*c^(1/4)*x + Sqrt[c]*x^2])/c^(1/4))/(32*a^(7/4))

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Maple [A]  time = 0.004, size = 188, normalized size = 0.8 \begin{align*}{\frac{dx}{4\,a \left ( c{x}^{4}+a \right ) }}+{\frac{3\,d\sqrt{2}}{32\,{a}^{2}}\sqrt [4]{{\frac{a}{c}}}\ln \left ({ \left ({x}^{2}+\sqrt [4]{{\frac{a}{c}}}x\sqrt{2}+\sqrt{{\frac{a}{c}}} \right ) \left ({x}^{2}-\sqrt [4]{{\frac{a}{c}}}x\sqrt{2}+\sqrt{{\frac{a}{c}}} \right ) ^{-1}} \right ) }+{\frac{3\,d\sqrt{2}}{16\,{a}^{2}}\sqrt [4]{{\frac{a}{c}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}+1 \right ) }+{\frac{3\,d\sqrt{2}}{16\,{a}^{2}}\sqrt [4]{{\frac{a}{c}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{c}}}}}}-1 \right ) }+{\frac{e{x}^{2}}{4\,a \left ( c{x}^{4}+a \right ) }}+{\frac{e}{4\,a}\arctan \left ({x}^{2}\sqrt{{\frac{c}{a}}} \right ){\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*d*x/a/(c*x^4+a)+3/32*d/a^2*(a/c)^(1/4)*2^(1/2)*ln((x^2+(a/c)^(1/4)*x*2^(1/2)+(a/c)^(1/2))/(x^2-(a/c)^(1/4)
*x*2^(1/2)+(a/c)^(1/2)))+3/16*d/a^2*(a/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x+1)+3/16*d/a^2*(a/c)^(1/4)
*2^(1/2)*arctan(2^(1/2)/(a/c)^(1/4)*x-1)+1/4*e*x^2/a/(c*x^4+a)+1/4*e/a/(a*c)^(1/2)*arctan(x^2*(1/a*c)^(1/2))

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

[Out]

Timed out

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Sympy [A]  time = 1.28393, size = 155, normalized size = 0.64 \begin{align*} \operatorname{RootSum}{\left (65536 t^{4} a^{7} c^{2} + 2048 t^{2} a^{4} c e^{2} - 1152 t a^{2} c d^{2} e + 16 a e^{4} + 81 c d^{4}, \left ( t \mapsto t \log{\left (x + \frac{- 32768 t^{3} a^{6} c e^{2} - 4608 t^{2} a^{4} c d^{2} e - 512 t a^{3} e^{4} - 1296 t a^{2} c d^{4} + 360 a d^{2} e^{3}}{192 a d e^{4} - 243 c d^{5}} \right )} \right )\right )} + \frac{d x + e x^{2}}{4 a^{2} + 4 a c x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(65536*_t**4*a**7*c**2 + 2048*_t**2*a**4*c*e**2 - 1152*_t*a**2*c*d**2*e + 16*a*e**4 + 81*c*d**4, Lambda
(_t, _t*log(x + (-32768*_t**3*a**6*c*e**2 - 4608*_t**2*a**4*c*d**2*e - 512*_t*a**3*e**4 - 1296*_t*a**2*c*d**4
+ 360*a*d**2*e**3)/(192*a*d*e**4 - 243*c*d**5)))) + (d*x + e*x**2)/(4*a**2 + 4*a*c*x**4)

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Giac [A]  time = 1.11144, size = 325, normalized size = 1.35 \begin{align*} \frac{3 \, \sqrt{2} \left (a c^{3}\right )^{\frac{1}{4}} d \log \left (x^{2} + \sqrt{2} x \left (\frac{a}{c}\right )^{\frac{1}{4}} + \sqrt{\frac{a}{c}}\right )}{32 \, a^{2} c} - \frac{3 \, \sqrt{2} \left (a c^{3}\right )^{\frac{1}{4}} d \log \left (x^{2} - \sqrt{2} x \left (\frac{a}{c}\right )^{\frac{1}{4}} + \sqrt{\frac{a}{c}}\right )}{32 \, a^{2} c} + \frac{x^{2} e + d x}{4 \,{\left (c x^{4} + a\right )} a} + \frac{\sqrt{2}{\left (2 \, \sqrt{2} \sqrt{a c} c e + 3 \, \left (a c^{3}\right )^{\frac{1}{4}} c d\right )} \arctan \left (\frac{\sqrt{2}{\left (2 \, x + \sqrt{2} \left (\frac{a}{c}\right )^{\frac{1}{4}}\right )}}{2 \, \left (\frac{a}{c}\right )^{\frac{1}{4}}}\right )}{16 \, a^{2} c^{2}} + \frac{\sqrt{2}{\left (2 \, \sqrt{2} \sqrt{a c} c e + 3 \, \left (a c^{3}\right )^{\frac{1}{4}} c d\right )} \arctan \left (\frac{\sqrt{2}{\left (2 \, x - \sqrt{2} \left (\frac{a}{c}\right )^{\frac{1}{4}}\right )}}{2 \, \left (\frac{a}{c}\right )^{\frac{1}{4}}}\right )}{16 \, a^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/32*sqrt(2)*(a*c^3)^(1/4)*d*log(x^2 + sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(a^2*c) - 3/32*sqrt(2)*(a*c^3)^(1/4)
*d*log(x^2 - sqrt(2)*x*(a/c)^(1/4) + sqrt(a/c))/(a^2*c) + 1/4*(x^2*e + d*x)/((c*x^4 + a)*a) + 1/16*sqrt(2)*(2*
sqrt(2)*sqrt(a*c)*c*e + 3*(a*c^3)^(1/4)*c*d)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/c)^(1/4))/(a/c)^(1/4))/(a^2*
c^2) + 1/16*sqrt(2)*(2*sqrt(2)*sqrt(a*c)*c*e + 3*(a*c^3)^(1/4)*c*d)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/c)^(1
/4))/(a/c)^(1/4))/(a^2*c^2)