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

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

[Out]

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

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Rubi [A]  time = 0.201599, 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} \[ -\frac{3 c \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right )}{16 \sqrt{2} a^{7/4} \sqrt [4]{b}}+\frac{3 c \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right )}{16 \sqrt{2} a^{7/4} \sqrt [4]{b}}-\frac{3 c \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt{2} a^{7/4} \sqrt [4]{b}}+\frac{3 c \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt{2} a^{7/4} \sqrt [4]{b}}+\frac{d \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{4 a^{3/2} \sqrt{b}}+\frac{x (c+d x)}{4 a \left (a+b x^4\right )} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a + b*x^4)^2,x]

[Out]

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

Mathematica [A]  time = 0.204034, size = 224, normalized size = 0.93 \[ \frac{\frac{8 a^{3/4} x (c+d x)}{a+b x^4}-\frac{2 \left (4 \sqrt [4]{a} d+3 \sqrt{2} \sqrt [4]{b} c\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{\sqrt{b}}+\frac{2 \left (3 \sqrt{2} \sqrt [4]{b} c-4 \sqrt [4]{a} d\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{\sqrt{b}}-\frac{3 \sqrt{2} c \log \left (-\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right )}{\sqrt [4]{b}}+\frac{3 \sqrt{2} c \log \left (\sqrt{2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt{a}+\sqrt{b} x^2\right )}{\sqrt [4]{b}}}{32 a^{7/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)/(a + b*x^4)^2,x]

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*c*x/a/(b*x^4+a)+3/32*c/a^2*(1/b*a)^(1/4)*2^(1/2)*ln((x^2+(1/b*a)^(1/4)*x*2^(1/2)+(1/b*a)^(1/2))/(x^2-(1/b*
a)^(1/4)*x*2^(1/2)+(1/b*a)^(1/2)))+3/16*c/a^2*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x+1)+3/16*c/a
^2*(1/b*a)^(1/4)*2^(1/2)*arctan(2^(1/2)/(1/b*a)^(1/4)*x-1)+1/4*d*x^2/a/(b*x^4+a)+1/4*d/a/(a*b)^(1/2)*arctan(x^
2*(b/a)^(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((d*x+c)/(b*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((d*x+c)/(b*x^4+a)^2,x, algorithm="fricas")

[Out]

Timed out

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Sympy [A]  time = 1.25806, size = 155, normalized size = 0.64 \begin{align*} \operatorname{RootSum}{\left (65536 t^{4} a^{7} b^{2} + 2048 t^{2} a^{4} b d^{2} - 1152 t a^{2} b c^{2} d + 16 a d^{4} + 81 b c^{4}, \left ( t \mapsto t \log{\left (x + \frac{- 32768 t^{3} a^{6} b d^{2} - 4608 t^{2} a^{4} b c^{2} d - 512 t a^{3} d^{4} - 1296 t a^{2} b c^{4} + 360 a c^{2} d^{3}}{192 a c d^{4} - 243 b c^{5}} \right )} \right )\right )} + \frac{c x + d x^{2}}{4 a^{2} + 4 a b x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.07802, size = 321, normalized size = 1.33 \begin{align*} \frac{3 \, \sqrt{2} \left (a b^{3}\right )^{\frac{1}{4}} c \log \left (x^{2} + \sqrt{2} x \left (\frac{a}{b}\right )^{\frac{1}{4}} + \sqrt{\frac{a}{b}}\right )}{32 \, a^{2} b} - \frac{3 \, \sqrt{2} \left (a b^{3}\right )^{\frac{1}{4}} c \log \left (x^{2} - \sqrt{2} x \left (\frac{a}{b}\right )^{\frac{1}{4}} + \sqrt{\frac{a}{b}}\right )}{32 \, a^{2} b} + \frac{d x^{2} + c x}{4 \,{\left (b x^{4} + a\right )} a} + \frac{\sqrt{2}{\left (2 \, \sqrt{2} \sqrt{a b} b d + 3 \, \left (a b^{3}\right )^{\frac{1}{4}} b c\right )} \arctan \left (\frac{\sqrt{2}{\left (2 \, x + \sqrt{2} \left (\frac{a}{b}\right )^{\frac{1}{4}}\right )}}{2 \, \left (\frac{a}{b}\right )^{\frac{1}{4}}}\right )}{16 \, a^{2} b^{2}} + \frac{\sqrt{2}{\left (2 \, \sqrt{2} \sqrt{a b} b d + 3 \, \left (a b^{3}\right )^{\frac{1}{4}} b c\right )} \arctan \left (\frac{\sqrt{2}{\left (2 \, x - \sqrt{2} \left (\frac{a}{b}\right )^{\frac{1}{4}}\right )}}{2 \, \left (\frac{a}{b}\right )^{\frac{1}{4}}}\right )}{16 \, a^{2} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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