3.186 \(\int \frac{c+d x+e x^2+f x^3+g x^4+h x^5}{a-b x^4} \, dx\)

Optimal. Leaf size=165 \[ \frac{\tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (-\sqrt{a} \sqrt{b} e+a g+b c\right )}{2 a^{3/4} b^{5/4}}+\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (\sqrt{a} \sqrt{b} e+a g+b c\right )}{2 a^{3/4} b^{5/4}}+\frac{(a h+b d) \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{2 \sqrt{a} b^{3/2}}-\frac{f \log \left (a-b x^4\right )}{4 b}-\frac{g x}{b}-\frac{h x^2}{2 b} \]

[Out]

-((g*x)/b) - (h*x^2)/(2*b) + ((b*c - Sqrt[a]*Sqrt[b]*e + a*g)*ArcTan[(b^(1/4)*x)/a^(1/4)])/(2*a^(3/4)*b^(5/4))
+ ((b*c + Sqrt[a]*Sqrt[b]*e + a*g)*ArcTanh[(b^(1/4)*x)/a^(1/4)])/(2*a^(3/4)*b^(5/4)) + ((b*d + a*h)*ArcTanh[(
Sqrt[b]*x^2)/Sqrt[a]])/(2*Sqrt[a]*b^(3/2)) - (f*Log[a - b*x^4])/(4*b)

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Rubi [A]  time = 0.260628, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1885, 1887, 1167, 205, 208, 1819, 1810, 635, 260} \[ \frac{\tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (-\sqrt{a} \sqrt{b} e+a g+b c\right )}{2 a^{3/4} b^{5/4}}+\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (\sqrt{a} \sqrt{b} e+a g+b c\right )}{2 a^{3/4} b^{5/4}}+\frac{(a h+b d) \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{2 \sqrt{a} b^{3/2}}-\frac{f \log \left (a-b x^4\right )}{4 b}-\frac{g x}{b}-\frac{h x^2}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5)/(a - b*x^4),x]

[Out]

-((g*x)/b) - (h*x^2)/(2*b) + ((b*c - Sqrt[a]*Sqrt[b]*e + a*g)*ArcTan[(b^(1/4)*x)/a^(1/4)])/(2*a^(3/4)*b^(5/4))
+ ((b*c + Sqrt[a]*Sqrt[b]*e + a*g)*ArcTanh[(b^(1/4)*x)/a^(1/4)])/(2*a^(3/4)*b^(5/4)) + ((b*d + a*h)*ArcTanh[(
Sqrt[b]*x^2)/Sqrt[a]])/(2*Sqrt[a]*b^(3/2)) - (f*Log[a - b*x^4])/(4*b)

Rule 1885

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[Sum[x^j*Sum[Coeff[P
q, x, j + (k*n)/2]*x^((k*n)/2), {k, 0, (2*(q - j))/n + 1}]*(a + b*x^n)^p, {j, 0, n/2 - 1}], x]] /; FreeQ[{a, b
, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] &&  !PolyQ[Pq, x^(n/2)]

Rule 1887

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^n), x], x] /; FreeQ[{a, b}, x
] && PolyQ[Pq, x] && IntegerQ[n]

Rule 1167

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

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 208

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

Rule 1819

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1)
, Pq, x]*(a + b*x^Simplify[n/(m + 1)])^p, x], x, x^(m + 1)], x] /; FreeQ[{a, b, m, n, p}, x] && NeQ[m, -1] &&
IGtQ[Simplify[n/(m + 1)], 0] && PolyQ[Pq, x^(m + 1)]

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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{c+d x+e x^2+f x^3+g x^4+h x^5}{a-b x^4} \, dx &=\int \left (\frac{c+e x^2+g x^4}{a-b x^4}+\frac{x \left (d+f x^2+h x^4\right )}{a-b x^4}\right ) \, dx\\ &=\int \frac{c+e x^2+g x^4}{a-b x^4} \, dx+\int \frac{x \left (d+f x^2+h x^4\right )}{a-b x^4} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{d+f x+h x^2}{a-b x^2} \, dx,x,x^2\right )+\int \left (-\frac{g}{b}+\frac{b c+a g+b e x^2}{b \left (a-b x^4\right )}\right ) \, dx\\ &=-\frac{g x}{b}+\frac{1}{2} \operatorname{Subst}\left (\int \left (-\frac{h}{b}+\frac{b d+a h+b f x}{b \left (a-b x^2\right )}\right ) \, dx,x,x^2\right )+\frac{\int \frac{b c+a g+b e x^2}{a-b x^4} \, dx}{b}\\ &=-\frac{g x}{b}-\frac{h x^2}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{b d+a h+b f x}{a-b x^2} \, dx,x,x^2\right )}{2 b}+\frac{1}{2} \left (e-\frac{b c+a g}{\sqrt{a} \sqrt{b}}\right ) \int \frac{1}{-\sqrt{a} \sqrt{b}-b x^2} \, dx+\frac{1}{2} \left (e+\frac{b c+a g}{\sqrt{a} \sqrt{b}}\right ) \int \frac{1}{\sqrt{a} \sqrt{b}-b x^2} \, dx\\ &=-\frac{g x}{b}-\frac{h x^2}{2 b}+\frac{\left (b c-\sqrt{a} \sqrt{b} e+a g\right ) \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/4}}+\frac{\left (b c+\sqrt{a} \sqrt{b} e+a g\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/4}}+\frac{1}{2} f \operatorname{Subst}\left (\int \frac{x}{a-b x^2} \, dx,x,x^2\right )+\frac{(b d+a h) \operatorname{Subst}\left (\int \frac{1}{a-b x^2} \, dx,x,x^2\right )}{2 b}\\ &=-\frac{g x}{b}-\frac{h x^2}{2 b}+\frac{\left (b c-\sqrt{a} \sqrt{b} e+a g\right ) \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/4}}+\frac{\left (b c+\sqrt{a} \sqrt{b} e+a g\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/4}}+\frac{(b d+a h) \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{2 \sqrt{a} b^{3/2}}-\frac{f \log \left (a-b x^4\right )}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.257312, size = 256, normalized size = 1.55 \[ \frac{-\log \left (\sqrt [4]{a}-\sqrt [4]{b} x\right ) \left (a^{5/4} h+\sqrt{a} b^{3/4} e+\sqrt [4]{a} b d+a \sqrt [4]{b} g+b^{5/4} c\right )+\log \left (\sqrt [4]{a}+\sqrt [4]{b} x\right ) \left (a^{5/4} (-h)+\sqrt{a} b^{3/4} e-\sqrt [4]{a} b d+a \sqrt [4]{b} g+b^{5/4} c\right )-a^{3/4} \sqrt{b} f \log \left (a-b x^4\right )-4 a^{3/4} \sqrt{b} g x-2 a^{3/4} \sqrt{b} h x^2+2 \sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (-\sqrt{a} \sqrt{b} e+a g+b c\right )+\sqrt [4]{a} (a h+b d) \log \left (\sqrt{a}+\sqrt{b} x^2\right )}{4 a^{3/4} b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x + e*x^2 + f*x^3 + g*x^4 + h*x^5)/(a - b*x^4),x]

[Out]

(-4*a^(3/4)*Sqrt[b]*g*x - 2*a^(3/4)*Sqrt[b]*h*x^2 + 2*b^(1/4)*(b*c - Sqrt[a]*Sqrt[b]*e + a*g)*ArcTan[(b^(1/4)*
x)/a^(1/4)] - (b^(5/4)*c + a^(1/4)*b*d + Sqrt[a]*b^(3/4)*e + a*b^(1/4)*g + a^(5/4)*h)*Log[a^(1/4) - b^(1/4)*x]
+ (b^(5/4)*c - a^(1/4)*b*d + Sqrt[a]*b^(3/4)*e + a*b^(1/4)*g - a^(5/4)*h)*Log[a^(1/4) + b^(1/4)*x] + a^(1/4)*
(b*d + a*h)*Log[Sqrt[a] + Sqrt[b]*x^2] - a^(3/4)*Sqrt[b]*f*Log[a - b*x^4])/(4*a^(3/4)*b^(3/2))

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Maple [B]  time = 0.045, size = 296, normalized size = 1.8 \begin{align*} -{\frac{h{x}^{2}}{2\,b}}-{\frac{gx}{b}}+{\frac{g}{2\,b}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \right ) }+{\frac{c}{2\,a}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \right ) }+{\frac{g}{4\,b}\sqrt [4]{{\frac{a}{b}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{a}{b}}} \right ) \left ( x-\sqrt [4]{{\frac{a}{b}}} \right ) ^{-1}} \right ) }+{\frac{c}{4\,a}\sqrt [4]{{\frac{a}{b}}}\ln \left ({ \left ( x+\sqrt [4]{{\frac{a}{b}}} \right ) \left ( x-\sqrt [4]{{\frac{a}{b}}} \right ) ^{-1}} \right ) }-{\frac{ah}{4\,b}\ln \left ({ \left ( -a+{x}^{2}\sqrt{ab} \right ) \left ( -a-{x}^{2}\sqrt{ab} \right ) ^{-1}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{d}{4}\ln \left ({ \left ( -a+{x}^{2}\sqrt{ab} \right ) \left ( -a-{x}^{2}\sqrt{ab} \right ) ^{-1}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{e}{2\,b}\arctan \left ({x{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+{\frac{e}{4\,b}\ln \left ({ \left ( x+\sqrt [4]{{\frac{a}{b}}} \right ) \left ( x-\sqrt [4]{{\frac{a}{b}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-{\frac{f\ln \left ( b{x}^{4}-a \right ) }{4\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/(-b*x^4+a),x)

[Out]

-1/2*h*x^2/b-g*x/b+1/2/b*(1/b*a)^(1/4)*arctan(x/(1/b*a)^(1/4))*g+1/2*c*(1/b*a)^(1/4)/a*arctan(x/(1/b*a)^(1/4))
+1/4/b*(1/b*a)^(1/4)*ln((x+(1/b*a)^(1/4))/(x-(1/b*a)^(1/4)))*g+1/4*c*(1/b*a)^(1/4)/a*ln((x+(1/b*a)^(1/4))/(x-(
1/b*a)^(1/4)))-1/4/b/(a*b)^(1/2)*ln((-a+x^2*(a*b)^(1/2))/(-a-x^2*(a*b)^(1/2)))*a*h-1/4*d/(a*b)^(1/2)*ln((-a+x^
2*(a*b)^(1/2))/(-a-x^2*(a*b)^(1/2)))-1/2*e/b/(1/b*a)^(1/4)*arctan(x/(1/b*a)^(1/4))+1/4*e/b/(1/b*a)^(1/4)*ln((x
+(1/b*a)^(1/4))/(x-(1/b*a)^(1/4)))-1/4/b*f*ln(b*x^4-a)

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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((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/(-b*x^4+a),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((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/(-b*x^4+a),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((h*x**5+g*x**4+f*x**3+e*x**2+d*x+c)/(-b*x**4+a),x)

[Out]

Timed out

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Giac [B]  time = 1.08378, size = 539, normalized size = 3.27 \begin{align*} -\frac{f \log \left ({\left | b x^{4} - a \right |}\right )}{4 \, b} - \frac{b h x^{2} + 2 \, b g x}{2 \, b^{2}} + \frac{\sqrt{2}{\left (\sqrt{2} \sqrt{-a b} b^{2} d - \sqrt{2} \sqrt{-a b} a b h + \left (-a b^{3}\right )^{\frac{1}{4}} b^{2} c + \left (-a b^{3}\right )^{\frac{1}{4}} a b g + \left (-a b^{3}\right )^{\frac{3}{4}} e\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 )}{4 \, a b^{3}} + \frac{\sqrt{2}{\left (\sqrt{2} \sqrt{-a b} b^{2} d - \sqrt{2} \sqrt{-a b} a b h + \left (-a b^{3}\right )^{\frac{1}{4}} b^{2} c + \left (-a b^{3}\right )^{\frac{1}{4}} a b g + \left (-a b^{3}\right )^{\frac{3}{4}} e\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 )}{4 \, a b^{3}} + \frac{\sqrt{2}{\left (\left (-a b^{3}\right )^{\frac{1}{4}} b^{2} c + \left (-a b^{3}\right )^{\frac{1}{4}} a b g - \left (-a b^{3}\right )^{\frac{3}{4}} e\right )} \log \left (x^{2} + \sqrt{2} x \left (-\frac{a}{b}\right )^{\frac{1}{4}} + \sqrt{-\frac{a}{b}}\right )}{8 \, a b^{3}} - \frac{\sqrt{2}{\left (\left (-a b^{3}\right )^{\frac{1}{4}} b^{2} c + \left (-a b^{3}\right )^{\frac{1}{4}} a b g - \left (-a b^{3}\right )^{\frac{3}{4}} e\right )} \log \left (x^{2} - \sqrt{2} x \left (-\frac{a}{b}\right )^{\frac{1}{4}} + \sqrt{-\frac{a}{b}}\right )}{8 \, a b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x^5+g*x^4+f*x^3+e*x^2+d*x+c)/(-b*x^4+a),x, algorithm="giac")

[Out]

-1/4*f*log(abs(b*x^4 - a))/b - 1/2*(b*h*x^2 + 2*b*g*x)/b^2 + 1/4*sqrt(2)*(sqrt(2)*sqrt(-a*b)*b^2*d - sqrt(2)*s
qrt(-a*b)*a*b*h + (-a*b^3)^(1/4)*b^2*c + (-a*b^3)^(1/4)*a*b*g + (-a*b^3)^(3/4)*e)*arctan(1/2*sqrt(2)*(2*x + sq
rt(2)*(-a/b)^(1/4))/(-a/b)^(1/4))/(a*b^3) + 1/4*sqrt(2)*(sqrt(2)*sqrt(-a*b)*b^2*d - sqrt(2)*sqrt(-a*b)*a*b*h +
(-a*b^3)^(1/4)*b^2*c + (-a*b^3)^(1/4)*a*b*g + (-a*b^3)^(3/4)*e)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(-a/b)^(1/4
))/(-a/b)^(1/4))/(a*b^3) + 1/8*sqrt(2)*((-a*b^3)^(1/4)*b^2*c + (-a*b^3)^(1/4)*a*b*g - (-a*b^3)^(3/4)*e)*log(x^
2 + sqrt(2)*x*(-a/b)^(1/4) + sqrt(-a/b))/(a*b^3) - 1/8*sqrt(2)*((-a*b^3)^(1/4)*b^2*c + (-a*b^3)^(1/4)*a*b*g -
(-a*b^3)^(3/4)*e)*log(x^2 - sqrt(2)*x*(-a/b)^(1/4) + sqrt(-a/b))/(a*b^3)