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

Optimal. Leaf size=184 \[ \frac{\tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (-\sqrt{a} \sqrt{b} e-a g+3 b c\right )}{8 a^{7/4} b^{5/4}}+\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (\sqrt{a} \sqrt{b} e-a g+3 b c\right )}{8 a^{7/4} b^{5/4}}+\frac{(b d-a h) \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{4 a^{3/2} b^{3/2}}+\frac{x \left (x (a h+b d)+a g+b c+b e x^2+b f x^3\right )}{4 a b \left (a-b x^4\right )} \]

[Out]

(x*(b*c + a*g + (b*d + a*h)*x + b*e*x^2 + b*f*x^3))/(4*a*b*(a - b*x^4)) + ((3*b*c - Sqrt[a]*Sqrt[b]*e - a*g)*A
rcTan[(b^(1/4)*x)/a^(1/4)])/(8*a^(7/4)*b^(5/4)) + ((3*b*c + Sqrt[a]*Sqrt[b]*e - a*g)*ArcTanh[(b^(1/4)*x)/a^(1/
4)])/(8*a^(7/4)*b^(5/4)) + ((b*d - a*h)*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a]])/(4*a^(3/2)*b^(3/2))

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Rubi [A]  time = 0.203857, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1858, 1876, 275, 208, 1167, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (-\sqrt{a} \sqrt{b} e-a g+3 b c\right )}{8 a^{7/4} b^{5/4}}+\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (\sqrt{a} \sqrt{b} e-a g+3 b c\right )}{8 a^{7/4} b^{5/4}}+\frac{(b d-a h) \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{4 a^{3/2} b^{3/2}}+\frac{x \left (x (a h+b d)+a g+b c+b e x^2+b f x^3\right )}{4 a b \left (a-b x^4\right )} \]

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)^2,x]

[Out]

(x*(b*c + a*g + (b*d + a*h)*x + b*e*x^2 + b*f*x^3))/(4*a*b*(a - b*x^4)) + ((3*b*c - Sqrt[a]*Sqrt[b]*e - a*g)*A
rcTan[(b^(1/4)*x)/a^(1/4)])/(8*a^(7/4)*b^(5/4)) + ((3*b*c + Sqrt[a]*Sqrt[b]*e - a*g)*ArcTanh[(b^(1/4)*x)/a^(1/
4)])/(8*a^(7/4)*b^(5/4)) + ((b*d - a*h)*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a]])/(4*a^(3/2)*b^(3/2))

Rule 1858

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = PolynomialQuotient
[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x
]}, Dist[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), Int[(a + b*x^n)^(p + 1)*ExpandToSum[a*n*(p + 1)*Q + n*(p +
1)*R + D[x*R, x], x], x], x] - Simp[(x*R*(a + b*x^n)^(p + 1))/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), x]] /; G
eQ[q, n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -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 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 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 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]

Rubi steps

\begin{align*} \int \frac{c+d x+e x^2+f x^3+g x^4+h x^5}{\left (a-b x^4\right )^2} \, dx &=\frac{x \left (b c+a g+(b d+a h) x+b e x^2+b f x^3\right )}{4 a b \left (a-b x^4\right )}-\frac{\int \frac{-b (3 b c-a g)-2 b (b d-a h) x-b^2 e x^2}{a-b x^4} \, dx}{4 a b^2}\\ &=\frac{x \left (b c+a g+(b d+a h) x+b e x^2+b f x^3\right )}{4 a b \left (a-b x^4\right )}-\frac{\int \left (-\frac{2 b (b d-a h) x}{a-b x^4}+\frac{-b (3 b c-a g)-b^2 e x^2}{a-b x^4}\right ) \, dx}{4 a b^2}\\ &=\frac{x \left (b c+a g+(b d+a h) x+b e x^2+b f x^3\right )}{4 a b \left (a-b x^4\right )}-\frac{\int \frac{-b (3 b c-a g)-b^2 e x^2}{a-b x^4} \, dx}{4 a b^2}+\frac{(b d-a h) \int \frac{x}{a-b x^4} \, dx}{2 a b}\\ &=\frac{x \left (b c+a g+(b d+a h) x+b e x^2+b f x^3\right )}{4 a b \left (a-b x^4\right )}-\frac{\left (3 b c-\sqrt{a} \sqrt{b} e-a g\right ) \int \frac{1}{-\sqrt{a} \sqrt{b}-b x^2} \, dx}{8 a^{3/2} \sqrt{b}}+\frac{\left (3 b c+\sqrt{a} \sqrt{b} e-a g\right ) \int \frac{1}{\sqrt{a} \sqrt{b}-b x^2} \, dx}{8 a^{3/2} \sqrt{b}}+\frac{(b d-a h) \operatorname{Subst}\left (\int \frac{1}{a-b x^2} \, dx,x,x^2\right )}{4 a b}\\ &=\frac{x \left (b c+a g+(b d+a h) x+b e x^2+b f x^3\right )}{4 a b \left (a-b x^4\right )}+\frac{\left (3 b c-\sqrt{a} \sqrt{b} e-a g\right ) \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 a^{7/4} b^{5/4}}+\frac{\left (3 b c+\sqrt{a} \sqrt{b} e-a g\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 a^{7/4} b^{5/4}}+\frac{(b d-a h) \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{4 a^{3/2} b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.163229, size = 257, normalized size = 1.4 \[ \frac{\log \left (\sqrt [4]{a}-\sqrt [4]{b} x\right ) \left (2 a^{5/4} h-\sqrt{a} b^{3/4} e-2 \sqrt [4]{a} b d+a \sqrt [4]{b} g-3 b^{5/4} c\right )+\log \left (\sqrt [4]{a}+\sqrt [4]{b} x\right ) \left (2 a^{5/4} h+\sqrt{a} b^{3/4} e-2 \sqrt [4]{a} b d-a \sqrt [4]{b} g+3 b^{5/4} c\right )+\frac{4 a^{3/4} \sqrt{b} (a (f+x (g+h x))+b x (c+x (d+e x)))}{a-b x^4}-2 \sqrt [4]{b} \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (\sqrt{a} \sqrt{b} e+a g-3 b c\right )-2 \sqrt [4]{a} (a h-b d) \log \left (\sqrt{a}+\sqrt{b} x^2\right )}{16 a^{7/4} b^{3/2}} \]

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)^2,x]

[Out]

((4*a^(3/4)*Sqrt[b]*(b*x*(c + x*(d + e*x)) + a*(f + x*(g + h*x))))/(a - b*x^4) - 2*b^(1/4)*(-3*b*c + Sqrt[a]*S
qrt[b]*e + a*g)*ArcTan[(b^(1/4)*x)/a^(1/4)] + (-3*b^(5/4)*c - 2*a^(1/4)*b*d - Sqrt[a]*b^(3/4)*e + a*b^(1/4)*g
+ 2*a^(5/4)*h)*Log[a^(1/4) - b^(1/4)*x] + (3*b^(5/4)*c - 2*a^(1/4)*b*d + Sqrt[a]*b^(3/4)*e - a*b^(1/4)*g + 2*a
^(5/4)*h)*Log[a^(1/4) + b^(1/4)*x] - 2*a^(1/4)*(-(b*d) + a*h)*Log[Sqrt[a] + Sqrt[b]*x^2])/(16*a^(7/4)*b^(3/2))

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Maple [B]  time = 0.008, size = 340, normalized size = 1.9 \begin{align*}{\frac{1}{b{x}^{4}-a} \left ( -{\frac{e{x}^{3}}{4\,a}}-{\frac{ \left ( ah+bd \right ){x}^{2}}{4\,ab}}-{\frac{ \left ( ag+bc \right ) x}{4\,ab}}-{\frac{f}{4\,b}} \right ) }-{\frac{g}{8\,ab}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \right ) }+{\frac{3\,c}{8\,{a}^{2}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \right ) }-{\frac{g}{16\,ab}\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{3\,c}{16\,{a}^{2}}\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{h}{8\,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}{8\,a}\ln \left ({ \left ( -a+{x}^{2}\sqrt{ab} \right ) \left ( -a-{x}^{2}\sqrt{ab} \right ) ^{-1}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{e}{8\,ab}\arctan \left ({x{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+{\frac{e}{16\,ab}\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}}}}}} \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)^2,x)

[Out]

(-1/4/a*e*x^3-1/4*(a*h+b*d)/a/b*x^2-1/4*(a*g+b*c)/a/b*x-1/4*f/b)/(b*x^4-a)-1/8/b/a*(1/b*a)^(1/4)*arctan(x/(1/b
*a)^(1/4))*g+3/8*c/a^2*(1/b*a)^(1/4)*arctan(x/(1/b*a)^(1/4))-1/16/b/a*(1/b*a)^(1/4)*ln((x+(1/b*a)^(1/4))/(x-(1
/b*a)^(1/4)))*g+3/16*c/a^2*(1/b*a)^(1/4)*ln((x+(1/b*a)^(1/4))/(x-(1/b*a)^(1/4)))+1/8/b/(a*b)^(1/2)*ln((-a+x^2*
(a*b)^(1/2))/(-a-x^2*(a*b)^(1/2)))*h-1/8*d/a/(a*b)^(1/2)*ln((-a+x^2*(a*b)^(1/2))/(-a-x^2*(a*b)^(1/2)))-1/8*e/a
/b/(1/b*a)^(1/4)*arctan(x/(1/b*a)^(1/4))+1/16*e/a/b/(1/b*a)^(1/4)*ln((x+(1/b*a)^(1/4))/(x-(1/b*a)^(1/4)))

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

[Out]

Timed out

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Giac [B]  time = 1.07615, size = 574, normalized size = 3.12 \begin{align*} -\frac{b x^{3} e + b d x^{2} + a h x^{2} + b c x + a g x + a f}{4 \,{\left (b x^{4} - a\right )} a b} - \frac{\sqrt{2}{\left (2 \, \sqrt{2} \sqrt{-a b} b^{2} d - 2 \, \sqrt{2} \sqrt{-a b} a b h - 3 \, \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 )}{16 \, a^{2} b^{3}} - \frac{\sqrt{2}{\left (2 \, \sqrt{2} \sqrt{-a b} b^{2} d - 2 \, \sqrt{2} \sqrt{-a b} a b h - 3 \, \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 )}{16 \, a^{2} b^{3}} + \frac{\sqrt{2}{\left (3 \, \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 )}{32 \, a^{2} b^{3}} - \frac{\sqrt{2}{\left (3 \, \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 )}{32 \, a^{2} 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)^2,x, algorithm="giac")

[Out]

-1/4*(b*x^3*e + b*d*x^2 + a*h*x^2 + b*c*x + a*g*x + a*f)/((b*x^4 - a)*a*b) - 1/16*sqrt(2)*(2*sqrt(2)*sqrt(-a*b
)*b^2*d - 2*sqrt(2)*sqrt(-a*b)*a*b*h - 3*(-a*b^3)^(1/4)*b^2*c + (-a*b^3)^(1/4)*a*b*g - (-a*b^3)^(3/4)*e)*arcta
n(1/2*sqrt(2)*(2*x + sqrt(2)*(-a/b)^(1/4))/(-a/b)^(1/4))/(a^2*b^3) - 1/16*sqrt(2)*(2*sqrt(2)*sqrt(-a*b)*b^2*d
- 2*sqrt(2)*sqrt(-a*b)*a*b*h - 3*(-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*sq
rt(2)*(2*x - sqrt(2)*(-a/b)^(1/4))/(-a/b)^(1/4))/(a^2*b^3) + 1/32*sqrt(2)*(3*(-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^2*b^3) - 1/32*sqrt(2)*(3*(-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^
2*b^3)