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

Optimal. Leaf size=221 \[ \frac{\tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (-5 \sqrt{a} \sqrt{b} e-3 a g+21 b c\right )}{64 a^{11/4} b^{5/4}}+\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (5 \sqrt{a} \sqrt{b} e-3 a g+21 b c\right )}{64 a^{11/4} b^{5/4}}+\frac{x \left (-a g+7 b c+6 b d x+5 b e x^2\right )+4 a f}{32 a^2 b \left (a-b x^4\right )}+\frac{3 d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{16 a^{5/2} \sqrt{b}}+\frac{x \left (a g+b c+b d x+b e x^2+b f x^3\right )}{8 a b \left (a-b x^4\right )^2} \]

[Out]

(x*(b*c + a*g + b*d*x + b*e*x^2 + b*f*x^3))/(8*a*b*(a - b*x^4)^2) + (4*a*f + x*(7*b*c - a*g + 6*b*d*x + 5*b*e*
x^2))/(32*a^2*b*(a - b*x^4)) + ((21*b*c - 5*Sqrt[a]*Sqrt[b]*e - 3*a*g)*ArcTan[(b^(1/4)*x)/a^(1/4)])/(64*a^(11/
4)*b^(5/4)) + ((21*b*c + 5*Sqrt[a]*Sqrt[b]*e - 3*a*g)*ArcTanh[(b^(1/4)*x)/a^(1/4)])/(64*a^(11/4)*b^(5/4)) + (3
*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a]])/(16*a^(5/2)*Sqrt[b])

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Rubi [A]  time = 0.263298, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {1858, 1854, 1876, 275, 208, 1167, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (-5 \sqrt{a} \sqrt{b} e-3 a g+21 b c\right )}{64 a^{11/4} b^{5/4}}+\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (5 \sqrt{a} \sqrt{b} e-3 a g+21 b c\right )}{64 a^{11/4} b^{5/4}}+\frac{x \left (-a g+7 b c+6 b d x+5 b e x^2\right )+4 a f}{32 a^2 b \left (a-b x^4\right )}+\frac{3 d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{16 a^{5/2} \sqrt{b}}+\frac{x \left (a g+b c+b d x+b e x^2+b f x^3\right )}{8 a b \left (a-b x^4\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(b*c + a*g + b*d*x + b*e*x^2 + b*f*x^3))/(8*a*b*(a - b*x^4)^2) + (4*a*f + x*(7*b*c - a*g + 6*b*d*x + 5*b*e*
x^2))/(32*a^2*b*(a - b*x^4)) + ((21*b*c - 5*Sqrt[a]*Sqrt[b]*e - 3*a*g)*ArcTan[(b^(1/4)*x)/a^(1/4)])/(64*a^(11/
4)*b^(5/4)) + ((21*b*c + 5*Sqrt[a]*Sqrt[b]*e - 3*a*g)*ArcTanh[(b^(1/4)*x)/a^(1/4)])/(64*a^(11/4)*b^(5/4)) + (3
*d*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a]])/(16*a^(5/2)*Sqrt[b])

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 1854

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], i}, Simp[((a*Coeff[Pq, x, q] -
b*x*ExpandToSum[Pq - Coeff[Pq, x, q]*x^q, x])*(a + b*x^n)^(p + 1))/(a*b*n*(p + 1)), x] + Dist[1/(a*n*(p + 1))
, Int[Sum[(n*(p + 1) + i + 1)*Coeff[Pq, x, i]*x^i, {i, 0, q - 1}]*(a + b*x^n)^(p + 1), x], x] /; q == n - 1] /
; 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}{\left (a-b x^4\right )^3} \, dx &=\frac{x \left (b c+a g+b d x+b e x^2+b f x^3\right )}{8 a b \left (a-b x^4\right )^2}+\frac{\int \frac{7 b c-a g+6 b d x+5 b e x^2+4 b f x^3}{\left (a-b x^4\right )^2} \, dx}{8 a b}\\ &=\frac{x \left (b c+a g+b d x+b e x^2+b f x^3\right )}{8 a b \left (a-b x^4\right )^2}+\frac{4 a f+x \left (7 b c-a g+6 b d x+5 b e x^2\right )}{32 a^2 b \left (a-b x^4\right )}-\frac{\int \frac{-3 (7 b c-a g)-12 b d x-5 b e x^2}{a-b x^4} \, dx}{32 a^2 b}\\ &=\frac{x \left (b c+a g+b d x+b e x^2+b f x^3\right )}{8 a b \left (a-b x^4\right )^2}+\frac{4 a f+x \left (7 b c-a g+6 b d x+5 b e x^2\right )}{32 a^2 b \left (a-b x^4\right )}-\frac{\int \left (-\frac{12 b d x}{a-b x^4}+\frac{-3 (7 b c-a g)-5 b e x^2}{a-b x^4}\right ) \, dx}{32 a^2 b}\\ &=\frac{x \left (b c+a g+b d x+b e x^2+b f x^3\right )}{8 a b \left (a-b x^4\right )^2}+\frac{4 a f+x \left (7 b c-a g+6 b d x+5 b e x^2\right )}{32 a^2 b \left (a-b x^4\right )}-\frac{\int \frac{-3 (7 b c-a g)-5 b e x^2}{a-b x^4} \, dx}{32 a^2 b}+\frac{(3 d) \int \frac{x}{a-b x^4} \, dx}{8 a^2}\\ &=\frac{x \left (b c+a g+b d x+b e x^2+b f x^3\right )}{8 a b \left (a-b x^4\right )^2}+\frac{4 a f+x \left (7 b c-a g+6 b d x+5 b e x^2\right )}{32 a^2 b \left (a-b x^4\right )}+\frac{(3 d) \operatorname{Subst}\left (\int \frac{1}{a-b x^2} \, dx,x,x^2\right )}{16 a^2}-\frac{\left (21 b c-5 \sqrt{a} \sqrt{b} e-3 a g\right ) \int \frac{1}{-\sqrt{a} \sqrt{b}-b x^2} \, dx}{64 a^{5/2} \sqrt{b}}+\frac{\left (21 b c+5 \sqrt{a} \sqrt{b} e-3 a g\right ) \int \frac{1}{\sqrt{a} \sqrt{b}-b x^2} \, dx}{64 a^{5/2} \sqrt{b}}\\ &=\frac{x \left (b c+a g+b d x+b e x^2+b f x^3\right )}{8 a b \left (a-b x^4\right )^2}+\frac{4 a f+x \left (7 b c-a g+6 b d x+5 b e x^2\right )}{32 a^2 b \left (a-b x^4\right )}+\frac{\left (21 b c-5 \sqrt{a} \sqrt{b} e-3 a g\right ) \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{64 a^{11/4} b^{5/4}}+\frac{\left (21 b c+5 \sqrt{a} \sqrt{b} e-3 a g\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{64 a^{11/4} b^{5/4}}+\frac{3 d \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{16 a^{5/2} \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.491858, size = 263, normalized size = 1.19 \[ \frac{\frac{4 a^{3/4} \sqrt [4]{b} \left (a^2 (4 f+3 g x)+a b x \left (11 c+x \left (10 d+9 e x+g x^3\right )\right )-b^2 x^5 (7 c+x (6 d+5 e x))\right )}{\left (a-b x^4\right )^2}-\log \left (\sqrt [4]{a}-\sqrt [4]{b} x\right ) \left (12 \sqrt [4]{a} b^{3/4} d+5 \sqrt{a} \sqrt{b} e-3 a g+21 b c\right )+\log \left (\sqrt [4]{a}+\sqrt [4]{b} x\right ) \left (-12 \sqrt [4]{a} b^{3/4} d+5 \sqrt{a} \sqrt{b} e-3 a g+21 b c\right )+12 \sqrt [4]{a} b^{3/4} d \log \left (\sqrt{a}+\sqrt{b} x^2\right )+2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (-5 \sqrt{a} \sqrt{b} e-3 a g+21 b c\right )}{128 a^{11/4} b^{5/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a^(3/4)*b^(1/4)*(a^2*(4*f + 3*g*x) - b^2*x^5*(7*c + x*(6*d + 5*e*x)) + a*b*x*(11*c + x*(10*d + 9*e*x + g*x
^3))))/(a - b*x^4)^2 + 2*(21*b*c - 5*Sqrt[a]*Sqrt[b]*e - 3*a*g)*ArcTan[(b^(1/4)*x)/a^(1/4)] - (21*b*c + 12*a^(
1/4)*b^(3/4)*d + 5*Sqrt[a]*Sqrt[b]*e - 3*a*g)*Log[a^(1/4) - b^(1/4)*x] + (21*b*c - 12*a^(1/4)*b^(3/4)*d + 5*Sq
rt[a]*Sqrt[b]*e - 3*a*g)*Log[a^(1/4) + b^(1/4)*x] + 12*a^(1/4)*b^(3/4)*d*Log[Sqrt[a] + Sqrt[b]*x^2])/(128*a^(1
1/4)*b^(5/4))

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Maple [A]  time = 0.011, size = 328, normalized size = 1.5 \begin{align*} -{\frac{1}{ \left ( b{x}^{4}-a \right ) ^{2}} \left ({\frac{5\,be{x}^{7}}{32\,{a}^{2}}}+{\frac{3\,bd{x}^{6}}{16\,{a}^{2}}}-{\frac{ \left ( ag-7\,bc \right ){x}^{5}}{32\,{a}^{2}}}-{\frac{9\,e{x}^{3}}{32\,a}}-{\frac{5\,d{x}^{2}}{16\,a}}-{\frac{ \left ( 3\,ag+11\,bc \right ) x}{32\,ab}}-{\frac{f}{8\,b}} \right ) }-{\frac{3\,g}{64\,b{a}^{2}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \right ) }+{\frac{21\,c}{64\,{a}^{3}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \right ) }-{\frac{3\,g}{128\,b{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{21\,c}{128\,{a}^{3}}\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\,d}{32\,{a}^{2}}\ln \left ({ \left ( -a+{x}^{2}\sqrt{ab} \right ) \left ( -a-{x}^{2}\sqrt{ab} \right ) ^{-1}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{5\,e}{64\,b{a}^{2}}\arctan \left ({x{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+{\frac{5\,e}{128\,b{a}^{2}}\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((g*x^4+f*x^3+e*x^2+d*x+c)/(-b*x^4+a)^3,x)

[Out]

-(5/32/a^2*b*e*x^7+3/16/a^2*d*b*x^6-1/32*(a*g-7*b*c)/a^2*x^5-9/32/a*e*x^3-5/16*d/a*x^2-1/32*(3*a*g+11*b*c)/a/b
*x-1/8*f/b)/(b*x^4-a)^2-3/64/b/a^2*(1/b*a)^(1/4)*arctan(x/(1/b*a)^(1/4))*g+21/64*c/a^3*(1/b*a)^(1/4)*arctan(x/
(1/b*a)^(1/4))-3/128/b/a^2*(1/b*a)^(1/4)*ln((x+(1/b*a)^(1/4))/(x-(1/b*a)^(1/4)))*g+21/128*c/a^3*(1/b*a)^(1/4)*
ln((x+(1/b*a)^(1/4))/(x-(1/b*a)^(1/4)))-3/32*d/a^2/(a*b)^(1/2)*ln((-a+x^2*(a*b)^(1/2))/(-a-x^2*(a*b)^(1/2)))-5
/64*e/a^2/b/(1/b*a)^(1/4)*arctan(x/(1/b*a)^(1/4))+5/128*e/a^2/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((g*x^4+f*x^3+e*x^2+d*x+c)/(-b*x^4+a)^3,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((g*x^4+f*x^3+e*x^2+d*x+c)/(-b*x^4+a)^3,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((g*x**4+f*x**3+e*x**2+d*x+c)/(-b*x**4+a)**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.08181, size = 595, normalized size = 2.69 \begin{align*} -\frac{\sqrt{2}{\left (12 \, \sqrt{2} \sqrt{-a b} b^{2} d - 21 \, \left (-a b^{3}\right )^{\frac{1}{4}} b^{2} c + 3 \, \left (-a b^{3}\right )^{\frac{1}{4}} a b g - 5 \, \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 )}{128 \, a^{3} b^{3}} - \frac{\sqrt{2}{\left (12 \, \sqrt{2} \sqrt{-a b} b^{2} d - 21 \, \left (-a b^{3}\right )^{\frac{1}{4}} b^{2} c + 3 \, \left (-a b^{3}\right )^{\frac{1}{4}} a b g - 5 \, \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 )}{128 \, a^{3} b^{3}} + \frac{\sqrt{2}{\left (21 \, \left (-a b^{3}\right )^{\frac{1}{4}} b^{2} c - 3 \, \left (-a b^{3}\right )^{\frac{1}{4}} a b g - 5 \, \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 )}{256 \, a^{3} b^{3}} - \frac{\sqrt{2}{\left (21 \, \left (-a b^{3}\right )^{\frac{1}{4}} b^{2} c - 3 \, \left (-a b^{3}\right )^{\frac{1}{4}} a b g - 5 \, \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 )}{256 \, a^{3} b^{3}} - \frac{5 \, b^{2} x^{7} e + 6 \, b^{2} d x^{6} + 7 \, b^{2} c x^{5} - a b g x^{5} - 9 \, a b x^{3} e - 10 \, a b d x^{2} - 11 \, a b c x - 3 \, a^{2} g x - 4 \, a^{2} f}{32 \,{\left (b x^{4} - a\right )}^{2} a^{2} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128*sqrt(2)*(12*sqrt(2)*sqrt(-a*b)*b^2*d - 21*(-a*b^3)^(1/4)*b^2*c + 3*(-a*b^3)^(1/4)*a*b*g - 5*(-a*b^3)^(3
/4)*e)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(-a/b)^(1/4))/(-a/b)^(1/4))/(a^3*b^3) - 1/128*sqrt(2)*(12*sqrt(2)*sqr
t(-a*b)*b^2*d - 21*(-a*b^3)^(1/4)*b^2*c + 3*(-a*b^3)^(1/4)*a*b*g - 5*(-a*b^3)^(3/4)*e)*arctan(1/2*sqrt(2)*(2*x
- sqrt(2)*(-a/b)^(1/4))/(-a/b)^(1/4))/(a^3*b^3) + 1/256*sqrt(2)*(21*(-a*b^3)^(1/4)*b^2*c - 3*(-a*b^3)^(1/4)*a
*b*g - 5*(-a*b^3)^(3/4)*e)*log(x^2 + sqrt(2)*x*(-a/b)^(1/4) + sqrt(-a/b))/(a^3*b^3) - 1/256*sqrt(2)*(21*(-a*b^
3)^(1/4)*b^2*c - 3*(-a*b^3)^(1/4)*a*b*g - 5*(-a*b^3)^(3/4)*e)*log(x^2 - sqrt(2)*x*(-a/b)^(1/4) + sqrt(-a/b))/(
a^3*b^3) - 1/32*(5*b^2*x^7*e + 6*b^2*d*x^6 + 7*b^2*c*x^5 - a*b*g*x^5 - 9*a*b*x^3*e - 10*a*b*d*x^2 - 11*a*b*c*x
- 3*a^2*g*x - 4*a^2*f)/((b*x^4 - a)^2*a^2*b)