3.1146 \(\int \frac{1}{x^9 (a+b x^4)^{5/4}} \, dx\)

Optimal. Leaf size=125 \[ \frac{45 b^2}{32 a^3 \sqrt [4]{a+b x^4}}+\frac{45 b^2 \tan ^{-1}\left (\frac{\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{13/4}}-\frac{45 b^2 \tanh ^{-1}\left (\frac{\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{13/4}}+\frac{9 b}{32 a^2 x^4 \sqrt [4]{a+b x^4}}-\frac{1}{8 a x^8 \sqrt [4]{a+b x^4}} \]

[Out]

(45*b^2)/(32*a^3*(a + b*x^4)^(1/4)) - 1/(8*a*x^8*(a + b*x^4)^(1/4)) + (9*b)/(32*a^2*x^4*(a + b*x^4)^(1/4)) + (
45*b^2*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)])/(64*a^(13/4)) - (45*b^2*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)])/(64*a^(1
3/4))

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Rubi [A]  time = 0.075702, antiderivative size = 122, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {266, 51, 63, 298, 203, 206} \[ \frac{45 b^2 \tan ^{-1}\left (\frac{\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{13/4}}-\frac{45 b^2 \tanh ^{-1}\left (\frac{\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{13/4}}+\frac{45 b \left (a+b x^4\right )^{3/4}}{32 a^3 x^4}-\frac{9 \left (a+b x^4\right )^{3/4}}{8 a^2 x^8}+\frac{1}{a x^8 \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^9*(a + b*x^4)^(5/4)),x]

[Out]

1/(a*x^8*(a + b*x^4)^(1/4)) - (9*(a + b*x^4)^(3/4))/(8*a^2*x^8) + (45*b*(a + b*x^4)^(3/4))/(32*a^3*x^4) + (45*
b^2*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)])/(64*a^(13/4)) - (45*b^2*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)])/(64*a^(13/4
))

Rule 266

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 298

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

Rule 203

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

Rule 206

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

Rubi steps

\begin{align*} \int \frac{1}{x^9 \left (a+b x^4\right )^{5/4}} \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{x^3 (a+b x)^{5/4}} \, dx,x,x^4\right )\\ &=\frac{1}{a x^8 \sqrt [4]{a+b x^4}}+\frac{9 \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt [4]{a+b x}} \, dx,x,x^4\right )}{4 a}\\ &=\frac{1}{a x^8 \sqrt [4]{a+b x^4}}-\frac{9 \left (a+b x^4\right )^{3/4}}{8 a^2 x^8}-\frac{(45 b) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt [4]{a+b x}} \, dx,x,x^4\right )}{32 a^2}\\ &=\frac{1}{a x^8 \sqrt [4]{a+b x^4}}-\frac{9 \left (a+b x^4\right )^{3/4}}{8 a^2 x^8}+\frac{45 b \left (a+b x^4\right )^{3/4}}{32 a^3 x^4}+\frac{\left (45 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt [4]{a+b x}} \, dx,x,x^4\right )}{128 a^3}\\ &=\frac{1}{a x^8 \sqrt [4]{a+b x^4}}-\frac{9 \left (a+b x^4\right )^{3/4}}{8 a^2 x^8}+\frac{45 b \left (a+b x^4\right )^{3/4}}{32 a^3 x^4}+\frac{(45 b) \operatorname{Subst}\left (\int \frac{x^2}{-\frac{a}{b}+\frac{x^4}{b}} \, dx,x,\sqrt [4]{a+b x^4}\right )}{32 a^3}\\ &=\frac{1}{a x^8 \sqrt [4]{a+b x^4}}-\frac{9 \left (a+b x^4\right )^{3/4}}{8 a^2 x^8}+\frac{45 b \left (a+b x^4\right )^{3/4}}{32 a^3 x^4}-\frac{\left (45 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a}-x^2} \, dx,x,\sqrt [4]{a+b x^4}\right )}{64 a^3}+\frac{\left (45 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a}+x^2} \, dx,x,\sqrt [4]{a+b x^4}\right )}{64 a^3}\\ &=\frac{1}{a x^8 \sqrt [4]{a+b x^4}}-\frac{9 \left (a+b x^4\right )^{3/4}}{8 a^2 x^8}+\frac{45 b \left (a+b x^4\right )^{3/4}}{32 a^3 x^4}+\frac{45 b^2 \tan ^{-1}\left (\frac{\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{13/4}}-\frac{45 b^2 \tanh ^{-1}\left (\frac{\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{13/4}}\\ \end{align*}

Mathematica [C]  time = 0.0080457, size = 36, normalized size = 0.29 \[ \frac{b^2 \, _2F_1\left (-\frac{1}{4},3;\frac{3}{4};\frac{b x^4}{a}+1\right )}{a^3 \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^9*(a + b*x^4)^(5/4)),x]

[Out]

(b^2*Hypergeometric2F1[-1/4, 3, 3/4, 1 + (b*x^4)/a])/(a^3*(a + b*x^4)^(1/4))

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Maple [F]  time = 0.06, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{9}} \left ( b{x}^{4}+a \right ) ^{-{\frac{5}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^9/(b*x^4+a)^(5/4),x)

[Out]

int(1/x^9/(b*x^4+a)^(5/4),x)

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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(1/x^9/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.61167, size = 651, normalized size = 5.21 \begin{align*} -\frac{180 \,{\left (a^{3} b x^{12} + a^{4} x^{8}\right )} \left (\frac{b^{8}}{a^{13}}\right )^{\frac{1}{4}} \arctan \left (-\frac{{\left (b x^{4} + a\right )}^{\frac{1}{4}} a^{3} b^{6} \left (\frac{b^{8}}{a^{13}}\right )^{\frac{1}{4}} - \sqrt{a^{7} b^{8} \sqrt{\frac{b^{8}}{a^{13}}} + \sqrt{b x^{4} + a} b^{12}} a^{3} \left (\frac{b^{8}}{a^{13}}\right )^{\frac{1}{4}}}{b^{8}}\right ) + 45 \,{\left (a^{3} b x^{12} + a^{4} x^{8}\right )} \left (\frac{b^{8}}{a^{13}}\right )^{\frac{1}{4}} \log \left (91125 \, a^{10} \left (\frac{b^{8}}{a^{13}}\right )^{\frac{3}{4}} + 91125 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}} b^{6}\right ) - 45 \,{\left (a^{3} b x^{12} + a^{4} x^{8}\right )} \left (\frac{b^{8}}{a^{13}}\right )^{\frac{1}{4}} \log \left (-91125 \, a^{10} \left (\frac{b^{8}}{a^{13}}\right )^{\frac{3}{4}} + 91125 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}} b^{6}\right ) - 4 \,{\left (45 \, b^{2} x^{8} + 9 \, a b x^{4} - 4 \, a^{2}\right )}{\left (b x^{4} + a\right )}^{\frac{3}{4}}}{128 \,{\left (a^{3} b x^{12} + a^{4} x^{8}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^9/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

-1/128*(180*(a^3*b*x^12 + a^4*x^8)*(b^8/a^13)^(1/4)*arctan(-((b*x^4 + a)^(1/4)*a^3*b^6*(b^8/a^13)^(1/4) - sqrt
(a^7*b^8*sqrt(b^8/a^13) + sqrt(b*x^4 + a)*b^12)*a^3*(b^8/a^13)^(1/4))/b^8) + 45*(a^3*b*x^12 + a^4*x^8)*(b^8/a^
13)^(1/4)*log(91125*a^10*(b^8/a^13)^(3/4) + 91125*(b*x^4 + a)^(1/4)*b^6) - 45*(a^3*b*x^12 + a^4*x^8)*(b^8/a^13
)^(1/4)*log(-91125*a^10*(b^8/a^13)^(3/4) + 91125*(b*x^4 + a)^(1/4)*b^6) - 4*(45*b^2*x^8 + 9*a*b*x^4 - 4*a^2)*(
b*x^4 + a)^(3/4))/(a^3*b*x^12 + a^4*x^8)

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Sympy [C]  time = 4.24266, size = 39, normalized size = 0.31 \begin{align*} - \frac{\Gamma \left (\frac{13}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{5}{4}, \frac{13}{4} \\ \frac{17}{4} \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{4}}} \right )}}{4 b^{\frac{5}{4}} x^{13} \Gamma \left (\frac{17}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**9/(b*x**4+a)**(5/4),x)

[Out]

-gamma(13/4)*hyper((5/4, 13/4), (17/4,), a*exp_polar(I*pi)/(b*x**4))/(4*b**(5/4)*x**13*gamma(17/4))

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Giac [B]  time = 1.12552, size = 324, normalized size = 2.59 \begin{align*} -\frac{1}{256} \, b^{2}{\left (\frac{90 \, \sqrt{2} \left (-a\right )^{\frac{3}{4}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (-a\right )^{\frac{1}{4}} + 2 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}\right )}}{2 \, \left (-a\right )^{\frac{1}{4}}}\right )}{a^{4}} + \frac{90 \, \sqrt{2} \left (-a\right )^{\frac{3}{4}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (-a\right )^{\frac{1}{4}} - 2 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}\right )}}{2 \, \left (-a\right )^{\frac{1}{4}}}\right )}{a^{4}} - \frac{45 \, \sqrt{2} \left (-a\right )^{\frac{3}{4}} \log \left (\sqrt{2}{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (-a\right )^{\frac{1}{4}} + \sqrt{b x^{4} + a} + \sqrt{-a}\right )}{a^{4}} + \frac{45 \, \sqrt{2} \left (-a\right )^{\frac{3}{4}} \log \left (-\sqrt{2}{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (-a\right )^{\frac{1}{4}} + \sqrt{b x^{4} + a} + \sqrt{-a}\right )}{a^{4}} - \frac{256}{{\left (b x^{4} + a\right )}^{\frac{1}{4}} a^{3}} - \frac{8 \,{\left (13 \,{\left (b x^{4} + a\right )}^{\frac{7}{4}} - 17 \,{\left (b x^{4} + a\right )}^{\frac{3}{4}} a\right )}}{a^{3} b^{2} x^{8}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^9/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

-1/256*b^2*(90*sqrt(2)*(-a)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/a^
4 + 90*sqrt(2)*(-a)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/a^4 - 45*
sqrt(2)*(-a)^(3/4)*log(sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/a^4 + 45*sqrt(2)*(-a
)^(3/4)*log(-sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/a^4 - 256/((b*x^4 + a)^(1/4)*a
^3) - 8*(13*(b*x^4 + a)^(7/4) - 17*(b*x^4 + a)^(3/4)*a)/(a^3*b^2*x^8))