3.1054 \(\int \frac {(a+b x^4)^{5/4}}{x^9} \, dx\)

Optimal. Leaf size=98 \[ -\frac {5 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{3/4}}-\frac {5 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{3/4}}-\frac {5 b \sqrt [4]{a+b x^4}}{32 x^4}-\frac {\left (a+b x^4\right )^{5/4}}{8 x^8} \]

[Out]

-5/32*b*(b*x^4+a)^(1/4)/x^4-1/8*(b*x^4+a)^(5/4)/x^8-5/64*b^2*arctan((b*x^4+a)^(1/4)/a^(1/4))/a^(3/4)-5/64*b^2*
arctanh((b*x^4+a)^(1/4)/a^(1/4))/a^(3/4)

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Rubi [A]  time = 0.06, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {266, 47, 63, 212, 206, 203} \[ -\frac {5 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{3/4}}-\frac {5 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{3/4}}-\frac {5 b \sqrt [4]{a+b x^4}}{32 x^4}-\frac {\left (a+b x^4\right )^{5/4}}{8 x^8} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(5/4)/x^9,x]

[Out]

(-5*b*(a + b*x^4)^(1/4))/(32*x^4) - (a + b*x^4)^(5/4)/(8*x^8) - (5*b^2*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)])/(64*
a^(3/4)) - (5*b^2*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)])/(64*a^(3/4))

Rule 47

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

Rule 212

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

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]]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 39, normalized size = 0.40 \[ -\frac {b^2 \left (a+b x^4\right )^{9/4} \, _2F_1\left (\frac {9}{4},3;\frac {13}{4};\frac {b x^4}{a}+1\right )}{9 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(5/4)/x^9,x]

[Out]

-1/9*(b^2*(a + b*x^4)^(9/4)*Hypergeometric2F1[9/4, 3, 13/4, 1 + (b*x^4)/a])/a^3

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fricas [B]  time = 0.81, size = 200, normalized size = 2.04 \[ \frac {20 \, \left (\frac {b^{8}}{a^{3}}\right )^{\frac {1}{4}} x^{8} \arctan \left (-\frac {\left (\frac {b^{8}}{a^{3}}\right )^{\frac {3}{4}} {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2} b^{2} - \left (\frac {b^{8}}{a^{3}}\right )^{\frac {3}{4}} \sqrt {\sqrt {b x^{4} + a} b^{4} + \sqrt {\frac {b^{8}}{a^{3}}} a^{2}} a^{2}}{b^{8}}\right ) - 5 \, \left (\frac {b^{8}}{a^{3}}\right )^{\frac {1}{4}} x^{8} \log \left (5 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{2} + 5 \, \left (\frac {b^{8}}{a^{3}}\right )^{\frac {1}{4}} a\right ) + 5 \, \left (\frac {b^{8}}{a^{3}}\right )^{\frac {1}{4}} x^{8} \log \left (5 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{2} - 5 \, \left (\frac {b^{8}}{a^{3}}\right )^{\frac {1}{4}} a\right ) - 4 \, {\left (9 \, b x^{4} + 4 \, a\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{128 \, x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(20*(b^8/a^3)^(1/4)*x^8*arctan(-((b^8/a^3)^(3/4)*(b*x^4 + a)^(1/4)*a^2*b^2 - (b^8/a^3)^(3/4)*sqrt(sqrt(b
*x^4 + a)*b^4 + sqrt(b^8/a^3)*a^2)*a^2)/b^8) - 5*(b^8/a^3)^(1/4)*x^8*log(5*(b*x^4 + a)^(1/4)*b^2 + 5*(b^8/a^3)
^(1/4)*a) + 5*(b^8/a^3)^(1/4)*x^8*log(5*(b*x^4 + a)^(1/4)*b^2 - 5*(b^8/a^3)^(1/4)*a) - 4*(9*b*x^4 + 4*a)*(b*x^
4 + a)^(1/4))/x^8

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giac [B]  time = 0.19, size = 232, normalized size = 2.37 \[ \frac {\frac {10 \, \sqrt {2} b^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}}} + \frac {10 \, \sqrt {2} b^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}}} + \frac {5 \, \sqrt {2} b^{3} \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {3}{4}}} + \frac {5 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{3} \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{a} - \frac {8 \, {\left (9 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} b^{3} - 5 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a b^{3}\right )}}{b^{2} x^{8}}}{256 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*(10*sqrt(2)*b^3*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/(-a)^(3/4) + 1
0*sqrt(2)*b^3*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/(-a)^(3/4) + 5*sqrt(2
)*b^3*log(sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/(-a)^(3/4) + 5*sqrt(2)*(-a)^(1/4)
*b^3*log(-sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/a - 8*(9*(b*x^4 + a)^(5/4)*b^3 -
5*(b*x^4 + a)^(1/4)*a*b^3)/(b^2*x^8))/b

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{4}+a \right )^{\frac {5}{4}}}{x^{9}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.89, size = 120, normalized size = 1.22 \[ -\frac {5 \, b^{2} \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{64 \, a^{\frac {3}{4}}} + \frac {5 \, b^{2} \log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{128 \, a^{\frac {3}{4}}} - \frac {9 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} b^{2} - 5 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a b^{2}}{32 \, {\left ({\left (b x^{4} + a\right )}^{2} - 2 \, {\left (b x^{4} + a\right )} a + a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/64*b^2*arctan((b*x^4 + a)^(1/4)/a^(1/4))/a^(3/4) + 5/128*b^2*log(((b*x^4 + a)^(1/4) - a^(1/4))/((b*x^4 + a)
^(1/4) + a^(1/4)))/a^(3/4) - 1/32*(9*(b*x^4 + a)^(5/4)*b^2 - 5*(b*x^4 + a)^(1/4)*a*b^2)/((b*x^4 + a)^2 - 2*(b*
x^4 + a)*a + a^2)

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mupad [B]  time = 1.50, size = 77, normalized size = 0.79 \[ \frac {5\,a\,{\left (b\,x^4+a\right )}^{1/4}}{32\,x^8}-\frac {5\,b^2\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{64\,a^{3/4}}-\frac {9\,{\left (b\,x^4+a\right )}^{5/4}}{32\,x^8}+\frac {b^2\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}\,1{}\mathrm {i}}{a^{1/4}}\right )\,5{}\mathrm {i}}{64\,a^{3/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^(5/4)/x^9,x)

[Out]

(b^2*atan(((a + b*x^4)^(1/4)*1i)/a^(1/4))*5i)/(64*a^(3/4)) - (5*b^2*atan((a + b*x^4)^(1/4)/a^(1/4)))/(64*a^(3/
4)) - (9*(a + b*x^4)^(5/4))/(32*x^8) + (5*a*(a + b*x^4)^(1/4))/(32*x^8)

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sympy [C]  time = 4.19, size = 41, normalized size = 0.42 \[ - \frac {b^{\frac {5}{4}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 x^{3} \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b**(5/4)*gamma(3/4)*hyper((-5/4, 3/4), (7/4,), a*exp_polar(I*pi)/(b*x**4))/(4*x**3*gamma(7/4))

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