3.12.46 \(\int \frac {1}{x^9 (a+b x^4)^{5/4}} \, dx\) [1146]

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

[Out]

45/32*b^2/a^3/(b*x^4+a)^(1/4)-1/8/a/x^8/(b*x^4+a)^(1/4)+9/32*b/a^2/x^4/(b*x^4+a)^(1/4)+45/64*b^2*arctan((b*x^4
+a)^(1/4)/a^(1/4))/a^(13/4)-45/64*b^2*arctanh((b*x^4+a)^(1/4)/a^(1/4))/a^(13/4)

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Rubi [A]
time = 0.06, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {272, 44, 53, 65, 304, 209, 212} \begin {gather*} \frac {45 b^2 \text {ArcTan}\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^2}{32 a^3 \sqrt [4]{a+b x^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}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^9*(a + b*x^4)^(5/4)),x]

[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))

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

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 65

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 209

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

Rule 212

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

Rule 272

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 304

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]

Rubi steps

\begin {align*} \int \frac {1}{x^9 \left (a+b x^4\right )^{5/4}} \, dx &=\frac {1}{4} \text {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 \text {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) \text {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 ) \text {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) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.18, size = 98, normalized size = 0.78 \begin {gather*} \frac {\frac {2 \sqrt [4]{a} \left (-4 a^2+9 a b x^4+45 b^2 x^8\right )}{x^8 \sqrt [4]{a+b x^4}}+45 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )-45 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{13/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{9} \left (b \,x^{4}+a \right )^{\frac {5}{4}}}\, dx\]

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 [A]
time = 0.50, size = 146, normalized size = 1.17 \begin {gather*} \frac {45 \, {\left (b x^{4} + a\right )}^{2} b^{2} - 81 \, {\left (b x^{4} + a\right )} a b^{2} + 32 \, a^{2} b^{2}}{32 \, {\left ({\left (b x^{4} + a\right )}^{\frac {9}{4}} a^{3} - 2 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{4} + {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{5}\right )}} + \frac {45 \, b^{2} {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\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 )}{a^{\frac {1}{4}}}\right )}}{128 \, a^{3}} \end {gather*}

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]

1/32*(45*(b*x^4 + a)^2*b^2 - 81*(b*x^4 + a)*a*b^2 + 32*a^2*b^2)/((b*x^4 + a)^(9/4)*a^3 - 2*(b*x^4 + a)^(5/4)*a
^4 + (b*x^4 + a)^(1/4)*a^5) + 45/128*b^2*(2*arctan((b*x^4 + a)^(1/4)/a^(1/4))/a^(1/4) + log(((b*x^4 + a)^(1/4)
- a^(1/4))/((b*x^4 + a)^(1/4) + a^(1/4)))/a^(1/4))/a^3

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 272 vs. \(2 (97) = 194\).
time = 0.40, size = 272, normalized size = 2.18 \begin {gather*} -\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 {gather*}

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] Result contains complex when optimal does not.
time = 2.63, size = 39, normalized size = 0.31 \begin {gather*} - \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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (97) = 194\).
time = 1.38, size = 255, normalized size = 2.04 \begin {gather*} \frac {45 \, \sqrt {2} b^{2} \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 )}{128 \, \left (-a\right )^{\frac {1}{4}} a^{3}} + \frac {45 \, \sqrt {2} b^{2} \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 )}{128 \, \left (-a\right )^{\frac {1}{4}} a^{3}} + \frac {45 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b^{2} \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 )}{256 \, a^{4}} + \frac {45 \, \sqrt {2} b^{2} \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 )}{256 \, \left (-a\right )^{\frac {1}{4}} a^{3}} + \frac {b^{2}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{3}} + \frac {13 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} b^{2} - 17 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a b^{2}}{32 \, a^{3} b^{2} x^{8}} \end {gather*}

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]

45/128*sqrt(2)*b^2*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/((-a)^(1/4)*a^3)
+ 45/128*sqrt(2)*b^2*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/((-a)^(1/4)*a^
3) + 45/256*sqrt(2)*(-a)^(3/4)*b^2*log(sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/a^4
+ 45/256*sqrt(2)*b^2*log(-sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/((-a)^(1/4)*a^3)
+ b^2/((b*x^4 + a)^(1/4)*a^3) + 1/32*(13*(b*x^4 + a)^(7/4)*b^2 - 17*(b*x^4 + a)^(3/4)*a*b^2)/(a^3*b^2*x^8)

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Mupad [B]
time = 1.51, size = 123, normalized size = 0.98 \begin {gather*} \frac {\frac {b^2}{a}-\frac {81\,b^2\,\left (b\,x^4+a\right )}{32\,a^2}+\frac {45\,b^2\,{\left (b\,x^4+a\right )}^2}{32\,a^3}}{{\left (b\,x^4+a\right )}^{9/4}-2\,a\,{\left (b\,x^4+a\right )}^{5/4}+a^2\,{\left (b\,x^4+a\right )}^{1/4}}+\frac {45\,b^2\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{64\,a^{13/4}}-\frac {45\,b^2\,\mathrm {atanh}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{64\,a^{13/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^9*(a + b*x^4)^(5/4)),x)

[Out]

(b^2/a - (81*b^2*(a + b*x^4))/(32*a^2) + (45*b^2*(a + b*x^4)^2)/(32*a^3))/((a + b*x^4)^(9/4) - 2*a*(a + b*x^4)
^(5/4) + a^2*(a + b*x^4)^(1/4)) + (45*b^2*atan((a + b*x^4)^(1/4)/a^(1/4)))/(64*a^(13/4)) - (45*b^2*atanh((a +
b*x^4)^(1/4)/a^(1/4)))/(64*a^(13/4))

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