3.1024 \(\int \frac {(a+b x^4)^{3/4}}{x^9} \, dx\)

Optimal. Leaf size=101 \[ -\frac {3 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{5/4}}+\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{5/4}}-\frac {3 b \left (a+b x^4\right )^{3/4}}{32 a x^4}-\frac {\left (a+b x^4\right )^{3/4}}{8 x^8} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {266, 47, 51, 63, 298, 203, 206} \[ -\frac {3 b^2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{5/4}}+\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a+b x^4}}{\sqrt [4]{a}}\right )}{64 a^{5/4}}-\frac {3 b \left (a+b x^4\right )^{3/4}}{32 a x^4}-\frac {\left (a+b x^4\right )^{3/4}}{8 x^8} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(3/4)/x^9,x]

[Out]

-(a + b*x^4)^(3/4)/(8*x^8) - (3*b*(a + b*x^4)^(3/4))/(32*a*x^4) - (3*b^2*ArcTan[(a + b*x^4)^(1/4)/a^(1/4)])/(6
4*a^(5/4)) + (3*b^2*ArcTanh[(a + b*x^4)^(1/4)/a^(1/4)])/(64*a^(5/4))

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 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 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 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 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]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 39, normalized size = 0.39 \[ -\frac {b^2 \left (a+b x^4\right )^{7/4} \, _2F_1\left (\frac {7}{4},3;\frac {11}{4};\frac {b x^4}{a}+1\right )}{7 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(3/4)/x^9,x]

[Out]

-1/7*(b^2*(a + b*x^4)^(7/4)*Hypergeometric2F1[7/4, 3, 11/4, 1 + (b*x^4)/a])/a^3

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fricas [B]  time = 0.94, size = 209, normalized size = 2.07 \[ \frac {12 \, \left (\frac {b^{8}}{a^{5}}\right )^{\frac {1}{4}} a x^{8} \arctan \left (-\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (\frac {b^{8}}{a^{5}}\right )^{\frac {1}{4}} a b^{6} - \sqrt {\sqrt {b x^{4} + a} b^{12} + \sqrt {\frac {b^{8}}{a^{5}}} a^{3} b^{8}} \left (\frac {b^{8}}{a^{5}}\right )^{\frac {1}{4}} a}{b^{8}}\right ) + 3 \, \left (\frac {b^{8}}{a^{5}}\right )^{\frac {1}{4}} a x^{8} \log \left (27 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{6} + 27 \, \left (\frac {b^{8}}{a^{5}}\right )^{\frac {3}{4}} a^{4}\right ) - 3 \, \left (\frac {b^{8}}{a^{5}}\right )^{\frac {1}{4}} a x^{8} \log \left (27 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{6} - 27 \, \left (\frac {b^{8}}{a^{5}}\right )^{\frac {3}{4}} a^{4}\right ) - 4 \, {\left (3 \, b x^{4} + 4 \, a\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{128 \, a x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(3/4)/x^9,x, algorithm="fricas")

[Out]

1/128*(12*(b^8/a^5)^(1/4)*a*x^8*arctan(-((b*x^4 + a)^(1/4)*(b^8/a^5)^(1/4)*a*b^6 - sqrt(sqrt(b*x^4 + a)*b^12 +
sqrt(b^8/a^5)*a^3*b^8)*(b^8/a^5)^(1/4)*a)/b^8) + 3*(b^8/a^5)^(1/4)*a*x^8*log(27*(b*x^4 + a)^(1/4)*b^6 + 27*(b
^8/a^5)^(3/4)*a^4) - 3*(b^8/a^5)^(1/4)*a*x^8*log(27*(b*x^4 + a)^(1/4)*b^6 - 27*(b^8/a^5)^(3/4)*a^4) - 4*(3*b*x
^4 + 4*a)*(b*x^4 + a)^(3/4))/(a*x^8)

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giac [B]  time = 0.18, size = 243, normalized size = 2.41 \[ \frac {\frac {6 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{2}} + \frac {6 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a^{2}} + \frac {3 \, \sqrt {2} b^{3} \log \left (\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{\left (-a\right )^{\frac {1}{4}} a} + \frac {3 \, \sqrt {2} \left (-a\right )^{\frac {3}{4}} b^{3} \log \left (-\sqrt {2} {\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (-a\right )^{\frac {1}{4}} + \sqrt {b x^{4} + a} + \sqrt {-a}\right )}{a^{2}} - \frac {8 \, {\left (3 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} b^{3} + {\left (b x^{4} + a\right )}^{\frac {3}{4}} a b^{3}\right )}}{a b^{2} x^{8}}}{256 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(3/4)/x^9,x, algorithm="giac")

[Out]

1/256*(6*sqrt(2)*(-a)^(3/4)*b^3*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/a^2
+ 6*sqrt(2)*(-a)^(3/4)*b^3*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) - 2*(b*x^4 + a)^(1/4))/(-a)^(1/4))/a^2 + 3*
sqrt(2)*b^3*log(sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/((-a)^(1/4)*a) + 3*sqrt(2)*
(-a)^(3/4)*b^3*log(-sqrt(2)*(b*x^4 + a)^(1/4)*(-a)^(1/4) + sqrt(b*x^4 + a) + sqrt(-a))/a^2 - 8*(3*(b*x^4 + a)^
(7/4)*b^3 + (b*x^4 + a)^(3/4)*a*b^3)/(a*b^2*x^8))/b

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}}}{x^{9}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^4+a)^(3/4)/x^9,x)

[Out]

int((b*x^4+a)^(3/4)/x^9,x)

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maxima [A]  time = 2.99, size = 125, normalized size = 1.24 \[ -\frac {3 \, b^{2} {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} - a^{\frac {1}{4}}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} + a^{\frac {1}{4}}}\right )}{a^{\frac {1}{4}}}\right )}}{128 \, a} - \frac {3 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} b^{2} + {\left (b x^{4} + a\right )}^{\frac {3}{4}} a b^{2}}{32 \, {\left ({\left (b x^{4} + a\right )}^{2} a - 2 \, {\left (b x^{4} + a\right )} a^{2} + a^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(3/4)/x^9,x, algorithm="maxima")

[Out]

-3/128*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 - 1/32*(3*(b*x^4 + a)^(7/4)*b^2 + (b*x^4 + a)^(3/4)*a*b^2)/((b*x^4 + a)^2*a - 2*(b*x^4
+ a)*a^2 + a^3)

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mupad [B]  time = 1.42, size = 79, normalized size = 0.78 \[ -\frac {{\left (b\,x^4+a\right )}^{3/4}}{32\,x^8}-\frac {3\,b^2\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}}{a^{1/4}}\right )}{64\,a^{5/4}}-\frac {3\,{\left (b\,x^4+a\right )}^{7/4}}{32\,a\,x^8}-\frac {b^2\,\mathrm {atan}\left (\frac {{\left (b\,x^4+a\right )}^{1/4}\,1{}\mathrm {i}}{a^{1/4}}\right )\,3{}\mathrm {i}}{64\,a^{5/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^(3/4)/x^9,x)

[Out]

- (a + b*x^4)^(3/4)/(32*x^8) - (3*b^2*atan((a + b*x^4)^(1/4)/a^(1/4)))/(64*a^(5/4)) - (b^2*atan(((a + b*x^4)^(
1/4)*1i)/a^(1/4))*3i)/(64*a^(5/4)) - (3*(a + b*x^4)^(7/4))/(32*a*x^8)

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sympy [C]  time = 2.98, size = 41, normalized size = 0.41 \[ - \frac {b^{\frac {3}{4}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{4 x^{5} \Gamma \left (\frac {9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**4+a)**(3/4)/x**9,x)

[Out]

-b**(3/4)*gamma(5/4)*hyper((-3/4, 5/4), (9/4,), a*exp_polar(I*pi)/(b*x**4))/(4*x**5*gamma(9/4))

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