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]

-(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))

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Rubi [A]  time = 0.0594983, antiderivative size = 101, normalized size of antiderivative = 1., 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 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 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 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{\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*}

Mathematica [C]  time = 0.0086075, 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]

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

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Maple [F]  time = 0.034, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{9}} \left ( b{x}^{4}+a \right ) ^{{\frac{3}{4}}}}\, dx \end{align*}

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.92516, size = 497, normalized size = 4.92 \begin{align*} \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}} \end{align*}

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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Sympy [C]  time = 2.51771, size = 41, normalized size = 0.41 \begin{align*} - \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 )} \end{align*}

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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Giac [B]  time = 1.15361, size = 304, normalized size = 3.01 \begin{align*} \frac{1}{256} \, b^{2}{\left (\frac{6 \, \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^{2}} + \frac{6 \, \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^{2}} - \frac{3 \, \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^{2}} + \frac{3 \, \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^{2}} - \frac{8 \,{\left (3 \,{\left (b x^{4} + a\right )}^{\frac{7}{4}} +{\left (b x^{4} + a\right )}^{\frac{3}{4}} a\right )}}{a b^{2} x^{8}}\right )} \end{align*}

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*b^2*(6*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^2
+ 6*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^2 - 3*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^2 + 3*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^2 - 8*(3*(b*x^4 + a)^(7/4) + (b*x
^4 + a)^(3/4)*a)/(a*b^2*x^8))