3.1045 \(\int \frac {(a+b x^4)^{3/4}}{x^{10}} \, dx\)

Optimal. Leaf size=126 \[ -\frac {2 b^{5/2} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{15 a^{3/2} \sqrt [4]{a+b x^4}}+\frac {2 b^2}{15 a x \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{9 x^9}-\frac {b \left (a+b x^4\right )^{3/4}}{15 a x^5} \]

[Out]

2/15*b^2/a/x/(b*x^4+a)^(1/4)-1/9*(b*x^4+a)^(3/4)/x^9-1/15*b*(b*x^4+a)^(3/4)/a/x^5-2/15*b^(5/2)*(1+a/b/x^4)^(1/
4)*x*(cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arc
cot(x^2*b^(1/2)/a^(1/2))),2^(1/2))/a^(3/2)/(b*x^4+a)^(1/4)

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Rubi [A]  time = 0.06, antiderivative size = 126, 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 = {277, 325, 312, 281, 335, 275, 196} \[ -\frac {2 b^{5/2} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{15 a^{3/2} \sqrt [4]{a+b x^4}}+\frac {2 b^2}{15 a x \sqrt [4]{a+b x^4}}-\frac {b \left (a+b x^4\right )^{3/4}}{15 a x^5}-\frac {\left (a+b x^4\right )^{3/4}}{9 x^9} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(3/4)/x^10,x]

[Out]

(2*b^2)/(15*a*x*(a + b*x^4)^(1/4)) - (a + b*x^4)^(3/4)/(9*x^9) - (b*(a + b*x^4)^(3/4))/(15*a*x^5) - (2*b^(5/2)
*(1 + a/(b*x^4))^(1/4)*x*EllipticE[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(15*a^(3/2)*(a + b*x^4)^(1/4))

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 281

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Dist[(x*(1 + a/(b*x^4))^(1/4))/(b*(a + b*x^4)^(1/4)), Int
[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 312

Int[1/((x_)^2*((a_) + (b_.)*(x_)^4)^(1/4)), x_Symbol] :> -Simp[(x*(a + b*x^4)^(1/4))^(-1), x] - Dist[b, Int[x^
2/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^4\right )^{3/4}}{x^{10}} \, dx &=-\frac {\left (a+b x^4\right )^{3/4}}{9 x^9}+\frac {1}{3} b \int \frac {1}{x^6 \sqrt [4]{a+b x^4}} \, dx\\ &=-\frac {\left (a+b x^4\right )^{3/4}}{9 x^9}-\frac {b \left (a+b x^4\right )^{3/4}}{15 a x^5}-\frac {\left (2 b^2\right ) \int \frac {1}{x^2 \sqrt [4]{a+b x^4}} \, dx}{15 a}\\ &=\frac {2 b^2}{15 a x \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{9 x^9}-\frac {b \left (a+b x^4\right )^{3/4}}{15 a x^5}+\frac {\left (2 b^3\right ) \int \frac {x^2}{\left (a+b x^4\right )^{5/4}} \, dx}{15 a}\\ &=\frac {2 b^2}{15 a x \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{9 x^9}-\frac {b \left (a+b x^4\right )^{3/4}}{15 a x^5}+\frac {\left (2 b^2 \sqrt [4]{1+\frac {a}{b x^4}} x\right ) \int \frac {1}{\left (1+\frac {a}{b x^4}\right )^{5/4} x^3} \, dx}{15 a \sqrt [4]{a+b x^4}}\\ &=\frac {2 b^2}{15 a x \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{9 x^9}-\frac {b \left (a+b x^4\right )^{3/4}}{15 a x^5}-\frac {\left (2 b^2 \sqrt [4]{1+\frac {a}{b x^4}} x\right ) \operatorname {Subst}\left (\int \frac {x}{\left (1+\frac {a x^4}{b}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )}{15 a \sqrt [4]{a+b x^4}}\\ &=\frac {2 b^2}{15 a x \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{9 x^9}-\frac {b \left (a+b x^4\right )^{3/4}}{15 a x^5}-\frac {\left (b^2 \sqrt [4]{1+\frac {a}{b x^4}} x\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{5/4}} \, dx,x,\frac {1}{x^2}\right )}{15 a \sqrt [4]{a+b x^4}}\\ &=\frac {2 b^2}{15 a x \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{9 x^9}-\frac {b \left (a+b x^4\right )^{3/4}}{15 a x^5}-\frac {2 b^{5/2} \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{15 a^{3/2} \sqrt [4]{a+b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 51, normalized size = 0.40 \[ -\frac {\left (a+b x^4\right )^{3/4} \, _2F_1\left (-\frac {9}{4},-\frac {3}{4};-\frac {5}{4};-\frac {b x^4}{a}\right )}{9 x^9 \left (\frac {b x^4}{a}+1\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(3/4)/x^10,x]

[Out]

-1/9*((a + b*x^4)^(3/4)*Hypergeometric2F1[-9/4, -3/4, -5/4, -((b*x^4)/a)])/(x^9*(1 + (b*x^4)/a)^(3/4))

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fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{x^{10}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(3/4)/x^10,x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(3/4)/x^10, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{x^{10}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(3/4)/x^10,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(3/4)/x^10, x)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}}}{x^{10}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^4+a)^(3/4)/x^10,x)

[Out]

int((b*x^4+a)^(3/4)/x^10,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{x^{10}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(3/4)/x^10,x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^(3/4)/x^10, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,x^4+a\right )}^{3/4}}{x^{10}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^(3/4)/x^10,x)

[Out]

int((a + b*x^4)^(3/4)/x^10, x)

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sympy [C]  time = 4.52, size = 31, normalized size = 0.25 \[ - \frac {b^{\frac {3}{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{4}}} \right )}}{6 x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**4+a)**(3/4)/x**10,x)

[Out]

-b**(3/4)*hyper((-3/4, 3/2), (5/2,), a*exp_polar(I*pi)/(b*x**4))/(6*x**6)

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