3.1030 \(\int \frac {(a+b x^4)^{3/4}}{x^{11}} \, dx\)

Optimal. Leaf size=149 \[ \frac {3 b^{5/2} \sqrt [4]{\frac {b x^4}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{40 a^{3/2} \sqrt [4]{a+b x^4}}-\frac {3 b^3 x^2}{40 a^2 \sqrt [4]{a+b x^4}}+\frac {3 b^2 \left (a+b x^4\right )^{3/4}}{40 a^2 x^2}-\frac {\left (a+b x^4\right )^{3/4}}{10 x^{10}}-\frac {b \left (a+b x^4\right )^{3/4}}{20 a x^6} \]

[Out]

-3/40*b^3*x^2/a^2/(b*x^4+a)^(1/4)-1/10*(b*x^4+a)^(3/4)/x^10-1/20*b*(b*x^4+a)^(3/4)/a/x^6+3/40*b^2*(b*x^4+a)^(3
/4)/a^2/x^2+3/40*b^(5/2)*(1+b*x^4/a)^(1/4)*(cos(1/2*arctan(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x^2*b
^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arctan(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 = 0.09, antiderivative size = 149, 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 = {275, 277, 325, 229, 227, 196} \[ -\frac {3 b^3 x^2}{40 a^2 \sqrt [4]{a+b x^4}}+\frac {3 b^2 \left (a+b x^4\right )^{3/4}}{40 a^2 x^2}+\frac {3 b^{5/2} \sqrt [4]{\frac {b x^4}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{40 a^{3/2} \sqrt [4]{a+b x^4}}-\frac {b \left (a+b x^4\right )^{3/4}}{20 a x^6}-\frac {\left (a+b x^4\right )^{3/4}}{10 x^{10}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(3/4)/x^11,x]

[Out]

(-3*b^3*x^2)/(40*a^2*(a + b*x^4)^(1/4)) - (a + b*x^4)^(3/4)/(10*x^10) - (b*(a + b*x^4)^(3/4))/(20*a*x^6) + (3*
b^2*(a + b*x^4)^(3/4))/(40*a^2*x^2) + (3*b^(5/2)*(1 + (b*x^4)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]
/2, 2])/(40*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 227

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

Rule 229

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x^4\right )^{3/4}}{x^{11}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (a+b x^2\right )^{3/4}}{x^6} \, dx,x,x^2\right )\\ &=-\frac {\left (a+b x^4\right )^{3/4}}{10 x^{10}}+\frac {1}{20} (3 b) \operatorname {Subst}\left (\int \frac {1}{x^4 \sqrt [4]{a+b x^2}} \, dx,x,x^2\right )\\ &=-\frac {\left (a+b x^4\right )^{3/4}}{10 x^{10}}-\frac {b \left (a+b x^4\right )^{3/4}}{20 a x^6}-\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt [4]{a+b x^2}} \, dx,x,x^2\right )}{40 a}\\ &=-\frac {\left (a+b x^4\right )^{3/4}}{10 x^{10}}-\frac {b \left (a+b x^4\right )^{3/4}}{20 a x^6}+\frac {3 b^2 \left (a+b x^4\right )^{3/4}}{40 a^2 x^2}-\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{a+b x^2}} \, dx,x,x^2\right )}{80 a^2}\\ &=-\frac {\left (a+b x^4\right )^{3/4}}{10 x^{10}}-\frac {b \left (a+b x^4\right )^{3/4}}{20 a x^6}+\frac {3 b^2 \left (a+b x^4\right )^{3/4}}{40 a^2 x^2}-\frac {\left (3 b^3 \sqrt [4]{1+\frac {b x^4}{a}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1+\frac {b x^2}{a}}} \, dx,x,x^2\right )}{80 a^2 \sqrt [4]{a+b x^4}}\\ &=-\frac {3 b^3 x^2}{40 a^2 \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{10 x^{10}}-\frac {b \left (a+b x^4\right )^{3/4}}{20 a x^6}+\frac {3 b^2 \left (a+b x^4\right )^{3/4}}{40 a^2 x^2}+\frac {\left (3 b^3 \sqrt [4]{1+\frac {b x^4}{a}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx,x,x^2\right )}{80 a^2 \sqrt [4]{a+b x^4}}\\ &=-\frac {3 b^3 x^2}{40 a^2 \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{10 x^{10}}-\frac {b \left (a+b x^4\right )^{3/4}}{20 a x^6}+\frac {3 b^2 \left (a+b x^4\right )^{3/4}}{40 a^2 x^2}+\frac {3 b^{5/2} \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{40 a^{3/2} \sqrt [4]{a+b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 51, normalized size = 0.34 \[ -\frac {\left (a+b x^4\right )^{3/4} \, _2F_1\left (-\frac {5}{2},-\frac {3}{4};-\frac {3}{2};-\frac {b x^4}{a}\right )}{10 x^{10} \left (\frac {b x^4}{a}+1\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(3/4)/x^11,x]

[Out]

-1/10*((a + b*x^4)^(3/4)*Hypergeometric2F1[-5/2, -3/4, -3/2, -((b*x^4)/a)])/(x^10*(1 + (b*x^4)/a)^(3/4))

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fricas [F]  time = 0.91, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{x^{11}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(3/4)/x^11,x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(3/4)/x^11, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{x^{11}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(3/4)/x^11,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(3/4)/x^11, x)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}}}{x^{11}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^4+a)^(3/4)/x^11,x)

[Out]

int((b*x^4+a)^(3/4)/x^11,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{x^{11}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(3/4)/x^11,x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^(3/4)/x^11, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,x^4+a\right )}^{3/4}}{x^{11}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^(3/4)/x^11,x)

[Out]

int((a + b*x^4)^(3/4)/x^11, x)

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sympy [C]  time = 2.22, size = 34, normalized size = 0.23 \[ - \frac {a^{\frac {3}{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, - \frac {3}{4} \\ - \frac {3}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{10 x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**4+a)**(3/4)/x**11,x)

[Out]

-a**(3/4)*hyper((-5/2, -3/4), (-3/2,), b*x**4*exp_polar(I*pi)/a)/(10*x**10)

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