3.1130 \(\int \frac {1}{x^{10} (a+b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=68 \[ -\frac {32 b^2 \sqrt [4]{a+b x^4}}{45 a^3 x}+\frac {8 b \sqrt [4]{a+b x^4}}{45 a^2 x^5}-\frac {\sqrt [4]{a+b x^4}}{9 a x^9} \]

[Out]

-1/9*(b*x^4+a)^(1/4)/a/x^9+8/45*b*(b*x^4+a)^(1/4)/a^2/x^5-32/45*b^2*(b*x^4+a)^(1/4)/a^3/x

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Rubi [A]  time = 0.02, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {271, 264} \[ -\frac {32 b^2 \sqrt [4]{a+b x^4}}{45 a^3 x}+\frac {8 b \sqrt [4]{a+b x^4}}{45 a^2 x^5}-\frac {\sqrt [4]{a+b x^4}}{9 a x^9} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^10*(a + b*x^4)^(3/4)),x]

[Out]

-(a + b*x^4)^(1/4)/(9*a*x^9) + (8*b*(a + b*x^4)^(1/4))/(45*a^2*x^5) - (32*b^2*(a + b*x^4)^(1/4))/(45*a^3*x)

Rule 264

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] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 271

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

Rubi steps

\begin {align*} \int \frac {1}{x^{10} \left (a+b x^4\right )^{3/4}} \, dx &=-\frac {\sqrt [4]{a+b x^4}}{9 a x^9}-\frac {(8 b) \int \frac {1}{x^6 \left (a+b x^4\right )^{3/4}} \, dx}{9 a}\\ &=-\frac {\sqrt [4]{a+b x^4}}{9 a x^9}+\frac {8 b \sqrt [4]{a+b x^4}}{45 a^2 x^5}+\frac {\left (32 b^2\right ) \int \frac {1}{x^2 \left (a+b x^4\right )^{3/4}} \, dx}{45 a^2}\\ &=-\frac {\sqrt [4]{a+b x^4}}{9 a x^9}+\frac {8 b \sqrt [4]{a+b x^4}}{45 a^2 x^5}-\frac {32 b^2 \sqrt [4]{a+b x^4}}{45 a^3 x}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 42, normalized size = 0.62 \[ -\frac {\sqrt [4]{a+b x^4} \left (5 a^2-8 a b x^4+32 b^2 x^8\right )}{45 a^3 x^9} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^10*(a + b*x^4)^(3/4)),x]

[Out]

-1/45*((a + b*x^4)^(1/4)*(5*a^2 - 8*a*b*x^4 + 32*b^2*x^8))/(a^3*x^9)

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fricas [A]  time = 0.92, size = 38, normalized size = 0.56 \[ -\frac {{\left (32 \, b^{2} x^{8} - 8 \, a b x^{4} + 5 \, a^{2}\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{45 \, a^{3} x^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^10/(b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

-1/45*(32*b^2*x^8 - 8*a*b*x^4 + 5*a^2)*(b*x^4 + a)^(1/4)/(a^3*x^9)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{10}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^10/(b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(3/4)*x^10), x)

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maple [A]  time = 0.01, size = 39, normalized size = 0.57 \[ -\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (32 b^{2} x^{8}-8 a b \,x^{4}+5 a^{2}\right )}{45 a^{3} x^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^10/(b*x^4+a)^(3/4),x)

[Out]

-1/45*(b*x^4+a)^(1/4)*(32*b^2*x^8-8*a*b*x^4+5*a^2)/a^3/x^9

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maxima [A]  time = 1.37, size = 52, normalized size = 0.76 \[ -\frac {\frac {45 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{2}}{x} - \frac {18 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} b}{x^{5}} + \frac {5 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}}}{x^{9}}}{45 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^10/(b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

-1/45*(45*(b*x^4 + a)^(1/4)*b^2/x - 18*(b*x^4 + a)^(5/4)*b/x^5 + 5*(b*x^4 + a)^(9/4)/x^9)/a^3

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mupad [B]  time = 1.27, size = 38, normalized size = 0.56 \[ -\frac {{\left (b\,x^4+a\right )}^{1/4}\,\left (5\,a^2-8\,a\,b\,x^4+32\,b^2\,x^8\right )}{45\,a^3\,x^9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^10*(a + b*x^4)^(3/4)),x)

[Out]

-((a + b*x^4)^(1/4)*(5*a^2 + 32*b^2*x^8 - 8*a*b*x^4))/(45*a^3*x^9)

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sympy [B]  time = 2.63, size = 406, normalized size = 5.97 \[ \frac {5 a^{4} b^{\frac {17}{4}} \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {9}{4}\right )}{64 a^{5} b^{4} x^{8} \Gamma \left (\frac {3}{4}\right ) + 128 a^{4} b^{5} x^{12} \Gamma \left (\frac {3}{4}\right ) + 64 a^{3} b^{6} x^{16} \Gamma \left (\frac {3}{4}\right )} + \frac {2 a^{3} b^{\frac {21}{4}} x^{4} \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {9}{4}\right )}{64 a^{5} b^{4} x^{8} \Gamma \left (\frac {3}{4}\right ) + 128 a^{4} b^{5} x^{12} \Gamma \left (\frac {3}{4}\right ) + 64 a^{3} b^{6} x^{16} \Gamma \left (\frac {3}{4}\right )} + \frac {21 a^{2} b^{\frac {25}{4}} x^{8} \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {9}{4}\right )}{64 a^{5} b^{4} x^{8} \Gamma \left (\frac {3}{4}\right ) + 128 a^{4} b^{5} x^{12} \Gamma \left (\frac {3}{4}\right ) + 64 a^{3} b^{6} x^{16} \Gamma \left (\frac {3}{4}\right )} + \frac {56 a b^{\frac {29}{4}} x^{12} \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {9}{4}\right )}{64 a^{5} b^{4} x^{8} \Gamma \left (\frac {3}{4}\right ) + 128 a^{4} b^{5} x^{12} \Gamma \left (\frac {3}{4}\right ) + 64 a^{3} b^{6} x^{16} \Gamma \left (\frac {3}{4}\right )} + \frac {32 b^{\frac {33}{4}} x^{16} \sqrt [4]{\frac {a}{b x^{4}} + 1} \Gamma \left (- \frac {9}{4}\right )}{64 a^{5} b^{4} x^{8} \Gamma \left (\frac {3}{4}\right ) + 128 a^{4} b^{5} x^{12} \Gamma \left (\frac {3}{4}\right ) + 64 a^{3} b^{6} x^{16} \Gamma \left (\frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**10/(b*x**4+a)**(3/4),x)

[Out]

5*a**4*b**(17/4)*(a/(b*x**4) + 1)**(1/4)*gamma(-9/4)/(64*a**5*b**4*x**8*gamma(3/4) + 128*a**4*b**5*x**12*gamma
(3/4) + 64*a**3*b**6*x**16*gamma(3/4)) + 2*a**3*b**(21/4)*x**4*(a/(b*x**4) + 1)**(1/4)*gamma(-9/4)/(64*a**5*b*
*4*x**8*gamma(3/4) + 128*a**4*b**5*x**12*gamma(3/4) + 64*a**3*b**6*x**16*gamma(3/4)) + 21*a**2*b**(25/4)*x**8*
(a/(b*x**4) + 1)**(1/4)*gamma(-9/4)/(64*a**5*b**4*x**8*gamma(3/4) + 128*a**4*b**5*x**12*gamma(3/4) + 64*a**3*b
**6*x**16*gamma(3/4)) + 56*a*b**(29/4)*x**12*(a/(b*x**4) + 1)**(1/4)*gamma(-9/4)/(64*a**5*b**4*x**8*gamma(3/4)
+ 128*a**4*b**5*x**12*gamma(3/4) + 64*a**3*b**6*x**16*gamma(3/4)) + 32*b**(33/4)*x**16*(a/(b*x**4) + 1)**(1/4
)*gamma(-9/4)/(64*a**5*b**4*x**8*gamma(3/4) + 128*a**4*b**5*x**12*gamma(3/4) + 64*a**3*b**6*x**16*gamma(3/4))

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