3.974 \(\int \frac {1}{(c x)^{11/2} (a+b x^2)^{3/4}} \, dx\)

Optimal. Leaf size=83 \[ -\frac {64 \left (a+b x^2\right )^{9/4}}{45 a^3 c (c x)^{9/2}}+\frac {16 \left (a+b x^2\right )^{5/4}}{5 a^2 c (c x)^{9/2}}-\frac {2 \sqrt [4]{a+b x^2}}{a c (c x)^{9/2}} \]

[Out]

-2*(b*x^2+a)^(1/4)/a/c/(c*x)^(9/2)+16/5*(b*x^2+a)^(5/4)/a^2/c/(c*x)^(9/2)-64/45*(b*x^2+a)^(9/4)/a^3/c/(c*x)^(9
/2)

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Rubi [A]  time = 0.02, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {273, 264} \[ -\frac {64 \left (a+b x^2\right )^{9/4}}{45 a^3 c (c x)^{9/2}}+\frac {16 \left (a+b x^2\right )^{5/4}}{5 a^2 c (c x)^{9/2}}-\frac {2 \sqrt [4]{a+b x^2}}{a c (c x)^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c*x)^(11/2)*(a + b*x^2)^(3/4)),x]

[Out]

(-2*(a + b*x^2)^(1/4))/(a*c*(c*x)^(9/2)) + (16*(a + b*x^2)^(5/4))/(5*a^2*c*(c*x)^(9/2)) - (64*(a + b*x^2)^(9/4
))/(45*a^3*c*(c*x)^(9/2))

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 273

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

Rubi steps

\begin {align*} \int \frac {1}{(c x)^{11/2} \left (a+b x^2\right )^{3/4}} \, dx &=-\frac {2 \sqrt [4]{a+b x^2}}{a c (c x)^{9/2}}-\frac {8 \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{11/2}} \, dx}{a}\\ &=-\frac {2 \sqrt [4]{a+b x^2}}{a c (c x)^{9/2}}+\frac {16 \left (a+b x^2\right )^{5/4}}{5 a^2 c (c x)^{9/2}}+\frac {32 \int \frac {\left (a+b x^2\right )^{5/4}}{(c x)^{11/2}} \, dx}{5 a^2}\\ &=-\frac {2 \sqrt [4]{a+b x^2}}{a c (c x)^{9/2}}+\frac {16 \left (a+b x^2\right )^{5/4}}{5 a^2 c (c x)^{9/2}}-\frac {64 \left (a+b x^2\right )^{9/4}}{45 a^3 c (c x)^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 52, normalized size = 0.63 \[ -\frac {2 \sqrt {c x} \sqrt [4]{a+b x^2} \left (5 a^2-8 a b x^2+32 b^2 x^4\right )}{45 a^3 c^6 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x)^(11/2)*(a + b*x^2)^(3/4)),x]

[Out]

(-2*Sqrt[c*x]*(a + b*x^2)^(1/4)*(5*a^2 - 8*a*b*x^2 + 32*b^2*x^4))/(45*a^3*c^6*x^5)

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fricas [A]  time = 0.61, size = 46, normalized size = 0.55 \[ -\frac {2 \, {\left (32 \, b^{2} x^{4} - 8 \, a b x^{2} + 5 \, a^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \sqrt {c x}}{45 \, a^{3} c^{6} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(11/2)/(b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

-2/45*(32*b^2*x^4 - 8*a*b*x^2 + 5*a^2)*(b*x^2 + a)^(1/4)*sqrt(c*x)/(a^3*c^6*x^5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \left (c x\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(11/2)/(b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(3/4)*(c*x)^(11/2)), x)

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maple [A]  time = 0.00, size = 42, normalized size = 0.51 \[ -\frac {2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (32 b^{2} x^{4}-8 a b \,x^{2}+5 a^{2}\right ) x}{45 \left (c x \right )^{\frac {11}{2}} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(11/2)/(b*x^2+a)^(3/4),x)

[Out]

-2/45*x*(b*x^2+a)^(1/4)*(32*b^2*x^4-8*a*b*x^2+5*a^2)/a^3/(c*x)^(11/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \left (c x\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(11/2)/(b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(3/4)*(c*x)^(11/2)), x)

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mupad [B]  time = 5.17, size = 54, normalized size = 0.65 \[ -\frac {{\left (b\,x^2+a\right )}^{1/4}\,\left (\frac {2}{9\,a\,c^5}-\frac {16\,b\,x^2}{45\,a^2\,c^5}+\frac {64\,b^2\,x^4}{45\,a^3\,c^5}\right )}{x^4\,\sqrt {c\,x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c*x)^(11/2)*(a + b*x^2)^(3/4)),x)

[Out]

-((a + b*x^2)^(1/4)*(2/(9*a*c^5) - (16*b*x^2)/(45*a^2*c^5) + (64*b^2*x^4)/(45*a^3*c^5)))/(x^4*(c*x)^(1/2))

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sympy [B]  time = 172.06, size = 483, normalized size = 5.82 \[ \frac {5 a^{4} b^{\frac {17}{4}} \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (- \frac {9}{4}\right )}{32 a^{5} b^{4} c^{\frac {11}{2}} x^{4} \Gamma \left (\frac {3}{4}\right ) + 64 a^{4} b^{5} c^{\frac {11}{2}} x^{6} \Gamma \left (\frac {3}{4}\right ) + 32 a^{3} b^{6} c^{\frac {11}{2}} x^{8} \Gamma \left (\frac {3}{4}\right )} + \frac {2 a^{3} b^{\frac {21}{4}} x^{2} \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (- \frac {9}{4}\right )}{32 a^{5} b^{4} c^{\frac {11}{2}} x^{4} \Gamma \left (\frac {3}{4}\right ) + 64 a^{4} b^{5} c^{\frac {11}{2}} x^{6} \Gamma \left (\frac {3}{4}\right ) + 32 a^{3} b^{6} c^{\frac {11}{2}} x^{8} \Gamma \left (\frac {3}{4}\right )} + \frac {21 a^{2} b^{\frac {25}{4}} x^{4} \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (- \frac {9}{4}\right )}{32 a^{5} b^{4} c^{\frac {11}{2}} x^{4} \Gamma \left (\frac {3}{4}\right ) + 64 a^{4} b^{5} c^{\frac {11}{2}} x^{6} \Gamma \left (\frac {3}{4}\right ) + 32 a^{3} b^{6} c^{\frac {11}{2}} x^{8} \Gamma \left (\frac {3}{4}\right )} + \frac {56 a b^{\frac {29}{4}} x^{6} \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (- \frac {9}{4}\right )}{32 a^{5} b^{4} c^{\frac {11}{2}} x^{4} \Gamma \left (\frac {3}{4}\right ) + 64 a^{4} b^{5} c^{\frac {11}{2}} x^{6} \Gamma \left (\frac {3}{4}\right ) + 32 a^{3} b^{6} c^{\frac {11}{2}} x^{8} \Gamma \left (\frac {3}{4}\right )} + \frac {32 b^{\frac {33}{4}} x^{8} \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (- \frac {9}{4}\right )}{32 a^{5} b^{4} c^{\frac {11}{2}} x^{4} \Gamma \left (\frac {3}{4}\right ) + 64 a^{4} b^{5} c^{\frac {11}{2}} x^{6} \Gamma \left (\frac {3}{4}\right ) + 32 a^{3} b^{6} c^{\frac {11}{2}} x^{8} \Gamma \left (\frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)**(11/2)/(b*x**2+a)**(3/4),x)

[Out]

5*a**4*b**(17/4)*(a/(b*x**2) + 1)**(1/4)*gamma(-9/4)/(32*a**5*b**4*c**(11/2)*x**4*gamma(3/4) + 64*a**4*b**5*c*
*(11/2)*x**6*gamma(3/4) + 32*a**3*b**6*c**(11/2)*x**8*gamma(3/4)) + 2*a**3*b**(21/4)*x**2*(a/(b*x**2) + 1)**(1
/4)*gamma(-9/4)/(32*a**5*b**4*c**(11/2)*x**4*gamma(3/4) + 64*a**4*b**5*c**(11/2)*x**6*gamma(3/4) + 32*a**3*b**
6*c**(11/2)*x**8*gamma(3/4)) + 21*a**2*b**(25/4)*x**4*(a/(b*x**2) + 1)**(1/4)*gamma(-9/4)/(32*a**5*b**4*c**(11
/2)*x**4*gamma(3/4) + 64*a**4*b**5*c**(11/2)*x**6*gamma(3/4) + 32*a**3*b**6*c**(11/2)*x**8*gamma(3/4)) + 56*a*
b**(29/4)*x**6*(a/(b*x**2) + 1)**(1/4)*gamma(-9/4)/(32*a**5*b**4*c**(11/2)*x**4*gamma(3/4) + 64*a**4*b**5*c**(
11/2)*x**6*gamma(3/4) + 32*a**3*b**6*c**(11/2)*x**8*gamma(3/4)) + 32*b**(33/4)*x**8*(a/(b*x**2) + 1)**(1/4)*ga
mma(-9/4)/(32*a**5*b**4*c**(11/2)*x**4*gamma(3/4) + 64*a**4*b**5*c**(11/2)*x**6*gamma(3/4) + 32*a**3*b**6*c**(
11/2)*x**8*gamma(3/4))

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