3.989 \(\int \frac {1}{(c x)^{9/2} (a+b x^2)^{5/4}} \, dx\)

Optimal. Leaf size=83 \[ \frac {64 \left (a+b x^2\right )^{7/4}}{21 a^3 c (c x)^{7/2}}-\frac {16 \left (a+b x^2\right )^{3/4}}{3 a^2 c (c x)^{7/2}}+\frac {2}{a c (c x)^{7/2} \sqrt [4]{a+b x^2}} \]

[Out]

2/a/c/(c*x)^(7/2)/(b*x^2+a)^(1/4)-16/3*(b*x^2+a)^(3/4)/a^2/c/(c*x)^(7/2)+64/21*(b*x^2+a)^(7/4)/a^3/c/(c*x)^(7/
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} \[ \frac {64 \left (a+b x^2\right )^{7/4}}{21 a^3 c (c x)^{7/2}}-\frac {16 \left (a+b x^2\right )^{3/4}}{3 a^2 c (c x)^{7/2}}+\frac {2}{a c (c x)^{7/2} \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c*x)^(9/2)*(a + b*x^2)^(5/4)),x]

[Out]

2/(a*c*(c*x)^(7/2)*(a + b*x^2)^(1/4)) - (16*(a + b*x^2)^(3/4))/(3*a^2*c*(c*x)^(7/2)) + (64*(a + b*x^2)^(7/4))/
(21*a^3*c*(c*x)^(7/2))

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)^{9/2} \left (a+b x^2\right )^{5/4}} \, dx &=\frac {2}{a c (c x)^{7/2} \sqrt [4]{a+b x^2}}+\frac {8 \int \frac {1}{(c x)^{9/2} \sqrt [4]{a+b x^2}} \, dx}{a}\\ &=\frac {2}{a c (c x)^{7/2} \sqrt [4]{a+b x^2}}-\frac {16 \left (a+b x^2\right )^{3/4}}{3 a^2 c (c x)^{7/2}}-\frac {32 \int \frac {\left (a+b x^2\right )^{3/4}}{(c x)^{9/2}} \, dx}{3 a^2}\\ &=\frac {2}{a c (c x)^{7/2} \sqrt [4]{a+b x^2}}-\frac {16 \left (a+b x^2\right )^{3/4}}{3 a^2 c (c x)^{7/2}}+\frac {64 \left (a+b x^2\right )^{7/4}}{21 a^3 c (c x)^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 47, normalized size = 0.57 \[ -\frac {2 x \left (3 a^2-8 a b x^2-32 b^2 x^4\right )}{21 a^3 (c x)^{9/2} \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x)^(9/2)*(a + b*x^2)^(5/4)),x]

[Out]

(-2*x*(3*a^2 - 8*a*b*x^2 - 32*b^2*x^4))/(21*a^3*(c*x)^(9/2)*(a + b*x^2)^(1/4))

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fricas [A]  time = 0.55, size = 61, normalized size = 0.73 \[ \frac {2 \, {\left (32 \, b^{2} x^{4} + 8 \, a b x^{2} - 3 \, a^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x}}{21 \, {\left (a^{3} b c^{5} x^{6} + a^{4} c^{5} x^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(9/2)/(b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

2/21*(32*b^2*x^4 + 8*a*b*x^2 - 3*a^2)*(b*x^2 + a)^(3/4)*sqrt(c*x)/(a^3*b*c^5*x^6 + a^4*c^5*x^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (c x\right )^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(9/2)/(b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(5/4)*(c*x)^(9/2)), x)

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maple [A]  time = 0.01, size = 42, normalized size = 0.51 \[ -\frac {2 \left (-32 b^{2} x^{4}-8 a b \,x^{2}+3 a^{2}\right ) x}{21 \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (c x \right )^{\frac {9}{2}} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(9/2)/(b*x^2+a)^(5/4),x)

[Out]

-2/21*x*(-32*b^2*x^4-8*a*b*x^2+3*a^2)/(b*x^2+a)^(1/4)/a^3/(c*x)^(9/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (c x\right )^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(9/2)/(b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(5/4)*(c*x)^(9/2)), x)

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mupad [B]  time = 5.15, size = 70, normalized size = 0.84 \[ \frac {{\left (b\,x^2+a\right )}^{3/4}\,\left (\frac {16\,x^2}{21\,a^2\,c^4}-\frac {2}{7\,a\,b\,c^4}+\frac {64\,b\,x^4}{21\,a^3\,c^4}\right )}{x^5\,\sqrt {c\,x}+\frac {a\,x^3\,\sqrt {c\,x}}{b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c*x)^(9/2)*(a + b*x^2)^(5/4)),x)

[Out]

((a + b*x^2)^(3/4)*((16*x^2)/(21*a^2*c^4) - 2/(7*a*b*c^4) + (64*b*x^4)/(21*a^3*c^4)))/(x^5*(c*x)^(1/2) + (a*x^
3*(c*x)^(1/2))/b)

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sympy [B]  time = 109.04, size = 384, normalized size = 4.63 \[ - \frac {3 a^{3} b^{\frac {19}{4}} \left (\frac {a}{b x^{2}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{32 a^{5} b^{4} c^{\frac {9}{2}} x^{2} \Gamma \left (\frac {5}{4}\right ) + 64 a^{4} b^{5} c^{\frac {9}{2}} x^{4} \Gamma \left (\frac {5}{4}\right ) + 32 a^{3} b^{6} c^{\frac {9}{2}} x^{6} \Gamma \left (\frac {5}{4}\right )} + \frac {5 a^{2} b^{\frac {23}{4}} x^{2} \left (\frac {a}{b x^{2}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{32 a^{5} b^{4} c^{\frac {9}{2}} x^{2} \Gamma \left (\frac {5}{4}\right ) + 64 a^{4} b^{5} c^{\frac {9}{2}} x^{4} \Gamma \left (\frac {5}{4}\right ) + 32 a^{3} b^{6} c^{\frac {9}{2}} x^{6} \Gamma \left (\frac {5}{4}\right )} + \frac {40 a b^{\frac {27}{4}} x^{4} \left (\frac {a}{b x^{2}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{32 a^{5} b^{4} c^{\frac {9}{2}} x^{2} \Gamma \left (\frac {5}{4}\right ) + 64 a^{4} b^{5} c^{\frac {9}{2}} x^{4} \Gamma \left (\frac {5}{4}\right ) + 32 a^{3} b^{6} c^{\frac {9}{2}} x^{6} \Gamma \left (\frac {5}{4}\right )} + \frac {32 b^{\frac {31}{4}} x^{6} \left (\frac {a}{b x^{2}} + 1\right )^{\frac {3}{4}} \Gamma \left (- \frac {7}{4}\right )}{32 a^{5} b^{4} c^{\frac {9}{2}} x^{2} \Gamma \left (\frac {5}{4}\right ) + 64 a^{4} b^{5} c^{\frac {9}{2}} x^{4} \Gamma \left (\frac {5}{4}\right ) + 32 a^{3} b^{6} c^{\frac {9}{2}} x^{6} \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)**(9/2)/(b*x**2+a)**(5/4),x)

[Out]

-3*a**3*b**(19/4)*(a/(b*x**2) + 1)**(3/4)*gamma(-7/4)/(32*a**5*b**4*c**(9/2)*x**2*gamma(5/4) + 64*a**4*b**5*c*
*(9/2)*x**4*gamma(5/4) + 32*a**3*b**6*c**(9/2)*x**6*gamma(5/4)) + 5*a**2*b**(23/4)*x**2*(a/(b*x**2) + 1)**(3/4
)*gamma(-7/4)/(32*a**5*b**4*c**(9/2)*x**2*gamma(5/4) + 64*a**4*b**5*c**(9/2)*x**4*gamma(5/4) + 32*a**3*b**6*c*
*(9/2)*x**6*gamma(5/4)) + 40*a*b**(27/4)*x**4*(a/(b*x**2) + 1)**(3/4)*gamma(-7/4)/(32*a**5*b**4*c**(9/2)*x**2*
gamma(5/4) + 64*a**4*b**5*c**(9/2)*x**4*gamma(5/4) + 32*a**3*b**6*c**(9/2)*x**6*gamma(5/4)) + 32*b**(31/4)*x**
6*(a/(b*x**2) + 1)**(3/4)*gamma(-7/4)/(32*a**5*b**4*c**(9/2)*x**2*gamma(5/4) + 64*a**4*b**5*c**(9/2)*x**4*gamm
a(5/4) + 32*a**3*b**6*c**(9/2)*x**6*gamma(5/4))

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