3.948 \(\int \frac {1}{(c x)^{13/2} \sqrt [4]{a+b x^2}} \, dx\)

Optimal. Leaf size=85 \[ -\frac {64 \left (a+b x^2\right )^{11/4}}{231 a^3 c (c x)^{11/2}}+\frac {16 \left (a+b x^2\right )^{7/4}}{21 a^2 c (c x)^{11/2}}-\frac {2 \left (a+b x^2\right )^{3/4}}{3 a c (c x)^{11/2}} \]

[Out]

-2/3*(b*x^2+a)^(3/4)/a/c/(c*x)^(11/2)+16/21*(b*x^2+a)^(7/4)/a^2/c/(c*x)^(11/2)-64/231*(b*x^2+a)^(11/4)/a^3/c/(
c*x)^(11/2)

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Rubi [A]  time = 0.03, antiderivative size = 85, 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} \[ -\frac {64 \left (a+b x^2\right )^{11/4}}{231 a^3 c (c x)^{11/2}}+\frac {16 \left (a+b x^2\right )^{7/4}}{21 a^2 c (c x)^{11/2}}-\frac {2 \left (a+b x^2\right )^{3/4}}{3 a c (c x)^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c*x)^(13/2)*(a + b*x^2)^(1/4)),x]

[Out]

(-2*(a + b*x^2)^(3/4))/(3*a*c*(c*x)^(11/2)) + (16*(a + b*x^2)^(7/4))/(21*a^2*c*(c*x)^(11/2)) - (64*(a + b*x^2)
^(11/4))/(231*a^3*c*(c*x)^(11/2))

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)^{13/2} \sqrt [4]{a+b x^2}} \, dx &=-\frac {2 \left (a+b x^2\right )^{3/4}}{3 a c (c x)^{11/2}}-\frac {8 \int \frac {\left (a+b x^2\right )^{3/4}}{(c x)^{13/2}} \, dx}{3 a}\\ &=-\frac {2 \left (a+b x^2\right )^{3/4}}{3 a c (c x)^{11/2}}+\frac {16 \left (a+b x^2\right )^{7/4}}{21 a^2 c (c x)^{11/2}}+\frac {32 \int \frac {\left (a+b x^2\right )^{7/4}}{(c x)^{13/2}} \, dx}{21 a^2}\\ &=-\frac {2 \left (a+b x^2\right )^{3/4}}{3 a c (c x)^{11/2}}+\frac {16 \left (a+b x^2\right )^{7/4}}{21 a^2 c (c x)^{11/2}}-\frac {64 \left (a+b x^2\right )^{11/4}}{231 a^3 c (c x)^{11/2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 52, normalized size = 0.61 \[ -\frac {2 \sqrt {c x} \left (a+b x^2\right )^{3/4} \left (21 a^2-24 a b x^2+32 b^2 x^4\right )}{231 a^3 c^7 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x)^(13/2)*(a + b*x^2)^(1/4)),x]

[Out]

(-2*Sqrt[c*x]*(a + b*x^2)^(3/4)*(21*a^2 - 24*a*b*x^2 + 32*b^2*x^4))/(231*a^3*c^7*x^6)

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fricas [A]  time = 0.77, size = 46, normalized size = 0.54 \[ -\frac {2 \, {\left (32 \, b^{2} x^{4} - 24 \, a b x^{2} + 21 \, a^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x}}{231 \, a^{3} c^{7} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(13/2)/(b*x^2+a)^(1/4),x, algorithm="fricas")

[Out]

-2/231*(32*b^2*x^4 - 24*a*b*x^2 + 21*a^2)*(b*x^2 + a)^(3/4)*sqrt(c*x)/(a^3*c^7*x^6)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (c x\right )^{\frac {13}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(13/2)/(b*x^2+a)^(1/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(1/4)*(c*x)^(13/2)), x)

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maple [A]  time = 0.01, size = 42, normalized size = 0.49 \[ -\frac {2 \left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (32 b^{2} x^{4}-24 a b \,x^{2}+21 a^{2}\right ) x}{231 \left (c x \right )^{\frac {13}{2}} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(13/2)/(b*x^2+a)^(1/4),x)

[Out]

-2/231*x*(b*x^2+a)^(3/4)*(32*b^2*x^4-24*a*b*x^2+21*a^2)/a^3/(c*x)^(13/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (c x\right )^{\frac {13}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(13/2)/(b*x^2+a)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(1/4)*(c*x)^(13/2)), x)

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mupad [B]  time = 5.00, size = 54, normalized size = 0.64 \[ -\frac {{\left (b\,x^2+a\right )}^{3/4}\,\left (\frac {2}{11\,a\,c^6}-\frac {16\,b\,x^2}{77\,a^2\,c^6}+\frac {64\,b^2\,x^4}{231\,a^3\,c^6}\right )}{x^5\,\sqrt {c\,x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c*x)^(13/2)*(a + b*x^2)^(1/4)),x)

[Out]

-((a + b*x^2)^(3/4)*(2/(11*a*c^6) - (16*b*x^2)/(77*a^2*c^6) + (64*b^2*x^4)/(231*a^3*c^6)))/(x^5*(c*x)^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)**(13/2)/(b*x**2+a)**(1/4),x)

[Out]

Timed out

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