3.923 \(\int \frac {\sqrt [4]{a+b x^2}}{(c x)^{9/2}} \, dx\)

Optimal. Leaf size=123 \[ \frac {4 b^{5/2} (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{21 a^{3/2} c^6 \left (a+b x^2\right )^{3/4}}-\frac {2 b \sqrt [4]{a+b x^2}}{21 a c^3 (c x)^{3/2}}-\frac {2 \sqrt [4]{a+b x^2}}{7 c (c x)^{7/2}} \]

[Out]

-2/7*(b*x^2+a)^(1/4)/c/(c*x)^(7/2)-2/21*b*(b*x^2+a)^(1/4)/a/c^3/(c*x)^(3/2)+4/21*b^(5/2)*(1+a/b/x^2)^(3/4)*(c*
x)^(3/2)*(cos(1/2*arccot(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arc
cot(x*b^(1/2)/a^(1/2))),2^(1/2))/a^(3/2)/c^6/(b*x^2+a)^(3/4)

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Rubi [A]  time = 0.09, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {277, 325, 329, 237, 335, 275, 231} \[ \frac {4 b^{5/2} (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{21 a^{3/2} c^6 \left (a+b x^2\right )^{3/4}}-\frac {2 b \sqrt [4]{a+b x^2}}{21 a c^3 (c x)^{3/2}}-\frac {2 \sqrt [4]{a+b x^2}}{7 c (c x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(1/4)/(c*x)^(9/2),x]

[Out]

(-2*(a + b*x^2)^(1/4))/(7*c*(c*x)^(7/2)) - (2*b*(a + b*x^2)^(1/4))/(21*a*c^3*(c*x)^(3/2)) + (4*b^(5/2)*(1 + a/
(b*x^2))^(3/4)*(c*x)^(3/2)*EllipticF[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(21*a^(3/2)*c^6*(a + b*x^2)^(3/4))

Rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(3/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 237

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Dist[(x^3*(1 + a/(b*x^4))^(3/4))/(a + b*x^4)^(3/4), Int[1/(x^3*
(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ[{a, b}, x]

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]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\sqrt [4]{a+b x^2}}{(c x)^{9/2}} \, dx &=-\frac {2 \sqrt [4]{a+b x^2}}{7 c (c x)^{7/2}}+\frac {b \int \frac {1}{(c x)^{5/2} \left (a+b x^2\right )^{3/4}} \, dx}{7 c^2}\\ &=-\frac {2 \sqrt [4]{a+b x^2}}{7 c (c x)^{7/2}}-\frac {2 b \sqrt [4]{a+b x^2}}{21 a c^3 (c x)^{3/2}}-\frac {\left (2 b^2\right ) \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{3/4}} \, dx}{21 a c^4}\\ &=-\frac {2 \sqrt [4]{a+b x^2}}{7 c (c x)^{7/2}}-\frac {2 b \sqrt [4]{a+b x^2}}{21 a c^3 (c x)^{3/2}}-\frac {\left (4 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a+\frac {b x^4}{c^2}\right )^{3/4}} \, dx,x,\sqrt {c x}\right )}{21 a c^5}\\ &=-\frac {2 \sqrt [4]{a+b x^2}}{7 c (c x)^{7/2}}-\frac {2 b \sqrt [4]{a+b x^2}}{21 a c^3 (c x)^{3/2}}-\frac {\left (4 b^2 \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {a c^2}{b x^4}\right )^{3/4} x^3} \, dx,x,\sqrt {c x}\right )}{21 a c^5 \left (a+b x^2\right )^{3/4}}\\ &=-\frac {2 \sqrt [4]{a+b x^2}}{7 c (c x)^{7/2}}-\frac {2 b \sqrt [4]{a+b x^2}}{21 a c^3 (c x)^{3/2}}+\frac {\left (4 b^2 \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {x}{\left (1+\frac {a c^2 x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{\sqrt {c x}}\right )}{21 a c^5 \left (a+b x^2\right )^{3/4}}\\ &=-\frac {2 \sqrt [4]{a+b x^2}}{7 c (c x)^{7/2}}-\frac {2 b \sqrt [4]{a+b x^2}}{21 a c^3 (c x)^{3/2}}+\frac {\left (2 b^2 \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {a c^2 x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{c x}\right )}{21 a c^5 \left (a+b x^2\right )^{3/4}}\\ &=-\frac {2 \sqrt [4]{a+b x^2}}{7 c (c x)^{7/2}}-\frac {2 b \sqrt [4]{a+b x^2}}{21 a c^3 (c x)^{3/2}}+\frac {4 b^{5/2} \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{21 a^{3/2} c^6 \left (a+b x^2\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 56, normalized size = 0.46 \[ -\frac {2 x \sqrt [4]{a+b x^2} \, _2F_1\left (-\frac {7}{4},-\frac {1}{4};-\frac {3}{4};-\frac {b x^2}{a}\right )}{7 (c x)^{9/2} \sqrt [4]{\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x*(a + b*x^2)^(1/4)*Hypergeometric2F1[-7/4, -1/4, -3/4, -((b*x^2)/a)])/(7*(c*x)^(9/2)*(1 + (b*x^2)/a)^(1/4
))

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}} \sqrt {c x}}{c^{5} x^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^(1/4)*sqrt(c*x)/(c^5*x^5), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{2}+a \right )^{\frac {1}{4}}}{\left (c x \right )^{\frac {9}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,x^2+a\right )}^{1/4}}{{\left (c\,x\right )}^{9/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/4)/(c*x)^(9/2),x)

[Out]

int((a + b*x^2)^(1/4)/(c*x)^(9/2), x)

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sympy [C]  time = 27.11, size = 36, normalized size = 0.29 \[ - \frac {\sqrt [4]{b} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{3 c^{\frac {9}{2}} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b**(1/4)*hyper((-1/4, 3/2), (5/2,), a*exp_polar(I*pi)/(b*x**2))/(3*c**(9/2)*x**3)

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