3.963 \(\int \frac {1}{(c x)^{7/2} \sqrt [4]{a-b x^2}} \, dx\)

Optimal. Leaf size=100 \[ -\frac {4 b^{3/2} \sqrt {c x} \sqrt [4]{1-\frac {a}{b x^2}} E\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{3/2} c^4 \sqrt [4]{a-b x^2}}-\frac {2 \left (a-b x^2\right )^{3/4}}{5 a c (c x)^{5/2}} \]

[Out]

-2/5*(-b*x^2+a)^(3/4)/a/c/(c*x)^(5/2)-4/5*b^(3/2)*(1-a/b/x^2)^(1/4)*(cos(1/2*arccsc(x*b^(1/2)/a^(1/2)))^2)^(1/
2)/cos(1/2*arccsc(x*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arccsc(x*b^(1/2)/a^(1/2))),2^(1/2))*(c*x)^(1/2)/a^(3/2
)/c^4/(-b*x^2+a)^(1/4)

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Rubi [A]  time = 0.04, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {325, 317, 335, 228} \[ -\frac {4 b^{3/2} \sqrt {c x} \sqrt [4]{1-\frac {a}{b x^2}} E\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{3/2} c^4 \sqrt [4]{a-b x^2}}-\frac {2 \left (a-b x^2\right )^{3/4}}{5 a c (c x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c*x)^(7/2)*(a - b*x^2)^(1/4)),x]

[Out]

(-2*(a - b*x^2)^(3/4))/(5*a*c*(c*x)^(5/2)) - (4*b^(3/2)*(1 - a/(b*x^2))^(1/4)*Sqrt[c*x]*EllipticE[ArcCsc[(Sqrt
[b]*x)/Sqrt[a]]/2, 2])/(5*a^(3/2)*c^4*(a - b*x^2)^(1/4))

Rule 228

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(1/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rule 317

Int[1/(((c_.)*(x_))^(3/2)*((a_) + (b_.)*(x_)^2)^(1/4)), x_Symbol] :> Dist[(Sqrt[c*x]*(1 + a/(b*x^2))^(1/4))/(c
^2*(a + b*x^2)^(1/4)), Int[1/(x^2*(1 + a/(b*x^2))^(1/4)), x], x] /; FreeQ[{a, b, c}, x] && NegQ[b/a]

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

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Mathematica [C]  time = 0.01, size = 57, normalized size = 0.57 \[ -\frac {2 x \sqrt [4]{1-\frac {b x^2}{a}} \, _2F_1\left (-\frac {5}{4},\frac {1}{4};-\frac {1}{4};\frac {b x^2}{a}\right )}{5 (c x)^{7/2} \sqrt [4]{a-b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x)^(7/2)*(a - b*x^2)^(1/4)),x]

[Out]

(-2*x*(1 - (b*x^2)/a)^(1/4)*Hypergeometric2F1[-5/4, 1/4, -1/4, (b*x^2)/a])/(5*(c*x)^(7/2)*(a - b*x^2)^(1/4))

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (-b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x}}{b c^{4} x^{6} - a c^{4} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(7/2)/(-b*x^2+a)^(1/4),x, algorithm="fricas")

[Out]

integral(-(-b*x^2 + a)^(3/4)*sqrt(c*x)/(b*c^4*x^6 - a*c^4*x^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{2} + a\right )}^{\frac {1}{4}} \left (c x\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(7/2)/(-b*x^2+a)^(1/4),x, algorithm="giac")

[Out]

integrate(1/((-b*x^2 + a)^(1/4)*(c*x)^(7/2)), x)

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maple [F]  time = 0.30, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (c x \right )^{\frac {7}{2}} \left (-b \,x^{2}+a \right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(7/2)/(-b*x^2+a)^(1/4),x)

[Out]

int(1/(c*x)^(7/2)/(-b*x^2+a)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-b x^{2} + a\right )}^{\frac {1}{4}} \left (c x\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(7/2)/(-b*x^2+a)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((-b*x^2 + a)^(1/4)*(c*x)^(7/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (c\,x\right )}^{7/2}\,{\left (a-b\,x^2\right )}^{1/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c*x)^(7/2)*(a - b*x^2)^(1/4)),x)

[Out]

int(1/((c*x)^(7/2)*(a - b*x^2)^(1/4)), x)

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sympy [C]  time = 12.20, size = 39, normalized size = 0.39 \[ - \frac {i e^{- \frac {3 i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {a}{b x^{2}}} \right )}}{3 \sqrt [4]{b} c^{\frac {7}{2}} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)**(7/2)/(-b*x**2+a)**(1/4),x)

[Out]

-I*exp(-3*I*pi/4)*hyper((1/4, 3/2), (5/2,), a/(b*x**2))/(3*b**(1/4)*c**(7/2)*x**3)

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