3.960 \(\int \frac {(c x)^{5/2}}{\sqrt [4]{a-b x^2}} \, dx\)

Optimal. Leaf size=128 \[ \frac {a^{3/2} c^2 \sqrt {c x} \sqrt [4]{1-\frac {a}{b x^2}} E\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 b^{3/2} \sqrt [4]{a-b x^2}}-\frac {a c^3 \left (a-b x^2\right )^{3/4}}{2 b^2 \sqrt {c x}}-\frac {c (c x)^{3/2} \left (a-b x^2\right )^{3/4}}{3 b} \]

[Out]

-1/3*c*(c*x)^(3/2)*(-b*x^2+a)^(3/4)/b-1/2*a*c^3*(-b*x^2+a)^(3/4)/b^2/(c*x)^(1/2)+1/2*a^(3/2)*c^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)/b^(3/2)/(-b*x^2+a)^(1/4)

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Rubi [A]  time = 0.05, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {321, 315, 317, 335, 228} \[ \frac {a^{3/2} c^2 \sqrt {c x} \sqrt [4]{1-\frac {a}{b x^2}} E\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 b^{3/2} \sqrt [4]{a-b x^2}}-\frac {a c^3 \left (a-b x^2\right )^{3/4}}{2 b^2 \sqrt {c x}}-\frac {c (c x)^{3/2} \left (a-b x^2\right )^{3/4}}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(5/2)/(a - b*x^2)^(1/4),x]

[Out]

-(a*c^3*(a - b*x^2)^(3/4))/(2*b^2*Sqrt[c*x]) - (c*(c*x)^(3/2)*(a - b*x^2)^(3/4))/(3*b) + (a^(3/2)*c^2*(1 - a/(
b*x^2))^(1/4)*Sqrt[c*x]*EllipticE[ArcCsc[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(2*b^(3/2)*(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 315

Int[Sqrt[(c_)*(x_)]/((a_) + (b_.)*(x_)^2)^(1/4), x_Symbol] :> Simp[(c*(a + b*x^2)^(3/4))/(b*Sqrt[c*x]), x] + D
ist[(a*c^2)/(2*b), Int[1/((c*x)^(3/2)*(a + b*x^2)^(1/4)), x], x] /; FreeQ[{a, b, c}, x] && 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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 {(c x)^{5/2}}{\sqrt [4]{a-b x^2}} \, dx &=-\frac {c (c x)^{3/2} \left (a-b x^2\right )^{3/4}}{3 b}+\frac {\left (a c^2\right ) \int \frac {\sqrt {c x}}{\sqrt [4]{a-b x^2}} \, dx}{2 b}\\ &=-\frac {a c^3 \left (a-b x^2\right )^{3/4}}{2 b^2 \sqrt {c x}}-\frac {c (c x)^{3/2} \left (a-b x^2\right )^{3/4}}{3 b}-\frac {\left (a^2 c^4\right ) \int \frac {1}{(c x)^{3/2} \sqrt [4]{a-b x^2}} \, dx}{4 b^2}\\ &=-\frac {a c^3 \left (a-b x^2\right )^{3/4}}{2 b^2 \sqrt {c x}}-\frac {c (c x)^{3/2} \left (a-b x^2\right )^{3/4}}{3 b}-\frac {\left (a^2 c^2 \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}{4 b^2 \sqrt [4]{a-b x^2}}\\ &=-\frac {a c^3 \left (a-b x^2\right )^{3/4}}{2 b^2 \sqrt {c x}}-\frac {c (c x)^{3/2} \left (a-b x^2\right )^{3/4}}{3 b}+\frac {\left (a^2 c^2 \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 )}{4 b^2 \sqrt [4]{a-b x^2}}\\ &=-\frac {a c^3 \left (a-b x^2\right )^{3/4}}{2 b^2 \sqrt {c x}}-\frac {c (c x)^{3/2} \left (a-b x^2\right )^{3/4}}{3 b}+\frac {a^{3/2} c^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 )}{2 b^{3/2} \sqrt [4]{a-b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 71, normalized size = 0.55 \[ \frac {c (c x)^{3/2} \left (a \sqrt [4]{1-\frac {b x^2}{a}} \, _2F_1\left (\frac {1}{4},\frac {3}{4};\frac {7}{4};\frac {b x^2}{a}\right )-a+b x^2\right )}{3 b \sqrt [4]{a-b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (-b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} c^{2} x^{2}}{b x^{2} - a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.31, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x \right )^{\frac {5}{2}}}{\left (-b \,x^{2}+a \right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x\right )}^{5/2}}{{\left (a-b\,x^2\right )}^{1/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(5/2)/(a - b*x^2)^(1/4),x)

[Out]

int((c*x)^(5/2)/(a - b*x^2)^(1/4), x)

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sympy [C]  time = 6.78, size = 46, normalized size = 0.36 \[ \frac {c^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{2 \sqrt [4]{a} \Gamma \left (\frac {11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**(5/2)*x**(7/2)*gamma(7/4)*hyper((1/4, 7/4), (11/4,), b*x**2*exp_polar(2*I*pi)/a)/(2*a**(1/4)*gamma(11/4))

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