3.10.67 \(\int \frac {1}{(c x)^{5/2} (a+b x^2)^{3/4}} \, dx\) [967]

Optimal. Leaf size=97 \[ -\frac {2 \sqrt [4]{a+b x^2}}{3 a c (c x)^{3/2}}+\frac {4 b^{3/2} \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 a^{3/2} c^4 \left (a+b x^2\right )^{3/4}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {331, 335, 243, 342, 281, 237} \begin {gather*} \frac {4 b^{3/2} (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 )}{3 a^{3/2} c^4 \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt [4]{a+b x^2}}{3 a c (c x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((c*x)^(5/2)*(a + b*x^2)^(3/4)),x]

[Out]

(-2*(a + b*x^2)^(1/4))/(3*a*c*(c*x)^(3/2)) + (4*b^(3/2)*(1 + a/(b*x^2))^(3/4)*(c*x)^(3/2)*EllipticF[ArcCot[(Sq
rt[b]*x)/Sqrt[a]]/2, 2])/(3*a^(3/2)*c^4*(a + b*x^2)^(3/4))

Rule 237

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

Rule 243

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 281

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 331

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[((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 342

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.01, size = 56, normalized size = 0.58 \begin {gather*} -\frac {2 x \left (1+\frac {b x^2}{a}\right )^{3/4} \, _2F_1\left (-\frac {3}{4},\frac {3}{4};\frac {1}{4};-\frac {b x^2}{a}\right )}{3 (c x)^{5/2} \left (a+b x^2\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x)^(5/2)*(a + b*x^2)^(3/4)),x]

[Out]

(-2*x*(1 + (b*x^2)/a)^(3/4)*Hypergeometric2F1[-3/4, 3/4, 1/4, -((b*x^2)/a)])/(3*(c*x)^(5/2)*(a + b*x^2)^(3/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (c x \right )^{\frac {5}{2}} \left (b \,x^{2}+a \right )^{\frac {3}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(5/2)/(b*x^2+a)^(3/4),x)

[Out]

int(1/(c*x)^(5/2)/(b*x^2+a)^(3/4),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(5/2)/(b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(3/4)*(c*x)^(5/2)), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(5/2)/(b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(1/4)*sqrt(c*x)/(b*c^3*x^5 + a*c^3*x^3), x)

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Sympy [C] Result contains complex when optimal does not.
time = 4.83, size = 48, normalized size = 0.49 \begin {gather*} \frac {\Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} c^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)**(5/2)/(b*x**2+a)**(3/4),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(5/2)/(b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(3/4)*(c*x)^(5/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (c\,x\right )}^{5/2}\,{\left (b\,x^2+a\right )}^{3/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c*x)^(5/2)*(a + b*x^2)^(3/4)),x)

[Out]

int(1/((c*x)^(5/2)*(a + b*x^2)^(3/4)), x)

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