3.968 \(\int \frac {1}{(c x)^{9/2} (a+b x^2)^{3/4}} \, dx\)

Optimal. Leaf size=126 \[ -\frac {8 b^{5/2} (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{7 a^{5/2} c^6 \left (a+b x^2\right )^{3/4}}+\frac {4 b \sqrt [4]{a+b x^2}}{7 a^2 c^3 (c x)^{3/2}}-\frac {2 \sqrt [4]{a+b x^2}}{7 a c (c x)^{7/2}} \]

[Out]

-2/7*(b*x^2+a)^(1/4)/a/c/(c*x)^(7/2)+4/7*b*(b*x^2+a)^(1/4)/a^2/c^3/(c*x)^(3/2)-8/7*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*a
rccot(x*b^(1/2)/a^(1/2))),2^(1/2))/a^(5/2)/c^6/(b*x^2+a)^(3/4)

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Rubi [A]  time = 0.09, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {325, 329, 237, 335, 275, 231} \[ -\frac {8 b^{5/2} (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{7 a^{5/2} c^6 \left (a+b x^2\right )^{3/4}}+\frac {4 b \sqrt [4]{a+b x^2}}{7 a^2 c^3 (c x)^{3/2}}-\frac {2 \sqrt [4]{a+b x^2}}{7 a c (c x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*x^2)^(1/4))/(7*a*c*(c*x)^(7/2)) + (4*b*(a + b*x^2)^(1/4))/(7*a^2*c^3*(c*x)^(3/2)) - (8*b^(5/2)*(1 +
a/(b*x^2))^(3/4)*(c*x)^(3/2)*EllipticF[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(7*a^(5/2)*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 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 {1}{(c x)^{9/2} \left (a+b x^2\right )^{3/4}} \, dx &=-\frac {2 \sqrt [4]{a+b x^2}}{7 a c (c x)^{7/2}}-\frac {(6 b) \int \frac {1}{(c x)^{5/2} \left (a+b x^2\right )^{3/4}} \, dx}{7 a c^2}\\ &=-\frac {2 \sqrt [4]{a+b x^2}}{7 a c (c x)^{7/2}}+\frac {4 b \sqrt [4]{a+b x^2}}{7 a^2 c^3 (c x)^{3/2}}+\frac {\left (4 b^2\right ) \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{3/4}} \, dx}{7 a^2 c^4}\\ &=-\frac {2 \sqrt [4]{a+b x^2}}{7 a c (c x)^{7/2}}+\frac {4 b \sqrt [4]{a+b x^2}}{7 a^2 c^3 (c x)^{3/2}}+\frac {\left (8 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 )}{7 a^2 c^5}\\ &=-\frac {2 \sqrt [4]{a+b x^2}}{7 a c (c x)^{7/2}}+\frac {4 b \sqrt [4]{a+b x^2}}{7 a^2 c^3 (c x)^{3/2}}+\frac {\left (8 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 )}{7 a^2 c^5 \left (a+b x^2\right )^{3/4}}\\ &=-\frac {2 \sqrt [4]{a+b x^2}}{7 a c (c x)^{7/2}}+\frac {4 b \sqrt [4]{a+b x^2}}{7 a^2 c^3 (c x)^{3/2}}-\frac {\left (8 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 )}{7 a^2 c^5 \left (a+b x^2\right )^{3/4}}\\ &=-\frac {2 \sqrt [4]{a+b x^2}}{7 a c (c x)^{7/2}}+\frac {4 b \sqrt [4]{a+b x^2}}{7 a^2 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 x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{c x}\right )}{7 a^2 c^5 \left (a+b x^2\right )^{3/4}}\\ &=-\frac {2 \sqrt [4]{a+b x^2}}{7 a c (c x)^{7/2}}+\frac {4 b \sqrt [4]{a+b x^2}}{7 a^2 c^3 (c x)^{3/2}}-\frac {8 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 )}{7 a^{5/2} 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 = 0.44 \[ -\frac {2 x \left (\frac {b x^2}{a}+1\right )^{3/4} \, _2F_1\left (-\frac {7}{4},\frac {3}{4};-\frac {3}{4};-\frac {b x^2}{a}\right )}{7 (c x)^{9/2} \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}} \sqrt {c x}}{b c^{5} x^{7} + a c^{5} x^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(9/2)/(b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(1/4)*sqrt(c*x)/(b*c^5*x^7 + a*c^5*x^5), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \left (c x\right )^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(9/2)/(b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(3/4)*(c*x)^(9/2)), x)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (c x \right )^{\frac {9}{2}} \left (b \,x^{2}+a \right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(9/2)/(b*x^2+a)^(3/4),x)

[Out]

int(1/(c*x)^(9/2)/(b*x^2+a)^(3/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \left (c x\right )^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(9/2)/(b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(3/4)*(c*x)^(9/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (c\,x\right )}^{9/2}\,{\left (b\,x^2+a\right )}^{3/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c*x)^(9/2)*(a + b*x^2)^(3/4)),x)

[Out]

int(1/((c*x)^(9/2)*(a + b*x^2)^(3/4)), x)

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sympy [C]  time = 66.19, size = 34, normalized size = 0.27 \[ - \frac {{{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{5 b^{\frac {3}{4}} c^{\frac {9}{2}} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)**(9/2)/(b*x**2+a)**(3/4),x)

[Out]

-hyper((3/4, 5/2), (7/2,), a*exp_polar(I*pi)/(b*x**2))/(5*b**(3/4)*c**(9/2)*x**5)

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