3.10.54 \(\int \frac {1}{(c x)^{11/2} \sqrt [4]{a+b x^2}} \, dx\) [954]

Optimal. Leaf size=157 \[ -\frac {8 b^2}{15 a^2 c^5 \sqrt {c x} \sqrt [4]{a+b x^2}}-\frac {2 \left (a+b x^2\right )^{3/4}}{9 a c (c x)^{9/2}}+\frac {4 b \left (a+b x^2\right )^{3/4}}{15 a^2 c^3 (c x)^{5/2}}+\frac {8 b^{5/2} \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {c x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 a^{5/2} c^6 \sqrt [4]{a+b x^2}} \]

[Out]

-2/9*(b*x^2+a)^(3/4)/a/c/(c*x)^(9/2)+4/15*b*(b*x^2+a)^(3/4)/a^2/c^3/(c*x)^(5/2)-8/15*b^2/a^2/c^5/(b*x^2+a)^(1/
4)/(c*x)^(1/2)+8/15*b^(5/2)*(1+a/b/x^2)^(1/4)*(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)))*EllipticE(sin(1/2*arccot(x*b^(1/2)/a^(1/2))),2^(1/2))*(c*x)^(1/2)/a^(5/2)/c^6/(b*x^2+a)^(1/4)

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Rubi [A]
time = 0.05, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {331, 322, 290, 342, 202} \begin {gather*} \frac {8 b^{5/2} \sqrt {c x} \sqrt [4]{\frac {a}{b x^2}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 a^{5/2} c^6 \sqrt [4]{a+b x^2}}-\frac {8 b^2}{15 a^2 c^5 \sqrt {c x} \sqrt [4]{a+b x^2}}+\frac {4 b \left (a+b x^2\right )^{3/4}}{15 a^2 c^3 (c x)^{5/2}}-\frac {2 \left (a+b x^2\right )^{3/4}}{9 a c (c x)^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((c*x)^(11/2)*(a + b*x^2)^(1/4)),x]

[Out]

(-8*b^2)/(15*a^2*c^5*Sqrt[c*x]*(a + b*x^2)^(1/4)) - (2*(a + b*x^2)^(3/4))/(9*a*c*(c*x)^(9/2)) + (4*b*(a + b*x^
2)^(3/4))/(15*a^2*c^3*(c*x)^(5/2)) + (8*b^(5/2)*(1 + a/(b*x^2))^(1/4)*Sqrt[c*x]*EllipticE[ArcCot[(Sqrt[b]*x)/S
qrt[a]]/2, 2])/(15*a^(5/2)*c^6*(a + b*x^2)^(1/4))

Rule 202

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]))*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 290

Int[Sqrt[(c_.)*(x_)]/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Dist[Sqrt[c*x]*((1 + a/(b*x^2))^(1/4)/(b*(a + b
*x^2)^(1/4))), Int[1/(x^2*(1 + a/(b*x^2))^(5/4)), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a]

Rule 322

Int[1/(((c_.)*(x_))^(3/2)*((a_) + (b_.)*(x_)^2)^(1/4)), x_Symbol] :> Simp[-2/(c*Sqrt[c*x]*(a + b*x^2)^(1/4)),
x] - Dist[b/c^2, Int[Sqrt[c*x]/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a]

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 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)^{11/2} \sqrt [4]{a+b x^2}} \, dx &=-\frac {2 \left (a+b x^2\right )^{3/4}}{9 a c (c x)^{9/2}}-\frac {(2 b) \int \frac {1}{(c x)^{7/2} \sqrt [4]{a+b x^2}} \, dx}{3 a c^2}\\ &=-\frac {2 \left (a+b x^2\right )^{3/4}}{9 a c (c x)^{9/2}}+\frac {4 b \left (a+b x^2\right )^{3/4}}{15 a^2 c^3 (c x)^{5/2}}+\frac {\left (4 b^2\right ) \int \frac {1}{(c x)^{3/2} \sqrt [4]{a+b x^2}} \, dx}{15 a^2 c^4}\\ &=-\frac {8 b^2}{15 a^2 c^5 \sqrt {c x} \sqrt [4]{a+b x^2}}-\frac {2 \left (a+b x^2\right )^{3/4}}{9 a c (c x)^{9/2}}+\frac {4 b \left (a+b x^2\right )^{3/4}}{15 a^2 c^3 (c x)^{5/2}}-\frac {\left (4 b^3\right ) \int \frac {\sqrt {c x}}{\left (a+b x^2\right )^{5/4}} \, dx}{15 a^2 c^6}\\ &=-\frac {8 b^2}{15 a^2 c^5 \sqrt {c x} \sqrt [4]{a+b x^2}}-\frac {2 \left (a+b x^2\right )^{3/4}}{9 a c (c x)^{9/2}}+\frac {4 b \left (a+b x^2\right )^{3/4}}{15 a^2 c^3 (c x)^{5/2}}-\frac {\left (4 b^2 \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {c x}\right ) \int \frac {1}{\left (1+\frac {a}{b x^2}\right )^{5/4} x^2} \, dx}{15 a^2 c^6 \sqrt [4]{a+b x^2}}\\ &=-\frac {8 b^2}{15 a^2 c^5 \sqrt {c x} \sqrt [4]{a+b x^2}}-\frac {2 \left (a+b x^2\right )^{3/4}}{9 a c (c x)^{9/2}}+\frac {4 b \left (a+b x^2\right )^{3/4}}{15 a^2 c^3 (c x)^{5/2}}+\frac {\left (4 b^2 \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {c x}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )}{15 a^2 c^6 \sqrt [4]{a+b x^2}}\\ &=-\frac {8 b^2}{15 a^2 c^5 \sqrt {c x} \sqrt [4]{a+b x^2}}-\frac {2 \left (a+b x^2\right )^{3/4}}{9 a c (c x)^{9/2}}+\frac {4 b \left (a+b x^2\right )^{3/4}}{15 a^2 c^3 (c x)^{5/2}}+\frac {8 b^{5/2} \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {c x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 a^{5/2} c^6 \sqrt [4]{a+b x^2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.02, size = 56, normalized size = 0.36 \begin {gather*} -\frac {2 x \sqrt [4]{1+\frac {b x^2}{a}} \, _2F_1\left (-\frac {9}{4},\frac {1}{4};-\frac {5}{4};-\frac {b x^2}{a}\right )}{9 (c x)^{11/2} \sqrt [4]{a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x)^(11/2)*(a + b*x^2)^(1/4)),x]

[Out]

(-2*x*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[-9/4, 1/4, -5/4, -((b*x^2)/a)])/(9*(c*x)^(11/2)*(a + b*x^2)^(1/4
))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (c x \right )^{\frac {11}{2}} \left (b \,x^{2}+a \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(11/2)/(b*x^2+a)^(1/4),x)

[Out]

int(1/(c*x)^(11/2)/(b*x^2+a)^(1/4),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)^(11/2)/(b*x^2+a)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(1/4)*(c*x)^(11/2)), 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)^(11/2)/(b*x^2+a)^(1/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(3/4)*sqrt(c*x)/(b*c^6*x^8 + a*c^6*x^6), x)

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Sympy [C] Result contains complex when optimal does not.
time = 73.09, size = 34, normalized size = 0.22 \begin {gather*} - \frac {{{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{5 \sqrt [4]{b} c^{\frac {11}{2}} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)**(11/2)/(b*x**2+a)**(1/4),x)

[Out]

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

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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)^(11/2)/(b*x^2+a)^(1/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(1/4)*(c*x)^(11/2)), 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 )}^{11/2}\,{\left (b\,x^2+a\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c*x)^(11/2)*(a + b*x^2)^(1/4)),x)

[Out]

int(1/((c*x)^(11/2)*(a + b*x^2)^(1/4)), x)

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