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

Optimal. Leaf size=126 \[ -\frac{4 b^{3/2} \sqrt{c x} \sqrt [4]{\frac{a}{b x^2}+1} E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{5 a^{3/2} c^4 \sqrt [4]{a+b x^2}}+\frac{4 b}{5 a c^3 \sqrt{c x} \sqrt [4]{a+b x^2}}-\frac{2 \left (a+b x^2\right )^{3/4}}{5 a c (c x)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.0490704, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {325, 316, 284, 335, 196} \[ -\frac{4 b^{3/2} \sqrt{c x} \sqrt [4]{\frac{a}{b x^2}+1} E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{5 a^{3/2} c^4 \sqrt [4]{a+b x^2}}+\frac{4 b}{5 a c^3 \sqrt{c x} \sqrt [4]{a+b x^2}}-\frac{2 \left (a+b x^2\right )^{3/4}}{5 a c (c x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 316

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 284

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 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]

Rule 196

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

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

Mathematica [C]  time = 0.0122788, size = 56, normalized size = 0.44 \[ -\frac{2 x \sqrt [4]{\frac{b x^2}{a}+1} \, _2F_1\left (-\frac{5}{4},\frac{1}{4};-\frac{1}{4};-\frac{b x^2}{a}\right )}{5 (c x)^{7/2} \sqrt [4]{a+b x^2}} \]

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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Maple [F]  time = 0.036, size = 0, normalized size = 0. \begin{align*} \int{ \left ( cx \right ) ^{-{\frac{7}{2}}}{\frac{1}{\sqrt [4]{b{x}^{2}+a}}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{1}{4}} \left (c x\right )^{\frac{7}{2}}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [C]  time = 96.643, size = 34, normalized size = 0.27 \begin{align*} - \frac{{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{3}{2} \\ \frac{5}{2} \end{matrix}\middle |{\frac{a e^{i \pi }}{b x^{2}}} \right )}}{3 \sqrt [4]{b} c^{\frac{7}{2}} x^{3}} \end{align*}

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]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{1}{4}} \left (c x\right )^{\frac{7}{2}}}\,{d x} \end{align*}

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)