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

Optimal. Leaf size=93 \[ \frac {4 \sqrt {b} \sqrt {c x} \sqrt [4]{\frac {a}{b x^2}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{a^{3/2} c^2 \sqrt [4]{a+b x^2}}-\frac {2}{a c \sqrt {c x} \sqrt [4]{a+b x^2}} \]

[Out]

-2/a/c/(b*x^2+a)^(1/4)/(c*x)^(1/2)+4*(1+a/b/x^2)^(1/4)*(cos(1/2*arccot(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*ar
ccot(x*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arccot(x*b^(1/2)/a^(1/2))),2^(1/2))*b^(1/2)*(c*x)^(1/2)/a^(3/2)/c^2
/(b*x^2+a)^(1/4)

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Rubi [A]  time = 0.03, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {286, 284, 335, 196} \[ \frac {4 \sqrt {b} \sqrt {c x} \sqrt [4]{\frac {a}{b x^2}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{a^{3/2} c^2 \sqrt [4]{a+b x^2}}-\frac {2}{a c \sqrt {c x} \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 286

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

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

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Mathematica [C]  time = 0.01, size = 57, normalized size = 0.61 \[ -\frac {2 x \sqrt [4]{\frac {b x^2}{a}+1} \, _2F_1\left (-\frac {1}{4},\frac {5}{4};\frac {3}{4};-\frac {b x^2}{a}\right )}{a (c x)^{3/2} \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x}}{b^{2} c^{2} x^{6} + 2 \, a b c^{2} x^{4} + a^{2} c^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.33, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (c x \right )^{\frac {3}{2}} \left (b \,x^{2}+a \right )^{\frac {5}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 6.01, size = 48, normalized size = 0.52 \[ \frac {\Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} c^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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