3.1.73 \(\int \frac {1}{x (a x+b x^3)^{3/2}} \, dx\) [73]

Optimal. Leaf size=139 \[ \frac {1}{a x \sqrt {a x+b x^3}}-\frac {5 \sqrt {a x+b x^3}}{3 a^2 x^2}-\frac {5 b^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{6 a^{9/4} \sqrt {a x+b x^3}} \]

[Out]

1/a/x/(b*x^3+a*x)^(1/2)-5/3*(b*x^3+a*x)^(1/2)/a^2/x^2-5/6*b^(3/4)*(cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))^2)^(
1/2)/cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*x^(1/2)/a^(1/4))),1/2*2^(1/2))*(a^(
1/2)+x*b^(1/2))*x^(1/2)*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)/a^(9/4)/(b*x^3+a*x)^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2048, 2050, 2036, 335, 226} \begin {gather*} -\frac {5 b^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{6 a^{9/4} \sqrt {a x+b x^3}}-\frac {5 \sqrt {a x+b x^3}}{3 a^2 x^2}+\frac {1}{a x \sqrt {a x+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(a*x + b*x^3)^(3/2)),x]

[Out]

1/(a*x*Sqrt[a*x + b*x^3]) - (5*Sqrt[a*x + b*x^3])/(3*a^2*x^2) - (5*b^(3/4)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[
(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(6*a^(9/4)*Sqrt[a*x
+ b*x^3])

Rule 226

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

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 2036

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2048

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-c^(j - 1))*(c*x)^(m - j
+ 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j)*(p + 1))), x] + Dist[c^j*((m + n*p + n - j + 1)/(a*(n - j)*(p + 1)))
, Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n]
&& (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1]

Rule 2050

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(j - 1)*(c*x)^(m - j +
1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {1}{x \left (a x+b x^3\right )^{3/2}} \, dx &=\frac {1}{a x \sqrt {a x+b x^3}}+\frac {5 \int \frac {1}{x^2 \sqrt {a x+b x^3}} \, dx}{2 a}\\ &=\frac {1}{a x \sqrt {a x+b x^3}}-\frac {5 \sqrt {a x+b x^3}}{3 a^2 x^2}-\frac {(5 b) \int \frac {1}{\sqrt {a x+b x^3}} \, dx}{6 a^2}\\ &=\frac {1}{a x \sqrt {a x+b x^3}}-\frac {5 \sqrt {a x+b x^3}}{3 a^2 x^2}-\frac {\left (5 b \sqrt {x} \sqrt {a+b x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x^2}} \, dx}{6 a^2 \sqrt {a x+b x^3}}\\ &=\frac {1}{a x \sqrt {a x+b x^3}}-\frac {5 \sqrt {a x+b x^3}}{3 a^2 x^2}-\frac {\left (5 b \sqrt {x} \sqrt {a+b x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\sqrt {x}\right )}{3 a^2 \sqrt {a x+b x^3}}\\ &=\frac {1}{a x \sqrt {a x+b x^3}}-\frac {5 \sqrt {a x+b x^3}}{3 a^2 x^2}-\frac {5 b^{3/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{6 a^{9/4} \sqrt {a x+b x^3}}\\ \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 = 0.40 \begin {gather*} -\frac {2 \sqrt {1+\frac {b x^2}{a}} \, _2F_1\left (-\frac {3}{4},\frac {3}{2};\frac {1}{4};-\frac {b x^2}{a}\right )}{3 a x \sqrt {x \left (a+b x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.41, size = 150, normalized size = 1.08

method	result	size
default	\(-\frac {b x}{a^{2} \sqrt {\left (x^{2}+\frac {a}{b}\right ) b x}}-\frac {2 \sqrt {b \,x^{3}+a x}}{3 a^{2} x^{2}}-\frac {5 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{6 a^{2} \sqrt {b \,x^{3}+a x}}\)	\(150\)
elliptic	\(-\frac {b x}{a^{2} \sqrt {\left (x^{2}+\frac {a}{b}\right ) b x}}-\frac {2 \sqrt {b \,x^{3}+a x}}{3 a^{2} x^{2}}-\frac {5 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{6 a^{2} \sqrt {b \,x^{3}+a x}}\)	\(150\)
risch	\(-\frac {2 \left (b \,x^{2}+a \right )}{3 a^{2} x \sqrt {x \left (b \,x^{2}+a \right )}}-\frac {b \left (\frac {\sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b \sqrt {b \,x^{3}+a x}}+3 a \left (\frac {x}{a \sqrt {\left (x^{2}+\frac {a}{b}\right ) b x}}+\frac {\sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{2 a b \sqrt {b \,x^{3}+a x}}\right )\right )}{3 a^{2}}\)	\(276\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(b*x^3+a*x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-b*x/a^2/((x^2+a/b)*b*x)^(1/2)-2/3*(b*x^3+a*x)^(1/2)/a^2/x^2-5/6/a^2*(-a*b)^(1/2)*((x+1/b*(-a*b)^(1/2))*b/(-a*
b)^(1/2))^(1/2)*(-2*(x-1/b*(-a*b)^(1/2))*b/(-a*b)^(1/2))^(1/2)*(-x*b/(-a*b)^(1/2))^(1/2)/(b*x^3+a*x)^(1/2)*Ell
ipticF(((x+1/b*(-a*b)^(1/2))*b/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))

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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/x/(b*x^3+a*x)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.41, size = 68, normalized size = 0.49 \begin {gather*} -\frac {5 \, {\left (b x^{4} + a x^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) + \sqrt {b x^{3} + a x} {\left (5 \, b x^{2} + 2 \, a\right )}}{3 \, {\left (a^{2} b x^{4} + a^{3} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(5*(b*x^4 + a*x^2)*sqrt(b)*weierstrassPInverse(-4*a/b, 0, x) + sqrt(b*x^3 + a*x)*(5*b*x^2 + 2*a))/(a^2*b*
x^4 + a^3*x^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \left (x \left (a + b x^{2}\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x*(x*(a + b*x**2))**(3/2)), x)

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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/x/(b*x^3+a*x)^(3/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a*x + b*x^3)^(3/2)),x)

[Out]

int(1/(x*(a*x + b*x^3)^(3/2)), x)

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