3.11.44 \(\int \frac {2 b+a x^2}{x (b^2+a^2 x^2)^{3/4}} \, dx\)

Optimal. Leaf size=79 \[ \frac {2 \sqrt [4]{a^2 x^2+b^2}}{a}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{a^2 x^2+b^2}}{\sqrt {b}}\right )}{\sqrt {b}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{a^2 x^2+b^2}}{\sqrt {b}}\right )}{\sqrt {b}} \]

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Rubi [A]  time = 0.07, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {446, 80, 63, 212, 206, 203} \begin {gather*} \frac {2 \sqrt [4]{a^2 x^2+b^2}}{a}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{a^2 x^2+b^2}}{\sqrt {b}}\right )}{\sqrt {b}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{a^2 x^2+b^2}}{\sqrt {b}}\right )}{\sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b^2 + a^2*x^2)^(1/4))/a - (2*ArcTan[(b^2 + a^2*x^2)^(1/4)/Sqrt[b]])/Sqrt[b] - (2*ArcTanh[(b^2 + a^2*x^2)^(
1/4)/Sqrt[b]])/Sqrt[b]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 203

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {2 b+a x^2}{x \left (b^2+a^2 x^2\right )^{3/4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {2 b+a x}{x \left (b^2+a^2 x\right )^{3/4}} \, dx,x,x^2\right )\\ &=\frac {2 \sqrt [4]{b^2+a^2 x^2}}{a}+b \operatorname {Subst}\left (\int \frac {1}{x \left (b^2+a^2 x\right )^{3/4}} \, dx,x,x^2\right )\\ &=\frac {2 \sqrt [4]{b^2+a^2 x^2}}{a}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {1}{-\frac {b^2}{a^2}+\frac {x^4}{a^2}} \, dx,x,\sqrt [4]{b^2+a^2 x^2}\right )}{a^2}\\ &=\frac {2 \sqrt [4]{b^2+a^2 x^2}}{a}-2 \operatorname {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt [4]{b^2+a^2 x^2}\right )-2 \operatorname {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt [4]{b^2+a^2 x^2}\right )\\ &=\frac {2 \sqrt [4]{b^2+a^2 x^2}}{a}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{b^2+a^2 x^2}}{\sqrt {b}}\right )}{\sqrt {b}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{b^2+a^2 x^2}}{\sqrt {b}}\right )}{\sqrt {b}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 79, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt [4]{a^2 x^2+b^2}}{a}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{a^2 x^2+b^2}}{\sqrt {b}}\right )}{\sqrt {b}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{a^2 x^2+b^2}}{\sqrt {b}}\right )}{\sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b^2 + a^2*x^2)^(1/4))/a - (2*ArcTan[(b^2 + a^2*x^2)^(1/4)/Sqrt[b]])/Sqrt[b] - (2*ArcTanh[(b^2 + a^2*x^2)^(
1/4)/Sqrt[b]])/Sqrt[b]

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IntegrateAlgebraic [A]  time = 0.07, size = 79, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt [4]{b^2+a^2 x^2}}{a}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{b^2+a^2 x^2}}{\sqrt {b}}\right )}{\sqrt {b}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{b^2+a^2 x^2}}{\sqrt {b}}\right )}{\sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(2*b + a*x^2)/(x*(b^2 + a^2*x^2)^(3/4)),x]

[Out]

(2*(b^2 + a^2*x^2)^(1/4))/a - (2*ArcTan[(b^2 + a^2*x^2)^(1/4)/Sqrt[b]])/Sqrt[b] - (2*ArcTanh[(b^2 + a^2*x^2)^(
1/4)/Sqrt[b]])/Sqrt[b]

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fricas [A]  time = 0.53, size = 264, normalized size = 3.34 \begin {gather*} \left [-\frac {2 \, a \sqrt {b} \arctan \left (\frac {{\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{\sqrt {b}}\right ) - a \sqrt {b} \log \left (\frac {a^{2} x^{2} + 2 \, b^{2} - 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}} b^{\frac {3}{2}} + 2 \, \sqrt {a^{2} x^{2} + b^{2}} b - 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {3}{4}} \sqrt {b}}{x^{2}}\right ) - 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}} b}{a b}, \frac {2 \, a \sqrt {-b} \arctan \left (\frac {{\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}} \sqrt {-b}}{b}\right ) - a \sqrt {-b} \log \left (\frac {a^{2} x^{2} + 2 \, b^{2} - 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}} \sqrt {-b} b - 2 \, \sqrt {a^{2} x^{2} + b^{2}} b + 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {3}{4}} \sqrt {-b}}{x^{2}}\right ) + 2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}} b}{a b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(2*a*sqrt(b)*arctan((a^2*x^2 + b^2)^(1/4)/sqrt(b)) - a*sqrt(b)*log((a^2*x^2 + 2*b^2 - 2*(a^2*x^2 + b^2)^(1/4
)*b^(3/2) + 2*sqrt(a^2*x^2 + b^2)*b - 2*(a^2*x^2 + b^2)^(3/4)*sqrt(b))/x^2) - 2*(a^2*x^2 + b^2)^(1/4)*b)/(a*b)
, (2*a*sqrt(-b)*arctan((a^2*x^2 + b^2)^(1/4)*sqrt(-b)/b) - a*sqrt(-b)*log((a^2*x^2 + 2*b^2 - 2*(a^2*x^2 + b^2)
^(1/4)*sqrt(-b)*b - 2*sqrt(a^2*x^2 + b^2)*b + 2*(a^2*x^2 + b^2)^(3/4)*sqrt(-b))/x^2) + 2*(a^2*x^2 + b^2)^(1/4)
*b)/(a*b)]

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giac [A]  time = 0.35, size = 69, normalized size = 0.87 \begin {gather*} \frac {2 \, \arctan \left (\frac {{\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{\sqrt {-b}}\right )}{\sqrt {-b}} - \frac {2 \, \arctan \left (\frac {{\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{\sqrt {b}}\right )}{\sqrt {b}} + \frac {2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan((a^2*x^2 + b^2)^(1/4)/sqrt(-b))/sqrt(-b) - 2*arctan((a^2*x^2 + b^2)^(1/4)/sqrt(b))/sqrt(b) + 2*(a^2*x
^2 + b^2)^(1/4)/a

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {a \,x^{2}+2 b}{x \left (a^{2} x^{2}+b^{2}\right )^{\frac {3}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 92, normalized size = 1.16 \begin {gather*} -b {\left (\frac {2 \, \arctan \left (\frac {{\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{\sqrt {b}}\right )}{b^{\frac {3}{2}}} - \frac {\log \left (-\frac {\sqrt {b} - {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{\sqrt {b} + {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}\right )}{b^{\frac {3}{2}}}\right )} + \frac {2 \, {\left (a^{2} x^{2} + b^{2}\right )}^{\frac {1}{4}}}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*(2*arctan((a^2*x^2 + b^2)^(1/4)/sqrt(b))/b^(3/2) - log(-(sqrt(b) - (a^2*x^2 + b^2)^(1/4))/(sqrt(b) + (a^2*x
^2 + b^2)^(1/4)))/b^(3/2)) + 2*(a^2*x^2 + b^2)^(1/4)/a

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mupad [B]  time = 1.14, size = 65, normalized size = 0.82 \begin {gather*} \frac {2\,{\left (a^2\,x^2+b^2\right )}^{1/4}}{a}-\frac {2\,\mathrm {atanh}\left (\frac {{\left (a^2\,x^2+b^2\right )}^{1/4}}{\sqrt {b}}\right )}{\sqrt {b}}-\frac {2\,\mathrm {atan}\left (\frac {{\left (a^2\,x^2+b^2\right )}^{1/4}}{\sqrt {b}}\right )}{\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(b^2 + a^2*x^2)^(1/4))/a - (2*atanh((b^2 + a^2*x^2)^(1/4)/b^(1/2)))/b^(1/2) - (2*atan((b^2 + a^2*x^2)^(1/4)
/b^(1/2)))/b^(1/2)

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sympy [A]  time = 7.86, size = 76, normalized size = 0.96 \begin {gather*} a \left (\begin {cases} \frac {x^{2}}{2 \left (b^{2}\right )^{\frac {3}{4}}} & \text {for}\: a^{2} = 0 \\\frac {2 \sqrt [4]{a^{2} x^{2} + b^{2}}}{a^{2}} & \text {otherwise} \end {cases}\right ) - \frac {b \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b^{2} e^{i \pi }}{a^{2} x^{2}}} \right )}}{a^{\frac {3}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*Piecewise((x**2/(2*(b**2)**(3/4)), Eq(a**2, 0)), (2*(a**2*x**2 + b**2)**(1/4)/a**2, True)) - b*gamma(3/4)*hy
per((3/4, 3/4), (7/4,), b**2*exp_polar(I*pi)/(a**2*x**2))/(a**(3/2)*x**(3/2)*gamma(7/4))

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