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

Optimal. Leaf size=78 \[ \frac {3 a \tan ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )}{4 b^{7/4}}+\frac {3 a \tanh ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )}{4 b^{7/4}}-\frac {\sqrt [4]{a x^2+b}}{2 b x^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {266, 51, 63, 212, 206, 203} \begin {gather*} \frac {3 a \tan ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )}{4 b^{7/4}}+\frac {3 a \tanh ^{-1}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )}{4 b^{7/4}}-\frac {\sqrt [4]{a x^2+b}}{2 b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(b + a*x^2)^(3/4)),x]

[Out]

-1/2*(b + a*x^2)^(1/4)/(b*x^2) + (3*a*ArcTan[(b + a*x^2)^(1/4)/b^(1/4)])/(4*b^(7/4)) + (3*a*ArcTanh[(b + a*x^2
)^(1/4)/b^(1/4)])/(4*b^(7/4))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 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 266

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

Rubi steps

\begin {align*} \int \frac {1}{x^3 \left (b+a x^2\right )^{3/4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^2 (b+a x)^{3/4}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt [4]{b+a x^2}}{2 b x^2}-\frac {(3 a) \operatorname {Subst}\left (\int \frac {1}{x (b+a x)^{3/4}} \, dx,x,x^2\right )}{8 b}\\ &=-\frac {\sqrt [4]{b+a x^2}}{2 b x^2}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{-\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{2 b}\\ &=-\frac {\sqrt [4]{b+a x^2}}{2 b x^2}+\frac {(3 a) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b}-x^2} \, dx,x,\sqrt [4]{b+a x^2}\right )}{4 b^{3/2}}+\frac {(3 a) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b}+x^2} \, dx,x,\sqrt [4]{b+a x^2}\right )}{4 b^{3/2}}\\ &=-\frac {\sqrt [4]{b+a x^2}}{2 b x^2}+\frac {3 a \tan ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )}{4 b^{7/4}}+\frac {3 a \tanh ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )}{4 b^{7/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 35, normalized size = 0.45 \begin {gather*} \frac {2 a \sqrt [4]{a x^2+b} \, _2F_1\left (\frac {1}{4},2;\frac {5}{4};\frac {a x^2}{b}+1\right )}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.16, size = 78, normalized size = 1.00 \begin {gather*} -\frac {\sqrt [4]{b+a x^2}}{2 b x^2}+\frac {3 a \tan ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )}{4 b^{7/4}}+\frac {3 a \tanh ^{-1}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )}{4 b^{7/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(b + a*x^2)^(1/4)/(b*x^2) + (3*a*ArcTan[(b + a*x^2)^(1/4)/b^(1/4)])/(4*b^(7/4)) + (3*a*ArcTanh[(b + a*x^2
)^(1/4)/b^(1/4)])/(4*b^(7/4))

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fricas [B]  time = 0.53, size = 194, normalized size = 2.49 \begin {gather*} -\frac {12 \, b x^{2} \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (a x^{2} + b\right )}^{\frac {1}{4}} a b^{5} \left (\frac {a^{4}}{b^{7}}\right )^{\frac {3}{4}} - \sqrt {b^{4} \sqrt {\frac {a^{4}}{b^{7}}} + \sqrt {a x^{2} + b} a^{2}} b^{5} \left (\frac {a^{4}}{b^{7}}\right )^{\frac {3}{4}}}{a^{4}}\right ) - 3 \, b x^{2} \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \log \left (3 \, b^{2} \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} + 3 \, {\left (a x^{2} + b\right )}^{\frac {1}{4}} a\right ) + 3 \, b x^{2} \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \log \left (-3 \, b^{2} \left (\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} + 3 \, {\left (a x^{2} + b\right )}^{\frac {1}{4}} a\right ) + 4 \, {\left (a x^{2} + b\right )}^{\frac {1}{4}}}{8 \, b x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(12*b*x^2*(a^4/b^7)^(1/4)*arctan(-((a*x^2 + b)^(1/4)*a*b^5*(a^4/b^7)^(3/4) - sqrt(b^4*sqrt(a^4/b^7) + sqr
t(a*x^2 + b)*a^2)*b^5*(a^4/b^7)^(3/4))/a^4) - 3*b*x^2*(a^4/b^7)^(1/4)*log(3*b^2*(a^4/b^7)^(1/4) + 3*(a*x^2 + b
)^(1/4)*a) + 3*b*x^2*(a^4/b^7)^(1/4)*log(-3*b^2*(a^4/b^7)^(1/4) + 3*(a*x^2 + b)^(1/4)*a) + 4*(a*x^2 + b)^(1/4)
)/(b*x^2)

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giac [B]  time = 0.36, size = 221, normalized size = 2.83 \begin {gather*} \frac {\frac {6 \, \sqrt {2} a^{2} \left (-b\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} + 2 \, {\left (a x^{2} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right )}{b^{2}} + \frac {6 \, \sqrt {2} a^{2} \left (-b\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} - 2 \, {\left (a x^{2} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right )}{b^{2}} + \frac {3 \, \sqrt {2} a^{2} \left (-b\right )^{\frac {1}{4}} \log \left (\sqrt {2} {\left (a x^{2} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{2} + b} + \sqrt {-b}\right )}{b^{2}} + \frac {3 \, \sqrt {2} a^{2} \log \left (-\sqrt {2} {\left (a x^{2} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{2} + b} + \sqrt {-b}\right )}{\left (-b\right )^{\frac {3}{4}} b} - \frac {8 \, {\left (a x^{2} + b\right )}^{\frac {1}{4}} a}{b x^{2}}}{16 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(6*sqrt(2)*a^2*(-b)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) + 2*(a*x^2 + b)^(1/4))/(-b)^(1/4))/b^2 +
6*sqrt(2)*a^2*(-b)^(1/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) - 2*(a*x^2 + b)^(1/4))/(-b)^(1/4))/b^2 + 3*s
qrt(2)*a^2*(-b)^(1/4)*log(sqrt(2)*(a*x^2 + b)^(1/4)*(-b)^(1/4) + sqrt(a*x^2 + b) + sqrt(-b))/b^2 + 3*sqrt(2)*a
^2*log(-sqrt(2)*(a*x^2 + b)^(1/4)*(-b)^(1/4) + sqrt(a*x^2 + b) + sqrt(-b))/((-b)^(3/4)*b) - 8*(a*x^2 + b)^(1/4
)*a/(b*x^2))/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 94, normalized size = 1.21 \begin {gather*} -\frac {{\left (a x^{2} + b\right )}^{\frac {1}{4}} a}{2 \, {\left ({\left (a x^{2} + b\right )} b - b^{2}\right )}} + \frac {3 \, {\left (\frac {2 \, a \arctan \left (\frac {{\left (a x^{2} + b\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} - \frac {a \log \left (\frac {{\left (a x^{2} + b\right )}^{\frac {1}{4}} - b^{\frac {1}{4}}}{{\left (a x^{2} + b\right )}^{\frac {1}{4}} + b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}}\right )}}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a*x^2 + b)^(1/4)*a/((a*x^2 + b)*b - b^2) + 3/8*(2*a*arctan((a*x^2 + b)^(1/4)/b^(1/4))/b^(3/4) - a*log(((
a*x^2 + b)^(1/4) - b^(1/4))/((a*x^2 + b)^(1/4) + b^(1/4)))/b^(3/4))/b

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mupad [B]  time = 0.99, size = 58, normalized size = 0.74 \begin {gather*} \frac {3\,a\,\mathrm {atan}\left (\frac {{\left (a\,x^2+b\right )}^{1/4}}{b^{1/4}}\right )}{4\,b^{7/4}}-\frac {{\left (a\,x^2+b\right )}^{1/4}}{2\,b\,x^2}+\frac {3\,a\,\mathrm {atanh}\left (\frac {{\left (a\,x^2+b\right )}^{1/4}}{b^{1/4}}\right )}{4\,b^{7/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(b + a*x^2)^(3/4)),x)

[Out]

(3*a*atan((b + a*x^2)^(1/4)/b^(1/4)))/(4*b^(7/4)) - (b + a*x^2)^(1/4)/(2*b*x^2) + (3*a*atanh((b + a*x^2)^(1/4)
/b^(1/4)))/(4*b^(7/4))

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sympy [C]  time = 1.44, size = 41, normalized size = 0.53 \begin {gather*} - \frac {\Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{2}}} \right )}}{2 a^{\frac {3}{4}} x^{\frac {7}{2}} \Gamma \left (\frac {11}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(7/4)*hyper((3/4, 7/4), (11/4,), b*exp_polar(I*pi)/(a*x**2))/(2*a**(3/4)*x**(7/2)*gamma(11/4))

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