Integrand size = 15, antiderivative size = 75 \[ \int \frac {\left (b+a x^2\right )^{3/4}}{x^3} \, dx=-\frac {\left (b+a x^2\right )^{3/4}}{2 x^2}+\frac {3 a \arctan \left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )}{4 \sqrt [4]{b}}-\frac {3 a \text {arctanh}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )}{4 \sqrt [4]{b}} \]
-1/2*(a*x^2+b)^(3/4)/x^2+3/4*a*arctan((a*x^2+b)^(1/4)/b^(1/4))/b^(1/4)-3/4 *a*arctanh((a*x^2+b)^(1/4)/b^(1/4))/b^(1/4)
Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int \frac {\left (b+a x^2\right )^{3/4}}{x^3} \, dx=-\frac {\left (b+a x^2\right )^{3/4}}{2 x^2}+\frac {3 a \arctan \left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )}{4 \sqrt [4]{b}}-\frac {3 a \text {arctanh}\left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right )}{4 \sqrt [4]{b}} \]
-1/2*(b + a*x^2)^(3/4)/x^2 + (3*a*ArcTan[(b + a*x^2)^(1/4)/b^(1/4)])/(4*b^ (1/4)) - (3*a*ArcTanh[(b + a*x^2)^(1/4)/b^(1/4)])/(4*b^(1/4))
Time = 0.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {243, 51, 73, 25, 27, 827, 216, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (a x^2+b\right )^{3/4}}{x^3} \, dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {\left (a x^2+b\right )^{3/4}}{x^4}dx^2\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{4} a \int \frac {1}{x^2 \sqrt [4]{a x^2+b}}dx^2-\frac {\left (a x^2+b\right )^{3/4}}{x^2}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (3 \int -\frac {a x^4}{b-x^8}d\sqrt [4]{a x^2+b}-\frac {\left (a x^2+b\right )^{3/4}}{x^2}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (-3 \int \frac {a x^4}{b-x^8}d\sqrt [4]{a x^2+b}-\frac {\left (a x^2+b\right )^{3/4}}{x^2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (-3 a \int \frac {x^4}{b-x^8}d\sqrt [4]{a x^2+b}-\frac {\left (a x^2+b\right )^{3/4}}{x^2}\right )\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {1}{2} \left (-3 a \left (\frac {1}{2} \int \frac {1}{\sqrt {b}-x^4}d\sqrt [4]{a x^2+b}-\frac {1}{2} \int \frac {1}{x^4+\sqrt {b}}d\sqrt [4]{a x^2+b}\right )-\frac {\left (a x^2+b\right )^{3/4}}{x^2}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \left (-3 a \left (\frac {1}{2} \int \frac {1}{\sqrt {b}-x^4}d\sqrt [4]{a x^2+b}-\frac {\arctan \left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )}{2 \sqrt [4]{b}}\right )-\frac {\left (a x^2+b\right )^{3/4}}{x^2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (-3 a \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )}{2 \sqrt [4]{b}}-\frac {\arctan \left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right )}{2 \sqrt [4]{b}}\right )-\frac {\left (a x^2+b\right )^{3/4}}{x^2}\right )\) |
(-((b + a*x^2)^(3/4)/x^2) - 3*a*(-1/2*ArcTan[(b + a*x^2)^(1/4)/b^(1/4)]/b^ (1/4) + ArcTanh[(b + a*x^2)^(1/4)/b^(1/4)]/(2*b^(1/4))))/2
3.10.83.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Time = 1.18 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.09
method | result | size |
pseudoelliptic | \(\frac {6 \arctan \left (\frac {\left (a \,x^{2}+b \right )^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) a \,x^{2}-3 \ln \left (\frac {\left (a \,x^{2}+b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{\left (a \,x^{2}+b \right )^{\frac {1}{4}}-b^{\frac {1}{4}}}\right ) a \,x^{2}-4 \left (a \,x^{2}+b \right )^{\frac {3}{4}} b^{\frac {1}{4}}}{8 x^{2} b^{\frac {1}{4}}}\) | \(82\) |
1/8*(6*arctan((a*x^2+b)^(1/4)/b^(1/4))*a*x^2-3*ln(((a*x^2+b)^(1/4)+b^(1/4) )/((a*x^2+b)^(1/4)-b^(1/4)))*a*x^2-4*(a*x^2+b)^(3/4)*b^(1/4))/x^2/b^(1/4)
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.47 \[ \int \frac {\left (b+a x^2\right )^{3/4}}{x^3} \, dx=-\frac {3 \, \left (\frac {a^{4}}{b}\right )^{\frac {1}{4}} x^{2} \log \left (27 \, {\left (a x^{2} + b\right )}^{\frac {1}{4}} a^{3} + 27 \, \left (\frac {a^{4}}{b}\right )^{\frac {3}{4}} b\right ) - 3 i \, \left (\frac {a^{4}}{b}\right )^{\frac {1}{4}} x^{2} \log \left (27 \, {\left (a x^{2} + b\right )}^{\frac {1}{4}} a^{3} + 27 i \, \left (\frac {a^{4}}{b}\right )^{\frac {3}{4}} b\right ) + 3 i \, \left (\frac {a^{4}}{b}\right )^{\frac {1}{4}} x^{2} \log \left (27 \, {\left (a x^{2} + b\right )}^{\frac {1}{4}} a^{3} - 27 i \, \left (\frac {a^{4}}{b}\right )^{\frac {3}{4}} b\right ) - 3 \, \left (\frac {a^{4}}{b}\right )^{\frac {1}{4}} x^{2} \log \left (27 \, {\left (a x^{2} + b\right )}^{\frac {1}{4}} a^{3} - 27 \, \left (\frac {a^{4}}{b}\right )^{\frac {3}{4}} b\right ) + 4 \, {\left (a x^{2} + b\right )}^{\frac {3}{4}}}{8 \, x^{2}} \]
-1/8*(3*(a^4/b)^(1/4)*x^2*log(27*(a*x^2 + b)^(1/4)*a^3 + 27*(a^4/b)^(3/4)* b) - 3*I*(a^4/b)^(1/4)*x^2*log(27*(a*x^2 + b)^(1/4)*a^3 + 27*I*(a^4/b)^(3/ 4)*b) + 3*I*(a^4/b)^(1/4)*x^2*log(27*(a*x^2 + b)^(1/4)*a^3 - 27*I*(a^4/b)^ (3/4)*b) - 3*(a^4/b)^(1/4)*x^2*log(27*(a*x^2 + b)^(1/4)*a^3 - 27*(a^4/b)^( 3/4)*b) + 4*(a*x^2 + b)^(3/4))/x^2
Result contains complex when optimal does not.
Time = 0.84 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.56 \[ \int \frac {\left (b+a x^2\right )^{3/4}}{x^3} \, dx=- \frac {a^{\frac {3}{4}} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{2}}} \right )}}{2 \sqrt {x} \Gamma \left (\frac {5}{4}\right )} \]
-a**(3/4)*gamma(1/4)*hyper((-3/4, 1/4), (5/4,), b*exp_polar(I*pi)/(a*x**2) )/(2*sqrt(x)*gamma(5/4))
Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.99 \[ \int \frac {\left (b+a x^2\right )^{3/4}}{x^3} \, dx=\frac {3}{8} \, a {\left (\frac {2 \, \arctan \left (\frac {{\left (a x^{2} + b\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (\frac {{\left (a x^{2} + b\right )}^{\frac {1}{4}} - b^{\frac {1}{4}}}{{\left (a x^{2} + b\right )}^{\frac {1}{4}} + b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}}\right )} - \frac {{\left (a x^{2} + b\right )}^{\frac {3}{4}}}{2 \, x^{2}} \]
3/8*a*(2*arctan((a*x^2 + b)^(1/4)/b^(1/4))/b^(1/4) + log(((a*x^2 + b)^(1/4 ) - b^(1/4))/((a*x^2 + b)^(1/4) + b^(1/4)))/b^(1/4)) - 1/2*(a*x^2 + b)^(3/ 4)/x^2
Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (55) = 110\).
Time = 0.28 (sec) , antiderivative size = 209, normalized size of antiderivative = 2.79 \[ \int \frac {\left (b+a x^2\right )^{3/4}}{x^3} \, dx=\frac {\frac {6 \, \sqrt {2} a^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} + 2 \, {\left (a x^{2} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right )}{\left (-b\right )^{\frac {1}{4}}} + \frac {6 \, \sqrt {2} a^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} - 2 \, {\left (a x^{2} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right )}{\left (-b\right )^{\frac {1}{4}}} + \frac {3 \, \sqrt {2} a^{2} \left (-b\right )^{\frac {3}{4}} \log \left (\sqrt {2} {\left (a x^{2} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{2} + b} + \sqrt {-b}\right )}{b} + \frac {3 \, \sqrt {2} a^{2} \log \left (-\sqrt {2} {\left (a x^{2} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{2} + b} + \sqrt {-b}\right )}{\left (-b\right )^{\frac {1}{4}}} - \frac {8 \, {\left (a x^{2} + b\right )}^{\frac {3}{4}} a}{x^{2}}}{16 \, a} \]
1/16*(6*sqrt(2)*a^2*arctan(1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) + 2*(a*x^2 + b) ^(1/4))/(-b)^(1/4))/(-b)^(1/4) + 6*sqrt(2)*a^2*arctan(-1/2*sqrt(2)*(sqrt(2 )*(-b)^(1/4) - 2*(a*x^2 + b)^(1/4))/(-b)^(1/4))/(-b)^(1/4) + 3*sqrt(2)*a^2 *(-b)^(3/4)*log(sqrt(2)*(a*x^2 + b)^(1/4)*(-b)^(1/4) + sqrt(a*x^2 + b) + s qrt(-b))/b + 3*sqrt(2)*a^2*log(-sqrt(2)*(a*x^2 + b)^(1/4)*(-b)^(1/4) + sqr t(a*x^2 + b) + sqrt(-b))/(-b)^(1/4) - 8*(a*x^2 + b)^(3/4)*a/x^2)/a
Time = 5.91 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.73 \[ \int \frac {\left (b+a x^2\right )^{3/4}}{x^3} \, dx=\frac {3\,a\,\mathrm {atan}\left (\frac {{\left (a\,x^2+b\right )}^{1/4}}{b^{1/4}}\right )}{4\,b^{1/4}}-\frac {{\left (a\,x^2+b\right )}^{3/4}}{2\,x^2}-\frac {3\,a\,\mathrm {atanh}\left (\frac {{\left (a\,x^2+b\right )}^{1/4}}{b^{1/4}}\right )}{4\,b^{1/4}} \]