\(\int \frac {b+a x^2}{(b-a x^2) \sqrt [4]{b x^3+a x^5}} \, dx\) [2807]

Optimal result

Integrand size = 33, antiderivative size = 275 \[ \int \frac {b+a x^2}{\left (b-a x^2\right ) \sqrt [4]{b x^3+a x^5}} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [8]{a} \sqrt [8]{b} x}{\sqrt [4]{b x^3+a x^5}}\right )}{\sqrt [4]{2} \sqrt [8]{a} \sqrt [8]{b}}-\frac {\arctan \left (\frac {2^{3/4} \sqrt [8]{a} \sqrt [8]{b} x \sqrt [4]{b x^3+a x^5}}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2-\sqrt {b x^3+a x^5}}\right )}{2^{3/4} \sqrt [8]{a} \sqrt [8]{b}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [8]{a} \sqrt [8]{b} x}{\sqrt [4]{b x^3+a x^5}}\right )}{\sqrt [4]{2} \sqrt [8]{a} \sqrt [8]{b}}+\frac {\text {arctanh}\left (\frac {\frac {\sqrt [8]{a} \sqrt [8]{b} x^2}{\sqrt [4]{2}}+\frac {\sqrt {b x^3+a x^5}}{2^{3/4} \sqrt [8]{a} \sqrt [8]{b}}}{x \sqrt [4]{b x^3+a x^5}}\right )}{2^{3/4} \sqrt [8]{a} \sqrt [8]{b}} \]

[Out]

1/2*arctan(2^(1/4)*a^(1/8)*b^(1/8)*x/(a*x^5+b*x^3)^(1/4))*2^(3/4)/a^(1/8)/b^(1/8)-1/2*arctan(2^(3/4)*a^(1/8)*b
^(1/8)*x*(a*x^5+b*x^3)^(1/4)/(2^(1/2)*a^(1/4)*b^(1/4)*x^2-(a*x^5+b*x^3)^(1/2)))*2^(1/4)/a^(1/8)/b^(1/8)+1/2*ar
ctanh(2^(1/4)*a^(1/8)*b^(1/8)*x/(a*x^5+b*x^3)^(1/4))*2^(3/4)/a^(1/8)/b^(1/8)+1/2*arctanh((1/2*a^(1/8)*b^(1/8)*
x^2*2^(3/4)+1/2*(a*x^5+b*x^3)^(1/2)*2^(1/4)/a^(1/8)/b^(1/8))/x/(a*x^5+b*x^3)^(1/4))*2^(1/4)/a^(1/8)/b^(1/8)

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.25, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {2081, 477, 441, 440} \[ \int \frac {b+a x^2}{\left (b-a x^2\right ) \sqrt [4]{b x^3+a x^5}} \, dx=\frac {4 x \left (a x^2+b\right ) \operatorname {AppellF1}\left (\frac {1}{8},1,-\frac {3}{4},\frac {9}{8},\frac {a x^2}{b},-\frac {a x^2}{b}\right )}{b \left (\frac {a x^2}{b}+1\right )^{3/4} \sqrt [4]{a x^5+b x^3}} \]

[In]

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

[Out]

(4*x*(b + a*x^2)*AppellF1[1/8, 1, -3/4, 9/8, (a*x^2)/b, -((a*x^2)/b)])/(b*(1 + (a*x^2)/b)^(3/4)*(b*x^3 + a*x^5
)^(1/4))

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 441

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^F
racPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 477

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/e^n))^p*(c + d*(x^(k*n)/e^n))^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (x^{3/4} \sqrt [4]{b+a x^2}\right ) \int \frac {\left (b+a x^2\right )^{3/4}}{x^{3/4} \left (b-a x^2\right )} \, dx}{\sqrt [4]{b x^3+a x^5}} \\ & = \frac {\left (4 x^{3/4} \sqrt [4]{b+a x^2}\right ) \text {Subst}\left (\int \frac {\left (b+a x^8\right )^{3/4}}{b-a x^8} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{b x^3+a x^5}} \\ & = \frac {\left (4 x^{3/4} \left (b+a x^2\right )\right ) \text {Subst}\left (\int \frac {\left (1+\frac {a x^8}{b}\right )^{3/4}}{b-a x^8} \, dx,x,\sqrt [4]{x}\right )}{\left (1+\frac {a x^2}{b}\right )^{3/4} \sqrt [4]{b x^3+a x^5}} \\ & = \frac {4 x \left (b+a x^2\right ) \operatorname {AppellF1}\left (\frac {1}{8},1,-\frac {3}{4},\frac {9}{8},\frac {a x^2}{b},-\frac {a x^2}{b}\right )}{b \left (1+\frac {a x^2}{b}\right )^{3/4} \sqrt [4]{b x^3+a x^5}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.23 (sec) , antiderivative size = 264, normalized size of antiderivative = 0.96 \[ \int \frac {b+a x^2}{\left (b-a x^2\right ) \sqrt [4]{b x^3+a x^5}} \, dx=\frac {x^{3/4} \sqrt [4]{b+a x^2} \left (\sqrt {2} \arctan \left (\frac {\sqrt [4]{2} \sqrt [8]{a} \sqrt [8]{b} \sqrt [4]{x}}{\sqrt [4]{b+a x^2}}\right )-\arctan \left (\frac {2^{3/4} \sqrt [8]{a} \sqrt [8]{b} \sqrt [4]{x} \sqrt [4]{b+a x^2}}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}-\sqrt {b+a x^2}}\right )+\sqrt {2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [8]{a} \sqrt [8]{b} \sqrt [4]{x}}{\sqrt [4]{b+a x^2}}\right )+\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b+a x^2}}{2^{3/4} \sqrt [8]{a} \sqrt [8]{b} \sqrt [4]{x} \sqrt [4]{b+a x^2}}\right )\right )}{2^{3/4} \sqrt [8]{a} \sqrt [8]{b} \sqrt [4]{x^3 \left (b+a x^2\right )}} \]

[In]

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

[Out]

(x^(3/4)*(b + a*x^2)^(1/4)*(Sqrt[2]*ArcTan[(2^(1/4)*a^(1/8)*b^(1/8)*x^(1/4))/(b + a*x^2)^(1/4)] - ArcTan[(2^(3
/4)*a^(1/8)*b^(1/8)*x^(1/4)*(b + a*x^2)^(1/4))/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] - Sqrt[b + a*x^2])] + Sqrt[2]*
ArcTanh[(2^(1/4)*a^(1/8)*b^(1/8)*x^(1/4))/(b + a*x^2)^(1/4)] + ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt
[b + a*x^2])/(2^(3/4)*a^(1/8)*b^(1/8)*x^(1/4)*(b + a*x^2)^(1/4))]))/(2^(3/4)*a^(1/8)*b^(1/8)*(x^3*(b + a*x^2))
^(1/4))

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.15

method	result	size
pseudoelliptic	\(-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-4 a b \right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\left (x^{3} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}}\right )}{2}\)	\(42\)

[In]

int((a*x^2+b)/(-a*x^2+b)/(a*x^5+b*x^3)^(1/4),x,method=_RETURNVERBOSE)

[Out]

-1/2*sum(ln((-_R*x+(x^3*(a*x^2+b))^(1/4))/x)/_R,_R=RootOf(_Z^8-4*a*b))

Fricas [F(-1)]

Timed out. \[ \int \frac {b+a x^2}{\left (b-a x^2\right ) \sqrt [4]{b x^3+a x^5}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

-Integral(b/(a*x**2*(a*x**5 + b*x**3)**(1/4) - b*(a*x**5 + b*x**3)**(1/4)), x) - Integral(a*x**2/(a*x**2*(a*x*
*5 + b*x**3)**(1/4) - b*(a*x**5 + b*x**3)**(1/4)), x)

Maxima [F]

\[ \int \frac {b+a x^2}{\left (b-a x^2\right ) \sqrt [4]{b x^3+a x^5}} \, dx=\int { -\frac {a x^{2} + b}{{\left (a x^{5} + b x^{3}\right )}^{\frac {1}{4}} {\left (a x^{2} - b\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {b+a x^2}{\left (b-a x^2\right ) \sqrt [4]{b x^3+a x^5}} \, dx=\int { -\frac {a x^{2} + b}{{\left (a x^{5} + b x^{3}\right )}^{\frac {1}{4}} {\left (a x^{2} - b\right )}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {b+a x^2}{\left (b-a x^2\right ) \sqrt [4]{b x^3+a x^5}} \, dx=\int \frac {a\,x^2+b}{\left (b-a\,x^2\right )\,{\left (a\,x^5+b\,x^3\right )}^{1/4}} \,d x \]

[In]

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

[Out]

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