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

Optimal. Leaf size=112 \[ -\frac {2^{3/4} a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^2}}\right )}{b}-\frac {2^{3/4} a^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4+b x^2}}\right )}{b}+\frac {2 \left (a x^4+b x^2\right )^{3/4}}{3 b x^3} \]

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Rubi [C]  time = 0.37, antiderivative size = 55, normalized size of antiderivative = 0.49, number of steps used = 4, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {2056, 466, 511, 510} \begin {gather*} \frac {2 \left (a x^2+b\right ) \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};\frac {2 a x^2}{a x^2+b}\right )}{3 b x \sqrt [4]{a x^4+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 466

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 510

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

Rule 511

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

Rule 2056

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*} \int \frac {b+a x^2}{x^2 \left (-b+a x^2\right ) \sqrt [4]{b x^2+a x^4}} \, dx &=\frac {\left (\sqrt {x} \sqrt [4]{b+a x^2}\right ) \int \frac {\left (b+a x^2\right )^{3/4}}{x^{5/2} \left (-b+a x^2\right )} \, dx}{\sqrt [4]{b x^2+a x^4}}\\ &=\frac {\left (2 \sqrt {x} \sqrt [4]{b+a x^2}\right ) \operatorname {Subst}\left (\int \frac {\left (b+a x^4\right )^{3/4}}{x^4 \left (-b+a x^4\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^4}}\\ &=\frac {\left (2 \sqrt {x} \left (b+a x^2\right )\right ) \operatorname {Subst}\left (\int \frac {\left (1+\frac {a x^4}{b}\right )^{3/4}}{x^4 \left (-b+a x^4\right )} \, dx,x,\sqrt {x}\right )}{\left (1+\frac {a x^2}{b}\right )^{3/4} \sqrt [4]{b x^2+a x^4}}\\ &=\frac {2 \left (b+a x^2\right ) \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};\frac {2 a x^2}{b+a x^2}\right )}{3 b x \sqrt [4]{b x^2+a x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 48, normalized size = 0.43 \begin {gather*} \frac {2 \left (x^2 \left (a x^2+b\right )\right )^{3/4} \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};\frac {2 a x^2}{a x^2+b}\right )}{3 b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.39, size = 112, normalized size = 1.00 \begin {gather*} \frac {2 \left (b x^2+a x^4\right )^{3/4}}{3 b x^3}-\frac {2^{3/4} a^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{b}-\frac {2^{3/4} a^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 78.00, size = 504, normalized size = 4.50 \begin {gather*} \frac {12 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} b x^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} \arctan \left (\frac {2 \, {\left (2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} {\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} a^{4} b x^{2} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} + 2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (a x^{4} + b x^{2}\right )}^{\frac {3}{4}} a^{2} b^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}} + {\left (2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {a x^{4} + b x^{2}} a^{2} b x \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} + \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (3 \, a b^{3} x^{3} + b^{4} x\right )} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}}\right )} \sqrt {\sqrt {\frac {1}{2}} a^{2} b^{2} \sqrt {\frac {a^{3}}{b^{4}}}}\right )}}{a^{5} x^{3} - a^{4} b x}\right ) - 3 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} b x^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} \log \left (\frac {4 \, \sqrt {\frac {1}{2}} {\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} a b^{2} x^{2} \sqrt {\frac {a^{3}}{b^{4}}} + 4 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} \sqrt {a x^{4} + b x^{2}} b^{3} x \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}} + 2 \, {\left (a x^{4} + b x^{2}\right )}^{\frac {3}{4}} a^{2} + \left (\frac {1}{2}\right )^{\frac {1}{4}} {\left (3 \, a^{2} b x^{3} + a b^{2} x\right )} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}}}{a x^{3} - b x}\right ) + 3 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} b x^{3} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}} \log \left (\frac {4 \, \sqrt {\frac {1}{2}} {\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} a b^{2} x^{2} \sqrt {\frac {a^{3}}{b^{4}}} - 4 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} \sqrt {a x^{4} + b x^{2}} b^{3} x \left (\frac {a^{3}}{b^{4}}\right )^{\frac {3}{4}} + 2 \, {\left (a x^{4} + b x^{2}\right )}^{\frac {3}{4}} a^{2} - \left (\frac {1}{2}\right )^{\frac {1}{4}} {\left (3 \, a^{2} b x^{3} + a b^{2} x\right )} \left (\frac {a^{3}}{b^{4}}\right )^{\frac {1}{4}}}{a x^{3} - b x}\right ) + 4 \, {\left (a x^{4} + b x^{2}\right )}^{\frac {3}{4}}}{6 \, b x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(12*(1/2)^(1/4)*b*x^3*(a^3/b^4)^(1/4)*arctan(2*(2*(1/2)^(1/4)*(a*x^4 + b*x^2)^(1/4)*a^4*b*x^2*(a^3/b^4)^(1
/4) + 2*(1/2)^(3/4)*(a*x^4 + b*x^2)^(3/4)*a^2*b^3*(a^3/b^4)^(3/4) + (2*(1/2)^(1/4)*sqrt(a*x^4 + b*x^2)*a^2*b*x
*(a^3/b^4)^(1/4) + (1/2)^(3/4)*(3*a*b^3*x^3 + b^4*x)*(a^3/b^4)^(3/4))*sqrt(sqrt(1/2)*a^2*b^2*sqrt(a^3/b^4)))/(
a^5*x^3 - a^4*b*x)) - 3*(1/2)^(1/4)*b*x^3*(a^3/b^4)^(1/4)*log((4*sqrt(1/2)*(a*x^4 + b*x^2)^(1/4)*a*b^2*x^2*sqr
t(a^3/b^4) + 4*(1/2)^(3/4)*sqrt(a*x^4 + b*x^2)*b^3*x*(a^3/b^4)^(3/4) + 2*(a*x^4 + b*x^2)^(3/4)*a^2 + (1/2)^(1/
4)*(3*a^2*b*x^3 + a*b^2*x)*(a^3/b^4)^(1/4))/(a*x^3 - b*x)) + 3*(1/2)^(1/4)*b*x^3*(a^3/b^4)^(1/4)*log((4*sqrt(1
/2)*(a*x^4 + b*x^2)^(1/4)*a*b^2*x^2*sqrt(a^3/b^4) - 4*(1/2)^(3/4)*sqrt(a*x^4 + b*x^2)*b^3*x*(a^3/b^4)^(3/4) +
2*(a*x^4 + b*x^2)^(3/4)*a^2 - (1/2)^(1/4)*(3*a^2*b*x^3 + a*b^2*x)*(a^3/b^4)^(1/4))/(a*x^3 - b*x)) + 4*(a*x^4 +
b*x^2)^(3/4))/(b*x^3)

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giac [B]  time = 0.38, size = 208, normalized size = 1.86 \begin {gather*} -\frac {2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} \arctan \left (\frac {2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{b} - \frac {2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} \arctan \left (-\frac {2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{b} + \frac {2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} \log \left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {2} \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right )}{2 \, b} - \frac {2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} \log \left (-2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {2} \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right )}{2 \, b} + \frac {2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{4}}}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {a \,x^{2}+b}{x^{2} \left (a \,x^{2}-b \right ) \left (a \,x^{4}+b \,x^{2}\right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{2} + b}{{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{2} - b\right )} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {a\,x^2+b}{x^2\,\left (b-a\,x^2\right )\,{\left (a\,x^4+b\,x^2\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{2} + b}{x^{2} \sqrt [4]{x^{2} \left (a x^{2} + b\right )} \left (a x^{2} - b\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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