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

Optimal. Leaf size=275 \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [8]{a} \sqrt [8]{b} x}{\sqrt [4]{a x^5+b x^3}}\right )}{\sqrt [4]{2} \sqrt [8]{a} \sqrt [8]{b}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [8]{a} \sqrt [8]{b} x}{\sqrt [4]{a x^5+b x^3}}\right )}{\sqrt [4]{2} \sqrt [8]{a} \sqrt [8]{b}}-\frac {\tan ^{-1}\left (\frac {2^{3/4} \sqrt [8]{a} \sqrt [8]{b} x \sqrt [4]{a x^5+b x^3}}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x^2-\sqrt {a x^5+b x^3}}\right )}{2^{3/4} \sqrt [8]{a} \sqrt [8]{b}}+\frac {\tanh ^{-1}\left (\frac {\frac {\sqrt [8]{a} \sqrt [8]{b} x^2}{\sqrt [4]{2}}+\frac {\sqrt {a x^5+b x^3}}{2^{3/4} \sqrt [8]{a} \sqrt [8]{b}}}{x \sqrt [4]{a x^5+b x^3}}\right )}{2^{3/4} \sqrt [8]{a} \sqrt [8]{b}} \]

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Rubi [C]  time = 0.23, antiderivative size = 70, normalized size of antiderivative = 0.25, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {2056, 466, 430, 429} \begin {gather*} \frac {4 x \left (a x^2+b\right ) F_1\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}} \end {gather*}

Warning: Unable to verify antiderivative.

[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 429

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 430

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 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 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}{\left (b-a x^2\right ) \sqrt [4]{b x^3+a x^5}} \, dx &=\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 ) \operatorname {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 ) \operatorname {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 ) F_1\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*}

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Mathematica [C]  time = 0.05, size = 65, normalized size = 0.24 \begin {gather*} \frac {4 \left (x^3 \left (a x^2+b\right )\right )^{3/4} F_1\left (\frac {1}{8};-\frac {3}{4},1;\frac {9}{8};-\frac {a x^2}{b},\frac {a x^2}{b}\right )}{b x^2 \left (\frac {a x^2}{b}+1\right )^{3/4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.85, size = 318, normalized size = 1.16 \begin {gather*} \frac {\tan ^{-1}\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 {\tan ^{-1}\left (\frac {\sqrt [8]{a} \sqrt [8]{b} x}{\sqrt [8]{a} \sqrt [8]{b} x-\sqrt [4]{2} \sqrt [4]{b x^3+a x^5}}\right )}{2^{3/4} \sqrt [8]{a} \sqrt [8]{b}}+\frac {\tan ^{-1}\left (\frac {\sqrt [8]{a} \sqrt [8]{b} x}{\sqrt [8]{a} \sqrt [8]{b} x+\sqrt [4]{2} \sqrt [4]{b x^3+a x^5}}\right )}{2^{3/4} \sqrt [8]{a} \sqrt [8]{b}}+\frac {\tanh ^{-1}\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 {\tanh ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(2^(1/4)*a^(1/8)*b^(1/8)*x)/(b*x^3 + a*x^5)^(1/4)]/(2^(1/4)*a^(1/8)*b^(1/8)) - ArcTan[(a^(1/8)*b^(1/8)*
x)/(a^(1/8)*b^(1/8)*x - 2^(1/4)*(b*x^3 + a*x^5)^(1/4))]/(2^(3/4)*a^(1/8)*b^(1/8)) + ArcTan[(a^(1/8)*b^(1/8)*x)
/(a^(1/8)*b^(1/8)*x + 2^(1/4)*(b*x^3 + a*x^5)^(1/4))]/(2^(3/4)*a^(1/8)*b^(1/8)) + ArcTanh[(2^(1/4)*a^(1/8)*b^(
1/8)*x)/(b*x^3 + a*x^5)^(1/4)]/(2^(1/4)*a^(1/8)*b^(1/8)) + ArcTanh[((a^(1/8)*b^(1/8)*x^2)/2^(1/4) + Sqrt[b*x^3
+ a*x^5]/(2^(3/4)*a^(1/8)*b^(1/8)))/(x*(b*x^3 + a*x^5)^(1/4))]/(2^(3/4)*a^(1/8)*b^(1/8))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {a \,x^{2}+b}{\left (-a \,x^{2}+b \right ) \left (a \,x^{5}+b \,x^{3}\right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a*x^2+b)/(-a*x^2+b)/(a*x^5+b*x^3)^(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^{5} + b x^{3}\right )}^{\frac {1}{4}} {\left (a x^{2} - b\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {a\,x^2+b}{\left (b-a\,x^2\right )\,{\left (a\,x^5+b\,x^3\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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