3.23.21 \(\int \frac {(-b+a x^4) (b+a x^4)^{3/4}}{x^8 (b+2 a x^4)} \, dx\)

Optimal. Leaf size=165 \[ -\frac {3 a^{7/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} x \sqrt [4]{a x^4+b}}{\sqrt {a x^4+b}-\sqrt {a} x^2}\right )}{2 \sqrt {2} b}-\frac {3 a^{7/4} \tanh ^{-1}\left (\frac {\frac {\sqrt {a x^4+b}}{\sqrt {2} \sqrt [4]{a}}+\frac {\sqrt [4]{a} x^2}{\sqrt {2}}}{x \sqrt [4]{a x^4+b}}\right )}{2 \sqrt {2} b}+\frac {\left (b-6 a x^4\right ) \left (a x^4+b\right )^{3/4}}{7 b x^7} \]

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Rubi [A]  time = 0.29, antiderivative size = 255, normalized size of antiderivative = 1.55, number of steps used = 13, number of rules used = 10, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {580, 583, 12, 377, 211, 1165, 628, 1162, 617, 204} \begin {gather*} \frac {3 a^{7/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2 \sqrt {2} b}-\frac {3 a^{7/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}+1\right )}{2 \sqrt {2} b}+\frac {3 a^{7/4} \log \left (-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}+\frac {\sqrt {a} x^2}{\sqrt {a x^4+b}}+1\right )}{4 \sqrt {2} b}-\frac {3 a^{7/4} \log \left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}+\frac {\sqrt {a} x^2}{\sqrt {a x^4+b}}+1\right )}{4 \sqrt {2} b}+\frac {\left (a x^4+b\right )^{3/4}}{7 x^7}-\frac {6 a \left (a x^4+b\right )^{3/4}}{7 b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b + a*x^4)^(3/4)/(7*x^7) - (6*a*(b + a*x^4)^(3/4))/(7*b*x^3) + (3*a^(7/4)*ArcTan[1 - (Sqrt[2]*a^(1/4)*x)/(b +
a*x^4)^(1/4)])/(2*Sqrt[2]*b) - (3*a^(7/4)*ArcTan[1 + (Sqrt[2]*a^(1/4)*x)/(b + a*x^4)^(1/4)])/(2*Sqrt[2]*b) +
(3*a^(7/4)*Log[1 + (Sqrt[a]*x^2)/Sqrt[b + a*x^4] - (Sqrt[2]*a^(1/4)*x)/(b + a*x^4)^(1/4)])/(4*Sqrt[2]*b) - (3*
a^(7/4)*Log[1 + (Sqrt[a]*x^2)/Sqrt[b + a*x^4] + (Sqrt[2]*a^(1/4)*x)/(b + a*x^4)^(1/4)])/(4*Sqrt[2]*b)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 377

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

Rule 580

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*g*(m + 1)), x] - Dist[1/(a*g^n*(m + 1
)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f)*(m + 1) + e*n*(b*c*(p + 1) + a*d*q)
+ d*((b*e - a*f)*(m + 1) + b*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && IGtQ[n
, 0] && GtQ[q, 0] && LtQ[m, -1] &&  !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

\begin {align*} \int \frac {\left (-b+a x^4\right ) \left (b+a x^4\right )^{3/4}}{x^8 \left (b+2 a x^4\right )} \, dx &=\frac {\left (b+a x^4\right )^{3/4}}{7 x^7}+\frac {\int \frac {18 a b^2+15 a^2 b x^4}{x^4 \sqrt [4]{b+a x^4} \left (b+2 a x^4\right )} \, dx}{7 b}\\ &=\frac {\left (b+a x^4\right )^{3/4}}{7 x^7}-\frac {6 a \left (b+a x^4\right )^{3/4}}{7 b x^3}-\frac {\int \frac {63 a^2 b^3}{\sqrt [4]{b+a x^4} \left (b+2 a x^4\right )} \, dx}{21 b^3}\\ &=\frac {\left (b+a x^4\right )^{3/4}}{7 x^7}-\frac {6 a \left (b+a x^4\right )^{3/4}}{7 b x^3}-\left (3 a^2\right ) \int \frac {1}{\sqrt [4]{b+a x^4} \left (b+2 a x^4\right )} \, dx\\ &=\frac {\left (b+a x^4\right )^{3/4}}{7 x^7}-\frac {6 a \left (b+a x^4\right )^{3/4}}{7 b x^3}-\left (3 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{b+a b x^4} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )\\ &=\frac {\left (b+a x^4\right )^{3/4}}{7 x^7}-\frac {6 a \left (b+a x^4\right )^{3/4}}{7 b x^3}-\frac {1}{2} \left (3 a^2\right ) \operatorname {Subst}\left (\int \frac {1-\sqrt {a} x^2}{b+a b x^4} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )-\frac {1}{2} \left (3 a^2\right ) \operatorname {Subst}\left (\int \frac {1+\sqrt {a} x^2}{b+a b x^4} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )\\ &=\frac {\left (b+a x^4\right )^{3/4}}{7 x^7}-\frac {6 a \left (b+a x^4\right )^{3/4}}{7 b x^3}-\frac {\left (3 a^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{\sqrt {a}}-\frac {\sqrt {2} x}{\sqrt [4]{a}}+x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{4 b}-\frac {\left (3 a^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{\sqrt {a}}+\frac {\sqrt {2} x}{\sqrt [4]{a}}+x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{4 b}+\frac {\left (3 a^{7/4}\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2}}{\sqrt [4]{a}}+2 x}{-\frac {1}{\sqrt {a}}-\frac {\sqrt {2} x}{\sqrt [4]{a}}-x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{4 \sqrt {2} b}+\frac {\left (3 a^{7/4}\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2}}{\sqrt [4]{a}}-2 x}{-\frac {1}{\sqrt {a}}+\frac {\sqrt {2} x}{\sqrt [4]{a}}-x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{4 \sqrt {2} b}\\ &=\frac {\left (b+a x^4\right )^{3/4}}{7 x^7}-\frac {6 a \left (b+a x^4\right )^{3/4}}{7 b x^3}+\frac {3 a^{7/4} \log \left (1+\frac {\sqrt {a} x^2}{\sqrt {b+a x^4}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{4 \sqrt {2} b}-\frac {3 a^{7/4} \log \left (1+\frac {\sqrt {a} x^2}{\sqrt {b+a x^4}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{4 \sqrt {2} b}-\frac {\left (3 a^{7/4}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{2 \sqrt {2} b}+\frac {\left (3 a^{7/4}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{2 \sqrt {2} b}\\ &=\frac {\left (b+a x^4\right )^{3/4}}{7 x^7}-\frac {6 a \left (b+a x^4\right )^{3/4}}{7 b x^3}+\frac {3 a^{7/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{2 \sqrt {2} b}-\frac {3 a^{7/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{2 \sqrt {2} b}+\frac {3 a^{7/4} \log \left (1+\frac {\sqrt {a} x^2}{\sqrt {b+a x^4}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{4 \sqrt {2} b}-\frac {3 a^{7/4} \log \left (1+\frac {\sqrt {a} x^2}{\sqrt {b+a x^4}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{4 \sqrt {2} b}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 202, normalized size = 1.22 \begin {gather*} \left (\frac {1}{7 x^7}-\frac {6 a}{7 b x^3}\right ) \left (a x^4+b\right )^{3/4}-\frac {3 a^{7/4} \left (-2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a+b x^4}}\right )+2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a+b x^4}}+1\right )-\log \left (-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a+b x^4}}+\frac {\sqrt {a} x^2}{\sqrt {a+b x^4}}+1\right )+\log \left (\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{a+b x^4}}+\frac {\sqrt {a} x^2}{\sqrt {a+b x^4}}+1\right )\right )}{4 \sqrt {2} b} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(1/(7*x^7) - (6*a)/(7*b*x^3))*(b + a*x^4)^(3/4) - (3*a^(7/4)*(-2*ArcTan[1 - (Sqrt[2]*a^(1/4)*x)/(a + b*x^4)^(1
/4)] + 2*ArcTan[1 + (Sqrt[2]*a^(1/4)*x)/(a + b*x^4)^(1/4)] - Log[1 + (Sqrt[a]*x^2)/Sqrt[a + b*x^4] - (Sqrt[2]*
a^(1/4)*x)/(a + b*x^4)^(1/4)] + Log[1 + (Sqrt[a]*x^2)/Sqrt[a + b*x^4] + (Sqrt[2]*a^(1/4)*x)/(a + b*x^4)^(1/4)]
))/(4*Sqrt[2]*b)

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IntegrateAlgebraic [A]  time = 0.52, size = 165, normalized size = 1.00 \begin {gather*} \frac {\left (b-6 a x^4\right ) \left (b+a x^4\right )^{3/4}}{7 b x^7}-\frac {3 a^{7/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} x \sqrt [4]{b+a x^4}}{-\sqrt {a} x^2+\sqrt {b+a x^4}}\right )}{2 \sqrt {2} b}-\frac {3 a^{7/4} \tanh ^{-1}\left (\frac {\frac {\sqrt [4]{a} x^2}{\sqrt {2}}+\frac {\sqrt {b+a x^4}}{\sqrt {2} \sqrt [4]{a}}}{x \sqrt [4]{b+a x^4}}\right )}{2 \sqrt {2} b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b - 6*a*x^4)*(b + a*x^4)^(3/4))/(7*b*x^7) - (3*a^(7/4)*ArcTan[(Sqrt[2]*a^(1/4)*x*(b + a*x^4)^(1/4))/(-(Sqrt[
a]*x^2) + Sqrt[b + a*x^4])])/(2*Sqrt[2]*b) - (3*a^(7/4)*ArcTanh[((a^(1/4)*x^2)/Sqrt[2] + Sqrt[b + a*x^4]/(Sqrt
[2]*a^(1/4)))/(x*(b + a*x^4)^(1/4))])/(2*Sqrt[2]*b)

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fricas [B]  time = 75.87, size = 429, normalized size = 2.60 \begin {gather*} \frac {84 \, \left (-\frac {a^{7}}{b^{4}}\right )^{\frac {1}{4}} b x^{7} \arctan \left (\frac {{\left (\left (-\frac {a^{7}}{b^{4}}\right )^{\frac {3}{4}} a^{5} b^{3} x^{2} + \sqrt {\frac {a^{10}}{b^{2}}} \left (-\frac {a^{7}}{b^{4}}\right )^{\frac {3}{4}} b^{4} x^{2}\right )} {\left (a x^{4} + b\right )}^{\frac {3}{4}} - {\left (a x^{4} + b\right )}^{\frac {1}{4}} {\left ({\left (a^{4} b^{2} x^{4} + a^{3} b^{3}\right )} \sqrt {\frac {a^{10}}{b^{2}}} \left (-\frac {a^{7}}{b^{4}}\right )^{\frac {1}{4}} - {\left (a^{9} b x^{4} + a^{8} b^{2}\right )} \left (-\frac {a^{7}}{b^{4}}\right )^{\frac {1}{4}}\right )}}{2 \, {\left (a^{11} x^{5} + a^{10} b x\right )}}\right ) - 21 \, \left (-\frac {a^{7}}{b^{4}}\right )^{\frac {1}{4}} b x^{7} \log \left (-\frac {27 \, {\left (2 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}} \left (-\frac {a^{7}}{b^{4}}\right )^{\frac {1}{4}} a^{4} x^{3} + 2 \, \sqrt {a x^{4} + b} \sqrt {-\frac {a^{7}}{b^{4}}} a^{2} b x^{2} - a^{5} + 2 \, {\left (a x^{4} + b\right )}^{\frac {3}{4}} \left (-\frac {a^{7}}{b^{4}}\right )^{\frac {3}{4}} b^{2} x\right )}}{2 \, a x^{4} + b}\right ) + 21 \, \left (-\frac {a^{7}}{b^{4}}\right )^{\frac {1}{4}} b x^{7} \log \left (\frac {27 \, {\left (2 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}} \left (-\frac {a^{7}}{b^{4}}\right )^{\frac {1}{4}} a^{4} x^{3} - 2 \, \sqrt {a x^{4} + b} \sqrt {-\frac {a^{7}}{b^{4}}} a^{2} b x^{2} + a^{5} + 2 \, {\left (a x^{4} + b\right )}^{\frac {3}{4}} \left (-\frac {a^{7}}{b^{4}}\right )^{\frac {3}{4}} b^{2} x\right )}}{2 \, a x^{4} + b}\right ) - 8 \, {\left (6 \, a x^{4} - b\right )} {\left (a x^{4} + b\right )}^{\frac {3}{4}}}{56 \, b x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4-b)*(a*x^4+b)^(3/4)/x^8/(2*a*x^4+b),x, algorithm="fricas")

[Out]

1/56*(84*(-a^7/b^4)^(1/4)*b*x^7*arctan(1/2*(((-a^7/b^4)^(3/4)*a^5*b^3*x^2 + sqrt(a^10/b^2)*(-a^7/b^4)^(3/4)*b^
4*x^2)*(a*x^4 + b)^(3/4) - (a*x^4 + b)^(1/4)*((a^4*b^2*x^4 + a^3*b^3)*sqrt(a^10/b^2)*(-a^7/b^4)^(1/4) - (a^9*b
*x^4 + a^8*b^2)*(-a^7/b^4)^(1/4)))/(a^11*x^5 + a^10*b*x)) - 21*(-a^7/b^4)^(1/4)*b*x^7*log(-27*(2*(a*x^4 + b)^(
1/4)*(-a^7/b^4)^(1/4)*a^4*x^3 + 2*sqrt(a*x^4 + b)*sqrt(-a^7/b^4)*a^2*b*x^2 - a^5 + 2*(a*x^4 + b)^(3/4)*(-a^7/b
^4)^(3/4)*b^2*x)/(2*a*x^4 + b)) + 21*(-a^7/b^4)^(1/4)*b*x^7*log(27*(2*(a*x^4 + b)^(1/4)*(-a^7/b^4)^(1/4)*a^4*x
^3 - 2*sqrt(a*x^4 + b)*sqrt(-a^7/b^4)*a^2*b*x^2 + a^5 + 2*(a*x^4 + b)^(3/4)*(-a^7/b^4)^(3/4)*b^2*x)/(2*a*x^4 +
b)) - 8*(6*a*x^4 - b)*(a*x^4 + b)^(3/4))/(b*x^7)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (a x^{4} + b\right )}^{\frac {3}{4}} {\left (a x^{4} - b\right )}}{{\left (2 \, a x^{4} + b\right )} x^{8}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4-b)*(a*x^4+b)^(3/4)/x^8/(2*a*x^4+b),x, algorithm="giac")

[Out]

integrate((a*x^4 + b)^(3/4)*(a*x^4 - b)/((2*a*x^4 + b)*x^8), x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (a \,x^{4}-b \right ) \left (a \,x^{4}+b \right )^{\frac {3}{4}}}{x^{8} \left (2 a \,x^{4}+b \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^4-b)*(a*x^4+b)^(3/4)/x^8/(2*a*x^4+b),x)

[Out]

int((a*x^4-b)*(a*x^4+b)^(3/4)/x^8/(2*a*x^4+b),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (a x^{4} + b\right )}^{\frac {3}{4}} {\left (a x^{4} - b\right )}}{{\left (2 \, a x^{4} + b\right )} x^{8}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^4-b)*(a*x^4+b)^(3/4)/x^8/(2*a*x^4+b),x, algorithm="maxima")

[Out]

integrate((a*x^4 + b)^(3/4)*(a*x^4 - b)/((2*a*x^4 + b)*x^8), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {{\left (a\,x^4+b\right )}^{3/4}\,\left (b-a\,x^4\right )}{x^8\,\left (2\,a\,x^4+b\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((b + a*x^4)^(3/4)*(b - a*x^4))/(x^8*(b + 2*a*x^4)),x)

[Out]

int(-((b + a*x^4)^(3/4)*(b - a*x^4))/(x^8*(b + 2*a*x^4)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a x^{4} - b\right ) \left (a x^{4} + b\right )^{\frac {3}{4}}}{x^{8} \left (2 a x^{4} + b\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**4-b)*(a*x**4+b)**(3/4)/x**8/(2*a*x**4+b),x)

[Out]

Integral((a*x**4 - b)*(a*x**4 + b)**(3/4)/(x**8*(2*a*x**4 + b)), x)

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