3.30.49 \(\int \frac {(-q+p x^4) \sqrt {q+p x^4}}{b x^4+a (q+p x^4)^2} \, dx\)

Optimal. Leaf size=352 \[ \frac {1}{4} \text {RootSum}\left [\text {$\#$1}^8 a-8 \text {$\#$1}^6 a \sqrt {p} \sqrt {q}+24 \text {$\#$1}^4 a p q+16 \text {$\#$1}^4 b-32 \text {$\#$1}^2 a p^{3/2} q^{3/2}+16 a p^2 q^2\& ,\frac {-\text {$\#$1}^6 \log \left (-\text {$\#$1} x+\sqrt {p x^4+q}+\sqrt {p} x^2+\sqrt {q}\right )+\text {$\#$1}^6 \log (x)+2 \text {$\#$1}^4 \sqrt {p} \sqrt {q} \log \left (-\text {$\#$1} x+\sqrt {p x^4+q}+\sqrt {p} x^2+\sqrt {q}\right )-2 \text {$\#$1}^4 \sqrt {p} \sqrt {q} \log (x)+4 \text {$\#$1}^2 p q \log \left (-\text {$\#$1} x+\sqrt {p x^4+q}+\sqrt {p} x^2+\sqrt {q}\right )-4 \text {$\#$1}^2 p q \log (x)-8 p^{3/2} q^{3/2} \log \left (-\text {$\#$1} x+\sqrt {p x^4+q}+\sqrt {p} x^2+\sqrt {q}\right )+8 p^{3/2} q^{3/2} \log (x)}{\text {$\#$1}^7 (-a)+6 \text {$\#$1}^5 a \sqrt {p} \sqrt {q}-12 \text {$\#$1}^3 a p q-8 \text {$\#$1}^3 b+8 \text {$\#$1} a p^{3/2} q^{3/2}}\& \right ] \]

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Rubi [A]  time = 0.43, antiderivative size = 247, normalized size of antiderivative = 0.70, number of steps used = 10, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {6712, 211, 1165, 628, 1162, 617, 204} \begin {gather*} \frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a} \sqrt {p x^4+q}}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{b}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a} \sqrt {p x^4+q}}+1\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{b}}+\frac {\log \left (-\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {p x^4+q}}+\sqrt {a}+\frac {\sqrt {b} x^2}{p x^4+q}\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{b}}-\frac {\log \left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {p x^4+q}}+\sqrt {a}+\frac {\sqrt {b} x^2}{p x^4+q}\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/(a^(1/4)*Sqrt[q + p*x^4])]/(2*Sqrt[2]*a^(3/4)*b^(1/4)) - ArcTan[1 + (Sqrt[2]*b^
(1/4)*x)/(a^(1/4)*Sqrt[q + p*x^4])]/(2*Sqrt[2]*a^(3/4)*b^(1/4)) + Log[Sqrt[a] + (Sqrt[b]*x^2)/(q + p*x^4) - (S
qrt[2]*a^(1/4)*b^(1/4)*x)/Sqrt[q + p*x^4]]/(4*Sqrt[2]*a^(3/4)*b^(1/4)) - Log[Sqrt[a] + (Sqrt[b]*x^2)/(q + p*x^
4) + (Sqrt[2]*a^(1/4)*b^(1/4)*x)/Sqrt[q + p*x^4]]/(4*Sqrt[2]*a^(3/4)*b^(1/4))

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 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]

Rule 6712

Int[(u_)*(v_)^(r_.)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x
] - q*v*D[w, x])]}, -Dist[c*q, Subst[Int[(a + b*x^q)^m, x], x, v^(m*p + r + 1)*w], x] /; FreeQ[c, x]] /; FreeQ
[{a, b, m, p, q, r}, x] && EqQ[p + q*(m*p + r + 1), 0] && IntegerQ[q] && IntegerQ[m]

Rubi steps

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

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Mathematica [C]  time = 1.93, size = 420, normalized size = 1.19 \begin {gather*} -\frac {i \sqrt {\frac {p x^4}{q}+1} \left (-\Pi \left (-\frac {i \sqrt {2} \sqrt {q}}{\sqrt {p} \sqrt {-\frac {b-\sqrt {b+4 a p q} \sqrt {b}+2 a p q}{a p^2}}};\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {p}}{\sqrt {q}}} x\right )\right |-1\right )-\Pi \left (\frac {i \sqrt {2} \sqrt {q}}{\sqrt {p} \sqrt {-\frac {b-\sqrt {b+4 a p q} \sqrt {b}+2 a p q}{a p^2}}};\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {p}}{\sqrt {q}}} x\right )\right |-1\right )-\Pi \left (-\frac {i \sqrt {2} \sqrt {q}}{\sqrt {p} \sqrt {-\frac {b+\sqrt {b+4 a p q} \sqrt {b}+2 a p q}{a p^2}}};\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {p}}{\sqrt {q}}} x\right )\right |-1\right )-\Pi \left (\frac {i \sqrt {2} \sqrt {q}}{\sqrt {p} \sqrt {-\frac {b+\sqrt {b+4 a p q} \sqrt {b}+2 a p q}{a p^2}}};\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {p}}{\sqrt {q}}} x\right )\right |-1\right )+2 F\left (\left .i \sinh ^{-1}\left (\sqrt {\frac {i \sqrt {p}}{\sqrt {q}}} x\right )\right |-1\right )\right )}{2 a \sqrt {\frac {i \sqrt {p}}{\sqrt {q}}} \sqrt {p x^4+q}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/2*I)*Sqrt[1 + (p*x^4)/q]*(2*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[p])/Sqrt[q]]*x], -1] - EllipticPi[((-I)*Sqrt
[2]*Sqrt[q])/(Sqrt[p]*Sqrt[-((b + 2*a*p*q - Sqrt[b]*Sqrt[b + 4*a*p*q])/(a*p^2))]), I*ArcSinh[Sqrt[(I*Sqrt[p])/
Sqrt[q]]*x], -1] - EllipticPi[(I*Sqrt[2]*Sqrt[q])/(Sqrt[p]*Sqrt[-((b + 2*a*p*q - Sqrt[b]*Sqrt[b + 4*a*p*q])/(a
*p^2))]), I*ArcSinh[Sqrt[(I*Sqrt[p])/Sqrt[q]]*x], -1] - EllipticPi[((-I)*Sqrt[2]*Sqrt[q])/(Sqrt[p]*Sqrt[-((b +
2*a*p*q + Sqrt[b]*Sqrt[b + 4*a*p*q])/(a*p^2))]), I*ArcSinh[Sqrt[(I*Sqrt[p])/Sqrt[q]]*x], -1] - EllipticPi[(I*
Sqrt[2]*Sqrt[q])/(Sqrt[p]*Sqrt[-((b + 2*a*p*q + Sqrt[b]*Sqrt[b + 4*a*p*q])/(a*p^2))]), I*ArcSinh[Sqrt[(I*Sqrt[
p])/Sqrt[q]]*x], -1]))/(a*Sqrt[(I*Sqrt[p])/Sqrt[q]]*Sqrt[q + p*x^4])

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IntegrateAlgebraic [A]  time = 1.63, size = 171, normalized size = 0.49 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x \sqrt {q+p x^4}}{\sqrt {a} q-\sqrt {b} x^2+\sqrt {a} p x^4}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{b}}-\frac {\tanh ^{-1}\left (\frac {\frac {\sqrt [4]{a} q}{\sqrt {2} \sqrt [4]{b}}+\frac {\sqrt [4]{b} x^2}{\sqrt {2} \sqrt [4]{a}}+\frac {\sqrt [4]{a} p x^4}{\sqrt {2} \sqrt [4]{b}}}{x \sqrt {q+p x^4}}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*ArcTan[(Sqrt[2]*a^(1/4)*b^(1/4)*x*Sqrt[q + p*x^4])/(Sqrt[a]*q - Sqrt[b]*x^2 + Sqrt[a]*p*x^4)]/(Sqrt[2]*a^
(3/4)*b^(1/4)) - ArcTanh[((a^(1/4)*q)/(Sqrt[2]*b^(1/4)) + (b^(1/4)*x^2)/(Sqrt[2]*a^(1/4)) + (a^(1/4)*p*x^4)/(S
qrt[2]*b^(1/4)))/(x*Sqrt[q + p*x^4])]/(2*Sqrt[2]*a^(3/4)*b^(1/4))

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fricas [B]  time = 10.70, size = 496, normalized size = 1.41 \begin {gather*} -\frac {1}{2} \, \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}} \arctan \left (-\frac {2 \, {\left (a b x^{3} \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}} + {\left (a^{3} b p x^{5} + a^{3} b q x\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {3}{4}}\right )} \sqrt {p x^{4} + q} - {\left ({\left (a^{4} b p^{2} x^{8} + a^{4} b q^{2} + {\left (2 \, a^{4} b p q - a^{3} b^{2}\right )} x^{4}\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {3}{4}} + 2 \, {\left (a^{2} b p x^{6} + a^{2} b q x^{2}\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}}\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}}}{a p^{2} x^{8} + {\left (2 \, a p q + b\right )} x^{4} + a q^{2}}\right ) + \frac {1}{8} \, \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}} \log \left (\frac {2 \, {\left (a^{2} b p x^{6} + a^{2} b q x^{2}\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {3}{4}} + 2 \, {\left (p x^{5} - a b x^{3} \sqrt {-\frac {1}{a^{3} b}} + q x\right )} \sqrt {p x^{4} + q} - {\left (a p^{2} x^{8} + {\left (2 \, a p q - b\right )} x^{4} + a q^{2}\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}}}{2 \, {\left (a p^{2} x^{8} + {\left (2 \, a p q + b\right )} x^{4} + a q^{2}\right )}}\right ) - \frac {1}{8} \, \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}} \log \left (-\frac {2 \, {\left (a^{2} b p x^{6} + a^{2} b q x^{2}\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {3}{4}} - 2 \, {\left (p x^{5} - a b x^{3} \sqrt {-\frac {1}{a^{3} b}} + q x\right )} \sqrt {p x^{4} + q} - {\left (a p^{2} x^{8} + {\left (2 \, a p q - b\right )} x^{4} + a q^{2}\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}}}{2 \, {\left (a p^{2} x^{8} + {\left (2 \, a p q + b\right )} x^{4} + a q^{2}\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(-1/(a^3*b))^(1/4)*arctan(-(2*(a*b*x^3*(-1/(a^3*b))^(1/4) + (a^3*b*p*x^5 + a^3*b*q*x)*(-1/(a^3*b))^(3/4))
*sqrt(p*x^4 + q) - ((a^4*b*p^2*x^8 + a^4*b*q^2 + (2*a^4*b*p*q - a^3*b^2)*x^4)*(-1/(a^3*b))^(3/4) + 2*(a^2*b*p*
x^6 + a^2*b*q*x^2)*(-1/(a^3*b))^(1/4))*(-1/(a^3*b))^(1/4))/(a*p^2*x^8 + (2*a*p*q + b)*x^4 + a*q^2)) + 1/8*(-1/
(a^3*b))^(1/4)*log(1/2*(2*(a^2*b*p*x^6 + a^2*b*q*x^2)*(-1/(a^3*b))^(3/4) + 2*(p*x^5 - a*b*x^3*sqrt(-1/(a^3*b))
+ q*x)*sqrt(p*x^4 + q) - (a*p^2*x^8 + (2*a*p*q - b)*x^4 + a*q^2)*(-1/(a^3*b))^(1/4))/(a*p^2*x^8 + (2*a*p*q +
b)*x^4 + a*q^2)) - 1/8*(-1/(a^3*b))^(1/4)*log(-1/2*(2*(a^2*b*p*x^6 + a^2*b*q*x^2)*(-1/(a^3*b))^(3/4) - 2*(p*x^
5 - a*b*x^3*sqrt(-1/(a^3*b)) + q*x)*sqrt(p*x^4 + q) - (a*p^2*x^8 + (2*a*p*q - b)*x^4 + a*q^2)*(-1/(a^3*b))^(1/
4))/(a*p^2*x^8 + (2*a*p*q + b)*x^4 + a*q^2))

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

[Out]

Timed out

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maple [B]  time = 0.20, size = 191, normalized size = 0.54

method	result	size
default	\(\frac {\left (\frac {\ln \left (\frac {\frac {p \,x^{4}+q}{2 x^{2}}-\frac {\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {p \,x^{4}+q}\, \sqrt {2}}{2 x}+\frac {\sqrt {\frac {b}{a}}}{2}}{\frac {p \,x^{4}+q}{2 x^{2}}+\frac {\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {p \,x^{4}+q}\, \sqrt {2}}{2 x}+\frac {\sqrt {\frac {b}{a}}}{2}}\right )}{4 a \left (\frac {b}{a}\right )^{\frac {1}{4}}}+\frac {\arctan \left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}}{\left (\frac {b}{a}\right )^{\frac {1}{4}} x}+1\right )}{2 a \left (\frac {b}{a}\right )^{\frac {1}{4}}}+\frac {\arctan \left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}}{\left (\frac {b}{a}\right )^{\frac {1}{4}} x}-1\right )}{2 a \left (\frac {b}{a}\right )^{\frac {1}{4}}}\right ) \sqrt {2}}{2}\)	\(191\)
elliptic	\(\frac {\left (\frac {\ln \left (\frac {\frac {p \,x^{4}+q}{2 x^{2}}-\frac {\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {p \,x^{4}+q}\, \sqrt {2}}{2 x}+\frac {\sqrt {\frac {b}{a}}}{2}}{\frac {p \,x^{4}+q}{2 x^{2}}+\frac {\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {p \,x^{4}+q}\, \sqrt {2}}{2 x}+\frac {\sqrt {\frac {b}{a}}}{2}}\right )}{4 a \left (\frac {b}{a}\right )^{\frac {1}{4}}}+\frac {\arctan \left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}}{\left (\frac {b}{a}\right )^{\frac {1}{4}} x}+1\right )}{2 a \left (\frac {b}{a}\right )^{\frac {1}{4}}}+\frac {\arctan \left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}}{\left (\frac {b}{a}\right )^{\frac {1}{4}} x}-1\right )}{2 a \left (\frac {b}{a}\right )^{\frac {1}{4}}}\right ) \sqrt {2}}{2}\)	\(191\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(1/4/a/(b/a)^(1/4)*ln((1/2*(p*x^4+q)/x^2-1/2*(b/a)^(1/4)*(p*x^4+q)^(1/2)*2^(1/2)/x+1/2*(b/a)^(1/2))/(1/2*(
p*x^4+q)/x^2+1/2*(b/a)^(1/4)*(p*x^4+q)^(1/2)*2^(1/2)/x+1/2*(b/a)^(1/2)))+1/2/a/(b/a)^(1/4)*arctan(1/(b/a)^(1/4
)*(p*x^4+q)^(1/2)*2^(1/2)/x+1)+1/2/a/(b/a)^(1/4)*arctan(1/(b/a)^(1/4)*(p*x^4+q)^(1/2)*2^(1/2)/x-1))*2^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(p*x^4 + q)*(p*x^4 - q)/(b*x^4 + (p*x^4 + q)^2*a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [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((p*x**4-q)*(p*x**4+q)**(1/2)/(b*x**4+a*(p*x**4+q)**2),x)

[Out]

Timed out

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