\(\int \frac {(-2 q+p x^3) \sqrt {q+p x^3}}{b x^4+a (q+p x^3)^2} \, dx\) [2246]

Optimal result

Integrand size = 40, antiderivative size = 167 \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{b x^4+a \left (q+p x^3\right )^2} \, dx=-\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x \sqrt {q+p x^3}}{\sqrt {a} q-\sqrt {b} x^2+\sqrt {a} p x^3}\right )}{\sqrt {2} a^{3/4} \sqrt [4]{b}}-\frac {\text {arctanh}\left (\frac {\frac {\sqrt [4]{a} q}{\sqrt {2} \sqrt [4]{b}}+\frac {\sqrt [4]{b} x^2}{\sqrt {2} \sqrt [4]{a}}+\frac {\sqrt [4]{a} p x^3}{\sqrt {2} \sqrt [4]{b}}}{x \sqrt {q+p x^3}}\right )}{\sqrt {2} a^{3/4} \sqrt [4]{b}} \]

[Out]

-1/2*arctan(2^(1/2)*a^(1/4)*b^(1/4)*x*(p*x^3+q)^(1/2)/(a^(1/2)*q-b^(1/2)*x^2+a^(1/2)*p*x^3))*2^(1/2)/a^(3/4)/b
^(1/4)-1/2*arctanh((1/2*a^(1/4)*q*2^(1/2)/b^(1/4)+1/2*b^(1/4)*x^2*2^(1/2)/a^(1/4)+1/2*a^(1/4)*p*x^3*2^(1/2)/b^
(1/4))/x/(p*x^3+q)^(1/2))*2^(1/2)/a^(3/4)/b^(1/4)

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.45, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {6844, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{b x^4+a \left (q+p x^3\right )^2} \, dx=\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a} \sqrt {p x^3+q}}\right )}{\sqrt {2} a^{3/4} \sqrt [4]{b}}-\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a} \sqrt {p x^3+q}}+1\right )}{\sqrt {2} a^{3/4} \sqrt [4]{b}}+\frac {\log \left (-\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {p x^3+q}}+\sqrt {a}+\frac {\sqrt {b} x^2}{p x^3+q}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{b}}-\frac {\log \left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {p x^3+q}}+\sqrt {a}+\frac {\sqrt {b} x^2}{p x^3+q}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{b}} \]

[In]

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

[Out]

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

Rule 210

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

Rule 217

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 631

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 642

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 1176

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 1179

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 6844

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]] /; Fre
eQ[{a, b, m, p, q, r}, x] && EqQ[p + q*(m*p + r + 1), 0] && IntegerQ[q] && IntegerQ[m]

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

Mathematica [A] (verified)

Time = 1.13 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.76 \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{b x^4+a \left (q+p x^3\right )^2} \, dx=-\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x \sqrt {q+p x^3}}{-\sqrt {b} x^2+\sqrt {a} \left (q+p x^3\right )}\right )+\text {arctanh}\left (\frac {\sqrt {b} x^2+\sqrt {a} \left (q+p x^3\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x \sqrt {q+p x^3}}\right )}{\sqrt {2} a^{3/4} \sqrt [4]{b}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.86 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.05

method	result	size
default	\(\frac {\sqrt {2}\, \left (\ln \left (\frac {-\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {p \,x^{3}+q}\, \sqrt {2}\, x +p \,x^{3}+\sqrt {\frac {b}{a}}\, x^{2}+q}{\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {p \,x^{3}+q}\, \sqrt {2}\, x +p \,x^{3}+\sqrt {\frac {b}{a}}\, x^{2}+q}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {p \,x^{3}+q}+\left (\frac {b}{a}\right )^{\frac {1}{4}} x}{\left (\frac {b}{a}\right )^{\frac {1}{4}} x}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {p \,x^{3}+q}-\left (\frac {b}{a}\right )^{\frac {1}{4}} x}{\left (\frac {b}{a}\right )^{\frac {1}{4}} x}\right )\right )}{4 \left (\frac {b}{a}\right )^{\frac {1}{4}} a}\)	\(175\)
pseudoelliptic	\(\frac {\sqrt {2}\, \left (\ln \left (\frac {-\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {p \,x^{3}+q}\, \sqrt {2}\, x +p \,x^{3}+\sqrt {\frac {b}{a}}\, x^{2}+q}{\left (\frac {b}{a}\right )^{\frac {1}{4}} \sqrt {p \,x^{3}+q}\, \sqrt {2}\, x +p \,x^{3}+\sqrt {\frac {b}{a}}\, x^{2}+q}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {p \,x^{3}+q}+\left (\frac {b}{a}\right )^{\frac {1}{4}} x}{\left (\frac {b}{a}\right )^{\frac {1}{4}} x}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {p \,x^{3}+q}-\left (\frac {b}{a}\right )^{\frac {1}{4}} x}{\left (\frac {b}{a}\right )^{\frac {1}{4}} x}\right )\right )}{4 \left (\frac {b}{a}\right )^{\frac {1}{4}} a}\)	\(175\)
elliptic	\(\text {Expression too large to display}\)	\(1164\)

[In]

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

[Out]

1/4/(b/a)^(1/4)*2^(1/2)*(ln((-(b/a)^(1/4)*(p*x^3+q)^(1/2)*2^(1/2)*x+p*x^3+(b/a)^(1/2)*x^2+q)/((b/a)^(1/4)*(p*x
^3+q)^(1/2)*2^(1/2)*x+p*x^3+(b/a)^(1/2)*x^2+q))+2*arctan((2^(1/2)*(p*x^3+q)^(1/2)+(b/a)^(1/4)*x)/(b/a)^(1/4)/x
)+2*arctan((2^(1/2)*(p*x^3+q)^(1/2)-(b/a)^(1/4)*x)/(b/a)^(1/4)/x))/a

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.65 (sec) , antiderivative size = 639, normalized size of antiderivative = 3.83 \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{b x^4+a \left (q+p x^3\right )^2} \, dx=\frac {1}{4} \, \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}} \log \left (\frac {a p^{2} x^{6} + 2 \, a p q x^{3} - b x^{4} + a q^{2} + 2 \, {\left (a b x^{3} \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}} + {\left (a^{3} b p x^{4} + a^{3} b q x\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {3}{4}}\right )} \sqrt {p x^{3} + q} - 2 \, {\left (a^{2} b p x^{5} + a^{2} b q x^{2}\right )} \sqrt {-\frac {1}{a^{3} b}}}{a p^{2} x^{6} + 2 \, a p q x^{3} + b x^{4} + a q^{2}}\right ) - \frac {1}{4} \, \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}} \log \left (\frac {a p^{2} x^{6} + 2 \, a p q x^{3} - b x^{4} + a q^{2} - 2 \, {\left (a b x^{3} \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}} + {\left (a^{3} b p x^{4} + a^{3} b q x\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {3}{4}}\right )} \sqrt {p x^{3} + q} - 2 \, {\left (a^{2} b p x^{5} + a^{2} b q x^{2}\right )} \sqrt {-\frac {1}{a^{3} b}}}{a p^{2} x^{6} + 2 \, a p q x^{3} + b x^{4} + a q^{2}}\right ) - \frac {1}{4} i \, \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}} \log \left (\frac {a p^{2} x^{6} + 2 \, a p q x^{3} - b x^{4} + a q^{2} - 2 \, {\left (i \, a b x^{3} \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}} + {\left (-i \, a^{3} b p x^{4} - i \, a^{3} b q x\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {3}{4}}\right )} \sqrt {p x^{3} + q} + 2 \, {\left (a^{2} b p x^{5} + a^{2} b q x^{2}\right )} \sqrt {-\frac {1}{a^{3} b}}}{a p^{2} x^{6} + 2 \, a p q x^{3} + b x^{4} + a q^{2}}\right ) + \frac {1}{4} i \, \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}} \log \left (\frac {a p^{2} x^{6} + 2 \, a p q x^{3} - b x^{4} + a q^{2} - 2 \, {\left (-i \, a b x^{3} \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}} + {\left (i \, a^{3} b p x^{4} + i \, a^{3} b q x\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {3}{4}}\right )} \sqrt {p x^{3} + q} + 2 \, {\left (a^{2} b p x^{5} + a^{2} b q x^{2}\right )} \sqrt {-\frac {1}{a^{3} b}}}{a p^{2} x^{6} + 2 \, a p q x^{3} + b x^{4} + a q^{2}}\right ) \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q+p x^3}}{b x^4+a \left (q+p x^3\right )^2} \, dx=\int \frac {\left (p x^{3} - 2 q\right ) \sqrt {p x^{3} + q}}{a p^{2} x^{6} + 2 a p q x^{3} + a q^{2} + b x^{4}}\, dx \]

[In]

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

[Out]

Integral((p*x**3 - 2*q)*sqrt(p*x**3 + q)/(a*p**2*x**6 + 2*a*p*q*x**3 + a*q**2 + b*x**4), x)

Maxima [F]

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

[In]

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

[Out]

integrate(sqrt(p*x^3 + q)*(p*x^3 - 2*q)/(b*x^4 + (p*x^3 + q)^2*a), x)

Giac [F]

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

[In]

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

[Out]

integrate(sqrt(p*x^3 + q)*(p*x^3 - 2*q)/(b*x^4 + (p*x^3 + q)^2*a), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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