3.23.48 \(\int \frac {\sqrt {q+p x^5} (-2 q+3 p x^5)}{b x^4+a (q+p x^5)^2} \, dx\)

Optimal. Leaf size=167 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x \sqrt {p x^5+q}}{\sqrt {a} p x^5+\sqrt {a} q-\sqrt {b} x^2}\right )}{\sqrt {2} a^{3/4} \sqrt [4]{b}}-\frac {\tanh ^{-1}\left (\frac {\frac {\sqrt [4]{a} p x^5}{\sqrt {2} \sqrt [4]{b}}+\frac {\sqrt [4]{a} q}{\sqrt {2} \sqrt [4]{b}}+\frac {\sqrt [4]{b} x^2}{\sqrt {2} \sqrt [4]{a}}}{x \sqrt {p x^5+q}}\right )}{\sqrt {2} a^{3/4} \sqrt [4]{b}} \]

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Rubi [A]  time = 0.47, antiderivative size = 242, normalized size of antiderivative = 1.45, number of steps used = 10, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, 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^5+q}}\right )}{\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^5+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^5+q}}+\sqrt {a}+\frac {\sqrt {b} x^2}{p x^5+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^5+q}}+\sqrt {a}+\frac {\sqrt {b} x^2}{p x^5+q}\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[q + p*x^5]*(-2*q + 3*p*x^5))/(b*x^4 + a*(q + p*x^5)^2),x]

[Out]

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

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Mathematica [F]  time = 0.31, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right )}{b x^4+a \left (q+p x^5\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 2.13, size = 167, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x \sqrt {q+p x^5}}{\sqrt {a} q-\sqrt {b} x^2+\sqrt {a} p x^5}\right )}{\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^5}{\sqrt {2} \sqrt [4]{b}}}{x \sqrt {q+p x^5}}\right )}{\sqrt {2} a^{3/4} \sqrt [4]{b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTan[(Sqrt[2]*a^(1/4)*b^(1/4)*x*Sqrt[q + p*x^5])/(Sqrt[a]*q - Sqrt[b]*x^2 + Sqrt[a]*p*x^5)]/(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^5)/(Sqr
t[2]*b^(1/4)))/(x*Sqrt[q + p*x^5])]/(Sqrt[2]*a^(3/4)*b^(1/4))

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fricas [B]  time = 2.18, size = 354, normalized size = 2.12 \begin {gather*} \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}} \arctan \left (\frac {a^{2} b x \left (-\frac {1}{a^{3} b}\right )^{\frac {3}{4}}}{\sqrt {p x^{5} + q}}\right ) + \frac {1}{4} \, \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}} \log \left (\frac {a p^{2} x^{10} + 2 \, a p q x^{5} - b x^{4} + a q^{2} + 2 \, \sqrt {p x^{5} + q} {\left (a b x^{3} \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}} + {\left (a^{3} b p x^{6} + a^{3} b q x\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {3}{4}}\right )} - 2 \, {\left (a^{2} b p x^{7} + a^{2} b q x^{2}\right )} \sqrt {-\frac {1}{a^{3} b}}}{a p^{2} x^{10} + 2 \, a p q x^{5} + 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^{10} + 2 \, a p q x^{5} - b x^{4} + a q^{2} - 2 \, \sqrt {p x^{5} + q} {\left (a b x^{3} \left (-\frac {1}{a^{3} b}\right )^{\frac {1}{4}} + {\left (a^{3} b p x^{6} + a^{3} b q x\right )} \left (-\frac {1}{a^{3} b}\right )^{\frac {3}{4}}\right )} - 2 \, {\left (a^{2} b p x^{7} + a^{2} b q x^{2}\right )} \sqrt {-\frac {1}{a^{3} b}}}{a p^{2} x^{10} + 2 \, a p q x^{5} + b x^{4} + a q^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1/(a^3*b))^(1/4)*arctan(a^2*b*x*(-1/(a^3*b))^(3/4)/sqrt(p*x^5 + q)) + 1/4*(-1/(a^3*b))^(1/4)*log((a*p^2*x^10
+ 2*a*p*q*x^5 - b*x^4 + a*q^2 + 2*sqrt(p*x^5 + q)*(a*b*x^3*(-1/(a^3*b))^(1/4) + (a^3*b*p*x^6 + a^3*b*q*x)*(-1
/(a^3*b))^(3/4)) - 2*(a^2*b*p*x^7 + a^2*b*q*x^2)*sqrt(-1/(a^3*b)))/(a*p^2*x^10 + 2*a*p*q*x^5 + b*x^4 + a*q^2))
- 1/4*(-1/(a^3*b))^(1/4)*log((a*p^2*x^10 + 2*a*p*q*x^5 - b*x^4 + a*q^2 - 2*sqrt(p*x^5 + q)*(a*b*x^3*(-1/(a^3*
b))^(1/4) + (a^3*b*p*x^6 + a^3*b*q*x)*(-1/(a^3*b))^(3/4)) - 2*(a^2*b*p*x^7 + a^2*b*q*x^2)*sqrt(-1/(a^3*b)))/(a
*p^2*x^10 + 2*a*p*q*x^5 + 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^5+q)^(1/2)*(3*p*x^5-2*q)/(b*x^4+a*(p*x^5+q)^2),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {p \,x^{5}+q}\, \left (3 p \,x^{5}-2 q \right )}{b \,x^{4}+a \left (p \,x^{5}+q \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (3 \, p x^{5} - 2 \, q\right )} \sqrt {p x^{5} + q}}{b x^{4} + {\left (p x^{5} + q\right )}^{2} a}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((3*p*x^5 - 2*q)*sqrt(p*x^5 + q)/(b*x^4 + (p*x^5 + q)^2*a), x)

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mupad [B]  time = 150.61, size = 498, normalized size = 2.98 \begin {gather*} \frac {\ln \left (\frac {\left (2\,\sqrt {p\,x^5+q}\,{\left (-a^3\,b\right )}^{15/4}-a^{23/2}\,b^{7/2}\,p\,x^4+a^{19/2}\,b^{7/2}\,x\,\sqrt {-a^3\,b}\right )\,\left (2\,a^{27/2}\,b^{7/2}\,q\,\sqrt {p\,x^5+q}\,{\left (-a^3\,b\right )}^{15/4}-a^{24}\,b^8\,x^3+a^{25}\,b^7\,p\,x^4\,\left (p\,x^5+q\right )+a^{23}\,b^7\,q\,x\,\sqrt {-a^3\,b}+2\,a^{23}\,b^7\,x\,\left (p\,x^5+q\right )\,\sqrt {-a^3\,b}\right )}{\left (x^2\,\sqrt {-a^3\,b}+a^2\,q+a^2\,p\,x^5\right )\,\left (4\,q\,{\left (-a^3\,b\right )}^{15/2}+2\,p\,x^5\,{\left (-a^3\,b\right )}^{15/2}+a^{22}\,b^8\,x^2-a^{23}\,b^7\,p^2\,x^8\right )}\right )\,{\left (-a^3\,b\right )}^{1/4}}{2\,a^{3/2}\,\sqrt {b}}+\frac {\ln \left (\frac {\left (-2\,\sqrt {p\,x^5+q}\,{\left (-a^3\,b\right )}^{15/4}+a^{23/2}\,b^{7/2}\,p\,x^4\,1{}\mathrm {i}+a^{19/2}\,b^{7/2}\,x\,\sqrt {-a^3\,b}\,1{}\mathrm {i}\right )\,\left (a^{24}\,b^8\,x^3\,1{}\mathrm {i}-2\,a^{27/2}\,b^{7/2}\,q\,\sqrt {p\,x^5+q}\,{\left (-a^3\,b\right )}^{15/4}-a^{25}\,b^7\,p\,x^4\,\left (p\,x^5+q\right )\,1{}\mathrm {i}+a^{23}\,b^7\,q\,x\,\sqrt {-a^3\,b}\,1{}\mathrm {i}+a^{23}\,b^7\,x\,\left (p\,x^5+q\right )\,\sqrt {-a^3\,b}\,2{}\mathrm {i}\right )}{\left (a^2\,q-x^2\,\sqrt {-a^3\,b}+a^2\,p\,x^5\right )\,\left (4\,q\,{\left (-a^3\,b\right )}^{15/2}+2\,p\,x^5\,{\left (-a^3\,b\right )}^{15/2}-a^{22}\,b^8\,x^2+a^{23}\,b^7\,p^2\,x^8\right )}\right )\,{\left (-a^3\,b\right )}^{1/4}\,1{}\mathrm {i}}{2\,a^{3/2}\,\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(((2*(q + p*x^5)^(1/2)*(-a^3*b)^(15/4) - a^(23/2)*b^(7/2)*p*x^4 + a^(19/2)*b^(7/2)*x*(-a^3*b)^(1/2))*(2*a^
(27/2)*b^(7/2)*q*(q + p*x^5)^(1/2)*(-a^3*b)^(15/4) - a^24*b^8*x^3 + a^25*b^7*p*x^4*(q + p*x^5) + a^23*b^7*q*x*
(-a^3*b)^(1/2) + 2*a^23*b^7*x*(q + p*x^5)*(-a^3*b)^(1/2)))/((x^2*(-a^3*b)^(1/2) + a^2*q + a^2*p*x^5)*(4*q*(-a^
3*b)^(15/2) + 2*p*x^5*(-a^3*b)^(15/2) + a^22*b^8*x^2 - a^23*b^7*p^2*x^8)))*(-a^3*b)^(1/4))/(2*a^(3/2)*b^(1/2))
+ (log(((a^(23/2)*b^(7/2)*p*x^4*1i - 2*(q + p*x^5)^(1/2)*(-a^3*b)^(15/4) + a^(19/2)*b^(7/2)*x*(-a^3*b)^(1/2)*
1i)*(a^24*b^8*x^3*1i - 2*a^(27/2)*b^(7/2)*q*(q + p*x^5)^(1/2)*(-a^3*b)^(15/4) - a^25*b^7*p*x^4*(q + p*x^5)*1i
+ a^23*b^7*q*x*(-a^3*b)^(1/2)*1i + a^23*b^7*x*(q + p*x^5)*(-a^3*b)^(1/2)*2i))/((a^2*q - x^2*(-a^3*b)^(1/2) + a
^2*p*x^5)*(4*q*(-a^3*b)^(15/2) + 2*p*x^5*(-a^3*b)^(15/2) - a^22*b^8*x^2 + a^23*b^7*p^2*x^8)))*(-a^3*b)^(1/4)*1
i)/(2*a^(3/2)*b^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {p x^{5} + q} \left (3 p x^{5} - 2 q\right )}{a p^{2} x^{10} + 2 a p q x^{5} + a q^{2} + b x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((p*x**5+q)**(1/2)*(3*p*x**5-2*q)/(b*x**4+a*(p*x**5+q)**2),x)

[Out]

Integral(sqrt(p*x**5 + q)*(3*p*x**5 - 2*q)/(a*p**2*x**10 + 2*a*p*q*x**5 + a*q**2 + b*x**4), x)

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