\(\int \frac {(-q+p x^2) \sqrt {q^2+p^2 x^4} (b x^3+a (q+p x^2)^3)}{x^6} \, dx\) [1775]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (verified)
   Fricas [A] (verification not implemented)
   Sympy [C] (verification not implemented)
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 45, antiderivative size = 119 \[ \int \frac {\left (-q+p x^2\right ) \sqrt {q^2+p^2 x^4} \left (b x^3+a \left (q+p x^2\right )^3\right )}{x^6} \, dx=\frac {\sqrt {q^2+p^2 x^4} \left (6 a q^4+20 a p q^3 x^2+15 b q x^3+12 a p^2 q^2 x^4+15 b p x^5+20 a p^3 q x^6+6 a p^4 x^8\right )}{30 x^5}+b p q \log (x)-b p q \log \left (q+p x^2+\sqrt {q^2+p^2 x^4}\right ) \]

[Out]

1/30*(p^2*x^4+q^2)^(1/2)*(6*a*p^4*x^8+20*a*p^3*q*x^6+12*a*p^2*q^2*x^4+20*a*p*q^3*x^2+15*b*p*x^5+6*a*q^4+15*b*q
*x^3)/x^5+b*p*q*ln(x)-b*p*q*ln(q+p*x^2+(p^2*x^4+q^2)^(1/2))

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.37, number of steps used = 16, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {1847, 1598, 1266, 827, 858, 223, 212, 272, 65, 214, 1849, 1600, 460} \[ \int \frac {\left (-q+p x^2\right ) \sqrt {q^2+p^2 x^4} \left (b x^3+a \left (q+p x^2\right )^3\right )}{x^6} \, dx=\frac {a p^2 \left (p^2 x^4+q^2\right )^{3/2}}{5 x}+\frac {a q^2 \left (p^2 x^4+q^2\right )^{3/2}}{5 x^5}+\frac {2 a p q \left (p^2 x^4+q^2\right )^{3/2}}{3 x^3}-\frac {1}{2} b p q \text {arctanh}\left (\frac {\sqrt {p^2 x^4+q^2}}{q}\right )-\frac {1}{2} b p q \text {arctanh}\left (\frac {p x^2}{\sqrt {p^2 x^4+q^2}}\right )+\frac {b \left (p x^2+q\right ) \sqrt {p^2 x^4+q^2}}{2 x^2} \]

[In]

Int[((-q + p*x^2)*Sqrt[q^2 + p^2*x^4]*(b*x^3 + a*(q + p*x^2)^3))/x^6,x]

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 212

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

Rule 214

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

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 460

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

Rule 827

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
 + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*((a + c*x^2)^p/(e^2*(m + 1)*(m + 2*p + 2))), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 858

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

Rule 1266

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
 - 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

Rule 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
 p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1847

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[
Sum[((c*x)^(m + j)/c^j)*Sum[Coeff[Pq, x, j + k*(n/2)]*x^(k*(n/2)), {k, 0, 2*((q - j)/n) + 1}]*(a + b*x^n)^p, {
j, 0, n/2 - 1}], x]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] &&  !PolyQ[Pq, x^(n/2)]

Rule 1849

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{Pq0 = Coeff[Pq, x, 0]}, Simp[Pq0
*(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] + Dist[1/(2*a*c*(m + 1)), Int[(c*x)^(m + 1)*ExpandToSum
[2*a*(m + 1)*((Pq - Pq0)/x) - 2*b*Pq0*(m + n*(p + 1) + 1)*x^(n - 1), x]*(a + b*x^n)^p, x], x] /; NeQ[Pq0, 0]]
/; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[m, -1] && LeQ[n - 1, Expon[Pq, x]]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\left (-b q x^2+b p x^4\right ) \sqrt {q^2+p^2 x^4}}{x^5}+\frac {\sqrt {q^2+p^2 x^4} \left (-a q^4-2 a p q^3 x^2+2 a p^3 q x^6+a p^4 x^8\right )}{x^6}\right ) \, dx \\ & = \int \frac {\left (-b q x^2+b p x^4\right ) \sqrt {q^2+p^2 x^4}}{x^5} \, dx+\int \frac {\sqrt {q^2+p^2 x^4} \left (-a q^4-2 a p q^3 x^2+2 a p^3 q x^6+a p^4 x^8\right )}{x^6} \, dx \\ & = \frac {a q^2 \left (q^2+p^2 x^4\right )^{3/2}}{5 x^5}-\frac {\int \frac {\sqrt {q^2+p^2 x^4} \left (20 a p q^5 x+2 a p^2 q^4 x^3-20 a p^3 q^3 x^5-10 a p^4 q^2 x^7\right )}{x^5} \, dx}{10 q^2}+\int \frac {\left (-b q+b p x^2\right ) \sqrt {q^2+p^2 x^4}}{x^3} \, dx \\ & = \frac {a q^2 \left (q^2+p^2 x^4\right )^{3/2}}{5 x^5}+\frac {1}{2} \text {Subst}\left (\int \frac {(-b q+b p x) \sqrt {q^2+p^2 x^2}}{x^2} \, dx,x,x^2\right )-\frac {\int \frac {\sqrt {q^2+p^2 x^4} \left (20 a p q^5+2 a p^2 q^4 x^2-20 a p^3 q^3 x^4-10 a p^4 q^2 x^6\right )}{x^4} \, dx}{10 q^2} \\ & = \frac {b \left (q+p x^2\right ) \sqrt {q^2+p^2 x^4}}{2 x^2}+\frac {a q^2 \left (q^2+p^2 x^4\right )^{3/2}}{5 x^5}+\frac {2 a p q \left (q^2+p^2 x^4\right )^{3/2}}{3 x^3}-\frac {1}{4} \text {Subst}\left (\int \frac {-2 b p q^2+2 b p^2 q x}{x \sqrt {q^2+p^2 x^2}} \, dx,x,x^2\right )+\frac {\int \frac {\sqrt {q^2+p^2 x^4} \left (-12 a p^2 q^6 x+60 a p^4 q^4 x^5\right )}{x^3} \, dx}{60 q^4} \\ & = \frac {b \left (q+p x^2\right ) \sqrt {q^2+p^2 x^4}}{2 x^2}+\frac {a q^2 \left (q^2+p^2 x^4\right )^{3/2}}{5 x^5}+\frac {2 a p q \left (q^2+p^2 x^4\right )^{3/2}}{3 x^3}+\frac {\int \frac {\sqrt {q^2+p^2 x^4} \left (-12 a p^2 q^6+60 a p^4 q^4 x^4\right )}{x^2} \, dx}{60 q^4}-\frac {1}{2} \left (b p^2 q\right ) \text {Subst}\left (\int \frac {1}{\sqrt {q^2+p^2 x^2}} \, dx,x,x^2\right )+\frac {1}{2} \left (b p q^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {q^2+p^2 x^2}} \, dx,x,x^2\right ) \\ & = \frac {b \left (q+p x^2\right ) \sqrt {q^2+p^2 x^4}}{2 x^2}+\frac {a q^2 \left (q^2+p^2 x^4\right )^{3/2}}{5 x^5}+\frac {2 a p q \left (q^2+p^2 x^4\right )^{3/2}}{3 x^3}+\frac {a p^2 \left (q^2+p^2 x^4\right )^{3/2}}{5 x}-\frac {1}{2} \left (b p^2 q\right ) \text {Subst}\left (\int \frac {1}{1-p^2 x^2} \, dx,x,\frac {x^2}{\sqrt {q^2+p^2 x^4}}\right )+\frac {1}{4} \left (b p q^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {q^2+p^2 x}} \, dx,x,x^4\right ) \\ & = \frac {b \left (q+p x^2\right ) \sqrt {q^2+p^2 x^4}}{2 x^2}+\frac {a q^2 \left (q^2+p^2 x^4\right )^{3/2}}{5 x^5}+\frac {2 a p q \left (q^2+p^2 x^4\right )^{3/2}}{3 x^3}+\frac {a p^2 \left (q^2+p^2 x^4\right )^{3/2}}{5 x}-\frac {1}{2} b p q \text {arctanh}\left (\frac {p x^2}{\sqrt {q^2+p^2 x^4}}\right )+\frac {\left (b q^2\right ) \text {Subst}\left (\int \frac {1}{-\frac {q^2}{p^2}+\frac {x^2}{p^2}} \, dx,x,\sqrt {q^2+p^2 x^4}\right )}{2 p} \\ & = \frac {b \left (q+p x^2\right ) \sqrt {q^2+p^2 x^4}}{2 x^2}+\frac {a q^2 \left (q^2+p^2 x^4\right )^{3/2}}{5 x^5}+\frac {2 a p q \left (q^2+p^2 x^4\right )^{3/2}}{3 x^3}+\frac {a p^2 \left (q^2+p^2 x^4\right )^{3/2}}{5 x}-\frac {1}{2} b p q \text {arctanh}\left (\frac {p x^2}{\sqrt {q^2+p^2 x^4}}\right )-\frac {1}{2} b p q \text {arctanh}\left (\frac {\sqrt {q^2+p^2 x^4}}{q}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 2.35 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-q+p x^2\right ) \sqrt {q^2+p^2 x^4} \left (b x^3+a \left (q+p x^2\right )^3\right )}{x^6} \, dx=\frac {\sqrt {q^2+p^2 x^4} \left (6 a q^4+20 a p q^3 x^2+15 b q x^3+12 a p^2 q^2 x^4+15 b p x^5+20 a p^3 q x^6+6 a p^4 x^8\right )}{30 x^5}+b p q \log (x)-b p q \log \left (q+p x^2+\sqrt {q^2+p^2 x^4}\right ) \]

[In]

Integrate[((-q + p*x^2)*Sqrt[q^2 + p^2*x^4]*(b*x^3 + a*(q + p*x^2)^3))/x^6,x]

[Out]

(Sqrt[q^2 + p^2*x^4]*(6*a*q^4 + 20*a*p*q^3*x^2 + 15*b*q*x^3 + 12*a*p^2*q^2*x^4 + 15*b*p*x^5 + 20*a*p^3*q*x^6 +
 6*a*p^4*x^8))/(30*x^5) + b*p*q*Log[x] - b*p*q*Log[q + p*x^2 + Sqrt[q^2 + p^2*x^4]]

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 3.86 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.01

method result size
pseudoelliptic \(\frac {-5 b p q \ln \left (\frac {\sqrt {p^{2} x^{4}+q^{2}}+\left (p \,x^{2}+q \right ) \operatorname {csgn}\left (p \right )}{x}\right ) \operatorname {csgn}\left (p \right ) x^{5}+\sqrt {p^{2} x^{4}+q^{2}}\, \left (a \,p^{4} x^{8}+\frac {10}{3} a \,p^{3} q \,x^{6}+2 a \,p^{2} q^{2} x^{4}+\frac {10}{3} a p \,q^{3} x^{2}+\frac {5}{2} b p \,x^{5}+a \,q^{4}+\frac {5}{2} b q \,x^{3}\right )}{5 x^{5}}\) \(120\)
elliptic \(\frac {b \left (p \sqrt {p^{2} x^{4}+q^{2}}-\frac {p^{2} q \ln \left (\frac {p^{2} x^{2}}{\sqrt {p^{2}}}+\sqrt {p^{2} x^{4}+q^{2}}\right )}{\sqrt {p^{2}}}+\frac {q \sqrt {p^{2} x^{4}+q^{2}}}{x^{2}}-\frac {p \,q^{2} \ln \left (\frac {2 q^{2}+2 \sqrt {q^{2}}\, \sqrt {p^{2} x^{4}+q^{2}}}{x^{2}}\right )}{\sqrt {q^{2}}}\right )}{2}+4 a \left (\frac {p q \left (p^{2} x^{4}+q^{2}\right )^{\frac {3}{2}} \sqrt {2}}{12 x^{3}}+\frac {\left (p^{2} x^{4}+q^{2}\right )^{\frac {5}{2}} \sqrt {2}}{40 x^{5}}\right ) \sqrt {2}\) \(170\)
risch \(\frac {q \sqrt {p^{2} x^{4}+q^{2}}\, \left (12 a \,p^{2} q \,x^{4}+20 a p \,q^{2} x^{2}+6 a \,q^{3}+15 b \,x^{3}\right )}{30 x^{5}}+\frac {a \,p^{4} x^{3} \sqrt {p^{2} x^{4}+q^{2}}}{5}+\frac {p b \sqrt {p^{2} x^{4}+q^{2}}}{2}-\frac {p b \,q^{2} \ln \left (\frac {2 q^{2}+2 \sqrt {q^{2}}\, \sqrt {p^{2} x^{4}+q^{2}}}{x^{2}}\right )}{2 \sqrt {q^{2}}}+\frac {2 a \,p^{3} q x \sqrt {p^{2} x^{4}+q^{2}}}{3}-\frac {q b \,p^{2} \ln \left (\frac {p^{2} x^{2}}{\sqrt {p^{2}}}+\sqrt {p^{2} x^{4}+q^{2}}\right )}{2 \sqrt {p^{2}}}\) \(196\)
default \(a \,p^{4} \left (\frac {x^{3} \sqrt {p^{2} x^{4}+q^{2}}}{5}+\frac {2 i q^{3} \sqrt {1-\frac {i p \,x^{2}}{q}}\, \sqrt {1+\frac {i p \,x^{2}}{q}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i p}{q}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i p}{q}}, i\right )\right )}{5 \sqrt {\frac {i p}{q}}\, \sqrt {p^{2} x^{4}+q^{2}}\, p}\right )+p b \left (\frac {\sqrt {p^{2} x^{4}+q^{2}}}{2}-\frac {q^{2} \ln \left (\frac {2 q^{2}+2 \sqrt {q^{2}}\, \sqrt {p^{2} x^{4}+q^{2}}}{x^{2}}\right )}{2 \sqrt {q^{2}}}\right )+2 q a \,p^{3} \left (\frac {x \sqrt {p^{2} x^{4}+q^{2}}}{3}+\frac {2 q^{2} \sqrt {1-\frac {i p \,x^{2}}{q}}\, \sqrt {1+\frac {i p \,x^{2}}{q}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i p}{q}}, i\right )}{3 \sqrt {\frac {i p}{q}}\, \sqrt {p^{2} x^{4}+q^{2}}}\right )-q b \left (-\frac {\left (p^{2} x^{4}+q^{2}\right )^{\frac {3}{2}}}{2 q^{2} x^{2}}+\frac {p^{2} x^{2} \sqrt {p^{2} x^{4}+q^{2}}}{2 q^{2}}+\frac {p^{2} \ln \left (\frac {p^{2} x^{2}}{\sqrt {p^{2}}}+\sqrt {p^{2} x^{4}+q^{2}}\right )}{2 \sqrt {p^{2}}}\right )-a \,q^{4} \left (-\frac {\sqrt {p^{2} x^{4}+q^{2}}}{5 x^{5}}-\frac {2 p^{2} \sqrt {p^{2} x^{4}+q^{2}}}{5 q^{2} x}+\frac {2 i p^{3} \sqrt {1-\frac {i p \,x^{2}}{q}}\, \sqrt {1+\frac {i p \,x^{2}}{q}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i p}{q}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i p}{q}}, i\right )\right )}{5 q \sqrt {\frac {i p}{q}}\, \sqrt {p^{2} x^{4}+q^{2}}}\right )-2 q^{3} a p \left (-\frac {\sqrt {p^{2} x^{4}+q^{2}}}{3 x^{3}}+\frac {2 p^{2} \sqrt {1-\frac {i p \,x^{2}}{q}}\, \sqrt {1+\frac {i p \,x^{2}}{q}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i p}{q}}, i\right )}{3 \sqrt {\frac {i p}{q}}\, \sqrt {p^{2} x^{4}+q^{2}}}\right )\) \(590\)

[In]

int((p*x^2-q)*(p^2*x^4+q^2)^(1/2)*(b*x^3+a*(p*x^2+q)^3)/x^6,x,method=_RETURNVERBOSE)

[Out]

1/5*(-5*b*p*q*ln(((p^2*x^4+q^2)^(1/2)+(p*x^2+q)*csgn(p))/x)*csgn(p)*x^5+(p^2*x^4+q^2)^(1/2)*(a*p^4*x^8+10/3*a*
p^3*q*x^6+2*a*p^2*q^2*x^4+10/3*a*p*q^3*x^2+5/2*b*p*x^5+a*q^4+5/2*b*q*x^3))/x^5

Fricas [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.98 \[ \int \frac {\left (-q+p x^2\right ) \sqrt {q^2+p^2 x^4} \left (b x^3+a \left (q+p x^2\right )^3\right )}{x^6} \, dx=\frac {30 \, b p q x^{5} \log \left (\frac {p x^{2} + q - \sqrt {p^{2} x^{4} + q^{2}}}{x}\right ) + {\left (6 \, a p^{4} x^{8} + 20 \, a p^{3} q x^{6} + 12 \, a p^{2} q^{2} x^{4} + 20 \, a p q^{3} x^{2} + 15 \, b p x^{5} + 6 \, a q^{4} + 15 \, b q x^{3}\right )} \sqrt {p^{2} x^{4} + q^{2}}}{30 \, x^{5}} \]

[In]

integrate((p*x^2-q)*(p^2*x^4+q^2)^(1/2)*(b*x^3+a*(p*x^2+q)^3)/x^6,x, algorithm="fricas")

[Out]

1/30*(30*b*p*q*x^5*log((p*x^2 + q - sqrt(p^2*x^4 + q^2))/x) + (6*a*p^4*x^8 + 20*a*p^3*q*x^6 + 12*a*p^2*q^2*x^4
 + 20*a*p*q^3*x^2 + 15*b*p*x^5 + 6*a*q^4 + 15*b*q*x^3)*sqrt(p^2*x^4 + q^2))/x^5

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.76 (sec) , antiderivative size = 323, normalized size of antiderivative = 2.71 \[ \int \frac {\left (-q+p x^2\right ) \sqrt {q^2+p^2 x^4} \left (b x^3+a \left (q+p x^2\right )^3\right )}{x^6} \, dx=\frac {a p^{4} q x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {p^{2} x^{4} e^{i \pi }}{q^{2}}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {a p^{3} q^{2} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {p^{2} x^{4} e^{i \pi }}{q^{2}}} \right )}}{2 \Gamma \left (\frac {5}{4}\right )} - \frac {a p q^{4} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {p^{2} x^{4} e^{i \pi }}{q^{2}}} \right )}}{2 x^{3} \Gamma \left (\frac {1}{4}\right )} - \frac {a q^{5} \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {p^{2} x^{4} e^{i \pi }}{q^{2}}} \right )}}{4 x^{5} \Gamma \left (- \frac {1}{4}\right )} + \frac {b p^{2} x^{2}}{2 \sqrt {\frac {p^{2} x^{4}}{q^{2}} + 1}} + \frac {b p^{2} x^{2}}{2 \sqrt {1 + \frac {q^{2}}{p^{2} x^{4}}}} - \frac {b p q \operatorname {asinh}{\left (\frac {q}{p x^{2}} \right )}}{2} - \frac {b p q \operatorname {asinh}{\left (\frac {p x^{2}}{q} \right )}}{2} + \frac {b q^{2}}{2 x^{2} \sqrt {\frac {p^{2} x^{4}}{q^{2}} + 1}} + \frac {b q^{2}}{2 x^{2} \sqrt {1 + \frac {q^{2}}{p^{2} x^{4}}}} \]

[In]

integrate((p*x**2-q)*(p**2*x**4+q**2)**(1/2)*(b*x**3+a*(p*x**2+q)**3)/x**6,x)

[Out]

a*p**4*q*x**3*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), p**2*x**4*exp_polar(I*pi)/q**2)/(4*gamma(7/4)) + a*p**3*q*
*2*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), p**2*x**4*exp_polar(I*pi)/q**2)/(2*gamma(5/4)) - a*p*q**4*gamma(-3/
4)*hyper((-3/4, -1/2), (1/4,), p**2*x**4*exp_polar(I*pi)/q**2)/(2*x**3*gamma(1/4)) - a*q**5*gamma(-5/4)*hyper(
(-5/4, -1/2), (-1/4,), p**2*x**4*exp_polar(I*pi)/q**2)/(4*x**5*gamma(-1/4)) + b*p**2*x**2/(2*sqrt(p**2*x**4/q*
*2 + 1)) + b*p**2*x**2/(2*sqrt(1 + q**2/(p**2*x**4))) - b*p*q*asinh(q/(p*x**2))/2 - b*p*q*asinh(p*x**2/q)/2 +
b*q**2/(2*x**2*sqrt(p**2*x**4/q**2 + 1)) + b*q**2/(2*x**2*sqrt(1 + q**2/(p**2*x**4)))

Maxima [F]

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

[In]

integrate((p*x^2-q)*(p^2*x^4+q^2)^(1/2)*(b*x^3+a*(p*x^2+q)^3)/x^6,x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((p*x^2-q)*(p^2*x^4+q^2)^(1/2)*(b*x^3+a*(p*x^2+q)^3)/x^6,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

int(-((p^2*x^4 + q^2)^(1/2)*(q - p*x^2)*(a*(q + p*x^2)^3 + b*x^3))/x^6,x)

[Out]

-int(((p^2*x^4 + q^2)^(1/2)*(q - p*x^2)*(a*(q + p*x^2)^3 + b*x^3))/x^6, x)