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

Optimal. Leaf size=119 \[ \frac {\sqrt {p^2 x^4+q^2} \left (6 a p^4 x^8+20 a p^3 q x^6+12 a p^2 q^2 x^4+20 a p q^3 x^2+6 a q^4+15 b p x^5+15 b q x^3\right )}{30 x^5}-b p q \log \left (\sqrt {p^2 x^4+q^2}+p x^2+q\right )+b p q \log (x) \]

________________________________________________________________________________________

Rubi [A]  time = 0.49, antiderivative size = 163, normalized size of antiderivative = 1.37, number of steps used = 16, number of rules used = 13, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {1833, 1584, 1252, 813, 844, 217, 206, 266, 63, 208, 1835, 1586, 449} \begin {gather*} \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 \tanh ^{-1}\left (\frac {\sqrt {p^2 x^4+q^2}}{q}\right )+\frac {b \left (p x^2+q\right ) \sqrt {p^2 x^4+q^2}}{2 x^2}-\frac {1}{2} b p q \tanh ^{-1}\left (\frac {p x^2}{\sqrt {p^2 x^4+q^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[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 63

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 206

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

Rule 208

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

Rule 217

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 266

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 449

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 813

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 844

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 1252

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 1584

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 1586

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 1833

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)*Sum[Coeff[Pq, x, j + (k*n)/2]*x^((k*n)/2), {k, 0, (2*(q - j))/n + 1}]*(a + b*x^n)^p)/c^j, {
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 1835

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{Pq0 = Coeff[Pq, x, 0]}, Simp[(Pq
0*(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*} \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 \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} \operatorname {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} \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 \tanh ^{-1}\left (\frac {p x^2}{\sqrt {q^2+p^2 x^4}}\right )+\frac {\left (b q^2\right ) \operatorname {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 \tanh ^{-1}\left (\frac {p x^2}{\sqrt {q^2+p^2 x^4}}\right )-\frac {1}{2} b p q \tanh ^{-1}\left (\frac {\sqrt {q^2+p^2 x^4}}{q}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.27, size = 354, normalized size = 2.97 \begin {gather*} \frac {6 a q^4 \sqrt {p^2 x^4+q^2} \, _2F_1\left (-\frac {5}{4},-\frac {1}{2};-\frac {1}{4};-\frac {p^2 x^4}{q^2}\right )+20 a p q^3 x^2 \sqrt {p^2 x^4+q^2} \, _2F_1\left (-\frac {3}{4},-\frac {1}{2};\frac {1}{4};-\frac {p^2 x^4}{q^2}\right )+10 a p^4 x^8 \sqrt {p^2 x^4+q^2} \, _2F_1\left (-\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {p^2 x^4}{q^2}\right )+60 a p^3 q x^6 \sqrt {p^2 x^4+q^2} \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};-\frac {p^2 x^4}{q^2}\right )+15 b p x^5 \sqrt {p^2 x^4+q^2} \sqrt {\frac {p^2 x^4}{q^2}+1}-15 b p q x^5 \sqrt {\frac {p^2 x^4}{q^2}+1} \tanh ^{-1}\left (\frac {\sqrt {p^2 x^4+q^2}}{q}\right )+15 b q x^3 \sqrt {p^2 x^4+q^2} \sqrt {\frac {p^2 x^4}{q^2}+1}-15 b p x^5 \sqrt {p^2 x^4+q^2} \sinh ^{-1}\left (\frac {p x^2}{q}\right )}{30 x^5 \sqrt {\frac {p^2 x^4}{q^2}+1}} \end {gather*}

Antiderivative was successfully verified.

[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]

(15*b*q*x^3*Sqrt[q^2 + p^2*x^4]*Sqrt[1 + (p^2*x^4)/q^2] + 15*b*p*x^5*Sqrt[q^2 + p^2*x^4]*Sqrt[1 + (p^2*x^4)/q^
2] - 15*b*p*x^5*Sqrt[q^2 + p^2*x^4]*ArcSinh[(p*x^2)/q] - 15*b*p*q*x^5*Sqrt[1 + (p^2*x^4)/q^2]*ArcTanh[Sqrt[q^2
+ p^2*x^4]/q] + 6*a*q^4*Sqrt[q^2 + p^2*x^4]*Hypergeometric2F1[-5/4, -1/2, -1/4, -((p^2*x^4)/q^2)] + 20*a*p*q^
3*x^2*Sqrt[q^2 + p^2*x^4]*Hypergeometric2F1[-3/4, -1/2, 1/4, -((p^2*x^4)/q^2)] + 60*a*p^3*q*x^6*Sqrt[q^2 + p^2
*x^4]*Hypergeometric2F1[-1/2, 1/4, 5/4, -((p^2*x^4)/q^2)] + 10*a*p^4*x^8*Sqrt[q^2 + p^2*x^4]*Hypergeometric2F1
[-1/2, 3/4, 7/4, -((p^2*x^4)/q^2)])/(30*x^5*Sqrt[1 + (p^2*x^4)/q^2])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 3.58, size = 119, normalized size = 1.00 \begin {gather*} \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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-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]]

________________________________________________________________________________________

fricas [A]  time = 0.73, size = 117, normalized size = 0.98 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [A]  time = 0.25, size = 172, normalized size = 1.45

method	result	size
elliptic	\(\frac {p b \sqrt {p^{2} x^{4}+q^{2}}}{2}-\frac {b \,p^{2} q \ln \left (\frac {p^{2} x^{2}}{\sqrt {p^{2}}}+\sqrt {p^{2} x^{4}+q^{2}}\right )}{2 \sqrt {p^{2}}}+\frac {b q \sqrt {p^{2} x^{4}+q^{2}}}{2 x^{2}}-\frac {p b \,q^{2} \ln \left (\frac {2 q^{2}+2 \sqrt {q^{2}}\, \sqrt {p^{2} x^{4}+q^{2}}}{x^{2}}\right )}{2 \sqrt {q^{2}}}+4 a \left (\frac {p q \left (p^{2} x^{4}+q^{2}\right )^{\frac {3}{2}} \sqrt {2}}{12 x^{3}}+\frac {\left (p^{2} x^{4}+q^{2}\right )^{\frac {5}{2}} \sqrt {2}}{40 x^{5}}\right ) \sqrt {2}\)	\(172\)
risch	\(\frac {\sqrt {p^{2} x^{4}+q^{2}}\, q \left (12 a \,p^{2} q \,x^{4}+20 a p \,q^{2} x^{2}+6 a \,q^{3}+15 b \,x^{3}\right )}{30 x^{5}}+\frac {a \,p^{4} x^{3} \sqrt {p^{2} x^{4}+q^{2}}}{5}+\frac {2 a \,p^{3} q x \sqrt {p^{2} x^{4}+q^{2}}}{3}+\frac {p b \sqrt {p^{2} x^{4}+q^{2}}}{2}-\frac {b \,p^{2} q \ln \left (\frac {p^{2} x^{2}}{\sqrt {p^{2}}}+\sqrt {p^{2} x^{4}+q^{2}}\right )}{2 \sqrt {p^{2}}}-\frac {p b \,q^{2} \ln \left (\frac {2 q^{2}+2 \sqrt {q^{2}}\, \sqrt {p^{2} x^{4}+q^{2}}}{x^{2}}\right )}{2 \sqrt {q^{2}}}\)	\(196\)
default	\(a \,p^{4} \left (\frac {x^{3} \sqrt {p^{2} x^{4}+q^{2}}}{5}+\frac {2 i q^{3} \sqrt {1-\frac {i p \,x^{2}}{q}}\, \sqrt {1+\frac {i p \,x^{2}}{q}}\, \left (\EllipticF \left (x \sqrt {\frac {i p}{q}}, i\right )-\EllipticE \left (x \sqrt {\frac {i p}{q}}, i\right )\right )}{5 \sqrt {\frac {i p}{q}}\, \sqrt {p^{2} x^{4}+q^{2}}\, p}\right )+2 q a \,p^{3} \left (\frac {x \sqrt {p^{2} x^{4}+q^{2}}}{3}+\frac {2 q^{2} \sqrt {1-\frac {i p \,x^{2}}{q}}\, \sqrt {1+\frac {i p \,x^{2}}{q}}\, \EllipticF \left (x \sqrt {\frac {i p}{q}}, i\right )}{3 \sqrt {\frac {i p}{q}}\, \sqrt {p^{2} x^{4}+q^{2}}}\right )-2 q^{3} a p \left (-\frac {\sqrt {p^{2} x^{4}+q^{2}}}{3 x^{3}}+\frac {2 p^{2} \sqrt {1-\frac {i p \,x^{2}}{q}}\, \sqrt {1+\frac {i p \,x^{2}}{q}}\, \EllipticF \left (x \sqrt {\frac {i p}{q}}, i\right )}{3 \sqrt {\frac {i p}{q}}\, \sqrt {p^{2} x^{4}+q^{2}}}\right )+p b \left (\frac {\sqrt {p^{2} x^{4}+q^{2}}}{2}-\frac {q^{2} \ln \left (\frac {2 q^{2}+2 \sqrt {q^{2}}\, \sqrt {p^{2} x^{4}+q^{2}}}{x^{2}}\right )}{2 \sqrt {q^{2}}}\right )-q b \left (-\frac {\left (p^{2} x^{4}+q^{2}\right )^{\frac {3}{2}}}{2 q^{2} x^{2}}+\frac {p^{2} x^{2} \sqrt {p^{2} x^{4}+q^{2}}}{2 q^{2}}+\frac {p^{2} \ln \left (\frac {p^{2} x^{2}}{\sqrt {p^{2}}}+\sqrt {p^{2} x^{4}+q^{2}}\right )}{2 \sqrt {p^{2}}}\right )-a \,q^{4} \left (-\frac {\sqrt {p^{2} x^{4}+q^{2}}}{5 x^{5}}-\frac {2 p^{2} \sqrt {p^{2} x^{4}+q^{2}}}{5 q^{2} x}+\frac {2 i p^{3} \sqrt {1-\frac {i p \,x^{2}}{q}}\, \sqrt {1+\frac {i p \,x^{2}}{q}}\, \left (\EllipticF \left (x \sqrt {\frac {i p}{q}}, i\right )-\EllipticE \left (x \sqrt {\frac {i p}{q}}, i\right )\right )}{5 q \sqrt {\frac {i p}{q}}\, \sqrt {p^{2} x^{4}+q^{2}}}\right )\)	\(590\)

Verification of antiderivative is not currently implemented for this CAS.

[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]

1/2*p*b*(p^2*x^4+q^2)^(1/2)-1/2*b*p^2*q*ln(p^2*x^2/(p^2)^(1/2)+(p^2*x^4+q^2)^(1/2))/(p^2)^(1/2)+1/2*b*q/x^2*(p
^2*x^4+q^2)^(1/2)-1/2*p*b*q^2/(q^2)^(1/2)*ln((2*q^2+2*(q^2)^(1/2)*(p^2*x^4+q^2)^(1/2))/x^2)+4*a*(1/12*p*q*(p^2
*x^4+q^2)^(3/2)*2^(1/2)/x^3+1/40*(p^2*x^4+q^2)^(5/2)*2^(1/2)/x^5)*2^(1/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [C]  time = 8.30, size = 323, normalized size = 2.71 \begin {gather*} \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}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

________________________________________________________________________________________