3.8.31 \(\int \frac {2 (-2 q+p x^6) \sqrt {q+p x^6} (a q+b x^4+a p x^6)}{x^{11}} \, dx\)

Optimal. Leaf size=56 \[ \frac {2 \sqrt {p x^6+q} \left (3 a p^2 x^{12}+6 a p q x^6+3 a q^2+5 b p x^{10}+5 b q x^4\right )}{15 x^{10}} \]

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Rubi [A]  time = 0.14, antiderivative size = 39, normalized size of antiderivative = 0.70, number of steps used = 8, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {12, 1833, 1584, 446, 74, 1478, 449} \begin {gather*} \frac {2 a \left (p x^6+q\right )^{5/2}}{5 x^{10}}+\frac {2 b \left (p x^6+q\right )^{3/2}}{3 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*(-2*q + p*x^6)*Sqrt[q + p*x^6]*(a*q + b*x^4 + a*p*x^6))/x^11,x]

[Out]

(2*b*(q + p*x^6)^(3/2))/(3*x^6) + (2*a*(q + p*x^6)^(5/2))/(5*x^10)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && 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 1478

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))^(q_.)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_.), x_Sym
bol] :> Int[(f*x)^m*(d + e*x^n)^(q + p)*(a/d + (c*x^n)/e)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, q}, x] && Eq
Q[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

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 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)]

Rubi steps

\begin {align*} \int \frac {2 \left (-2 q+p x^6\right ) \sqrt {q+p x^6} \left (a q+b x^4+a p x^6\right )}{x^{11}} \, dx &=2 \int \frac {\left (-2 q+p x^6\right ) \sqrt {q+p x^6} \left (a q+b x^4+a p x^6\right )}{x^{11}} \, dx\\ &=2 \int \left (\frac {\sqrt {q+p x^6} \left (-2 b q x^3+b p x^9\right )}{x^{10}}+\frac {\sqrt {q+p x^6} \left (-2 a q^2-a p q x^6+a p^2 x^{12}\right )}{x^{11}}\right ) \, dx\\ &=2 \int \frac {\sqrt {q+p x^6} \left (-2 b q x^3+b p x^9\right )}{x^{10}} \, dx+2 \int \frac {\sqrt {q+p x^6} \left (-2 a q^2-a p q x^6+a p^2 x^{12}\right )}{x^{11}} \, dx\\ &=2 \int \frac {\left (q+p x^6\right )^{3/2} \left (-2 a q+a p x^6\right )}{x^{11}} \, dx+2 \int \frac {\sqrt {q+p x^6} \left (-2 b q+b p x^6\right )}{x^7} \, dx\\ &=\frac {2 a \left (q+p x^6\right )^{5/2}}{5 x^{10}}+\frac {1}{3} \operatorname {Subst}\left (\int \frac {\sqrt {q+p x} (-2 b q+b p x)}{x^2} \, dx,x,x^6\right )\\ &=\frac {2 b \left (q+p x^6\right )^{3/2}}{3 x^6}+\frac {2 a \left (q+p x^6\right )^{5/2}}{5 x^{10}}\\ \end {align*}

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Mathematica [C]  time = 0.58, size = 248, normalized size = 4.43 \begin {gather*} \frac {5 x^4 \left (6 a p^2 x^8 \left (p x^6+q\right ) \, _2F_1\left (-\frac {1}{2},\frac {1}{3};\frac {4}{3};-\frac {p x^6}{q}\right )+3 a p q x^2 \left (p x^6+q\right ) \, _2F_1\left (-\frac {2}{3},-\frac {1}{2};\frac {1}{3};-\frac {p x^6}{q}\right )-4 b p \sqrt {q} x^6 \sqrt {p x^6+q} \sqrt {\frac {p x^6}{q}+1} \tanh ^{-1}\left (\frac {\sqrt {p x^6+q}}{\sqrt {q}}\right )+4 b \left (p x^6+q\right ) \left (\left (p x^6+q\right ) \sqrt {\frac {p x^6}{q}+1}+p x^6 \tanh ^{-1}\left (\sqrt {\frac {p x^6}{q}+1}\right )\right )\right )+12 a q^2 \left (p x^6+q\right ) \, _2F_1\left (-\frac {5}{3},-\frac {1}{2};-\frac {2}{3};-\frac {p x^6}{q}\right )}{30 x^{10} \sqrt {p x^6+q} \sqrt {\frac {p x^6}{q}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*(-2*q + p*x^6)*Sqrt[q + p*x^6]*(a*q + b*x^4 + a*p*x^6))/x^11,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 18.79, size = 56, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {q+p x^6} \left (3 a q^2+5 b q x^4+6 a p q x^6+5 b p x^{10}+3 a p^2 x^{12}\right )}{15 x^{10}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(2*(-2*q + p*x^6)*Sqrt[q + p*x^6]*(a*q + b*x^4 + a*p*x^6))/x^11,x]

[Out]

(2*Sqrt[q + p*x^6]*(3*a*q^2 + 5*b*q*x^4 + 6*a*p*q*x^6 + 5*b*p*x^10 + 3*a*p^2*x^12))/(15*x^10)

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fricas [A]  time = 0.51, size = 52, normalized size = 0.93 \begin {gather*} \frac {2 \, {\left (3 \, a p^{2} x^{12} + 5 \, b p x^{10} + 6 \, a p q x^{6} + 5 \, b q x^{4} + 3 \, a q^{2}\right )} \sqrt {p x^{6} + q}}{15 \, x^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*(p*x^6-2*q)*(p*x^6+q)^(1/2)*(a*p*x^6+b*x^4+a*q)/x^11,x, algorithm="fricas")

[Out]

2/15*(3*a*p^2*x^12 + 5*b*p*x^10 + 6*a*p*q*x^6 + 5*b*q*x^4 + 3*a*q^2)*sqrt(p*x^6 + q)/x^10

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giac [A]  time = 0.66, size = 65, normalized size = 1.16 \begin {gather*} \frac {2}{15} \, \sqrt {p x^{6} + q} {\left (3 \, a p^{2} x^{2} + 5 \, b p\right )} + \frac {2}{15} \, {\left (6 \, a p q + \frac {5 \, b q + \frac {3 \, a q^{2}}{x^{4}}}{x^{2}}\right )} \sqrt {\frac {p}{x^{2}} + \frac {q}{x^{8}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*(p*x^6-2*q)*(p*x^6+q)^(1/2)*(a*p*x^6+b*x^4+a*q)/x^11,x, algorithm="giac")

[Out]

2/15*sqrt(p*x^6 + q)*(3*a*p^2*x^2 + 5*b*p) + 2/15*(6*a*p*q + (5*b*q + 3*a*q^2/x^4)/x^2)*sqrt(p/x^2 + q/x^8)

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maple [A]  time = 0.11, size = 33, normalized size = 0.59

method	result	size
gosper	\(\frac {2 \left (p \,x^{6}+q \right )^{\frac {3}{2}} \left (3 a p \,x^{6}+5 b \,x^{4}+3 a q \right )}{15 x^{10}}\)	\(33\)
trager	\(\frac {2 \sqrt {p \,x^{6}+q}\, \left (3 a \,p^{2} x^{12}+5 b p \,x^{10}+6 a p q \,x^{6}+5 b q \,x^{4}+3 a \,q^{2}\right )}{15 x^{10}}\)	\(53\)
risch	\(\frac {2 \sqrt {p \,x^{6}+q}\, \left (3 a \,p^{2} x^{12}+5 b p \,x^{10}+6 a p q \,x^{6}+5 b q \,x^{4}+3 a \,q^{2}\right )}{15 x^{10}}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*(p*x^6-2*q)*(p*x^6+q)^(1/2)*(a*p*x^6+b*x^4+a*q)/x^11,x,method=_RETURNVERBOSE)

[Out]

2/15*(p*x^6+q)^(3/2)*(3*a*p*x^6+5*b*x^4+3*a*q)/x^10

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maxima [A]  time = 0.37, size = 52, normalized size = 0.93 \begin {gather*} \frac {2 \, {\left (3 \, a p^{2} x^{12} + 5 \, b p x^{10} + 6 \, a p q x^{6} + 5 \, b q x^{4} + 3 \, a q^{2}\right )} \sqrt {p x^{6} + q}}{15 \, x^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*(p*x^6-2*q)*(p*x^6+q)^(1/2)*(a*p*x^6+b*x^4+a*q)/x^11,x, algorithm="maxima")

[Out]

2/15*(3*a*p^2*x^12 + 5*b*p*x^10 + 6*a*p*q*x^6 + 5*b*q*x^4 + 3*a*q^2)*sqrt(p*x^6 + q)/x^10

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mupad [B]  time = 1.69, size = 76, normalized size = 1.36 \begin {gather*} \sqrt {p\,x^6+q}\,\left (\frac {2\,a\,p^2\,x^2}{5}+\frac {2\,b\,p}{3}\right )+\frac {2\,a\,q^2\,\sqrt {p\,x^6+q}}{5\,x^{10}}+\frac {2\,b\,q\,\sqrt {p\,x^6+q}}{3\,x^6}+\frac {4\,a\,p\,q\,\sqrt {p\,x^6+q}}{5\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*(q + p*x^6)^(1/2)*(2*q - p*x^6)*(a*q + b*x^4 + a*p*x^6))/x^11,x)

[Out]

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

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sympy [C]  time = 10.91, size = 223, normalized size = 3.98 \begin {gather*} \frac {a p^{2} \sqrt {q} x^{2} \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {p x^{6} e^{i \pi }}{q}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} - \frac {a p q^{\frac {3}{2}} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {1}{2} \\ \frac {1}{3} \end {matrix}\middle | {\frac {p x^{6} e^{i \pi }}{q}} \right )}}{3 x^{4} \Gamma \left (\frac {1}{3}\right )} - \frac {2 a q^{\frac {5}{2}} \Gamma \left (- \frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{3}, - \frac {1}{2} \\ - \frac {2}{3} \end {matrix}\middle | {\frac {p x^{6} e^{i \pi }}{q}} \right )}}{3 x^{10} \Gamma \left (- \frac {2}{3}\right )} + \frac {2 b p^{\frac {3}{2}} x^{3}}{3 \sqrt {1 + \frac {q}{p x^{6}}}} + \frac {2 b \sqrt {p} q \sqrt {1 + \frac {q}{p x^{6}}}}{3 x^{3}} + \frac {2 b \sqrt {p} q}{3 x^{3} \sqrt {1 + \frac {q}{p x^{6}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*(p*x**6-2*q)*(p*x**6+q)**(1/2)*(a*p*x**6+b*x**4+a*q)/x**11,x)

[Out]

a*p**2*sqrt(q)*x**2*gamma(1/3)*hyper((-1/2, 1/3), (4/3,), p*x**6*exp_polar(I*pi)/q)/(3*gamma(4/3)) - a*p*q**(3
/2)*gamma(-2/3)*hyper((-2/3, -1/2), (1/3,), p*x**6*exp_polar(I*pi)/q)/(3*x**4*gamma(1/3)) - 2*a*q**(5/2)*gamma
(-5/3)*hyper((-5/3, -1/2), (-2/3,), p*x**6*exp_polar(I*pi)/q)/(3*x**10*gamma(-2/3)) + 2*b*p**(3/2)*x**3/(3*sqr
t(1 + q/(p*x**6))) + 2*b*sqrt(p)*q*sqrt(1 + q/(p*x**6))/(3*x**3) + 2*b*sqrt(p)*q/(3*x**3*sqrt(1 + q/(p*x**6)))

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