3.13.65 \(\int \frac {1}{x^7 (b+a x^3)^{3/4}} \, dx\)

Optimal. Leaf size=92 \[ -\frac {7 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{a x^3+b}}{\sqrt [4]{b}}\right )}{16 b^{11/4}}-\frac {7 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a x^3+b}}{\sqrt [4]{b}}\right )}{16 b^{11/4}}+\frac {\sqrt [4]{a x^3+b} \left (7 a x^3-4 b\right )}{24 b^2 x^6} \]

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Rubi [A]  time = 0.08, antiderivative size = 104, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {266, 51, 63, 212, 206, 203} \begin {gather*} -\frac {7 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{a x^3+b}}{\sqrt [4]{b}}\right )}{16 b^{11/4}}-\frac {7 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a x^3+b}}{\sqrt [4]{b}}\right )}{16 b^{11/4}}+\frac {7 a \sqrt [4]{a x^3+b}}{24 b^2 x^3}-\frac {\sqrt [4]{a x^3+b}}{6 b x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(b + a*x^3)^(3/4)),x]

[Out]

-1/6*(b + a*x^3)^(1/4)/(b*x^6) + (7*a*(b + a*x^3)^(1/4))/(24*b^2*x^3) - (7*a^2*ArcTan[(b + a*x^3)^(1/4)/b^(1/4
)])/(16*b^(11/4)) - (7*a^2*ArcTanh[(b + a*x^3)^(1/4)/b^(1/4)])/(16*b^(11/4))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 203

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

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 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 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]]

Rubi steps

\begin {align*} \int \frac {1}{x^7 \left (b+a x^3\right )^{3/4}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x^3 (b+a x)^{3/4}} \, dx,x,x^3\right )\\ &=-\frac {\sqrt [4]{b+a x^3}}{6 b x^6}-\frac {(7 a) \operatorname {Subst}\left (\int \frac {1}{x^2 (b+a x)^{3/4}} \, dx,x,x^3\right )}{24 b}\\ &=-\frac {\sqrt [4]{b+a x^3}}{6 b x^6}+\frac {7 a \sqrt [4]{b+a x^3}}{24 b^2 x^3}+\frac {\left (7 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{x (b+a x)^{3/4}} \, dx,x,x^3\right )}{32 b^2}\\ &=-\frac {\sqrt [4]{b+a x^3}}{6 b x^6}+\frac {7 a \sqrt [4]{b+a x^3}}{24 b^2 x^3}+\frac {(7 a) \operatorname {Subst}\left (\int \frac {1}{-\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{b+a x^3}\right )}{8 b^2}\\ &=-\frac {\sqrt [4]{b+a x^3}}{6 b x^6}+\frac {7 a \sqrt [4]{b+a x^3}}{24 b^2 x^3}-\frac {\left (7 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b}-x^2} \, dx,x,\sqrt [4]{b+a x^3}\right )}{16 b^{5/2}}-\frac {\left (7 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b}+x^2} \, dx,x,\sqrt [4]{b+a x^3}\right )}{16 b^{5/2}}\\ &=-\frac {\sqrt [4]{b+a x^3}}{6 b x^6}+\frac {7 a \sqrt [4]{b+a x^3}}{24 b^2 x^3}-\frac {7 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{b+a x^3}}{\sqrt [4]{b}}\right )}{16 b^{11/4}}-\frac {7 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{b+a x^3}}{\sqrt [4]{b}}\right )}{16 b^{11/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 39, normalized size = 0.42 \begin {gather*} -\frac {4 a^2 \sqrt [4]{a x^3+b} \, _2F_1\left (\frac {1}{4},3;\frac {5}{4};\frac {a x^3}{b}+1\right )}{3 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*(b + a*x^3)^(3/4)),x]

[Out]

(-4*a^2*(b + a*x^3)^(1/4)*Hypergeometric2F1[1/4, 3, 5/4, 1 + (a*x^3)/b])/(3*b^3)

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IntegrateAlgebraic [A]  time = 0.15, size = 92, normalized size = 1.00 \begin {gather*} \frac {\sqrt [4]{b+a x^3} \left (-4 b+7 a x^3\right )}{24 b^2 x^6}-\frac {7 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{b+a x^3}}{\sqrt [4]{b}}\right )}{16 b^{11/4}}-\frac {7 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{b+a x^3}}{\sqrt [4]{b}}\right )}{16 b^{11/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^7*(b + a*x^3)^(3/4)),x]

[Out]

((b + a*x^3)^(1/4)*(-4*b + 7*a*x^3))/(24*b^2*x^6) - (7*a^2*ArcTan[(b + a*x^3)^(1/4)/b^(1/4)])/(16*b^(11/4)) -
(7*a^2*ArcTanh[(b + a*x^3)^(1/4)/b^(1/4)])/(16*b^(11/4))

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fricas [B]  time = 0.53, size = 216, normalized size = 2.35 \begin {gather*} \frac {84 \, b^{2} x^{6} \left (\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (a x^{3} + b\right )}^{\frac {1}{4}} a^{2} b^{8} \left (\frac {a^{8}}{b^{11}}\right )^{\frac {3}{4}} - \sqrt {b^{6} \sqrt {\frac {a^{8}}{b^{11}}} + \sqrt {a x^{3} + b} a^{4}} b^{8} \left (\frac {a^{8}}{b^{11}}\right )^{\frac {3}{4}}}{a^{8}}\right ) - 21 \, b^{2} x^{6} \left (\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (7 \, b^{3} \left (\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} + 7 \, {\left (a x^{3} + b\right )}^{\frac {1}{4}} a^{2}\right ) + 21 \, b^{2} x^{6} \left (\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} \log \left (-7 \, b^{3} \left (\frac {a^{8}}{b^{11}}\right )^{\frac {1}{4}} + 7 \, {\left (a x^{3} + b\right )}^{\frac {1}{4}} a^{2}\right ) + 4 \, {\left (7 \, a x^{3} - 4 \, b\right )} {\left (a x^{3} + b\right )}^{\frac {1}{4}}}{96 \, b^{2} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(a*x^3+b)^(3/4),x, algorithm="fricas")

[Out]

1/96*(84*b^2*x^6*(a^8/b^11)^(1/4)*arctan(-((a*x^3 + b)^(1/4)*a^2*b^8*(a^8/b^11)^(3/4) - sqrt(b^6*sqrt(a^8/b^11
) + sqrt(a*x^3 + b)*a^4)*b^8*(a^8/b^11)^(3/4))/a^8) - 21*b^2*x^6*(a^8/b^11)^(1/4)*log(7*b^3*(a^8/b^11)^(1/4) +
7*(a*x^3 + b)^(1/4)*a^2) + 21*b^2*x^6*(a^8/b^11)^(1/4)*log(-7*b^3*(a^8/b^11)^(1/4) + 7*(a*x^3 + b)^(1/4)*a^2)
+ 4*(7*a*x^3 - 4*b)*(a*x^3 + b)^(1/4))/(b^2*x^6)

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giac [B]  time = 0.24, size = 244, normalized size = 2.65 \begin {gather*} \frac {\frac {42 \, \sqrt {2} a^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} + 2 \, {\left (a x^{3} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right )}{\left (-b\right )^{\frac {3}{4}} b^{2}} + \frac {42 \, \sqrt {2} a^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-b\right )^{\frac {1}{4}} - 2 \, {\left (a x^{3} + b\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-b\right )^{\frac {1}{4}}}\right )}{\left (-b\right )^{\frac {3}{4}} b^{2}} + \frac {21 \, \sqrt {2} a^{3} \log \left (\sqrt {2} {\left (a x^{3} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{3} + b} + \sqrt {-b}\right )}{\left (-b\right )^{\frac {3}{4}} b^{2}} + \frac {21 \, \sqrt {2} a^{3} \left (-b\right )^{\frac {1}{4}} \log \left (-\sqrt {2} {\left (a x^{3} + b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {1}{4}} + \sqrt {a x^{3} + b} + \sqrt {-b}\right )}{b^{3}} + \frac {8 \, {\left (7 \, {\left (a x^{3} + b\right )}^{\frac {5}{4}} a^{3} - 11 \, {\left (a x^{3} + b\right )}^{\frac {1}{4}} a^{3} b\right )}}{a^{2} b^{2} x^{6}}}{192 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(a*x^3+b)^(3/4),x, algorithm="giac")

[Out]

1/192*(42*sqrt(2)*a^3*arctan(1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) + 2*(a*x^3 + b)^(1/4))/(-b)^(1/4))/((-b)^(3/4)*b^
2) + 42*sqrt(2)*a^3*arctan(-1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) - 2*(a*x^3 + b)^(1/4))/(-b)^(1/4))/((-b)^(3/4)*b^2
) + 21*sqrt(2)*a^3*log(sqrt(2)*(a*x^3 + b)^(1/4)*(-b)^(1/4) + sqrt(a*x^3 + b) + sqrt(-b))/((-b)^(3/4)*b^2) + 2
1*sqrt(2)*a^3*(-b)^(1/4)*log(-sqrt(2)*(a*x^3 + b)^(1/4)*(-b)^(1/4) + sqrt(a*x^3 + b) + sqrt(-b))/b^3 + 8*(7*(a
*x^3 + b)^(5/4)*a^3 - 11*(a*x^3 + b)^(1/4)*a^3*b)/(a^2*b^2*x^6))/a

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maple [F]  time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{7} \left (a \,x^{3}+b \right )^{\frac {3}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(a*x^3+b)^(3/4),x)

[Out]

int(1/x^7/(a*x^3+b)^(3/4),x)

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maxima [A]  time = 0.50, size = 132, normalized size = 1.43 \begin {gather*} \frac {7 \, {\left (a x^{3} + b\right )}^{\frac {5}{4}} a^{2} - 11 \, {\left (a x^{3} + b\right )}^{\frac {1}{4}} a^{2} b}{24 \, {\left ({\left (a x^{3} + b\right )}^{2} b^{2} - 2 \, {\left (a x^{3} + b\right )} b^{3} + b^{4}\right )}} - \frac {7 \, {\left (\frac {2 \, a^{2} \arctan \left (\frac {{\left (a x^{3} + b\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} - \frac {a^{2} \log \left (\frac {{\left (a x^{3} + b\right )}^{\frac {1}{4}} - b^{\frac {1}{4}}}{{\left (a x^{3} + b\right )}^{\frac {1}{4}} + b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}}\right )}}{32 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(a*x^3+b)^(3/4),x, algorithm="maxima")

[Out]

1/24*(7*(a*x^3 + b)^(5/4)*a^2 - 11*(a*x^3 + b)^(1/4)*a^2*b)/((a*x^3 + b)^2*b^2 - 2*(a*x^3 + b)*b^3 + b^4) - 7/
32*(2*a^2*arctan((a*x^3 + b)^(1/4)/b^(1/4))/b^(3/4) - a^2*log(((a*x^3 + b)^(1/4) - b^(1/4))/((a*x^3 + b)^(1/4)
+ b^(1/4)))/b^(3/4))/b^2

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mupad [B]  time = 1.09, size = 82, normalized size = 0.89 \begin {gather*} \frac {7\,{\left (a\,x^3+b\right )}^{5/4}}{24\,b^2\,x^6}-\frac {11\,{\left (a\,x^3+b\right )}^{1/4}}{24\,b\,x^6}-\frac {7\,a^2\,\mathrm {atan}\left (\frac {{\left (a\,x^3+b\right )}^{1/4}}{b^{1/4}}\right )}{16\,b^{11/4}}+\frac {a^2\,\mathrm {atan}\left (\frac {{\left (a\,x^3+b\right )}^{1/4}\,1{}\mathrm {i}}{b^{1/4}}\right )\,7{}\mathrm {i}}{16\,b^{11/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^7*(b + a*x^3)^(3/4)),x)

[Out]

(a^2*atan(((b + a*x^3)^(1/4)*1i)/b^(1/4))*7i)/(16*b^(11/4)) - (7*a^2*atan((b + a*x^3)^(1/4)/b^(1/4)))/(16*b^(1
1/4)) - (11*(b + a*x^3)^(1/4))/(24*b*x^6) + (7*(b + a*x^3)^(5/4))/(24*b^2*x^6)

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sympy [C]  time = 1.47, size = 41, normalized size = 0.45 \begin {gather*} - \frac {\Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{3}}} \right )}}{3 a^{\frac {3}{4}} x^{\frac {33}{4}} \Gamma \left (\frac {15}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**7/(a*x**3+b)**(3/4),x)

[Out]

-gamma(11/4)*hyper((3/4, 11/4), (15/4,), b*exp_polar(I*pi)/(a*x**3))/(3*a**(3/4)*x**(33/4)*gamma(15/4))

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