3.3.100 \(\int \frac {1}{x^7 (8 c-d x^3)^2 (c+d x^3)^{3/2}} \, dx\)

Optimal. Leaf size=185 \[ \frac {13 d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{497664 c^{11/2}}-\frac {33 d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2048 c^{11/2}}+\frac {665 d^2}{41472 c^5 \sqrt {c+d x^3}}-\frac {71 d^2}{13824 c^4 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {17 d}{384 c^3 x^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {1}{48 c^2 x^6 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \]

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Rubi [A]  time = 0.16, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {446, 103, 151, 152, 156, 63, 208, 206} \begin {gather*} -\frac {71 d^2}{13824 c^4 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {665 d^2}{41472 c^5 \sqrt {c+d x^3}}+\frac {13 d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{497664 c^{11/2}}-\frac {33 d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2048 c^{11/2}}+\frac {17 d}{384 c^3 x^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {1}{48 c^2 x^6 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(8*c - d*x^3)^2*(c + d*x^3)^(3/2)),x]

[Out]

(665*d^2)/(41472*c^5*Sqrt[c + d*x^3]) - (71*d^2)/(13824*c^4*(8*c - d*x^3)*Sqrt[c + d*x^3]) - 1/(48*c^2*x^6*(8*
c - d*x^3)*Sqrt[c + d*x^3]) + (17*d)/(384*c^3*x^3*(8*c - d*x^3)*Sqrt[c + d*x^3]) + (13*d^2*ArcTanh[Sqrt[c + d*
x^3]/(3*Sqrt[c])])/(497664*c^(11/2)) - (33*d^2*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(2048*c^(11/2))

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 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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 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]]

Rubi steps

\begin {align*} \int \frac {1}{x^7 \left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x^3 (8 c-d x)^2 (c+d x)^{3/2}} \, dx,x,x^3\right )\\ &=-\frac {1}{48 c^2 x^6 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {\operatorname {Subst}\left (\int \frac {17 c d-\frac {7 d^2 x}{2}}{x^2 (8 c-d x)^2 (c+d x)^{3/2}} \, dx,x,x^3\right )}{48 c^2}\\ &=-\frac {1}{48 c^2 x^6 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {17 d}{384 c^3 x^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {\operatorname {Subst}\left (\int \frac {198 c^2 d^2-\frac {85}{2} c d^3 x}{x (8 c-d x)^2 (c+d x)^{3/2}} \, dx,x,x^3\right )}{384 c^4}\\ &=-\frac {71 d^2}{13824 c^4 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {1}{48 c^2 x^6 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {17 d}{384 c^3 x^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {\operatorname {Subst}\left (\int \frac {-1782 c^3 d^3+213 c^2 d^4 x}{x (8 c-d x) (c+d x)^{3/2}} \, dx,x,x^3\right )}{27648 c^6 d}\\ &=\frac {665 d^2}{41472 c^5 \sqrt {c+d x^3}}-\frac {71 d^2}{13824 c^4 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {1}{48 c^2 x^6 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {17 d}{384 c^3 x^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {\operatorname {Subst}\left (\int \frac {-8019 c^4 d^4+\frac {1995}{2} c^3 d^5 x}{x (8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{124416 c^8 d^2}\\ &=\frac {665 d^2}{41472 c^5 \sqrt {c+d x^3}}-\frac {71 d^2}{13824 c^4 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {1}{48 c^2 x^6 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {17 d}{384 c^3 x^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {\left (33 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^3\right )}{4096 c^5}+\frac {\left (13 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{331776 c^5}\\ &=\frac {665 d^2}{41472 c^5 \sqrt {c+d x^3}}-\frac {71 d^2}{13824 c^4 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {1}{48 c^2 x^6 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {17 d}{384 c^3 x^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {(33 d) \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^3}\right )}{2048 c^5}+\frac {\left (13 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{165888 c^5}\\ &=\frac {665 d^2}{41472 c^5 \sqrt {c+d x^3}}-\frac {71 d^2}{13824 c^4 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {1}{48 c^2 x^6 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {17 d}{384 c^3 x^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {13 d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{497664 c^{11/2}}-\frac {33 d^2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2048 c^{11/2}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 135, normalized size = 0.73 \begin {gather*} \frac {13 d^2 x^6 \left (d x^3-8 c\right ) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {d x^3+c}{9 c}\right )-3 \left (4 c \left (288 c^2-612 c d x^3+71 d^2 x^6\right )+891 d^2 x^6 \left (d x^3-8 c\right ) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {d x^3}{c}+1\right )\right )}{165888 c^5 x^6 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*(8*c - d*x^3)^2*(c + d*x^3)^(3/2)),x]

[Out]

(13*d^2*x^6*(-8*c + d*x^3)*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*x^3)/(9*c)] - 3*(4*c*(288*c^2 - 612*c*d*x^3
+ 71*d^2*x^6) + 891*d^2*x^6*(-8*c + d*x^3)*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (d*x^3)/c]))/(165888*c^5*x^6*(8
*c - d*x^3)*Sqrt[c + d*x^3])

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IntegrateAlgebraic [A]  time = 0.19, size = 208, normalized size = 1.12 \begin {gather*} \frac {x^6 \sqrt {c+d x^3} \left (\frac {13 d^2}{62208 c^{9/2}}-\frac {13 d^3 x^3}{497664 c^{11/2}}\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )+x^6 \sqrt {c+d x^3} \left (\frac {33 d^3 x^3}{2048 c^{11/2}}-\frac {33 d^2}{256 c^{9/2}}\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )-\frac {665 d^3 x^9}{41472 c^5}+\frac {5107 d^2 x^6}{41472 c^4}+\frac {17 d x^3}{384 c^3}-\frac {1}{48 c^2}}{x^6 \left (8 c \sqrt {c+d x^3}-d x^3 \sqrt {c+d x^3}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^7*(8*c - d*x^3)^2*(c + d*x^3)^(3/2)),x]

[Out]

(-1/48*1/c^2 + (17*d*x^3)/(384*c^3) + (5107*d^2*x^6)/(41472*c^4) - (665*d^3*x^9)/(41472*c^5) + x^6*Sqrt[c + d*
x^3]*((13*d^2)/(62208*c^(9/2)) - (13*d^3*x^3)/(497664*c^(11/2)))*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])] + x^6*Sq
rt[c + d*x^3]*((-33*d^2)/(256*c^(9/2)) + (33*d^3*x^3)/(2048*c^(11/2)))*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(x^6*
(8*c*Sqrt[c + d*x^3] - d*x^3*Sqrt[c + d*x^3]))

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fricas [A]  time = 0.90, size = 398, normalized size = 2.15 \begin {gather*} \left [\frac {13 \, {\left (d^{4} x^{12} - 7 \, c d^{3} x^{9} - 8 \, c^{2} d^{2} x^{6}\right )} \sqrt {c} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 8019 \, {\left (d^{4} x^{12} - 7 \, c d^{3} x^{9} - 8 \, c^{2} d^{2} x^{6}\right )} \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) + 24 \, {\left (665 \, c d^{3} x^{9} - 5107 \, c^{2} d^{2} x^{6} - 1836 \, c^{3} d x^{3} + 864 \, c^{4}\right )} \sqrt {d x^{3} + c}}{995328 \, {\left (c^{6} d^{2} x^{12} - 7 \, c^{7} d x^{9} - 8 \, c^{8} x^{6}\right )}}, \frac {8019 \, {\left (d^{4} x^{12} - 7 \, c d^{3} x^{9} - 8 \, c^{2} d^{2} x^{6}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) - 13 \, {\left (d^{4} x^{12} - 7 \, c d^{3} x^{9} - 8 \, c^{2} d^{2} x^{6}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + 12 \, {\left (665 \, c d^{3} x^{9} - 5107 \, c^{2} d^{2} x^{6} - 1836 \, c^{3} d x^{3} + 864 \, c^{4}\right )} \sqrt {d x^{3} + c}}{497664 \, {\left (c^{6} d^{2} x^{12} - 7 \, c^{7} d x^{9} - 8 \, c^{8} x^{6}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="fricas")

[Out]

[1/995328*(13*(d^4*x^12 - 7*c*d^3*x^9 - 8*c^2*d^2*x^6)*sqrt(c)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/
(d*x^3 - 8*c)) + 8019*(d^4*x^12 - 7*c*d^3*x^9 - 8*c^2*d^2*x^6)*sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c)
+ 2*c)/x^3) + 24*(665*c*d^3*x^9 - 5107*c^2*d^2*x^6 - 1836*c^3*d*x^3 + 864*c^4)*sqrt(d*x^3 + c))/(c^6*d^2*x^12
- 7*c^7*d*x^9 - 8*c^8*x^6), 1/497664*(8019*(d^4*x^12 - 7*c*d^3*x^9 - 8*c^2*d^2*x^6)*sqrt(-c)*arctan(sqrt(d*x^3
+ c)*sqrt(-c)/c) - 13*(d^4*x^12 - 7*c*d^3*x^9 - 8*c^2*d^2*x^6)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c
) + 12*(665*c*d^3*x^9 - 5107*c^2*d^2*x^6 - 1836*c^3*d*x^3 + 864*c^4)*sqrt(d*x^3 + c))/(c^6*d^2*x^12 - 7*c^7*d*
x^9 - 8*c^8*x^6)]

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giac [A]  time = 0.18, size = 149, normalized size = 0.81 \begin {gather*} \frac {33 \, d^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{2048 \, \sqrt {-c} c^{5}} - \frac {13 \, d^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{497664 \, \sqrt {-c} c^{5}} + \frac {341 \, {\left (d x^{3} + c\right )} d^{2} - 3072 \, c d^{2}}{41472 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} - 9 \, \sqrt {d x^{3} + c} c\right )} c^{5}} + \frac {3 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} d^{2} - 4 \, \sqrt {d x^{3} + c} c d^{2}}{384 \, c^{5} d^{2} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="giac")

[Out]

33/2048*d^2*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^5) - 13/497664*d^2*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c
))/(sqrt(-c)*c^5) + 1/41472*(341*(d*x^3 + c)*d^2 - 3072*c*d^2)/(((d*x^3 + c)^(3/2) - 9*sqrt(d*x^3 + c)*c)*c^5)
+ 1/384*(3*(d*x^3 + c)^(3/2)*d^2 - 4*sqrt(d*x^3 + c)*c*d^2)/(c^5*d^2*x^6)

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maple [C]  time = 0.21, size = 1106, normalized size = 5.98

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x)

[Out]

1/512/c^3*d^3*(-1/243*(d*x^3+c)^(1/2)/(d*x^3-8*c)/c^2/d-2/243/((x^3+c/d)*d)^(1/2)/c^2/d-1/1458*I/d^3/c^3*2^(1/
2)*sum((-c*d^2)^(1/3)*(1/2*I*(2*x+(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)*((x-(-
c*d^2)^(1/3)/d)/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3))*d)^(1/2)*(-1/2*I*(2*x+(I*3^(1/2)*(-c*d^2)^(1/3)+(
-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*(2*_alpha^2*d^2+I*(-c*d^2)^(1/3)*3^(1/2)*_alpha*d-(-
c*d^2)^(1/3)*_alpha*d-I*3^(1/2)*(-c*d^2)^(2/3)-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)
/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2),-1/18*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*
d+I*3^(1/2)*c*d-3*c*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha-3*(-c*d^2)^(2/3)*_alpha)/c/d,(I*3^(1/2)*(-c*d^2)^(1/3)/(
-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2)),_alpha=RootOf(_Z^3*d-8*c)))+1/256/c^3*d*(-1/3*
(d*x^3+c)^(1/2)/c^2/x^3-2/3/((x^3+c/d)*d)^(1/2)/c^2*d+d*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(5/2))+1/64/c^2*(-1
/6*(d*x^3+c)^(1/2)/c^2/x^6+7/12*(d*x^3+c)^(1/2)/c^3*d/x^3+2/3/((x^3+c/d)*d)^(1/2)/c^3*d^2-5/4*d^2*arctanh((d*x
^3+c)^(1/2)/c^(1/2))/c^(7/2))-3/4096/c^4*d^3*(2/27/((x^3+c/d)*d)^(1/2)/c/d+1/243*I/c^2/d^3*2^(1/2)*sum((-c*d^2
)^(1/3)*(1/2*I*(2*x+(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)*((x-(-c*d^2)^(1/3)/d
)/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3))*d)^(1/2)*(-1/2*I*(2*x+(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))
/d)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*(2*_alpha^2*d^2+I*(-c*d^2)^(1/3)*3^(1/2)*_alpha*d-(-c*d^2)^(1/3)*_
alpha*d-I*3^(1/2)*(-c*d^2)^(2/3)-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/
2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2),-1/18*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d+I*3^(1/2)*c*
d-3*c*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha-3*(-c*d^2)^(2/3)*_alpha)/c/d,(I*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^
(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2)),_alpha=RootOf(_Z^3*d-8*c)))+3/4096/c^4*d^2*(2/3/((x^3+c/d)*d
)^(1/2)/c-2/3*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(3/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )}^{2} x^{7}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((d*x^3 + c)^(3/2)*(d*x^3 - 8*c)^2*x^7), x)

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mupad [B]  time = 4.76, size = 171, normalized size = 0.92 \begin {gather*} \frac {\frac {2\,d^2}{9\,c^2}-\frac {10373\,d^2\,\left (d\,x^3+c\right )}{13824\,c^3}+\frac {3551\,d^2\,{\left (d\,x^3+c\right )}^2}{6912\,c^4}-\frac {665\,d^2\,{\left (d\,x^3+c\right )}^3}{13824\,c^5}}{33\,c\,{\left (d\,x^3+c\right )}^{5/2}-3\,{\left (d\,x^3+c\right )}^{7/2}+27\,c^3\,\sqrt {d\,x^3+c}-57\,c^2\,{\left (d\,x^3+c\right )}^{3/2}}+\frac {d^2\,\left (\mathrm {atanh}\left (\frac {c^5\,\sqrt {d\,x^3+c}}{\sqrt {c^{11}}}\right )\,1{}\mathrm {i}-\frac {\mathrm {atanh}\left (\frac {c^5\,\sqrt {d\,x^3+c}}{3\,\sqrt {c^{11}}}\right )\,13{}\mathrm {i}}{8019}\right )\,33{}\mathrm {i}}{2048\,\sqrt {c^{11}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^7*(c + d*x^3)^(3/2)*(8*c - d*x^3)^2),x)

[Out]

((2*d^2)/(9*c^2) - (10373*d^2*(c + d*x^3))/(13824*c^3) + (3551*d^2*(c + d*x^3)^2)/(6912*c^4) - (665*d^2*(c + d
*x^3)^3)/(13824*c^5))/(33*c*(c + d*x^3)^(5/2) - 3*(c + d*x^3)^(7/2) + 27*c^3*(c + d*x^3)^(1/2) - 57*c^2*(c + d
*x^3)^(3/2)) + (d^2*(atanh((c^5*(c + d*x^3)^(1/2))/(c^11)^(1/2))*1i - (atanh((c^5*(c + d*x^3)^(1/2))/(3*(c^11)
^(1/2)))*13i)/8019)*33i)/(2048*(c^11)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{7} \left (- 8 c + d x^{3}\right )^{2} \left (c + d x^{3}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**7/(-d*x**3+8*c)**2/(d*x**3+c)**(3/2),x)

[Out]

Integral(1/(x**7*(-8*c + d*x**3)**2*(c + d*x**3)**(3/2)), x)

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