3.447 \(\int \frac {1}{x^4 (8 c-d x^3)^2 (c+d x^3)^{3/2}} \, dx\)

Optimal. Leaf size=143 \[ \frac {5 d \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{31104 c^{9/2}}+\frac {5 d \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{384 c^{9/2}}-\frac {35 d}{2592 c^4 \sqrt {c+d x^3}}+\frac {5 d}{864 c^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {1}{24 c^2 x^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \]

[Out]

5/31104*d*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(9/2)+5/384*d*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(9/2)-35/259
2*d/c^4/(d*x^3+c)^(1/2)+5/864*d/c^3/(-d*x^3+8*c)/(d*x^3+c)^(1/2)-1/24/c^2/x^3/(-d*x^3+8*c)/(d*x^3+c)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 9, 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} \[ \frac {5 d}{864 c^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {35 d}{2592 c^4 \sqrt {c+d x^3}}-\frac {1}{24 c^2 x^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {5 d \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{31104 c^{9/2}}+\frac {5 d \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{384 c^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(8*c - d*x^3)^2*(c + d*x^3)^(3/2)),x]

[Out]

(-35*d)/(2592*c^4*Sqrt[c + d*x^3]) + (5*d)/(864*c^3*(8*c - d*x^3)*Sqrt[c + d*x^3]) - 1/(24*c^2*x^3*(8*c - d*x^
3)*Sqrt[c + d*x^3]) + (5*d*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(31104*c^(9/2)) + (5*d*ArcTanh[Sqrt[c + d*x^3
]/Sqrt[c]])/(384*c^(9/2))

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^4 \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^2 (8 c-d x)^2 (c+d x)^{3/2}} \, dx,x,x^3\right )\\ &=-\frac {1}{24 c^2 x^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {\operatorname {Subst}\left (\int \frac {10 c d-\frac {5 d^2 x}{2}}{x (8 c-d x)^2 (c+d x)^{3/2}} \, dx,x,x^3\right )}{24 c^2}\\ &=\frac {5 d}{864 c^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {1}{24 c^2 x^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {\operatorname {Subst}\left (\int \frac {-90 c^2 d^2+15 c d^3 x}{x (8 c-d x) (c+d x)^{3/2}} \, dx,x,x^3\right )}{1728 c^4 d}\\ &=-\frac {35 d}{2592 c^4 \sqrt {c+d x^3}}+\frac {5 d}{864 c^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {1}{24 c^2 x^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {\operatorname {Subst}\left (\int \frac {-405 c^3 d^3+\frac {105}{2} c^2 d^4 x}{x (8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{7776 c^6 d^2}\\ &=-\frac {35 d}{2592 c^4 \sqrt {c+d x^3}}+\frac {5 d}{864 c^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {1}{24 c^2 x^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {(5 d) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^3\right )}{768 c^4}+\frac {\left (5 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{20736 c^4}\\ &=-\frac {35 d}{2592 c^4 \sqrt {c+d x^3}}+\frac {5 d}{864 c^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {1}{24 c^2 x^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^3}\right )}{384 c^4}+\frac {(5 d) \operatorname {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{10368 c^4}\\ &=-\frac {35 d}{2592 c^4 \sqrt {c+d x^3}}+\frac {5 d}{864 c^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {1}{24 c^2 x^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {5 d \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{31104 c^{9/2}}+\frac {5 d \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{384 c^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 117, normalized size = 0.82 \[ \frac {5 d x^3 \left (d x^3-8 c\right ) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {d x^3+c}{9 c}\right )+135 d x^3 \left (d x^3-8 c\right ) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {d x^3}{c}+1\right )+12 c \left (5 d x^3-36 c\right )}{10368 c^4 x^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(8*c - d*x^3)^2*(c + d*x^3)^(3/2)),x]

[Out]

(12*c*(-36*c + 5*d*x^3) + 5*d*x^3*(-8*c + d*x^3)*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*x^3)/(9*c)] + 135*d*x^
3*(-8*c + d*x^3)*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (d*x^3)/c])/(10368*c^4*x^3*(8*c - d*x^3)*Sqrt[c + d*x^3])

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fricas [A]  time = 0.65, size = 368, normalized size = 2.57 \[ \left [\frac {5 \, {\left (d^{3} x^{9} - 7 \, c d^{2} x^{6} - 8 \, c^{2} d x^{3}\right )} \sqrt {c} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 405 \, {\left (d^{3} x^{9} - 7 \, c d^{2} x^{6} - 8 \, c^{2} d x^{3}\right )} \sqrt {c} \log \left (\frac {d x^{3} + 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 24 \, {\left (35 \, c d^{2} x^{6} - 265 \, c^{2} d x^{3} - 108 \, c^{3}\right )} \sqrt {d x^{3} + c}}{62208 \, {\left (c^{5} d^{2} x^{9} - 7 \, c^{6} d x^{6} - 8 \, c^{7} x^{3}\right )}}, -\frac {405 \, {\left (d^{3} x^{9} - 7 \, c d^{2} x^{6} - 8 \, c^{2} d x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) + 5 \, {\left (d^{3} x^{9} - 7 \, c d^{2} x^{6} - 8 \, c^{2} d x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + 12 \, {\left (35 \, c d^{2} x^{6} - 265 \, c^{2} d x^{3} - 108 \, c^{3}\right )} \sqrt {d x^{3} + c}}{31104 \, {\left (c^{5} d^{2} x^{9} - 7 \, c^{6} d x^{6} - 8 \, c^{7} x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="fricas")

[Out]

[1/62208*(5*(d^3*x^9 - 7*c*d^2*x^6 - 8*c^2*d*x^3)*sqrt(c)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^
3 - 8*c)) + 405*(d^3*x^9 - 7*c*d^2*x^6 - 8*c^2*d*x^3)*sqrt(c)*log((d*x^3 + 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^
3) - 24*(35*c*d^2*x^6 - 265*c^2*d*x^3 - 108*c^3)*sqrt(d*x^3 + c))/(c^5*d^2*x^9 - 7*c^6*d*x^6 - 8*c^7*x^3), -1/
31104*(405*(d^3*x^9 - 7*c*d^2*x^6 - 8*c^2*d*x^3)*sqrt(-c)*arctan(sqrt(d*x^3 + c)*sqrt(-c)/c) + 5*(d^3*x^9 - 7*
c*d^2*x^6 - 8*c^2*d*x^3)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) + 12*(35*c*d^2*x^6 - 265*c^2*d*x^3 -
108*c^3)*sqrt(d*x^3 + c))/(c^5*d^2*x^9 - 7*c^6*d*x^6 - 8*c^7*x^3)]

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giac [A]  time = 0.17, size = 129, normalized size = 0.90 \[ -\frac {5 \, d \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{384 \, \sqrt {-c} c^{4}} - \frac {5 \, d \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{31104 \, \sqrt {-c} c^{4}} - \frac {35 \, {\left (d x^{3} + c\right )}^{2} d - 335 \, {\left (d x^{3} + c\right )} c d + 192 \, c^{2} d}{2592 \, {\left ({\left (d x^{3} + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c + 9 \, \sqrt {d x^{3} + c} c^{2}\right )} c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="giac")

[Out]

-5/384*d*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^4) - 5/31104*d*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqr
t(-c)*c^4) - 1/2592*(35*(d*x^3 + c)^2*d - 335*(d*x^3 + c)*c*d + 192*c^2*d)/(((d*x^3 + c)^(5/2) - 10*(d*x^3 + c
)^(3/2)*c + 9*sqrt(d*x^3 + c)*c^2)*c^4)

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maple [C]  time = 0.20, size = 1019, normalized size = 7.13 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x)

[Out]

1/64/c^2*d^2*(-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)))+1/64/c^2*(-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))-1/256/c^3*d^2*(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)))+1/256/c^3*d*(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 \[ \int \frac {1}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )}^{2} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(-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^4), x)

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mupad [B]  time = 4.56, size = 133, normalized size = 0.93 \[ -\frac {\frac {2\,d}{9\,c^2}+\frac {35\,d\,{\left (d\,x^3+c\right )}^2}{864\,c^4}-\frac {335\,d\,\left (d\,x^3+c\right )}{864\,c^3}}{3\,{\left (d\,x^3+c\right )}^{5/2}-30\,c\,{\left (d\,x^3+c\right )}^{3/2}+27\,c^2\,\sqrt {d\,x^3+c}}-\frac {d\,\left (\mathrm {atanh}\left (\frac {c^4\,\sqrt {d\,x^3+c}}{\sqrt {c^9}}\right )\,1{}\mathrm {i}+\frac {\mathrm {atanh}\left (\frac {c^4\,\sqrt {d\,x^3+c}}{3\,\sqrt {c^9}}\right )\,1{}\mathrm {i}}{81}\right )\,5{}\mathrm {i}}{384\,\sqrt {c^9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(c + d*x^3)^(3/2)*(8*c - d*x^3)^2),x)

[Out]

- ((2*d)/(9*c^2) + (35*d*(c + d*x^3)^2)/(864*c^4) - (335*d*(c + d*x^3))/(864*c^3))/(3*(c + d*x^3)^(5/2) - 30*c
*(c + d*x^3)^(3/2) + 27*c^2*(c + d*x^3)^(1/2)) - (d*(atanh((c^4*(c + d*x^3)^(1/2))/(c^9)^(1/2))*1i + (atanh((c
^4*(c + d*x^3)^(1/2))/(3*(c^9)^(1/2)))*1i)/81)*5i)/(384*(c^9)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{4} \left (- 8 c + d x^{3}\right )^{2} \left (c + d x^{3}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(-d*x**3+8*c)**2/(d*x**3+c)**(3/2),x)

[Out]

Integral(1/(x**4*(-8*c + d*x**3)**2*(c + d*x**3)**(3/2)), x)

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