3.5.47 \(\int \frac {1}{x^4 (8 c-d x^3)^2 (c+d x^3)^{3/2}} \, dx\) [447]

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

[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 = 0.09, 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 = {457, 105, 156, 157, 162, 65, 214, 212} \begin {gather*} \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}} \end {gather*}

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 65

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 105

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] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

Rule 156

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] && ILtQ[m, -1]

Rule 157

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 162

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 212

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

Rule 214

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

Rule 457

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} \text {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 {\text {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 {\text {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 {\text {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) \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^3\right )}{768 c^4}+\frac {\left (5 d^2\right ) \text {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 \text {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) \text {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 [A]
time = 0.23, size = 109, normalized size = 0.76 \begin {gather*} \frac {-\frac {12 \sqrt {c} \left (108 c^2+265 c d x^3-35 d^2 x^6\right )}{x^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+5 d \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )+405 d \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{31104 c^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-12*Sqrt[c]*(108*c^2 + 265*c*d*x^3 - 35*d^2*x^6))/(x^3*(8*c - d*x^3)*Sqrt[c + d*x^3]) + 5*d*ArcTanh[Sqrt[c +
d*x^3]/(3*Sqrt[c])] + 405*d*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(31104*c^(9/2))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.50, size = 1020, normalized size = 7.13

method	result	size
risch	\(\text {Expression too large to display}\)	\(913\)
default	\(\text {Expression too large to display}\)	\(1020\)
elliptic	\(\text {Expression too large to display}\)	\(1569\)

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,method=_RETURNVERBOSE)

[Out]

1/64/c^2*d^2*(1/243/c^2/d*(d*x^3+c)^(1/2)/(-d*x^3+8*c)-2/243/d/c^2/((x^3+c/d)*d)^(1/2)-1/1458*I/c^3/d^3*2^(1/2
)*sum((-c*d^2)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)*(d*(x
-1/d*(-c*d^2)^(1/3))/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-c*d^2
)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-c*d^2)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-
c*d^2)^(2/3)+2*_alpha^2*d^2-(-c*d^2)^(1/3)*_alpha*d-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2
)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2),-1/18/d*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_a
lpha^2*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-c*d^2)^(2/3)*_alpha-3*c*d)/c,(I*3^(1/2)/d*(-c*d^2)^
(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*d-8*c)))-1/256/c^3*d^2
*(2/27/d/c/((x^3+c/d)*d)^(1/2)+1/243*I/d^3/c^2*2^(1/2)*sum((-c*d^2)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-c*d^
2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)*(d*(x-1/d*(-c*d^2)^(1/3))/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2
)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^
(1/2)*(I*(-c*d^2)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-c*d^2)^(2/3)+2*_alpha^2*d^2-(-c*d^2)^(1/3)*_alpha*d-(-c*d
^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2
)^(1/3))^(1/2),-1/18/d*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*
(-c*d^2)^(2/3)*_alpha-3*c*d)/c,(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/
3)))^(1/2)),_alpha=RootOf(_Z^3*d-8*c)))+1/64/c^2*(-2/3*d/c^2/((x^3+c/d)*d)^(1/2)-1/3*(d*x^3+c)^(1/2)/c^2/x^3+d
*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(5/2))+1/256/c^3*d*(2/3/c/((x^3+c/d)*d)^(1/2)-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*} \text {Failed to integrate} \end {gather*}

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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Fricas [A]
time = 3.33, size = 368, normalized size = 2.57 \begin {gather*} \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 ] \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{4} \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**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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Giac [A]
time = 1.32, size = 129, normalized size = 0.90 \begin {gather*} -\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}} \end {gather*}

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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Mupad [B]
time = 4.56, size = 133, normalized size = 0.93 \begin {gather*} -\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}} \end {gather*}

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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