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

Optimal. Leaf size=100 \[ -\frac{25 d}{216 c^3 \sqrt{c+d x^3}}-\frac{1}{24 c^2 x^3 \sqrt{c+d x^3}}+\frac{d \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{2592 c^{7/2}}+\frac{11 d \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{96 c^{7/2}} \]

[Out]

(-25*d)/(216*c^3*Sqrt[c + d*x^3]) - 1/(24*c^2*x^3*Sqrt[c + d*x^3]) + (d*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/
(2592*c^(7/2)) + (11*d*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(96*c^(7/2))

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Rubi [A]  time = 0.0950825, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {446, 103, 152, 156, 63, 208, 206} \[ -\frac{25 d}{216 c^3 \sqrt{c+d x^3}}-\frac{1}{24 c^2 x^3 \sqrt{c+d x^3}}+\frac{d \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{2592 c^{7/2}}+\frac{11 d \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{96 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-25*d)/(216*c^3*Sqrt[c + d*x^3]) - 1/(24*c^2*x^3*Sqrt[c + d*x^3]) + (d*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/
(2592*c^(7/2)) + (11*d*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(96*c^(7/2))

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 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 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 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 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 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])

Rubi steps

\begin{align*} \int \frac{1}{x^4 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{x^2 (8 c-d x) (c+d x)^{3/2}} \, dx,x,x^3\right )\\ &=-\frac{1}{24 c^2 x^3 \sqrt{c+d x^3}}-\frac{\operatorname{Subst}\left (\int \frac{11 c d-\frac{3 d^2 x}{2}}{x (8 c-d x) (c+d x)^{3/2}} \, dx,x,x^3\right )}{24 c^2}\\ &=-\frac{25 d}{216 c^3 \sqrt{c+d x^3}}-\frac{1}{24 c^2 x^3 \sqrt{c+d x^3}}-\frac{\operatorname{Subst}\left (\int \frac{\frac{99 c^2 d^2}{2}-\frac{25}{4} c d^3 x}{x (8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )}{108 c^4 d}\\ &=-\frac{25 d}{216 c^3 \sqrt{c+d x^3}}-\frac{1}{24 c^2 x^3 \sqrt{c+d x^3}}-\frac{(11 d) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^3\right )}{192 c^3}+\frac{d^2 \operatorname{Subst}\left (\int \frac{1}{(8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )}{1728 c^3}\\ &=-\frac{25 d}{216 c^3 \sqrt{c+d x^3}}-\frac{1}{24 c^2 x^3 \sqrt{c+d x^3}}-\frac{11 \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x^3}\right )}{96 c^3}+\frac{d \operatorname{Subst}\left (\int \frac{1}{9 c-x^2} \, dx,x,\sqrt{c+d x^3}\right )}{864 c^3}\\ &=-\frac{25 d}{216 c^3 \sqrt{c+d x^3}}-\frac{1}{24 c^2 x^3 \sqrt{c+d x^3}}+\frac{d \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{2592 c^{7/2}}+\frac{11 d \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{96 c^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0288534, size = 77, normalized size = 0.77 \[ \frac{-d x^3 \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{d x^3+c}{9 c}\right )-99 d x^3 \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{d x^3}{c}+1\right )-36 c}{864 c^3 x^3 \sqrt{c+d x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-36*c - d*x^3*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*x^3)/(9*c)] - 99*d*x^3*Hypergeometric2F1[-1/2, 1, 1/2, 1
+ (d*x^3)/c])/(864*c^3*x^3*Sqrt[c + d*x^3])

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Maple [C]  time = 0.024, size = 549, normalized size = 5.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*d^2/c^2*(2/27/d/c/((x^3+1/d*c)*d)^(1/2)+1/243*I/d^3/c^2*2^(1/2)*sum((-d^2*c)^(1/3)*(1/2*I*d*(2*x+1/d*(-I
*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(d*(x-1/d*(-d^2*c)^(1/3))/(-3*(-d^2*c)^(1/3)+I*
3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(
1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-d^2*c)^(2/3)+2*_alpha^2*d^2-(-d^2*c)^(1/3)
*_alpha*d-(-d^2*c)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^
(1/2)*d/(-d^2*c)^(1/3))^(1/2),-1/18/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_alpha^2*d-I*(-d^2*c)^(2/3)*3^(1/2)*_alpha+I
*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*d)/c,(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)
/d*(-d^2*c)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*d-8*c)))+1/8/c*(-2/3*d/c^2/((x^3+1/d*c)*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))+1/64*d/c^2*(2/3/c/((x^3+1/d*c)*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., size = 0, normalized size = 0. \begin{align*} -\int \frac{1}{{\left (d x^{3} + c\right )}^{\frac{3}{2}}{\left (d x^{3} - 8 \, c\right )} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.37141, size = 629, normalized size = 6.29 \begin{align*} \left [\frac{{\left (d^{2} x^{6} + c 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 ) + 297 \,{\left (d^{2} x^{6} + c 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 (25 \, c d x^{3} + 9 \, c^{2}\right )} \sqrt{d x^{3} + c}}{5184 \,{\left (c^{4} d x^{6} + c^{5} x^{3}\right )}}, -\frac{297 \,{\left (d^{2} x^{6} + c d x^{3}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{c}\right ) +{\left (d^{2} x^{6} + c d x^{3}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{3 \, c}\right ) + 12 \,{\left (25 \, c d x^{3} + 9 \, c^{2}\right )} \sqrt{d x^{3} + c}}{2592 \,{\left (c^{4} d x^{6} + c^{5} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/5184*((d^2*x^6 + c*d*x^3)*sqrt(c)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) + 297*(d^2*
x^6 + c*d*x^3)*sqrt(c)*log((d*x^3 + 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) - 24*(25*c*d*x^3 + 9*c^2)*sqrt(d*x^3
+ c))/(c^4*d*x^6 + c^5*x^3), -1/2592*(297*(d^2*x^6 + c*d*x^3)*sqrt(-c)*arctan(sqrt(d*x^3 + c)*sqrt(-c)/c) + (
d^2*x^6 + c*d*x^3)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) + 12*(25*c*d*x^3 + 9*c^2)*sqrt(d*x^3 + c))/
(c^4*d*x^6 + c^5*x^3)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.1405, size = 128, normalized size = 1.28 \begin{align*} -\frac{1}{2592} \, d{\left (\frac{297 \, \arctan \left (\frac{\sqrt{d x^{3} + c}}{\sqrt{-c}}\right )}{\sqrt{-c} c^{3}} + \frac{\arctan \left (\frac{\sqrt{d x^{3} + c}}{3 \, \sqrt{-c}}\right )}{\sqrt{-c} c^{3}} + \frac{12 \,{\left (25 \, d x^{3} + 9 \, c\right )}}{{\left ({\left (d x^{3} + c\right )}^{\frac{3}{2}} - \sqrt{d x^{3} + c} c\right )} c^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2592*d*(297*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^3) + arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c
)*c^3) + 12*(25*d*x^3 + 9*c)/(((d*x^3 + c)^(3/2) - sqrt(d*x^3 + c)*c)*c^3))