3.5.28 \(\int \frac {1}{x (8 c-d x^3)^2 \sqrt {c+d x^3}} \, dx\) [428]

Optimal. Leaf size=88 \[ \frac {\sqrt {c+d x^3}}{216 c^2 \left (8 c-d x^3\right )}+\frac {13 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2592 c^{5/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{96 c^{5/2}} \]

[Out]

13/2592*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(5/2)-1/96*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(5/2)+1/216*(d*x^
3+c)^(1/2)/c^2/(-d*x^3+8*c)

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Rubi [A]
time = 0.06, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {457, 105, 162, 65, 214, 212} \begin {gather*} \frac {13 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2592 c^{5/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{96 c^{5/2}}+\frac {\sqrt {c+d x^3}}{216 c^2 \left (8 c-d x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[c + d*x^3]/(216*c^2*(8*c - d*x^3)) + (13*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(2592*c^(5/2)) - ArcTanh[S
qrt[c + d*x^3]/Sqrt[c]]/(96*c^(5/2))

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 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 \left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {1}{x (8 c-d x)^2 \sqrt {c+d x}} \, dx,x,x^3\right )\\ &=\frac {\sqrt {c+d x^3}}{216 c^2 \left (8 c-d x^3\right )}-\frac {\text {Subst}\left (\int \frac {-9 c d-\frac {d^2 x}{2}}{x (8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{216 c^2 d}\\ &=\frac {\sqrt {c+d x^3}}{216 c^2 \left (8 c-d x^3\right )}+\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^3\right )}{192 c^2}+\frac {(13 d) \text {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{1728 c^2}\\ &=\frac {\sqrt {c+d x^3}}{216 c^2 \left (8 c-d x^3\right )}+\frac {13 \text {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{864 c^2}+\frac {\text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^3}\right )}{96 c^2 d}\\ &=\frac {\sqrt {c+d x^3}}{216 c^2 \left (8 c-d x^3\right )}+\frac {13 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2592 c^{5/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{96 c^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 83, normalized size = 0.94 \begin {gather*} \frac {\frac {12 \sqrt {c} \sqrt {c+d x^3}}{8 c-d x^3}+13 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )-27 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2592 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((12*Sqrt[c]*Sqrt[c + d*x^3])/(8*c - d*x^3) + 13*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])] - 27*ArcTanh[Sqrt[c + d*
x^3]/Sqrt[c]])/(2592*c^(5/2))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.41, size = 881, normalized size = 10.01

method	result	size
default	\(\text {Expression too large to display}\)	\(881\)
elliptic	\(\text {Expression too large to display}\)	\(1534\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*d/c*(1/27*(d*x^3+c)^(1/2)/c/d/(-d*x^3+8*c)-1/486*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)))-1/1728*I/d^2/c^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)*_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))-1/96*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(5/2)

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

[Out]

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

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Fricas [A]
time = 1.82, size = 226, normalized size = 2.57 \begin {gather*} \left [\frac {13 \, {\left (d x^{3} - 8 \, c\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 ) + 27 \, {\left (d x^{3} - 8 \, c\right )} \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 24 \, \sqrt {d x^{3} + c} c}{5184 \, {\left (c^{3} d x^{3} - 8 \, c^{4}\right )}}, \frac {27 \, {\left (d x^{3} - 8 \, c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) - 13 \, {\left (d x^{3} - 8 \, c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) - 12 \, \sqrt {d x^{3} + c} c}{2592 \, {\left (c^{3} d x^{3} - 8 \, c^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/5184*(13*(d*x^3 - 8*c)*sqrt(c)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) + 27*(d*x^3 -
8*c)*sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) - 24*sqrt(d*x^3 + c)*c)/(c^3*d*x^3 - 8*c^4), 1
/2592*(27*(d*x^3 - 8*c)*sqrt(-c)*arctan(sqrt(d*x^3 + c)*sqrt(-c)/c) - 13*(d*x^3 - 8*c)*sqrt(-c)*arctan(1/3*sqr
t(d*x^3 + c)*sqrt(-c)/c) - 12*sqrt(d*x^3 + c)*c)/(c^3*d*x^3 - 8*c^4)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.01, size = 79, normalized size = 0.90 \begin {gather*} \frac {\arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{96 \, \sqrt {-c} c^{2}} - \frac {13 \, \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{2592 \, \sqrt {-c} c^{2}} - \frac {\sqrt {d x^{3} + c}}{216 \, {\left (d x^{3} - 8 \, c\right )} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^2) - 13/2592*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*
c^2) - 1/216*sqrt(d*x^3 + c)/((d*x^3 - 8*c)*c^2)

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Mupad [B]
time = 4.01, size = 80, normalized size = 0.91 \begin {gather*} \frac {13\,\mathrm {atanh}\left (\frac {c^2\,\sqrt {d\,x^3+c}}{3\,\sqrt {c^5}}\right )}{2592\,\sqrt {c^5}}-\frac {\mathrm {atanh}\left (\frac {c^2\,\sqrt {d\,x^3+c}}{\sqrt {c^5}}\right )}{96\,\sqrt {c^5}}+\frac {\sqrt {d\,x^3+c}}{72\,c^2\,\left (24\,c-3\,d\,x^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(13*atanh((c^2*(c + d*x^3)^(1/2))/(3*(c^5)^(1/2))))/(2592*(c^5)^(1/2)) - atanh((c^2*(c + d*x^3)^(1/2))/(c^5)^(
1/2))/(96*(c^5)^(1/2)) + (c + d*x^3)^(1/2)/(72*c^2*(24*c - 3*d*x^3))

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