3.403 \(\int \frac {\sqrt {c+d x^3}}{x^4 (8 c-d x^3)^2} \, dx\)

Optimal. Leaf size=124 \[ \frac {7 d \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{1152 c^{5/2}}-\frac {d \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{128 c^{5/2}}+\frac {d \sqrt {c+d x^3}}{96 c^2 \left (8 c-d x^3\right )}-\frac {\sqrt {c+d x^3}}{24 c x^3 \left (8 c-d x^3\right )} \]

[Out]

7/1152*d*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(5/2)-1/128*d*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(5/2)+1/96*d*
(d*x^3+c)^(1/2)/c^2/(-d*x^3+8*c)-1/24*(d*x^3+c)^(1/2)/c/x^3/(-d*x^3+8*c)

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Rubi [A]  time = 0.10, antiderivative size = 124, normalized size of antiderivative = 1.00, 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, 99, 151, 156, 63, 208, 206} \[ \frac {d \sqrt {c+d x^3}}{96 c^2 \left (8 c-d x^3\right )}+\frac {7 d \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{1152 c^{5/2}}-\frac {d \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{128 c^{5/2}}-\frac {\sqrt {c+d x^3}}{24 c x^3 \left (8 c-d x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*Sqrt[c + d*x^3])/(96*c^2*(8*c - d*x^3)) - Sqrt[c + d*x^3]/(24*c*x^3*(8*c - d*x^3)) + (7*d*ArcTanh[Sqrt[c +
d*x^3]/(3*Sqrt[c])])/(1152*c^(5/2)) - (d*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(128*c^(5/2))

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 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

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 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 {\sqrt {c+d x^3}}{x^4 \left (8 c-d x^3\right )^2} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{x^2 (8 c-d x)^2} \, dx,x,x^3\right )\\ &=-\frac {\sqrt {c+d x^3}}{24 c x^3 \left (8 c-d x^3\right )}+\frac {\operatorname {Subst}\left (\int \frac {6 c d+\frac {3 d^2 x}{2}}{x (8 c-d x)^2 \sqrt {c+d x}} \, dx,x,x^3\right )}{24 c}\\ &=\frac {d \sqrt {c+d x^3}}{96 c^2 \left (8 c-d x^3\right )}-\frac {\sqrt {c+d x^3}}{24 c x^3 \left (8 c-d x^3\right )}-\frac {\operatorname {Subst}\left (\int \frac {-54 c^2 d^2-9 c d^3 x}{x (8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{1728 c^3 d}\\ &=\frac {d \sqrt {c+d x^3}}{96 c^2 \left (8 c-d x^3\right )}-\frac {\sqrt {c+d x^3}}{24 c x^3 \left (8 c-d x^3\right )}+\frac {d \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^3\right )}{256 c^2}+\frac {\left (7 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{768 c^2}\\ &=\frac {d \sqrt {c+d x^3}}{96 c^2 \left (8 c-d x^3\right )}-\frac {\sqrt {c+d x^3}}{24 c x^3 \left (8 c-d x^3\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^3}\right )}{128 c^2}+\frac {(7 d) \operatorname {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{384 c^2}\\ &=\frac {d \sqrt {c+d x^3}}{96 c^2 \left (8 c-d x^3\right )}-\frac {\sqrt {c+d x^3}}{24 c x^3 \left (8 c-d x^3\right )}+\frac {7 d \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{1152 c^{5/2}}-\frac {d \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{128 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 97, normalized size = 0.78 \[ \frac {7 d \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )-9 d \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )+\frac {12 \sqrt {c} \sqrt {c+d x^3} \left (4 c-d x^3\right )}{d x^6-8 c x^3}}{1152 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((12*Sqrt[c]*(4*c - d*x^3)*Sqrt[c + d*x^3])/(-8*c*x^3 + d*x^6) + 7*d*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])] - 9*
d*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(1152*c^(5/2))

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fricas [A]  time = 0.87, size = 278, normalized size = 2.24 \[ \left [\frac {7 \, {\left (d^{2} x^{6} - 8 \, c d x^{3}\right )} \sqrt {c} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 9 \, {\left (d^{2} x^{6} - 8 \, c d x^{3}\right )} \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 24 \, {\left (c d x^{3} - 4 \, c^{2}\right )} \sqrt {d x^{3} + c}}{2304 \, {\left (c^{3} d x^{6} - 8 \, c^{4} x^{3}\right )}}, \frac {9 \, {\left (d^{2} x^{6} - 8 \, c d x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) - 7 \, {\left (d^{2} x^{6} - 8 \, c d x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) - 12 \, {\left (c d x^{3} - 4 \, c^{2}\right )} \sqrt {d x^{3} + c}}{1152 \, {\left (c^{3} d x^{6} - 8 \, c^{4} x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2304*(7*(d^2*x^6 - 8*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)) + 9*(d^
2*x^6 - 8*c*d*x^3)*sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) - 24*(c*d*x^3 - 4*c^2)*sqrt(d*x^
3 + c))/(c^3*d*x^6 - 8*c^4*x^3), 1/1152*(9*(d^2*x^6 - 8*c*d*x^3)*sqrt(-c)*arctan(sqrt(d*x^3 + c)*sqrt(-c)/c) -
7*(d^2*x^6 - 8*c*d*x^3)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) - 12*(c*d*x^3 - 4*c^2)*sqrt(d*x^3 + c
))/(c^3*d*x^6 - 8*c^4*x^3)]

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giac [A]  time = 0.17, size = 113, normalized size = 0.91 \[ \frac {d \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{128 \, \sqrt {-c} c^{2}} - \frac {7 \, d \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{1152 \, \sqrt {-c} c^{2}} - \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}} d - 5 \, \sqrt {d x^{3} + c} c d}{96 \, {\left ({\left (d x^{3} + c\right )}^{2} - 10 \, {\left (d x^{3} + c\right )} c + 9 \, c^{2}\right )} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*d*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^2) - 7/1152*d*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(
-c)*c^2) - 1/96*((d*x^3 + c)^(3/2)*d - 5*sqrt(d*x^3 + c)*c*d)/(((d*x^3 + c)^2 - 10*(d*x^3 + c)*c + 9*c^2)*c^2)

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maple [C]  time = 0.21, size = 957, normalized size = 7.72 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64/c^2*d^2*(-1/3*(d*x^3+c)^(1/2)/(d*x^3-8*c)/d+1/54*I/d^3/c*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)/x^3-1/3*d*arctanh((d*x^3+c)^(1
/2)/c^(1/2))/c^(1/2))-1/256/c^3*d^2*(2/3*(d*x^3+c)^(1/2)/d+1/3*I/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*(d*x^3+c)^(1/2)-2/3*arctanh((d*x^3+c)
^(1/2)/c^(1/2))*c^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.21, size = 117, normalized size = 0.94 \[ \frac {\frac {5\,d\,\sqrt {d\,x^3+c}}{32\,c}-\frac {d\,{\left (d\,x^3+c\right )}^{3/2}}{32\,c^2}}{3\,{\left (d\,x^3+c\right )}^2-30\,c\,\left (d\,x^3+c\right )+27\,c^2}+\frac {d\,\left (\mathrm {atanh}\left (\frac {c^2\,\sqrt {d\,x^3+c}}{\sqrt {c^5}}\right )\,1{}\mathrm {i}-\frac {\mathrm {atanh}\left (\frac {c^2\,\sqrt {d\,x^3+c}}{3\,\sqrt {c^5}}\right )\,7{}\mathrm {i}}{9}\right )\,1{}\mathrm {i}}{128\,\sqrt {c^5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((5*d*(c + d*x^3)^(1/2))/(32*c) - (d*(c + d*x^3)^(3/2))/(32*c^2))/(3*(c + d*x^3)^2 - 30*c*(c + d*x^3) + 27*c^2
) + (d*(atanh((c^2*(c + d*x^3)^(1/2))/(c^5)^(1/2))*1i - (atanh((c^2*(c + d*x^3)^(1/2))/(3*(c^5)^(1/2)))*7i)/9)
*1i)/(128*(c^5)^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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