3.288 \(\int \frac{\sqrt{c+d x^3}}{x^7 (8 c-d x^3)} \, dx\)

Optimal. Leaf size=107 \[ \frac{d^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{256 c^{5/2}}+\frac{d^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{256 c^{5/2}}-\frac{d \sqrt{c+d x^3}}{64 c^2 x^3}-\frac{\sqrt{c+d x^3}}{48 c x^6} \]

[Out]

-Sqrt[c + d*x^3]/(48*c*x^6) - (d*Sqrt[c + d*x^3])/(64*c^2*x^3) + (d^2*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(2
56*c^(5/2)) + (d^2*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(256*c^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0940667, antiderivative size = 107, 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, 99, 151, 156, 63, 208, 206} \[ \frac{d^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{256 c^{5/2}}+\frac{d^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{256 c^{5/2}}-\frac{d \sqrt{c+d x^3}}{64 c^2 x^3}-\frac{\sqrt{c+d x^3}}{48 c x^6} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x^3]/(x^7*(8*c - d*x^3)),x]

[Out]

-Sqrt[c + d*x^3]/(48*c*x^6) - (d*Sqrt[c + d*x^3])/(64*c^2*x^3) + (d^2*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(2
56*c^(5/2)) + (d^2*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(256*c^(5/2))

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 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 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{\sqrt{c+d x^3}}{x^7 \left (8 c-d x^3\right )} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{x^3 (8 c-d x)} \, dx,x,x^3\right )\\ &=-\frac{\sqrt{c+d x^3}}{48 c x^6}+\frac{\operatorname{Subst}\left (\int \frac{6 c d+\frac{3 d^2 x}{2}}{x^2 (8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )}{48 c}\\ &=-\frac{\sqrt{c+d x^3}}{48 c x^6}-\frac{d \sqrt{c+d x^3}}{64 c^2 x^3}-\frac{\operatorname{Subst}\left (\int \frac{6 c^2 d^2-3 c d^3 x}{x (8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )}{384 c^3}\\ &=-\frac{\sqrt{c+d x^3}}{48 c x^6}-\frac{d \sqrt{c+d x^3}}{64 c^2 x^3}-\frac{d^2 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^3\right )}{512 c^2}+\frac{\left (3 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{(8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )}{512 c^2}\\ &=-\frac{\sqrt{c+d x^3}}{48 c x^6}-\frac{d \sqrt{c+d x^3}}{64 c^2 x^3}-\frac{d \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x^3}\right )}{256 c^2}+\frac{\left (3 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{9 c-x^2} \, dx,x,\sqrt{c+d x^3}\right )}{256 c^2}\\ &=-\frac{\sqrt{c+d x^3}}{48 c x^6}-\frac{d \sqrt{c+d x^3}}{64 c^2 x^3}+\frac{d^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{256 c^{5/2}}+\frac{d^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{256 c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0522811, size = 96, normalized size = 0.9 \[ \frac{3 d^2 x^6 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )+3 d^2 x^6 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )-4 \sqrt{c} \sqrt{c+d x^3} \left (4 c+3 d x^3\right )}{768 c^{5/2} x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*x^3]/(x^7*(8*c - d*x^3)),x]

[Out]

(-4*Sqrt[c]*Sqrt[c + d*x^3]*(4*c + 3*d*x^3) + 3*d^2*x^6*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])] + 3*d^2*x^6*ArcTa
nh[Sqrt[c + d*x^3]/Sqrt[c]])/(768*c^(5/2)*x^6)

________________________________________________________________________________________

Maple [C]  time = 0.024, size = 574, normalized size = 5.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^3+c)^(1/2)/x^7/(-d*x^3+8*c),x)

[Out]

-1/512*d^3/c^3*(2/3*(d*x^3+c)^(1/2)/d+1/3*I/d^3*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*(-1/6*(d*x^3+c)^(1/2)/x^6-1/12*d*(d*x^3+c)^(1/2)/c/x^3+1/12*d^2
*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(3/2))+1/64*d/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))+1/512*d^2/c^3*(2/3*(d*x^3+c)^(1/2)-2/3*arctanh((d*x^3+c)^(1/2)/c^(1/2))*c^(1/2))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{\sqrt{d x^{3} + c}}{{\left (d x^{3} - 8 \, c\right )} x^{7}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(1/2)/x^7/(-d*x^3+8*c),x, algorithm="maxima")

[Out]

-integrate(sqrt(d*x^3 + c)/((d*x^3 - 8*c)*x^7), x)

________________________________________________________________________________________

Fricas [A]  time = 1.51803, size = 439, normalized size = 4.1 \begin{align*} \left [\frac{3 \, \sqrt{c} d^{2} x^{6} \log \left (\frac{d^{2} x^{6} + 24 \, c d x^{3} + 8 \,{\left (d x^{3} + 4 \, c\right )} \sqrt{d x^{3} + c} \sqrt{c} + 32 \, c^{2}}{d x^{6} - 8 \, c x^{3}}\right ) - 8 \,{\left (3 \, c d x^{3} + 4 \, c^{2}\right )} \sqrt{d x^{3} + c}}{1536 \, c^{3} x^{6}}, -\frac{3 \, \sqrt{-c} d^{2} x^{6} \arctan \left (\frac{{\left (d x^{3} + 4 \, c\right )} \sqrt{d x^{3} + c} \sqrt{-c}}{4 \,{\left (c d x^{3} + c^{2}\right )}}\right ) + 4 \,{\left (3 \, c d x^{3} + 4 \, c^{2}\right )} \sqrt{d x^{3} + c}}{768 \, c^{3} x^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(1/2)/x^7/(-d*x^3+8*c),x, algorithm="fricas")

[Out]

[1/1536*(3*sqrt(c)*d^2*x^6*log((d^2*x^6 + 24*c*d*x^3 + 8*(d*x^3 + 4*c)*sqrt(d*x^3 + c)*sqrt(c) + 32*c^2)/(d*x^
6 - 8*c*x^3)) - 8*(3*c*d*x^3 + 4*c^2)*sqrt(d*x^3 + c))/(c^3*x^6), -1/768*(3*sqrt(-c)*d^2*x^6*arctan(1/4*(d*x^3
+ 4*c)*sqrt(d*x^3 + c)*sqrt(-c)/(c*d*x^3 + c^2)) + 4*(3*c*d*x^3 + 4*c^2)*sqrt(d*x^3 + c))/(c^3*x^6)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{\sqrt{c + d x^{3}}}{- 8 c x^{7} + d x^{10}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**3+c)**(1/2)/x**7/(-d*x**3+8*c),x)

[Out]

-Integral(sqrt(c + d*x**3)/(-8*c*x**7 + d*x**10), x)

________________________________________________________________________________________

Giac [A]  time = 1.11543, size = 126, normalized size = 1.18 \begin{align*} -\frac{1}{768} \, d^{2}{\left (\frac{3 \, \arctan \left (\frac{\sqrt{d x^{3} + c}}{\sqrt{-c}}\right )}{\sqrt{-c} c^{2}} + \frac{3 \, \arctan \left (\frac{\sqrt{d x^{3} + c}}{3 \, \sqrt{-c}}\right )}{\sqrt{-c} c^{2}} + \frac{4 \,{\left (3 \,{\left (d x^{3} + c\right )}^{\frac{3}{2}} + \sqrt{d x^{3} + c} c\right )}}{c^{2} d^{2} x^{6}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+c)^(1/2)/x^7/(-d*x^3+8*c),x, algorithm="giac")

[Out]

-1/768*d^2*(3*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^2) + 3*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-
c)*c^2) + 4*(3*(d*x^3 + c)^(3/2) + sqrt(d*x^3 + c)*c)/(c^2*d^2*x^6))