3.404 \(\int \frac{\sqrt{c+d x^3}}{x^7 (8 c-d x^3)^2} \, dx\)

Optimal. Leaf size=164 \[ \frac{5 d^2 \sqrt{c+d x^3}}{1536 c^3 \left (8 c-d x^3\right )}+\frac{23 d^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{18432 c^{7/2}}-\frac{d^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{2048 c^{7/2}}-\frac{7 d \sqrt{c+d x^3}}{384 c^2 x^3 \left (8 c-d x^3\right )}-\frac{\sqrt{c+d x^3}}{48 c x^6 \left (8 c-d x^3\right )} \]

[Out]

(5*d^2*Sqrt[c + d*x^3])/(1536*c^3*(8*c - d*x^3)) - Sqrt[c + d*x^3]/(48*c*x^6*(8*c - d*x^3)) - (7*d*Sqrt[c + d*
x^3])/(384*c^2*x^3*(8*c - d*x^3)) + (23*d^2*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(18432*c^(7/2)) - (d^2*ArcTa
nh[Sqrt[c + d*x^3]/Sqrt[c]])/(2048*c^(7/2))

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Rubi [A]  time = 0.131336, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ \frac{5 d^2 \sqrt{c+d x^3}}{1536 c^3 \left (8 c-d x^3\right )}+\frac{23 d^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{18432 c^{7/2}}-\frac{d^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{2048 c^{7/2}}-\frac{7 d \sqrt{c+d x^3}}{384 c^2 x^3 \left (8 c-d x^3\right )}-\frac{\sqrt{c+d x^3}}{48 c x^6 \left (8 c-d x^3\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x^3]/(x^7*(8*c - d*x^3)^2),x]

[Out]

(5*d^2*Sqrt[c + d*x^3])/(1536*c^3*(8*c - d*x^3)) - Sqrt[c + d*x^3]/(48*c*x^6*(8*c - d*x^3)) - (7*d*Sqrt[c + d*
x^3])/(384*c^2*x^3*(8*c - d*x^3)) + (23*d^2*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(18432*c^(7/2)) - (d^2*ArcTa
nh[Sqrt[c + d*x^3]/Sqrt[c]])/(2048*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 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 )^2} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{x^3 (8 c-d x)^2} \, dx,x,x^3\right )\\ &=-\frac{\sqrt{c+d x^3}}{48 c x^6 \left (8 c-d x^3\right )}+\frac{\operatorname{Subst}\left (\int \frac{7 c d+\frac{5 d^2 x}{2}}{x^2 (8 c-d x)^2 \sqrt{c+d x}} \, dx,x,x^3\right )}{48 c}\\ &=-\frac{\sqrt{c+d x^3}}{48 c x^6 \left (8 c-d x^3\right )}-\frac{7 d \sqrt{c+d x^3}}{384 c^2 x^3 \left (8 c-d x^3\right )}-\frac{\operatorname{Subst}\left (\int \frac{-6 c^2 d^2-\frac{21}{2} c d^3 x}{x (8 c-d x)^2 \sqrt{c+d x}} \, dx,x,x^3\right )}{384 c^3}\\ &=\frac{5 d^2 \sqrt{c+d x^3}}{1536 c^3 \left (8 c-d x^3\right )}-\frac{\sqrt{c+d x^3}}{48 c x^6 \left (8 c-d x^3\right )}-\frac{7 d \sqrt{c+d x^3}}{384 c^2 x^3 \left (8 c-d x^3\right )}+\frac{\operatorname{Subst}\left (\int \frac{54 c^3 d^3+45 c^2 d^4 x}{x (8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )}{27648 c^5 d}\\ &=\frac{5 d^2 \sqrt{c+d x^3}}{1536 c^3 \left (8 c-d x^3\right )}-\frac{\sqrt{c+d x^3}}{48 c x^6 \left (8 c-d x^3\right )}-\frac{7 d \sqrt{c+d x^3}}{384 c^2 x^3 \left (8 c-d x^3\right )}+\frac{d^2 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^3\right )}{4096 c^3}+\frac{\left (23 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{(8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )}{12288 c^3}\\ &=\frac{5 d^2 \sqrt{c+d x^3}}{1536 c^3 \left (8 c-d x^3\right )}-\frac{\sqrt{c+d x^3}}{48 c x^6 \left (8 c-d x^3\right )}-\frac{7 d \sqrt{c+d x^3}}{384 c^2 x^3 \left (8 c-d x^3\right )}+\frac{d \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x^3}\right )}{2048 c^3}+\frac{\left (23 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{9 c-x^2} \, dx,x,\sqrt{c+d x^3}\right )}{6144 c^3}\\ &=\frac{5 d^2 \sqrt{c+d x^3}}{1536 c^3 \left (8 c-d x^3\right )}-\frac{\sqrt{c+d x^3}}{48 c x^6 \left (8 c-d x^3\right )}-\frac{7 d \sqrt{c+d x^3}}{384 c^2 x^3 \left (8 c-d x^3\right )}+\frac{23 d^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{18432 c^{7/2}}-\frac{d^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{2048 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.171532, size = 112, normalized size = 0.68 \[ \frac{\frac{12 \sqrt{c} \sqrt{c+d x^3} \left (32 c^2+28 c d x^3-5 d^2 x^6\right )}{d x^9-8 c x^6}+23 d^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )-9 d^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{18432 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*x^3]/(x^7*(8*c - d*x^3)^2),x]

[Out]

((12*Sqrt[c]*Sqrt[c + d*x^3]*(32*c^2 + 28*c*d*x^3 - 5*d^2*x^6))/(-8*c*x^6 + d*x^9) + 23*d^2*ArcTanh[Sqrt[c + d
*x^3]/(3*Sqrt[c])] - 9*d^2*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(18432*c^(7/2))

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Maple [C]  time = 0.015, size = 1020, normalized size = 6.2 \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)^2,x)

[Out]

-3/4096*d^3/c^4*(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)))+1/512*d^3/c^3*(-1/3/d*(d*x^3+c)^(1/2)/(d*x^3-8*c)+1/54*I/d^3/c*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)))+1/64/c^2*(-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))+1/256/c
^3*d*(-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))+3/4096/c^4*d^2*(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., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d x^{3} + c}}{{\left (d x^{3} - 8 \, c\right )}^{2} 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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.03926, size = 706, normalized size = 4.3 \begin{align*} \left [\frac{23 \,{\left (d^{3} x^{9} - 8 \, c d^{2} x^{6}\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^{3} x^{9} - 8 \, c d^{2} x^{6}\right )} \sqrt{c} \log \left (\frac{d x^{3} - 2 \, \sqrt{d x^{3} + c} \sqrt{c} + 2 \, c}{x^{3}}\right ) - 24 \,{\left (5 \, c d^{2} x^{6} - 28 \, c^{2} d x^{3} - 32 \, c^{3}\right )} \sqrt{d x^{3} + c}}{36864 \,{\left (c^{4} d x^{9} - 8 \, c^{5} x^{6}\right )}}, \frac{9 \,{\left (d^{3} x^{9} - 8 \, c d^{2} x^{6}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{c}\right ) - 23 \,{\left (d^{3} x^{9} - 8 \, c d^{2} x^{6}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{3 \, c}\right ) - 12 \,{\left (5 \, c d^{2} x^{6} - 28 \, c^{2} d x^{3} - 32 \, c^{3}\right )} \sqrt{d x^{3} + c}}{18432 \,{\left (c^{4} d x^{9} - 8 \, c^{5} x^{6}\right )}}\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)^2,x, algorithm="fricas")

[Out]

[1/36864*(23*(d^3*x^9 - 8*c*d^2*x^6)*sqrt(c)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) + 9
*(d^3*x^9 - 8*c*d^2*x^6)*sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) - 24*(5*c*d^2*x^6 - 28*c^2
*d*x^3 - 32*c^3)*sqrt(d*x^3 + c))/(c^4*d*x^9 - 8*c^5*x^6), 1/18432*(9*(d^3*x^9 - 8*c*d^2*x^6)*sqrt(-c)*arctan(
sqrt(d*x^3 + c)*sqrt(-c)/c) - 23*(d^3*x^9 - 8*c*d^2*x^6)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) - 12*
(5*c*d^2*x^6 - 28*c^2*d*x^3 - 32*c^3)*sqrt(d*x^3 + c))/(c^4*d*x^9 - 8*c^5*x^6)]

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

[Out]

Timed out

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Giac [A]  time = 1.11387, size = 140, normalized size = 0.85 \begin{align*} \frac{1}{18432} \, d^{2}{\left (\frac{9 \, \arctan \left (\frac{\sqrt{d x^{3} + c}}{\sqrt{-c}}\right )}{\sqrt{-c} c^{3}} - \frac{23 \, \arctan \left (\frac{\sqrt{d x^{3} + c}}{3 \, \sqrt{-c}}\right )}{\sqrt{-c} c^{3}} - \frac{12 \, \sqrt{d x^{3} + c}}{{\left (d x^{3} - 8 \, c\right )} c^{3}} - \frac{48 \,{\left (d x^{3} + c\right )}^{\frac{3}{2}}}{c^{3} 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)^2,x, algorithm="giac")

[Out]

1/18432*d^2*(9*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^3) - 23*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt
(-c)*c^3) - 12*sqrt(d*x^3 + c)/((d*x^3 - 8*c)*c^3) - 48*(d*x^3 + c)^(3/2)/(c^3*d^2*x^6))