3.416 \(\int \frac{(c+d x^3)^{3/2}}{x^4 (8 c-d x^3)^2} \, dx\)
Optimal. Leaf size=121 \[ \frac{3 d \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{128 c^{3/2}}-\frac{7 d \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{384 c^{3/2}}+\frac{5 d \sqrt{c+d x^3}}{96 c \left (8 c-d x^3\right )}-\frac{\sqrt{c+d x^3}}{24 x^3 \left (8 c-d x^3\right )} \]
[Out]
(5*d*Sqrt[c + d*x^3])/(96*c*(8*c - d*x^3)) - Sqrt[c + d*x^3]/(24*x^3*(8*c - d*x^3)) + (3*d*ArcTanh[Sqrt[c + d*
x^3]/(3*Sqrt[c])])/(128*c^(3/2)) - (7*d*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(384*c^(3/2))
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Rubi [A] time = 0.103458, antiderivative size = 121, 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, 98, 151, 156, 63, 208, 206} \[ \frac{3 d \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{128 c^{3/2}}-\frac{7 d \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{384 c^{3/2}}+\frac{5 d \sqrt{c+d x^3}}{96 c \left (8 c-d x^3\right )}-\frac{\sqrt{c+d x^3}}{24 x^3 \left (8 c-d x^3\right )} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*x^3)^(3/2)/(x^4*(8*c - d*x^3)^2),x]
[Out]
(5*d*Sqrt[c + d*x^3])/(96*c*(8*c - d*x^3)) - Sqrt[c + d*x^3]/(24*x^3*(8*c - d*x^3)) + (3*d*ArcTanh[Sqrt[c + d*
x^3]/(3*Sqrt[c])])/(128*c^(3/2)) - (7*d*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(384*c^(3/2))
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 98
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (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{\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )^2} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{(c+d x)^{3/2}}{x^2 (8 c-d x)^2} \, dx,x,x^3\right )\\ &=-\frac{\sqrt{c+d x^3}}{24 x^3 \left (8 c-d x^3\right )}-\frac{\operatorname{Subst}\left (\int \frac{-14 c^2 d-\frac{19}{2} c d^2 x}{x (8 c-d x)^2 \sqrt{c+d x}} \, dx,x,x^3\right )}{24 c}\\ &=\frac{5 d \sqrt{c+d x^3}}{96 c \left (8 c-d x^3\right )}-\frac{\sqrt{c+d x^3}}{24 x^3 \left (8 c-d x^3\right )}+\frac{\operatorname{Subst}\left (\int \frac{126 c^3 d^2+45 c^2 d^3 x}{x (8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )}{1728 c^3 d}\\ &=\frac{5 d \sqrt{c+d x^3}}{96 c \left (8 c-d x^3\right )}-\frac{\sqrt{c+d x^3}}{24 x^3 \left (8 c-d x^3\right )}+\frac{(7 d) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^3\right )}{768 c}+\frac{\left (9 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{(8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )}{256 c}\\ &=\frac{5 d \sqrt{c+d x^3}}{96 c \left (8 c-d x^3\right )}-\frac{\sqrt{c+d x^3}}{24 x^3 \left (8 c-d x^3\right )}+\frac{7 \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x^3}\right )}{384 c}+\frac{(9 d) \operatorname{Subst}\left (\int \frac{1}{9 c-x^2} \, dx,x,\sqrt{c+d x^3}\right )}{128 c}\\ &=\frac{5 d \sqrt{c+d x^3}}{96 c \left (8 c-d x^3\right )}-\frac{\sqrt{c+d x^3}}{24 x^3 \left (8 c-d x^3\right )}+\frac{3 d \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{128 c^{3/2}}-\frac{7 d \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{384 c^{3/2}}\\ \end{align*}
Mathematica [A] time = 0.132811, size = 97, normalized size = 0.8 \[ \frac{\frac{4 \sqrt{c} \sqrt{c+d x^3} \left (4 c-5 d x^3\right )}{d x^6-8 c x^3}+9 d \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )-7 d \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{384 c^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*x^3)^(3/2)/(x^4*(8*c - d*x^3)^2),x]
[Out]
((4*Sqrt[c]*(4*c - 5*d*x^3)*Sqrt[c + d*x^3])/(-8*c*x^3 + d*x^6) + 9*d*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])] - 7
*d*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(384*c^(3/2))
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Maple [C] time = 0.015, size = 1014, normalized size = 8.4 \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)^(3/2)/x^4/(-d*x^3+8*c)^2,x)
[Out]
-1/256*d^2/c^3*(2/9*x^3*(d*x^3+c)^(1/2)+56/9*c*(d*x^3+c)^(1/2)/d+3*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*d^2/c^2*(-3/d*c*(d*x^3+c)^(1/2)/(d*x^
3-8*c)+2/3*(d*x^3+c)^(1/2)/d+1/2*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)))+1/64/c^2*(-1/3*c*(d*x^3+c)^(1/2)/x^3+2/3*d*(d*x^3+c)^(1/2)-c^(1/2)*d*arctanh((
d*x^3+c)^(1/2)/c^(1/2)))+1/256/c^3*d*(2/9*d*x^3*(d*x^3+c)^(1/2)+8/9*c*(d*x^3+c)^(1/2)-2/3*c^(3/2)*arctanh((d*x
^3+c)^(1/2)/c^(1/2)))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{3} + c\right )}^{\frac{3}{2}}}{{\left (d x^{3} - 8 \, c\right )}^{2} x^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x^3+c)^(3/2)/x^4/(-d*x^3+8*c)^2,x, algorithm="maxima")
[Out]
integrate((d*x^3 + c)^(3/2)/((d*x^3 - 8*c)^2*x^4), x)
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Fricas [A] time = 1.64422, size = 636, normalized size = 5.26 \begin{align*} \left [\frac{9 \,{\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 ) + 7 \,{\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 ) - 8 \,{\left (5 \, c d x^{3} - 4 \, c^{2}\right )} \sqrt{d x^{3} + c}}{768 \,{\left (c^{2} d x^{6} - 8 \, c^{3} x^{3}\right )}}, \frac{7 \,{\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 ) - 9 \,{\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 ) - 4 \,{\left (5 \, c d x^{3} - 4 \, c^{2}\right )} \sqrt{d x^{3} + c}}{384 \,{\left (c^{2} d x^{6} - 8 \, c^{3} x^{3}\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x^3+c)^(3/2)/x^4/(-d*x^3+8*c)^2,x, algorithm="fricas")
[Out]
[1/768*(9*(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)) + 7*(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) - 8*(5*c*d*x^3 - 4*c^2)*sqrt(d*x^
3 + c))/(c^2*d*x^6 - 8*c^3*x^3), 1/384*(7*(d^2*x^6 - 8*c*d*x^3)*sqrt(-c)*arctan(sqrt(d*x^3 + c)*sqrt(-c)/c) -
9*(d^2*x^6 - 8*c*d*x^3)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) - 4*(5*c*d*x^3 - 4*c^2)*sqrt(d*x^3 + c
))/(c^2*d*x^6 - 8*c^3*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((d*x**3+c)**(3/2)/x**4/(-d*x**3+8*c)**2,x)
[Out]
Timed out
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Giac [A] time = 1.1618, size = 153, normalized size = 1.26 \begin{align*} \frac{1}{384} \, d{\left (\frac{7 \, \arctan \left (\frac{\sqrt{d x^{3} + c}}{\sqrt{-c}}\right )}{\sqrt{-c} c} - \frac{9 \, \arctan \left (\frac{\sqrt{d x^{3} + c}}{3 \, \sqrt{-c}}\right )}{\sqrt{-c} c} - \frac{4 \,{\left (5 \,{\left (d x^{3} + c\right )}^{\frac{3}{2}} - 9 \, \sqrt{d x^{3} + c} c\right )}}{{\left ({\left (d x^{3} + c\right )}^{2} - 10 \,{\left (d x^{3} + c\right )} c + 9 \, c^{2}\right )} c}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x^3+c)^(3/2)/x^4/(-d*x^3+8*c)^2,x, algorithm="giac")
[Out]
1/384*d*(7*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c) - 9*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c)
- 4*(5*(d*x^3 + c)^(3/2) - 9*sqrt(d*x^3 + c)*c)/(((d*x^3 + c)^2 - 10*(d*x^3 + c)*c + 9*c^2)*c))