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

Optimal. Leaf size=121 \[ \frac {3 d \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{128 c^{3/2}}-\frac {7 d \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{384 c^{3/2}}+\frac {5 d \sqrt {c+d x^3}}{96 c \left (8 c-d x^3\right )}-\frac {\sqrt {c+d x^3}}{24 x^3 \left (8 c-d x^3\right )} \]

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Rubi [A]  time = 0.10, antiderivative size = 121, 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, 98, 151, 156, 63, 208, 206} \begin {gather*} \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 )} \end {gather*}

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 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 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 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 {\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*}

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Mathematica [A]  time = 0.16, size = 97, normalized size = 0.80 \begin {gather*} \frac {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 )+\frac {4 \sqrt {c} \sqrt {c+d x^3} \left (4 c-5 d x^3\right )}{d x^6-8 c x^3}}{384 c^{3/2}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.12, size = 103, normalized size = 0.85 \begin {gather*} \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 {\sqrt {c+d x^3} \left (5 d x^3-4 c\right )}{96 c x^3 \left (8 c-d x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^3]*(-4*c + 5*d*x^3))/(96*c*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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fricas [A]  time = 0.45, size = 280, normalized size = 2.31 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.19, size = 114, normalized size = 0.94 \begin {gather*} \frac {7 \, d \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{384 \, \sqrt {-c} c} - \frac {3 \, d \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{128 \, \sqrt {-c} c} - \frac {5 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} d - 9 \, \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} \end {gather*}

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]

7/384*d*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c) - 3/128*d*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)
*c) - 1/96*(5*(d*x^3 + c)^(3/2)*d - 9*sqrt(d*x^3 + c)*c*d)/(((d*x^3 + c)^2 - 10*(d*x^3 + c)*c + 9*c^2)*c)

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maple [C]  time = 0.19, size = 1014, normalized size = 8.38

result too large to display

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]

1/64/c^2*d^2*(-3*(d*x^3+c)^(1/2)/(d*x^3-8*c)*c/d+2/3*(d*x^3+c)^(1/2)/d+1/2*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)))+1/64/c^2*(-1/3*(d*x^3+c)^(1/2)*c/x^3+2/3*(
d*x^3+c)^(1/2)*d-c^(1/2)*d*arctanh((d*x^3+c)^(1/2)/c^(1/2)))-1/256/c^3*d^2*(2/9*(d*x^3+c)^(1/2)*x^3+56/9*(d*x^
3+c)^(1/2)*c/d+3*I*c/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*(2/9*(d*x^3+c)^(1/2)*d*x^3+8/9*(d*x^3+c)^(1/2)*c-2/3*c^(3/2)*arctanh((d*x^3+c)^(1/2)/c
^(1/2)))

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

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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mupad [B]  time = 4.24, size = 110, normalized size = 0.91 \begin {gather*} \frac {\frac {9\,d\,\sqrt {d\,x^3+c}}{32}-\frac {5\,d\,{\left (d\,x^3+c\right )}^{3/2}}{32\,c}}{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\,\sqrt {d\,x^3+c}}{\sqrt {c^3}}\right )\,1{}\mathrm {i}-\frac {\mathrm {atanh}\left (\frac {c\,\sqrt {d\,x^3+c}}{3\,\sqrt {c^3}}\right )\,9{}\mathrm {i}}{7}\right )\,7{}\mathrm {i}}{384\,\sqrt {c^3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((9*d*(c + d*x^3)^(1/2))/32 - (5*d*(c + d*x^3)^(3/2))/(32*c))/(3*(c + d*x^3)^2 - 30*c*(c + d*x^3) + 27*c^2) +
(d*(atanh((c*(c + d*x^3)^(1/2))/(c^3)^(1/2))*1i - (atanh((c*(c + d*x^3)^(1/2))/(3*(c^3)^(1/2)))*9i)/7)*7i)/(38
4*(c^3)^(1/2))

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

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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