3.3.72 \(\int \frac {1}{x^4 (a+b x^3) (c+d x^3)^{3/2}} \, dx\)

Optimal. Leaf size=158 \[ -\frac {2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^2 (b c-a d)^{3/2}}+\frac {(3 a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2 c^{5/2}}-\frac {d (b c-3 a d)}{3 a c^2 \sqrt {c+d x^3} (b c-a d)}-\frac {1}{3 a c x^3 \sqrt {c+d x^3}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.22, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {446, 103, 152, 156, 63, 208} \begin {gather*} -\frac {2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^2 (b c-a d)^{3/2}}+\frac {(3 a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2 c^{5/2}}-\frac {d (b c-3 a d)}{3 a c^2 \sqrt {c+d x^3} (b c-a d)}-\frac {1}{3 a c x^3 \sqrt {c+d x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^3)*(c + d*x^3)^(3/2)),x]

[Out]

-(d*(b*c - 3*a*d))/(3*a*c^2*(b*c - a*d)*Sqrt[c + d*x^3]) - 1/(3*a*c*x^3*Sqrt[c + d*x^3]) + ((2*b*c + 3*a*d)*Ar
cTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(3*a^2*c^(5/2)) - (2*b^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]
])/(3*a^2*(b*c - a*d)^(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 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(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*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 152

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] && IntegersQ[2*m, 2*n, 2*p]

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 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 {1}{x^4 \left (a+b x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x^2 (a+b x) (c+d x)^{3/2}} \, dx,x,x^3\right )\\ &=-\frac {1}{3 a c x^3 \sqrt {c+d x^3}}-\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} (2 b c+3 a d)+\frac {3 b d x}{2}}{x (a+b x) (c+d x)^{3/2}} \, dx,x,x^3\right )}{3 a c}\\ &=-\frac {d (b c-3 a d)}{3 a c^2 (b c-a d) \sqrt {c+d x^3}}-\frac {1}{3 a c x^3 \sqrt {c+d x^3}}+\frac {2 \operatorname {Subst}\left (\int \frac {-\frac {1}{4} (b c-a d) (2 b c+3 a d)-\frac {1}{4} b d (b c-3 a d) x}{x (a+b x) \sqrt {c+d x}} \, dx,x,x^3\right )}{3 a c^2 (b c-a d)}\\ &=-\frac {d (b c-3 a d)}{3 a c^2 (b c-a d) \sqrt {c+d x^3}}-\frac {1}{3 a c x^3 \sqrt {c+d x^3}}+\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^3\right )}{3 a^2 (b c-a d)}-\frac {(2 b c+3 a d) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^3\right )}{6 a^2 c^2}\\ &=-\frac {d (b c-3 a d)}{3 a c^2 (b c-a d) \sqrt {c+d x^3}}-\frac {1}{3 a c x^3 \sqrt {c+d x^3}}+\frac {\left (2 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^3}\right )}{3 a^2 d (b c-a d)}-\frac {(2 b c+3 a d) \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^3}\right )}{3 a^2 c^2 d}\\ &=-\frac {d (b c-3 a d)}{3 a c^2 (b c-a d) \sqrt {c+d x^3}}-\frac {1}{3 a c x^3 \sqrt {c+d x^3}}+\frac {(2 b c+3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2 c^{5/2}}-\frac {2 b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^2 (b c-a d)^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.06, size = 117, normalized size = 0.74 \begin {gather*} \frac {2 b^2 c^2 x^3 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b \left (d x^3+c\right )}{b c-a d}\right )+(a d-b c) \left (x^3 (3 a d+2 b c) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {d x^3}{c}+1\right )+a c\right )}{3 a^2 c^2 x^3 \sqrt {c+d x^3} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*x^3)*(c + d*x^3)^(3/2)),x]

[Out]

(2*b^2*c^2*x^3*Hypergeometric2F1[-1/2, 1, 1/2, (b*(c + d*x^3))/(b*c - a*d)] + (-(b*c) + a*d)*(a*c + (2*b*c + 3
*a*d)*x^3*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (d*x^3)/c]))/(3*a^2*c^2*(b*c - a*d)*x^3*Sqrt[c + d*x^3])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.39, size = 165, normalized size = 1.04 \begin {gather*} \frac {2 b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3} \sqrt {a d-b c}}{b c-a d}\right )}{3 a^2 (a d-b c)^{3/2}}+\frac {(3 a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2 c^{5/2}}+\frac {-a c d-3 a d^2 x^3+b c^2+b c d x^3}{3 a c^2 x^3 \sqrt {c+d x^3} (a d-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^4*(a + b*x^3)*(c + d*x^3)^(3/2)),x]

[Out]

(b*c^2 - a*c*d + b*c*d*x^3 - 3*a*d^2*x^3)/(3*a*c^2*(-(b*c) + a*d)*x^3*Sqrt[c + d*x^3]) + (2*b^(5/2)*ArcTan[(Sq
rt[b]*Sqrt[-(b*c) + a*d]*Sqrt[c + d*x^3])/(b*c - a*d)])/(3*a^2*(-(b*c) + a*d)^(3/2)) + ((2*b*c + 3*a*d)*ArcTan
h[Sqrt[c + d*x^3]/Sqrt[c]])/(3*a^2*c^(5/2))

________________________________________________________________________________________

fricas [B]  time = 0.87, size = 1120, normalized size = 7.09 \begin {gather*} \left [-\frac {2 \, {\left (b^{2} c^{3} d x^{6} + b^{2} c^{4} x^{3}\right )} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x^{3} + 2 \, b c - a d + 2 \, \sqrt {d x^{3} + c} {\left (b c - a d\right )} \sqrt {\frac {b}{b c - a d}}}{b x^{3} + a}\right ) - {\left ({\left (2 \, b^{2} c^{2} d + a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{6} + {\left (2 \, b^{2} c^{3} + a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{3}\right )} \sqrt {c} \log \left (\frac {d x^{3} + 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) + 2 \, {\left (a b c^{3} - a^{2} c^{2} d + {\left (a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{6 \, {\left ({\left (a^{2} b c^{4} d - a^{3} c^{3} d^{2}\right )} x^{6} + {\left (a^{2} b c^{5} - a^{3} c^{4} d\right )} x^{3}\right )}}, -\frac {4 \, {\left (b^{2} c^{3} d x^{6} + b^{2} c^{4} x^{3}\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (-\frac {\sqrt {d x^{3} + c} {\left (b c - a d\right )} \sqrt {-\frac {b}{b c - a d}}}{b d x^{3} + b c}\right ) - {\left ({\left (2 \, b^{2} c^{2} d + a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{6} + {\left (2 \, b^{2} c^{3} + a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{3}\right )} \sqrt {c} \log \left (\frac {d x^{3} + 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) + 2 \, {\left (a b c^{3} - a^{2} c^{2} d + {\left (a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{6 \, {\left ({\left (a^{2} b c^{4} d - a^{3} c^{3} d^{2}\right )} x^{6} + {\left (a^{2} b c^{5} - a^{3} c^{4} d\right )} x^{3}\right )}}, -\frac {{\left ({\left (2 \, b^{2} c^{2} d + a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{6} + {\left (2 \, b^{2} c^{3} + a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) + {\left (b^{2} c^{3} d x^{6} + b^{2} c^{4} x^{3}\right )} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x^{3} + 2 \, b c - a d + 2 \, \sqrt {d x^{3} + c} {\left (b c - a d\right )} \sqrt {\frac {b}{b c - a d}}}{b x^{3} + a}\right ) + {\left (a b c^{3} - a^{2} c^{2} d + {\left (a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{3 \, {\left ({\left (a^{2} b c^{4} d - a^{3} c^{3} d^{2}\right )} x^{6} + {\left (a^{2} b c^{5} - a^{3} c^{4} d\right )} x^{3}\right )}}, -\frac {2 \, {\left (b^{2} c^{3} d x^{6} + b^{2} c^{4} x^{3}\right )} \sqrt {-\frac {b}{b c - a d}} \arctan \left (-\frac {\sqrt {d x^{3} + c} {\left (b c - a d\right )} \sqrt {-\frac {b}{b c - a d}}}{b d x^{3} + b c}\right ) + {\left ({\left (2 \, b^{2} c^{2} d + a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{6} + {\left (2 \, b^{2} c^{3} + a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) + {\left (a b c^{3} - a^{2} c^{2} d + {\left (a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{3 \, {\left ({\left (a^{2} b c^{4} d - a^{3} c^{3} d^{2}\right )} x^{6} + {\left (a^{2} b c^{5} - a^{3} c^{4} d\right )} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(2*(b^2*c^3*d*x^6 + b^2*c^4*x^3)*sqrt(b/(b*c - a*d))*log((b*d*x^3 + 2*b*c - a*d + 2*sqrt(d*x^3 + c)*(b*c
- a*d)*sqrt(b/(b*c - a*d)))/(b*x^3 + a)) - ((2*b^2*c^2*d + a*b*c*d^2 - 3*a^2*d^3)*x^6 + (2*b^2*c^3 + a*b*c^2*
d - 3*a^2*c*d^2)*x^3)*sqrt(c)*log((d*x^3 + 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) + 2*(a*b*c^3 - a^2*c^2*d + (a
*b*c^2*d - 3*a^2*c*d^2)*x^3)*sqrt(d*x^3 + c))/((a^2*b*c^4*d - a^3*c^3*d^2)*x^6 + (a^2*b*c^5 - a^3*c^4*d)*x^3),
-1/6*(4*(b^2*c^3*d*x^6 + b^2*c^4*x^3)*sqrt(-b/(b*c - a*d))*arctan(-sqrt(d*x^3 + c)*(b*c - a*d)*sqrt(-b/(b*c -
a*d))/(b*d*x^3 + b*c)) - ((2*b^2*c^2*d + a*b*c*d^2 - 3*a^2*d^3)*x^6 + (2*b^2*c^3 + a*b*c^2*d - 3*a^2*c*d^2)*x
^3)*sqrt(c)*log((d*x^3 + 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) + 2*(a*b*c^3 - a^2*c^2*d + (a*b*c^2*d - 3*a^2*c
*d^2)*x^3)*sqrt(d*x^3 + c))/((a^2*b*c^4*d - a^3*c^3*d^2)*x^6 + (a^2*b*c^5 - a^3*c^4*d)*x^3), -1/3*(((2*b^2*c^2
*d + a*b*c*d^2 - 3*a^2*d^3)*x^6 + (2*b^2*c^3 + a*b*c^2*d - 3*a^2*c*d^2)*x^3)*sqrt(-c)*arctan(sqrt(d*x^3 + c)*s
qrt(-c)/c) + (b^2*c^3*d*x^6 + b^2*c^4*x^3)*sqrt(b/(b*c - a*d))*log((b*d*x^3 + 2*b*c - a*d + 2*sqrt(d*x^3 + c)*
(b*c - a*d)*sqrt(b/(b*c - a*d)))/(b*x^3 + a)) + (a*b*c^3 - a^2*c^2*d + (a*b*c^2*d - 3*a^2*c*d^2)*x^3)*sqrt(d*x
^3 + c))/((a^2*b*c^4*d - a^3*c^3*d^2)*x^6 + (a^2*b*c^5 - a^3*c^4*d)*x^3), -1/3*(2*(b^2*c^3*d*x^6 + b^2*c^4*x^3
)*sqrt(-b/(b*c - a*d))*arctan(-sqrt(d*x^3 + c)*(b*c - a*d)*sqrt(-b/(b*c - a*d))/(b*d*x^3 + b*c)) + ((2*b^2*c^2
*d + a*b*c*d^2 - 3*a^2*d^3)*x^6 + (2*b^2*c^3 + a*b*c^2*d - 3*a^2*c*d^2)*x^3)*sqrt(-c)*arctan(sqrt(d*x^3 + c)*s
qrt(-c)/c) + (a*b*c^3 - a^2*c^2*d + (a*b*c^2*d - 3*a^2*c*d^2)*x^3)*sqrt(d*x^3 + c))/((a^2*b*c^4*d - a^3*c^3*d^
2)*x^6 + (a^2*b*c^5 - a^3*c^4*d)*x^3)]

________________________________________________________________________________________

giac [A]  time = 0.17, size = 173, normalized size = 1.09 \begin {gather*} \frac {2 \, b^{3} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, {\left (a^{2} b c - a^{3} d\right )} \sqrt {-b^{2} c + a b d}} - \frac {{\left (d x^{3} + c\right )} b c d - 3 \, {\left (d x^{3} + c\right )} a d^{2} + 2 \, a c d^{2}}{3 \, {\left (a b c^{3} - a^{2} c^{2} d\right )} {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} - \sqrt {d x^{3} + c} c\right )}} - \frac {{\left (2 \, b c + 3 \, a d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{3 \, a^{2} \sqrt {-c} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*b^3*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/((a^2*b*c - a^3*d)*sqrt(-b^2*c + a*b*d)) - 1/3*((d*x^3
+ c)*b*c*d - 3*(d*x^3 + c)*a*d^2 + 2*a*c*d^2)/((a*b*c^3 - a^2*c^2*d)*((d*x^3 + c)^(3/2) - sqrt(d*x^3 + c)*c))
- 1/3*(2*b*c + 3*a*d)*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(a^2*sqrt(-c)*c^2)

________________________________________________________________________________________

maple [C]  time = 0.41, size = 575, normalized size = 3.64 \begin {gather*} \frac {\left (-\frac {i b \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {\left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (2 \RootOf \left (\textit {\_Z}^{3} b +a \right )^{2} d^{2}+i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (\textit {\_Z}^{3} b +a \right ) d -\left (-c \,d^{2}\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{3} b +a \right ) d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}-\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, \frac {\left (2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (\textit {\_Z}^{3} b +a \right )^{2} d +i \sqrt {3}\, c d -3 c d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \RootOf \left (\textit {\_Z}^{3} b +a \right )-3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \RootOf \left (\textit {\_Z}^{3} b +a \right )\right ) b}{2 \left (a d -b c \right ) d}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{\left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) d}}\right )}{3 d^{2} \left (-a d +b c \right ) \left (a d -b c \right ) \sqrt {d \,x^{3}+c}}-\frac {2}{3 \left (a d -b c \right ) \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}\right ) b^{2}}{a^{2}}+\frac {\frac {d \arctanh \left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{c^{\frac {5}{2}}}-\frac {2 d}{3 \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}\, c^{2}}-\frac {\sqrt {d \,x^{3}+c}}{3 c^{2} x^{3}}}{a}-\frac {\left (-\frac {2 \arctanh \left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 c^{\frac {3}{2}}}+\frac {2}{3 \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}\, c}\right ) b}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(b*x^3+a)/(d*x^3+c)^(3/2),x)

[Out]

1/a*(-1/3*(d*x^3+c)^(1/2)/c^2/x^3-2/3/((x^3+c/d)*d)^(1/2)/c^2*d+d*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(5/2))+1/
a^2*b^2*(-2/3/(a*d-b*c)/((x^3+c/d)*d)^(1/2)-1/3*I*b/d^2*2^(1/2)*sum(1/(-a*d+b*c)/(a*d-b*c)*(-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),1/2*(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)/(a*d-b*c)*b/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*b+a)))-1/a^2*b*(2/3/((x^3+c/d)*d)^(1/2)/c-2/3
*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(3/2))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x^{3} + a\right )} {\left (d x^{3} + c\right )}^{\frac {3}{2}} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((b*x^3 + a)*(d*x^3 + c)^(3/2)*x^4), x)

________________________________________________________________________________________

mupad [B]  time = 10.47, size = 597, normalized size = 3.78 \begin {gather*} \frac {\ln \left (\frac {\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )\,{\left (\sqrt {d\,x^3+c}+\sqrt {c}\right )}^3}{x^6}\right )\,\left (3\,a\,d+2\,b\,c\right )}{6\,a^2\,c^{5/2}}-\frac {\sqrt {d\,x^3+c}}{3\,a\,c^2\,x^3}-\frac {\frac {c\,\left (\frac {c\,\left (\frac {c\,\left (\frac {3\,a^2\,d^4+24\,a\,b\,c\,d^3+15\,b^2\,c^2\,d^2}{8\,a^3\,c^5}+\frac {c\,\left (\frac {c\,\left (\frac {3\,b^2\,d^4}{8\,a^3\,c^5}+\frac {b^2\,d^4\,\left (5\,a\,d-3\,b\,c\right )}{8\,a^3\,c^4\,\left (b\,c^2-a\,c\,d\right )}-\frac {b\,d^4\,\left (a\,d+2\,b\,c\right )\,\left (5\,a\,d-3\,b\,c\right )}{4\,a^3\,c^5\,\left (b\,c^2-a\,c\,d\right )}\right )}{d}-\frac {3\,b\,d^3\,\left (a\,d+2\,b\,c\right )}{4\,a^3\,c^5}+\frac {d\,\left (5\,a\,d-3\,b\,c\right )\,\left (3\,a^2\,d^4+24\,a\,b\,c\,d^3+15\,b^2\,c^2\,d^2\right )}{24\,a^3\,c^5\,\left (b\,c^2-a\,c\,d\right )}\right )}{d}-\frac {d^2\,\left (5\,a\,d-3\,b\,c\right )\,\left (6\,a^2\,d^2+14\,a\,b\,c\,d+3\,b^2\,c^2\right )}{12\,a^3\,c^4\,\left (b\,c^2-a\,c\,d\right )}\right )}{d}-\frac {d\,\left (6\,a^2\,d^2+14\,a\,b\,c\,d+3\,b^2\,c^2\right )}{4\,a^3\,c^4}+\frac {d^2\,\left (5\,a\,d-3\,b\,c\right )\,\left (13\,a\,d+18\,b\,c\right )}{24\,a^2\,c^3\,\left (b\,c^2-a\,c\,d\right )}\right )}{d}+\frac {d\,\left (13\,a\,d+18\,b\,c\right )}{8\,a^2\,c^3}-\frac {d\,\left (3\,a\,d+2\,b\,c\right )\,\left (5\,a\,d-3\,b\,c\right )}{6\,a^2\,c^2\,\left (b\,c^2-a\,c\,d\right )}\right )}{d}-\frac {3\,a\,d+2\,b\,c}{2\,a^2\,c^2}}{\sqrt {d\,x^3+c}}+\frac {b^{5/2}\,\ln \left (\frac {a\,d-2\,b\,c-b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,1{}\mathrm {i}}{3\,a^2\,{\left (a\,d-b\,c\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b*x^3)*(c + d*x^3)^(3/2)),x)

[Out]

(log((((c + d*x^3)^(1/2) - c^(1/2))*((c + d*x^3)^(1/2) + c^(1/2))^3)/x^6)*(3*a*d + 2*b*c))/(6*a^2*c^(5/2)) - (
c + d*x^3)^(1/2)/(3*a*c^2*x^3) - ((c*((c*((c*((3*a^2*d^4 + 15*b^2*c^2*d^2 + 24*a*b*c*d^3)/(8*a^3*c^5) + (c*((c
*((3*b^2*d^4)/(8*a^3*c^5) + (b^2*d^4*(5*a*d - 3*b*c))/(8*a^3*c^4*(b*c^2 - a*c*d)) - (b*d^4*(a*d + 2*b*c)*(5*a*
d - 3*b*c))/(4*a^3*c^5*(b*c^2 - a*c*d))))/d - (3*b*d^3*(a*d + 2*b*c))/(4*a^3*c^5) + (d*(5*a*d - 3*b*c)*(3*a^2*
d^4 + 15*b^2*c^2*d^2 + 24*a*b*c*d^3))/(24*a^3*c^5*(b*c^2 - a*c*d))))/d - (d^2*(5*a*d - 3*b*c)*(6*a^2*d^2 + 3*b
^2*c^2 + 14*a*b*c*d))/(12*a^3*c^4*(b*c^2 - a*c*d))))/d - (d*(6*a^2*d^2 + 3*b^2*c^2 + 14*a*b*c*d))/(4*a^3*c^4)
+ (d^2*(5*a*d - 3*b*c)*(13*a*d + 18*b*c))/(24*a^2*c^3*(b*c^2 - a*c*d))))/d + (d*(13*a*d + 18*b*c))/(8*a^2*c^3)
- (d*(3*a*d + 2*b*c)*(5*a*d - 3*b*c))/(6*a^2*c^2*(b*c^2 - a*c*d))))/d - (3*a*d + 2*b*c)/(2*a^2*c^2))/(c + d*x
^3)^(1/2) + (b^(5/2)*log((a*d - 2*b*c + b^(1/2)*(c + d*x^3)^(1/2)*(a*d - b*c)^(1/2)*2i - b*d*x^3)/(a + b*x^3))
*1i)/(3*a^2*(a*d - b*c)^(3/2))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{4} \left (a + b x^{3}\right ) \left (c + d x^{3}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(b*x**3+a)/(d*x**3+c)**(3/2),x)

[Out]

Integral(1/(x**4*(a + b*x**3)*(c + d*x**3)**(3/2)), x)

________________________________________________________________________________________