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

Optimal. Leaf size=170 \[ \frac {\sqrt {c} (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^3}-\frac {\sqrt {b c-a d} (4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^3 \sqrt {b}}-\frac {\sqrt {c+d x^3} (2 b c-a d)}{3 a^2 \left (a+b x^3\right )}-\frac {c \sqrt {c+d x^3}}{3 a x^3 \left (a+b x^3\right )} \]

[Out]

1/3*(-3*a*d+4*b*c)*arctanh((d*x^3+c)^(1/2)/c^(1/2))*c^(1/2)/a^3-1/3*(-a*d+4*b*c)*arctanh(b^(1/2)*(d*x^3+c)^(1/
2)/(-a*d+b*c)^(1/2))*(-a*d+b*c)^(1/2)/a^3/b^(1/2)-1/3*(-a*d+2*b*c)*(d*x^3+c)^(1/2)/a^2/(b*x^3+a)-1/3*c*(d*x^3+
c)^(1/2)/a/x^3/(b*x^3+a)

________________________________________________________________________________________

Rubi [A]  time = 0.26, antiderivative size = 170, 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, 98, 151, 156, 63, 208} \[ -\frac {\sqrt {c+d x^3} (2 b c-a d)}{3 a^2 \left (a+b x^3\right )}+\frac {\sqrt {c} (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^3}-\frac {\sqrt {b c-a d} (4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^3 \sqrt {b}}-\frac {c \sqrt {c+d x^3}}{3 a x^3 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*b*c - a*d)*Sqrt[c + d*x^3])/(3*a^2*(a + b*x^3)) - (c*Sqrt[c + d*x^3])/(3*a*x^3*(a + b*x^3)) + (Sqrt[c]*(4
*b*c - 3*a*d)*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(3*a^3) - (Sqrt[b*c - a*d]*(4*b*c - a*d)*ArcTanh[(Sqrt[b]*Sqrt
[c + d*x^3])/Sqrt[b*c - a*d]])/(3*a^3*Sqrt[b])

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

________________________________________________________________________________________

Mathematica [A]  time = 0.33, size = 142, normalized size = 0.84 \[ \frac {\frac {a \sqrt {c+d x^3} \left (-a c+a d x^3-2 b c x^3\right )}{x^3 \left (a+b x^3\right )}+\sqrt {c} (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )-\frac {\sqrt {b c-a d} (4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{\sqrt {b}}}{3 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*Sqrt[c + d*x^3]*(-(a*c) - 2*b*c*x^3 + a*d*x^3))/(x^3*(a + b*x^3)) + Sqrt[c]*(4*b*c - 3*a*d)*ArcTanh[Sqrt[c
+ d*x^3]/Sqrt[c]] - (Sqrt[b*c - a*d]*(4*b*c - a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/Sqrt[b
])/(3*a^3)

________________________________________________________________________________________

fricas [A]  time = 1.17, size = 838, normalized size = 4.93 \[ \left [-\frac {{\left ({\left (4 \, b^{2} c - a b d\right )} x^{6} + {\left (4 \, a b c - a^{2} d\right )} x^{3}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x^{3} + 2 \, b c - a d + 2 \, \sqrt {d x^{3} + c} b \sqrt {\frac {b c - a d}{b}}}{b x^{3} + a}\right ) + {\left ({\left (4 \, b^{2} c - 3 \, a b d\right )} x^{6} + {\left (4 \, a b c - 3 \, a^{2} d\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 ({\left (2 \, a b c - a^{2} d\right )} x^{3} + a^{2} c\right )} \sqrt {d x^{3} + c}}{6 \, {\left (a^{3} b x^{6} + a^{4} x^{3}\right )}}, -\frac {2 \, {\left ({\left (4 \, b^{2} c - a b d\right )} x^{6} + {\left (4 \, a b c - a^{2} d\right )} x^{3}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x^{3} + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left ({\left (4 \, b^{2} c - 3 \, a b d\right )} x^{6} + {\left (4 \, a b c - 3 \, a^{2} d\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 ({\left (2 \, a b c - a^{2} d\right )} x^{3} + a^{2} c\right )} \sqrt {d x^{3} + c}}{6 \, {\left (a^{3} b x^{6} + a^{4} x^{3}\right )}}, -\frac {2 \, {\left ({\left (4 \, b^{2} c - 3 \, a b d\right )} x^{6} + {\left (4 \, a b c - 3 \, a^{2} d\right )} x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) + {\left ({\left (4 \, b^{2} c - a b d\right )} x^{6} + {\left (4 \, a b c - a^{2} d\right )} x^{3}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x^{3} + 2 \, b c - a d + 2 \, \sqrt {d x^{3} + c} b \sqrt {\frac {b c - a d}{b}}}{b x^{3} + a}\right ) + 2 \, {\left ({\left (2 \, a b c - a^{2} d\right )} x^{3} + a^{2} c\right )} \sqrt {d x^{3} + c}}{6 \, {\left (a^{3} b x^{6} + a^{4} x^{3}\right )}}, -\frac {{\left ({\left (4 \, b^{2} c - a b d\right )} x^{6} + {\left (4 \, a b c - a^{2} d\right )} x^{3}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x^{3} + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left ({\left (4 \, b^{2} c - 3 \, a b d\right )} x^{6} + {\left (4 \, a b c - 3 \, a^{2} d\right )} x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) + {\left ({\left (2 \, a b c - a^{2} d\right )} x^{3} + a^{2} c\right )} \sqrt {d x^{3} + c}}{3 \, {\left (a^{3} b x^{6} + a^{4} x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.22, size = 216, normalized size = 1.27 \[ \frac {{\left (4 \, b^{2} c^{2} - 5 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, \sqrt {-b^{2} c + a b d} a^{3}} - \frac {{\left (4 \, b c^{2} - 3 \, a c d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{3 \, a^{3} \sqrt {-c}} - \frac {2 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} b c d - 2 \, \sqrt {d x^{3} + c} b c^{2} d - {\left (d x^{3} + c\right )}^{\frac {3}{2}} a d^{2} + 2 \, \sqrt {d x^{3} + c} a c d^{2}}{3 \, {\left ({\left (d x^{3} + c\right )}^{2} b - 2 \, {\left (d x^{3} + c\right )} b c + b c^{2} + {\left (d x^{3} + c\right )} a d - a c d\right )} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(4*b^2*c^2 - 5*a*b*c*d + a^2*d^2)*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*a^3
) - 1/3*(4*b*c^2 - 3*a*c*d)*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(a^3*sqrt(-c)) - 1/3*(2*(d*x^3 + c)^(3/2)*b*c*d -
2*sqrt(d*x^3 + c)*b*c^2*d - (d*x^3 + c)^(3/2)*a*d^2 + 2*sqrt(d*x^3 + c)*a*c*d^2)/(((d*x^3 + c)^2*b - 2*(d*x^3
+ c)*b*c + b*c^2 + (d*x^3 + c)*a*d - a*c*d)*a^2)

________________________________________________________________________________________

maple [C]  time = 0.25, size = 1093, normalized size = 6.43 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^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)))+2/a^3*b^2*
(2/9*(d*x^3+c)^(1/2)/b*d*x^3+2/3*(-2/3/b*c*d-(a*d-2*b*c)/b^2*d)*(d*x^3+c)^(1/2)/d+1/3*I/b^2/d^2*2^(1/2)*sum((-
a^2*d^2+2*a*b*c*d-b^2*c^2)/(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=R
ootOf(_Z^3*b+a)))+1/a^2*b^2*(1/3*(a*d-b*c)*(d*x^3+c)^(1/2)/(b*x^3+a)/b^2+2/3*(d*x^3+c)^(1/2)/b^2*d+1/2*I/d/b^2
*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)*_alph
a*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)*_alph
a^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)))-2*b/a^
3*(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)))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 10.82, size = 531, normalized size = 3.12 \[ \frac {\sqrt {c}\,\ln \left (\frac {{\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}^3\,\left (\sqrt {d\,x^3+c}+\sqrt {c}\right )}{x^6}\right )\,\left (3\,a\,d-4\,b\,c\right )}{6\,a^3}-\frac {c\,\sqrt {d\,x^3+c}}{3\,a^2\,x^3}-\frac {\sqrt {d\,x^3+c}\,\left (\frac {3\,a\,d-4\,b\,c}{2\,a^2}-\frac {a\,\left (\frac {a\,\left (\frac {a\,\left (\frac {b\,d^2\,\left (a\,d+b\,c\right )}{a^3\,c^2}-\frac {a\,\left (\frac {b^2\,d^3}{2\,a^3\,c^2}-\frac {b^2\,d^3\,\left (3\,a\,d-4\,b\,c\right )}{6\,a^2\,c^2\,\left (a^2\,d-a\,b\,c\right )}+\frac {b^2\,d^2\,\left (a\,d+b\,c\right )\,\left (3\,a\,d-4\,b\,c\right )}{3\,a^3\,c^2\,\left (a^2\,d-a\,b\,c\right )}\right )}{b}+\frac {b\,\left (3\,a\,d-4\,b\,c\right )\,\left (a^2\,d^3+4\,a\,b\,c\,d^2-b^2\,c^2\,d\right )}{6\,a^3\,c^2\,\left (a^2\,d-a\,b\,c\right )}\right )}{b}-\frac {a^2\,d^3+4\,a\,b\,c\,d^2-b^2\,c^2\,d}{2\,a^3\,c^2}+\frac {b\,\left (3\,a\,d-4\,b\,c\right )\,\left (-4\,a^2\,c\,d^2+2\,a\,b\,c^2\,d+2\,b^2\,c^3\right )}{6\,a^3\,c^2\,\left (a^2\,d-a\,b\,c\right )}\right )}{b}-\frac {-4\,a^2\,c\,d^2+2\,a\,b\,c^2\,d+2\,b^2\,c^3}{2\,a^3\,c^2}+\frac {b\,{\left (3\,a\,d-4\,b\,c\right )}^2}{6\,a^2\,\left (a^2\,d-a\,b\,c\right )}\right )}{b}\right )}{b\,x^3+a}+\frac {\ln \left (\frac {2\,b\,c-a\,d+b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,\sqrt {a\,d-b\,c}\,\left (a\,d-4\,b\,c\right )\,1{}\mathrm {i}}{6\,a^3\,\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^(1/2)*log((((c + d*x^3)^(1/2) - c^(1/2))^3*((c + d*x^3)^(1/2) + c^(1/2)))/x^6)*(3*a*d - 4*b*c))/(6*a^3) - (
c*(c + d*x^3)^(1/2))/(3*a^2*x^3) - ((c + d*x^3)^(1/2)*((3*a*d - 4*b*c)/(2*a^2) - (a*((a*((a*((b*d^2*(a*d + b*c
))/(a^3*c^2) - (a*((b^2*d^3)/(2*a^3*c^2) - (b^2*d^3*(3*a*d - 4*b*c))/(6*a^2*c^2*(a^2*d - a*b*c)) + (b^2*d^2*(a
*d + b*c)*(3*a*d - 4*b*c))/(3*a^3*c^2*(a^2*d - a*b*c))))/b + (b*(3*a*d - 4*b*c)*(a^2*d^3 - b^2*c^2*d + 4*a*b*c
*d^2))/(6*a^3*c^2*(a^2*d - a*b*c))))/b - (a^2*d^3 - b^2*c^2*d + 4*a*b*c*d^2)/(2*a^3*c^2) + (b*(3*a*d - 4*b*c)*
(2*b^2*c^3 - 4*a^2*c*d^2 + 2*a*b*c^2*d))/(6*a^3*c^2*(a^2*d - a*b*c))))/b - (2*b^2*c^3 - 4*a^2*c*d^2 + 2*a*b*c^
2*d)/(2*a^3*c^2) + (b*(3*a*d - 4*b*c)^2)/(6*a^2*(a^2*d - a*b*c))))/b))/(a + b*x^3) + (log((2*b*c - a*d + b^(1/
2)*(c + d*x^3)^(1/2)*(a*d - b*c)^(1/2)*2i + b*d*x^3)/(a + b*x^3))*(a*d - b*c)^(1/2)*(a*d - 4*b*c)*1i)/(6*a^3*b
^(1/2))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________