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

Optimal. Leaf size=161 \[ -\frac {\sqrt {b} (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^3 \sqrt {b c-a d}}+\frac {(4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^3 \sqrt {c}}-\frac {2 b \sqrt {c+d x^3}}{3 a^2 \left (a+b x^3\right )}-\frac {\sqrt {c+d x^3}}{3 a x^3 \left (a+b x^3\right )} \]

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Rubi [A]  time = 0.22, antiderivative size = 161, 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, 99, 151, 156, 63, 208} \begin {gather*} -\frac {2 b \sqrt {c+d x^3}}{3 a^2 \left (a+b x^3\right )}-\frac {\sqrt {b} (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^3 \sqrt {b c-a d}}+\frac {(4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^3 \sqrt {c}}-\frac {\sqrt {c+d x^3}}{3 a x^3 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.28, size = 190, normalized size = 1.18 \begin {gather*} \frac {\sqrt {c} \left (a \left (a+2 b x^3\right ) \sqrt {c+d x^3} (b c-a d)+\sqrt {b} x^3 \left (a+b x^3\right ) (4 b c-3 a d) \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )\right )-x^3 \left (a+b x^3\right ) \left (a^2 d^2-5 a b c d+4 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^3 \sqrt {c} x^3 \left (a+b x^3\right ) (a d-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((4*b^2*c^2 - 5*a*b*c*d + a^2*d^2)*x^3*(a + b*x^3)*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]]) + Sqrt[c]*(a*(b*c - a*d
)*(a + 2*b*x^3)*Sqrt[c + d*x^3] + Sqrt[b]*(4*b*c - 3*a*d)*Sqrt[b*c - a*d]*x^3*(a + b*x^3)*ArcTanh[(Sqrt[b]*Sqr
t[c + d*x^3])/Sqrt[b*c - a*d]]))/(3*a^3*Sqrt[c]*(-(b*c) + a*d)*x^3*(a + b*x^3))

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IntegrateAlgebraic [A]  time = 0.54, size = 157, normalized size = 0.98 \begin {gather*} \frac {\left (3 a \sqrt {b} d-4 b^{3/2} c\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3} \sqrt {a d-b c}}{b c-a d}\right )}{3 a^3 \sqrt {a d-b c}}+\frac {(4 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^3 \sqrt {c}}+\frac {\left (-a-2 b x^3\right ) \sqrt {c+d x^3}}{3 a^2 x^3 \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.11, size = 870, normalized size = 5.40 \begin {gather*} \left [-\frac {{\left ({\left (4 \, b^{2} c^{2} - 3 \, a b c d\right )} x^{6} + {\left (4 \, a b c^{2} - 3 \, a^{2} c d\right )} 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 (4 \, b^{2} c - a b d\right )} x^{6} + {\left (4 \, a b c - 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 (2 \, a b c x^{3} + a^{2} c\right )} \sqrt {d x^{3} + c}}{6 \, {\left (a^{3} b c x^{6} + a^{4} c x^{3}\right )}}, -\frac {2 \, {\left ({\left (4 \, b^{2} c^{2} - 3 \, a b c d\right )} x^{6} + {\left (4 \, a b c^{2} - 3 \, a^{2} c d\right )} 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 (4 \, b^{2} c - a b d\right )} x^{6} + {\left (4 \, a b c - 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 (2 \, a b c x^{3} + a^{2} c\right )} \sqrt {d x^{3} + c}}{6 \, {\left (a^{3} b c x^{6} + a^{4} c 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 {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) + {\left ({\left (4 \, b^{2} c^{2} - 3 \, a b c d\right )} x^{6} + {\left (4 \, a b c^{2} - 3 \, a^{2} c d\right )} 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 ) + 2 \, {\left (2 \, a b c x^{3} + a^{2} c\right )} \sqrt {d x^{3} + c}}{6 \, {\left (a^{3} b c x^{6} + a^{4} c x^{3}\right )}}, -\frac {{\left ({\left (4 \, b^{2} c^{2} - 3 \, a b c d\right )} x^{6} + {\left (4 \, a b c^{2} - 3 \, a^{2} c d\right )} 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 (4 \, b^{2} c - a b d\right )} x^{6} + {\left (4 \, a b c - a^{2} d\right )} x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) + {\left (2 \, a b c x^{3} + a^{2} c\right )} \sqrt {d x^{3} + c}}{3 \, {\left (a^{3} b c x^{6} + a^{4} c x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(((4*b^2*c^2 - 3*a*b*c*d)*x^6 + (4*a*b*c^2 - 3*a^2*c*d)*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)) + ((4*b^2*c - a*b*d)*x^6 + (4*a*b*c - 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*x^3 + a^2*c)*sqrt(d*x^3 + c)
)/(a^3*b*c*x^6 + a^4*c*x^3), -1/6*(2*((4*b^2*c^2 - 3*a*b*c*d)*x^6 + (4*a*b*c^2 - 3*a^2*c*d)*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)) + ((4*b^2*c - a*b*d)*x^6 + (
4*a*b*c - 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*x^3 + a^2*c)*sqr
t(d*x^3 + c))/(a^3*b*c*x^6 + a^4*c*x^3), -1/6*(2*((4*b^2*c - a*b*d)*x^6 + (4*a*b*c - a^2*d)*x^3)*sqrt(-c)*arct
an(sqrt(d*x^3 + c)*sqrt(-c)/c) + ((4*b^2*c^2 - 3*a*b*c*d)*x^6 + (4*a*b*c^2 - 3*a^2*c*d)*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*(2*a*b*c*x
^3 + a^2*c)*sqrt(d*x^3 + c))/(a^3*b*c*x^6 + a^4*c*x^3), -1/3*(((4*b^2*c^2 - 3*a*b*c*d)*x^6 + (4*a*b*c^2 - 3*a^
2*c*d)*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)) + (
(4*b^2*c - a*b*d)*x^6 + (4*a*b*c - a^2*d)*x^3)*sqrt(-c)*arctan(sqrt(d*x^3 + c)*sqrt(-c)/c) + (2*a*b*c*x^3 + a^
2*c)*sqrt(d*x^3 + c))/(a^3*b*c*x^6 + a^4*c*x^3)]

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giac [A]  time = 0.22, size = 183, normalized size = 1.14 \begin {gather*} \frac {{\left (4 \, b^{2} c - 3 \, a b d\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 - a 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 d - 2 \, \sqrt {d x^{3} + c} b c d + \sqrt {d x^{3} + c} a 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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.26, size = 978, normalized size = 6.07

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^2*(-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))+2/a^3*b^2*(2/3*(d*x^3+c)^(1/2)
/b+1/3*I/b/d^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)*_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*(-1/3*(d*x^3+c)^(1/2)/(b*x^3+a)/b-1/6*I/d/b*2^(1/2)*sum(1/(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)))-2*b/a^3*(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {d x^{3} + c}}{{\left (b x^{3} + a\right )}^{2} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 9.69, size = 438, normalized size = 2.72 \begin {gather*} \frac {\left (\frac {a\,\left (\frac {a\,\left (\frac {a\,\left (\frac {b^2\,d^2}{2\,a^3\,c^2}-\frac {b^2\,d^2\,\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\,\left (2\,a\,d-b\,c\right )\,\left (3\,a\,d-4\,b\,c\right )}{6\,a^3\,c^2\,\left (a^2\,d-a\,b\,c\right )}\right )}{b}-\frac {b\,d\,\left (2\,a\,d-b\,c\right )}{2\,a^3\,c^2}+\frac {b\,\left (3\,a\,d-4\,b\,c\right )\,\left (-a^2\,d^2+2\,a\,b\,c\,d+2\,b^2\,c^2\right )}{6\,a^3\,c^2\,\left (a^2\,d-a\,b\,c\right )}\right )}{b}-\frac {-a^2\,d^2+2\,a\,b\,c\,d+2\,b^2\,c^2}{2\,a^3\,c^2}+\frac {b\,\left (a\,d-4\,b\,c\right )\,\left (3\,a\,d-4\,b\,c\right )}{6\,a^2\,c\,\left (a^2\,d-a\,b\,c\right )}\right )}{b}-\frac {a\,d-4\,b\,c}{2\,a^2\,c}\right )\,\sqrt {d\,x^3+c}}{b\,x^3+a}-\frac {\sqrt {d\,x^3+c}}{3\,a^2\,x^3}+\frac {\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 (a\,d-4\,b\,c\right )}{6\,a^3\,\sqrt {c}}+\frac {\sqrt {b}\,\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 )\,\left (3\,a\,d-4\,b\,c\right )\,1{}\mathrm {i}}{6\,a^3\,\sqrt {a\,d-b\,c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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