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

Optimal. Leaf size=185 \[ -\frac {b^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^3 (b c-a d)^{3/2}}+\frac {(a d+4 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^3 c^{3/2}}-\frac {b \sqrt {c+d x^3} (2 b c-a d)}{3 a^2 c \left (a+b x^3\right ) (b c-a d)}-\frac {\sqrt {c+d x^3}}{3 a c x^3 \left (a+b x^3\right )} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 185, 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, 151, 156, 63, 208} \[ -\frac {b^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^3 (b c-a d)^{3/2}}+\frac {(a d+4 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^3 c^{3/2}}-\frac {b \sqrt {c+d x^3} (2 b c-a d)}{3 a^2 c \left (a+b x^3\right ) (b c-a d)}-\frac {\sqrt {c+d x^3}}{3 a c x^3 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.61, size = 163, normalized size = 0.88 \[ \frac {\frac {a \sqrt {c+d x^3} \left (a^2 d+a b \left (d x^3-c\right )-2 b^2 c x^3\right )}{x^3 \left (a+b x^3\right ) (b c-a d)}+\frac {b^{3/2} c (5 a d-4 b c) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{3/2}}+\frac {(a d+4 b c) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{\sqrt {c}}}{3 a^3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.28, size = 1236, normalized size = 6.68 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 257, normalized size = 1.39 \[ \frac {{\left (4 \, b^{3} c - 5 \, a b^{2} d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, {\left (a^{3} b c - a^{4} d\right )} \sqrt {-b^{2} c + a b d}} - \frac {2 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} b^{2} c d - 2 \, \sqrt {d x^{3} + c} b^{2} c^{2} d - {\left (d x^{3} + c\right )}^{\frac {3}{2}} a b d^{2} + 2 \, \sqrt {d x^{3} + c} a b c d^{2} - \sqrt {d x^{3} + c} a^{2} d^{3}}{3 \, {\left (a^{2} b c^{2} - a^{3} c d\right )} {\left ({\left (d x^{3} + c\right )}^{2} b - 2 \, {\left (d x^{3} + c\right )} b c + b c^{2} + {\left (d x^{3} + c\right )} a d - a c d\right )}} - \frac {{\left (4 \, b c + a d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{3 \, a^{3} \sqrt {-c} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.29, size = 961, normalized size = 5.19 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^2*(-1/3*(d*x^3+c)^(1/2)/c/x^3+1/3*d*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(3/2))-2/3*I/a^3*b^2/d^2*2^(1/2)*su
m(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)*_alp
ha*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)*_alp
ha^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/(a*d-b*c)*(d*x^3+c)^(1/2)/(b*x^3+a)-1/6*I/d*2^(1/2)*sum(1/(a*d-b*c)^2*(-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)))+4/3*b/a^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 \[ \int \frac {1}{{\left (b x^{3} + a\right )}^{2} \sqrt {d x^{3} + c} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 11.55, size = 355, normalized size = 1.92 \[ \frac {\sqrt {d\,x^3+c}\,\left (\frac {d\,a^2+4\,b\,c\,a}{2\,a^3\,c^2}-\frac {a\,\left (\frac {2\,c\,b^2+2\,a\,d\,b}{2\,a^3\,c^2}-\frac {a\,\left (\frac {b^2\,d}{2\,a^3\,c^2}+\frac {b\,\left (2\,c\,b^2+2\,a\,d\,b\right )\,\left (3\,a\,d-4\,b\,c\right )}{6\,a^3\,c^2\,\left (a^2\,d-a\,b\,c\right )}-\frac {b^2\,d\,\left (3\,a\,d-4\,b\,c\right )}{6\,a^2\,c^2\,\left (a^2\,d-a\,b\,c\right )}\right )}{b}+\frac {b\,\left (d\,a^2+4\,b\,c\,a\right )\,\left (3\,a\,d-4\,b\,c\right )}{6\,a^3\,c^2\,\left (a^2\,d-a\,b\,c\right )}\right )}{b}\right )}{b\,x^3+a}-\frac {\sqrt {d\,x^3+c}}{3\,a^2\,c\,x^3}+\frac {\ln \left (\frac {\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )\,{\left (\sqrt {d\,x^3+c}+\sqrt {c}\right )}^3}{x^6}\right )\,\left (a\,d+4\,b\,c\right )}{6\,a^3\,c^{3/2}}+\frac {b^{3/2}\,\ln \left (\frac {2\,b\,c-a\,d+b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,\left (5\,a\,d-4\,b\,c\right )\,1{}\mathrm {i}}{6\,a^3\,{\left (a\,d-b\,c\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c + d*x^3)^(1/2)*((a^2*d + 4*a*b*c)/(2*a^3*c^2) - (a*((2*b^2*c + 2*a*b*d)/(2*a^3*c^2) - (a*((b^2*d)/(2*a^3*c
^2) + (b*(2*b^2*c + 2*a*b*d)*(3*a*d - 4*b*c))/(6*a^3*c^2*(a^2*d - a*b*c)) - (b^2*d*(3*a*d - 4*b*c))/(6*a^2*c^2
*(a^2*d - a*b*c))))/b + (b*(a^2*d + 4*a*b*c)*(3*a*d - 4*b*c))/(6*a^3*c^2*(a^2*d - a*b*c))))/b))/(a + b*x^3) -
(c + d*x^3)^(1/2)/(3*a^2*c*x^3) + (log((((c + d*x^3)^(1/2) - c^(1/2))*((c + d*x^3)^(1/2) + c^(1/2))^3)/x^6)*(a
*d + 4*b*c))/(6*a^3*c^(3/2)) + (b^(3/2)*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))*(5*a*d - 4*b*c)*1i)/(6*a^3*(a*d - b*c)^(3/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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