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

Optimal. Leaf size=132 \[ \frac {\sqrt {b} (2 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^2 (b c-a d)^{3/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2 \sqrt {c}}+\frac {b \sqrt {c+d x^3}}{3 a \left (a+b x^3\right ) (b c-a d)} \]

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Rubi [A]  time = 0.14, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {446, 103, 156, 63, 208} \begin {gather*} \frac {\sqrt {b} (2 b c-3 a d) \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 {2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2 \sqrt {c}}+\frac {b \sqrt {c+d x^3}}{3 a \left (a+b x^3\right ) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.26, size = 123, normalized size = 0.93 \begin {gather*} \frac {\frac {a b \sqrt {c+d x^3}}{\left (a+b x^3\right ) (b c-a d)}+\frac {\sqrt {b} (2 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{3/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{\sqrt {c}}}{3 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.31, size = 146, normalized size = 1.11 \begin {gather*} \frac {\left (3 a \sqrt {b} d-2 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^2 (a d-b c)^{3/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2 \sqrt {c}}-\frac {b \sqrt {c+d x^3}}{3 a \left (a+b x^3\right ) (a d-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.93, size = 862, normalized size = 6.53 \begin {gather*} \left [\frac {2 \, \sqrt {d x^{3} + c} a b c + {\left (2 \, a b c^{2} - 3 \, a^{2} c d + {\left (2 \, b^{2} c^{2} - 3 \, a b 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 ({\left (b^{2} c - a b d\right )} x^{3} + a b c - a^{2} d\right )} \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right )}{6 \, {\left (a^{3} b c^{2} - a^{4} c d + {\left (a^{2} b^{2} c^{2} - a^{3} b c d\right )} x^{3}\right )}}, \frac {\sqrt {d x^{3} + c} a b c + {\left (2 \, a b c^{2} - 3 \, a^{2} c d + {\left (2 \, b^{2} c^{2} - 3 \, a b 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 (b^{2} c - a b d\right )} x^{3} + a b c - a^{2} d\right )} \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right )}{3 \, {\left (a^{3} b c^{2} - a^{4} c d + {\left (a^{2} b^{2} c^{2} - a^{3} b c d\right )} x^{3}\right )}}, \frac {2 \, \sqrt {d x^{3} + c} a b c + 4 \, {\left ({\left (b^{2} c - a b d\right )} x^{3} + a b c - a^{2} d\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) + {\left (2 \, a b c^{2} - 3 \, a^{2} c d + {\left (2 \, b^{2} c^{2} - 3 \, a b 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 )}{6 \, {\left (a^{3} b c^{2} - a^{4} c d + {\left (a^{2} b^{2} c^{2} - a^{3} b c d\right )} x^{3}\right )}}, \frac {\sqrt {d x^{3} + c} a b c + {\left (2 \, a b c^{2} - 3 \, a^{2} c d + {\left (2 \, b^{2} c^{2} - 3 \, a b 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 ) + 2 \, {\left ({\left (b^{2} c - a b d\right )} x^{3} + a b c - a^{2} d\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right )}{3 \, {\left (a^{3} b c^{2} - a^{4} c d + {\left (a^{2} b^{2} c^{2} - a^{3} b c d\right )} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 139, normalized size = 1.05 \begin {gather*} \frac {\sqrt {d x^{3} + c} b d}{3 \, {\left (a b c - a^{2} d\right )} {\left ({\left (d x^{3} + c\right )} b - b c + a d\right )}} - \frac {{\left (2 \, 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 \, {\left (a^{2} b c - a^{3} d\right )} \sqrt {-b^{2} c + a b d}} + \frac {2 \, \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{3 \, a^{2} \sqrt {-c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.28, size = 915, normalized size = 6.93

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*I/a^2*b/d^2*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)),_alph
a=RootOf(_Z^3*b+a))-1/a*b*(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)*_al
pha*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)))-2/3*arctanh((d*x^3+c)^(1/2)/
c^(1/2))/a^2/c^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(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), x)

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mupad [B]  time = 9.65, size = 162, normalized size = 1.23 \begin {gather*} \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 )}{3\,a^2\,\sqrt {c}}+\frac {b^2\,\sqrt {d\,x^3+c}}{3\,a\,\left (b\,x^3+a\right )\,\left (b^2\,c-a\,b\,d\right )}+\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-2\,b\,c\right )\,1{}\mathrm {i}}{6\,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*(a + b*x^3)^2*(c + d*x^3)^(1/2)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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