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

Optimal. Leaf size=116 \[ -\frac {2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^2 \sqrt {b}}+\frac {\sqrt {c} (2 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2}-\frac {c \sqrt {c+d x^3}}{3 a x^3} \]

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Rubi [A]  time = 0.15, antiderivative size = 116, 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, 98, 156, 63, 208} \begin {gather*} -\frac {2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^2 \sqrt {b}}+\frac {\sqrt {c} (2 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2}-\frac {c \sqrt {c+d x^3}}{3 a x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.13, size = 108, normalized size = 0.93 \begin {gather*} \frac {-\frac {2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{\sqrt {b}}+\sqrt {c} (2 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )-\frac {a c \sqrt {c+d x^3}}{x^3}}{3 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.25, size = 130, normalized size = 1.12 \begin {gather*} \frac {\left (2 b c^{3/2}-3 a \sqrt {c} d\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2}-\frac {2 (a d-b c)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3} \sqrt {a d-b c}}{b c-a d}\right )}{3 a^2 \sqrt {b}}-\frac {c \sqrt {c+d x^3}}{3 a x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.88, size = 538, normalized size = 4.64 \begin {gather*} \left [-\frac {2 \, {\left (b c - a d\right )} x^{3} \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 (2 \, b c - 3 \, a d\right )} \sqrt {c} x^{3} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) + 2 \, \sqrt {d x^{3} + c} a c}{6 \, a^{2} x^{3}}, -\frac {4 \, {\left (b c - a d\right )} x^{3} \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 (2 \, b c - 3 \, a d\right )} \sqrt {c} x^{3} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) + 2 \, \sqrt {d x^{3} + c} a c}{6 \, a^{2} x^{3}}, -\frac {{\left (2 \, b c - 3 \, a d\right )} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) + {\left (b c - a d\right )} x^{3} \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 ) + \sqrt {d x^{3} + c} a c}{3 \, a^{2} x^{3}}, -\frac {2 \, {\left (b c - a d\right )} x^{3} \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 (2 \, b c - 3 \, a d\right )} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) + \sqrt {d x^{3} + c} a c}{3 \, a^{2} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 121, normalized size = 1.04 \begin {gather*} \frac {2 \, {\left (b^{2} c^{2} - 2 \, 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^{2}} - \frac {{\left (2 \, b c^{2} - 3 \, a c d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{3 \, a^{2} \sqrt {-c}} - \frac {\sqrt {d x^{3} + c} c}{3 \, a x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.28, size = 620, normalized size = 5.34 \begin {gather*} \frac {\left (\frac {2 \sqrt {d \,x^{3}+c}\, d \,x^{3}}{9 b}+\frac {2 \left (-\frac {2 c d}{3 b}-\frac {\left (a d -2 b c \right ) d}{b^{2}}\right ) \sqrt {d \,x^{3}+c}}{3 d}+\frac {i \left (-a^{2} d^{2}+2 a b c d -b^{2} c^{2}\right ) \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 b^{2} d^{2} \left (a d -b c \right ) \sqrt {d \,x^{3}+c}}\right ) b^{2}}{a^{2}}+\frac {-\sqrt {c}\, d \arctanh \left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )+\frac {2 \sqrt {d \,x^{3}+c}\, d}{3}-\frac {\sqrt {d \,x^{3}+c}\, c}{3 x^{3}}}{a}-\frac {\left (\frac {2 \sqrt {d \,x^{3}+c}\, d \,x^{3}}{9}-\frac {2 c^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3}+\frac {8 \sqrt {d \,x^{3}+c}\, c}{9}\right ) b}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a*(-1/3*c*(d*x^3+c)^(1/2)/x^3+2/3*d*(d*x^3+c)^(1/2)-c^(1/2)*d*arctanh((d*x^3+c)^(1/2)/c^(1/2)))+1/a^2*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=Roo
tOf(_Z^3*b+a)))-1/a^2*b*(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)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 9.52, size = 167, normalized size = 1.44 \begin {gather*} \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-2\,b\,c\right )}{6\,a^2}-\frac {c\,\sqrt {d\,x^3+c}}{3\,a\,x^3}+\frac {\ln \left (\frac {a^2\,d^2+2\,b^2\,c^2-a\,b\,d^2\,x^3+b^2\,c\,d\,x^3-3\,a\,b\,c\,d-\sqrt {b}\,\sqrt {d\,x^3+c}\,{\left (a\,d-b\,c\right )}^{3/2}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,{\left (a\,d-b\,c\right )}^{3/2}\,1{}\mathrm {i}}{3\,a^2\,\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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