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

Optimal. Leaf size=234 \[ -\frac {\log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 b^{2/3} d}-\frac {\tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3} d}-\frac {c^{2/3} \log \left (c+d x^3\right )}{6 d (b c-a d)^{2/3}}+\frac {c^{2/3} \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 d (b c-a d)^{2/3}}+\frac {c^{2/3} \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} d (b c-a d)^{2/3}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.31, antiderivative size = 346, normalized size of antiderivative = 1.48, number of steps used = 14, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {494, 481, 292, 31, 634, 617, 204, 628} \begin {gather*} -\frac {\log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{2/3} d}+\frac {\log \left (\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )}{6 b^{2/3} d}-\frac {\tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3} d}+\frac {c^{2/3} \log \left (\sqrt [3]{c}-\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}\right )}{3 d (b c-a d)^{2/3}}-\frac {c^{2/3} \log \left (\frac {x^2 (b c-a d)^{2/3}}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}+c^{2/3}\right )}{6 d (b c-a d)^{2/3}}+\frac {c^{2/3} \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}+\sqrt [3]{c}}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} d (b c-a d)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*b^(2/3)*d)) + (c^(2/3)*ArcTan[(c^(1/3) + (2*(
b*c - a*d)^(1/3)*x)/(a + b*x^3)^(1/3))/(Sqrt[3]*c^(1/3))])/(Sqrt[3]*d*(b*c - a*d)^(2/3)) - Log[1 - (b^(1/3)*x)
/(a + b*x^3)^(1/3)]/(3*b^(2/3)*d) + Log[1 + (b^(2/3)*x^2)/(a + b*x^3)^(2/3) + (b^(1/3)*x)/(a + b*x^3)^(1/3)]/(
6*b^(2/3)*d) + (c^(2/3)*Log[c^(1/3) - ((b*c - a*d)^(1/3)*x)/(a + b*x^3)^(1/3)])/(3*d*(b*c - a*d)^(2/3)) - (c^(
2/3)*Log[c^(2/3) + ((b*c - a*d)^(2/3)*x^2)/(a + b*x^3)^(2/3) + (c^(1/3)*(b*c - a*d)^(1/3)*x)/(a + b*x^3)^(1/3)
])/(6*d*(b*c - a*d)^(2/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 481

Int[((e_.)*(x_))^(m_.)/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> -Dist[(a*e^n)/(b*c -
a*d), Int[(e*x)^(m - n)/(a + b*x^n), x], x] + Dist[(c*e^n)/(b*c - a*d), Int[(e*x)^(m - n)/(c + d*x^n), x], x]
/; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LeQ[n, m, 2*n - 1]

Rule 494

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = Denominato
r[p]}, Dist[(k*a^(p + (m + 1)/n))/n, Subst[Int[(x^((k*(m + 1))/n - 1)*(c - (b*c - a*d)*x^k)^q)/(1 - b*x^k)^(p
+ q + (m + 1)/n + 1), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && Ration
alQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 0]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 0.04, size = 65, normalized size = 0.28 \begin {gather*} \frac {x^5 \left (\frac {a+b x^3}{a}\right )^{2/3} F_1\left (\frac {5}{3};\frac {2}{3},1;\frac {8}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{5 c \left (a+b x^3\right )^{2/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x^5*((a + b*x^3)/a)^(2/3)*AppellF1[5/3, 2/3, 1, 8/3, -((b*x^3)/a), -((d*x^3)/c)])/(5*c*(a + b*x^3)^(2/3))

________________________________________________________________________________________

IntegrateAlgebraic [C]  time = 3.42, size = 475, normalized size = 2.03 \begin {gather*} -\frac {\log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{3 b^{2/3} d}-\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{2 \sqrt [3]{a+b x^3}+\sqrt [3]{b} x}\right )}{\sqrt {3} b^{2/3} d}+\frac {\log \left (\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}+b^{2/3} x^2\right )}{6 b^{2/3} d}+\frac {i \left (\sqrt {3} c^{2/3}+i c^{2/3}\right ) \log \left (2 x \sqrt [3]{b c-a d}+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{6 d (b c-a d)^{2/3}}-\frac {\sqrt {\frac {1}{6} \left (-1-i \sqrt {3}\right )} c^{2/3} \tan ^{-1}\left (\frac {3 x \sqrt [3]{b c-a d}}{\sqrt {3} x \sqrt [3]{b c-a d}-\sqrt {3} \sqrt [3]{c} \sqrt [3]{a+b x^3}-3 i \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{d (b c-a d)^{2/3}}+\frac {\left (c^{2/3}-i \sqrt {3} c^{2/3}\right ) \log \left (\left (\sqrt {3}+i\right ) c^{2/3} \left (a+b x^3\right )^{2/3}+\sqrt [3]{c} \left (-\sqrt {3} x+i x\right ) \sqrt [3]{a+b x^3} \sqrt [3]{b c-a d}-2 i x^2 (b c-a d)^{2/3}\right )}{12 d (b c-a d)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))]/(Sqrt[3]*b^(2/3)*d)) - (Sqrt[(-1 - I*Sqrt[3])/
6]*c^(2/3)*ArcTan[(3*(b*c - a*d)^(1/3)*x)/(Sqrt[3]*(b*c - a*d)^(1/3)*x - (3*I)*c^(1/3)*(a + b*x^3)^(1/3) - Sqr
t[3]*c^(1/3)*(a + b*x^3)^(1/3))])/(d*(b*c - a*d)^(2/3)) - Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)]/(3*b^(2/3)*d)
+ ((I/6)*(I*c^(2/3) + Sqrt[3]*c^(2/3))*Log[2*(b*c - a*d)^(1/3)*x + (1 + I*Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3)])
/(d*(b*c - a*d)^(2/3)) + Log[b^(2/3)*x^2 + b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)]/(6*b^(2/3)*d) + ((
c^(2/3) - I*Sqrt[3]*c^(2/3))*Log[(-2*I)*(b*c - a*d)^(2/3)*x^2 + c^(1/3)*(b*c - a*d)^(1/3)*(I*x - Sqrt[3]*x)*(a
+ b*x^3)^(1/3) + (I + Sqrt[3])*c^(2/3)*(a + b*x^3)^(2/3)])/(12*d*(b*c - a*d)^(2/3))

________________________________________________________________________________________

fricas [B]  time = 0.55, size = 530, normalized size = 2.26 \begin {gather*} \frac {2 \, \sqrt {3} b^{2} \left (\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} \arctan \left (-\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )} \left (\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {2}{3}} + \sqrt {3} c x}{3 \, c x}\right ) + 2 \, b^{2} \left (\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} \log \left (-\frac {{\left (b c - a d\right )} \left (\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} c}{x}\right ) - b^{2} \left (\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} \log \left (\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \left (\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {2}{3}} c^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c^{2} - a c d\right )} \left (\frac {c^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}\right )^{\frac {1}{3}} x}{x^{2}}\right ) + 2 \, \sqrt {3} b \sqrt {-\left (-b^{2}\right )^{\frac {1}{3}}} \arctan \left (-\frac {{\left (\sqrt {3} \left (-b^{2}\right )^{\frac {1}{3}} b x - 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}}\right )} \sqrt {-\left (-b^{2}\right )^{\frac {1}{3}}}}{3 \, b^{2} x}\right ) - 2 \, \left (-b^{2}\right )^{\frac {2}{3}} \log \left (-\frac {\left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) + \left (-b^{2}\right )^{\frac {2}{3}} \log \left (-\frac {\left (-b^{2}\right )^{\frac {1}{3}} b x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right )}{6 \, b^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.64, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (d \,x^{3}+c \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^4}{{\left (b\,x^3+a\right )}^{2/3}\,\left (d\,x^3+c\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________