3.1.42 \(\int \frac {1}{(c+d x)^3 (a+b x^3)^{2/3}} \, dx\) [42]

Optimal. Leaf size=1357 \[ \frac {3 c^4 d^2 \sqrt [3]{a+b x^3}}{2 \left (b c^3-a d^3\right ) \left (c^3+d^3 x^3\right )^2}+\frac {3 c^2 d^4 x^2 \sqrt [3]{a+b x^3}}{2 \left (b c^3-a d^3\right ) \left (c^3+d^3 x^3\right )^2}+\frac {5 b c^4 d^2 \sqrt [3]{a+b x^3}}{3 \left (b c^3-a d^3\right )^2 \left (c^3+d^3 x^3\right )}-\frac {c d^2 \left (b c^3-6 a d^3\right ) \sqrt [3]{a+b x^3}}{6 \left (b c^3-a d^3\right )^2 \left (c^3+d^3 x^3\right )}+\frac {d^4 \left (9 b c^3-4 a d^3\right ) x^2 \sqrt [3]{a+b x^3}}{6 c \left (b c^3-a d^3\right )^2 \left (c^3+d^3 x^3\right )}+\frac {d^4 \left (3 b c^3+2 a d^3\right ) x^2 \sqrt [3]{a+b x^3}}{3 c \left (b c^3-a d^3\right )^2 \left (c^3+d^3 x^3\right )}+\frac {x \left (1+\frac {b x^3}{a}\right )^{2/3} F_1\left (\frac {1}{3};\frac {2}{3},3;\frac {4}{3};-\frac {b x^3}{a},-\frac {d^3 x^3}{c^3}\right )}{c^3 \left (a+b x^3\right )^{2/3}}-\frac {7 d^3 x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} F_1\left (\frac {4}{3};\frac {2}{3},3;\frac {7}{3};-\frac {b x^3}{a},-\frac {d^3 x^3}{c^3}\right )}{4 c^6 \left (a+b x^3\right )^{2/3}}+\frac {d^6 x^7 \left (1+\frac {b x^3}{a}\right )^{2/3} F_1\left (\frac {7}{3};\frac {2}{3},3;\frac {10}{3};-\frac {b x^3}{a},-\frac {d^3 x^3}{c^3}\right )}{7 c^9 \left (a+b x^3\right )^{2/3}}+\frac {2 a d^4 \left (6 b c^3-a d^3\right ) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b c^3-a d^3} x}{c \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} c^2 \left (b c^3-a d^3\right )^{8/3}}+\frac {d \left (9 b^2 c^6-6 a b c^3 d^3+2 a^2 d^6\right ) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b c^3-a d^3} x}{c \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} c^2 \left (b c^3-a d^3\right )^{8/3}}-\frac {10 b^2 c^4 d \tan ^{-1}\left (\frac {1-\frac {2 d \sqrt [3]{a+b x^3}}{\sqrt [3]{b c^3-a d^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} \left (b c^3-a d^3\right )^{8/3}}+\frac {b c d \left (b c^3-6 a d^3\right ) \tan ^{-1}\left (\frac {1-\frac {2 d \sqrt [3]{a+b x^3}}{\sqrt [3]{b c^3-a d^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} \left (b c^3-a d^3\right )^{8/3}}-\frac {5 b^2 c^4 d \log \left (c^3+d^3 x^3\right )}{9 \left (b c^3-a d^3\right )^{8/3}}+\frac {b c d \left (b c^3-6 a d^3\right ) \log \left (c^3+d^3 x^3\right )}{18 \left (b c^3-a d^3\right )^{8/3}}-\frac {a d^4 \left (6 b c^3-a d^3\right ) \log \left (c^3+d^3 x^3\right )}{9 c^2 \left (b c^3-a d^3\right )^{8/3}}-\frac {d \left (9 b^2 c^6-6 a b c^3 d^3+2 a^2 d^6\right ) \log \left (c^3+d^3 x^3\right )}{18 c^2 \left (b c^3-a d^3\right )^{8/3}}+\frac {a d^4 \left (6 b c^3-a d^3\right ) \log \left (\frac {\sqrt [3]{b c^3-a d^3} x}{c}-\sqrt [3]{a+b x^3}\right )}{3 c^2 \left (b c^3-a d^3\right )^{8/3}}+\frac {d \left (9 b^2 c^6-6 a b c^3 d^3+2 a^2 d^6\right ) \log \left (\frac {\sqrt [3]{b c^3-a d^3} x}{c}-\sqrt [3]{a+b x^3}\right )}{6 c^2 \left (b c^3-a d^3\right )^{8/3}}+\frac {5 b^2 c^4 d \log \left (\sqrt [3]{b c^3-a d^3}+d \sqrt [3]{a+b x^3}\right )}{3 \left (b c^3-a d^3\right )^{8/3}}-\frac {b c d \left (b c^3-6 a d^3\right ) \log \left (\sqrt [3]{b c^3-a d^3}+d \sqrt [3]{a+b x^3}\right )}{6 \left (b c^3-a d^3\right )^{8/3}} \]

[Out]

3/2*c^4*d^2*(b*x^3+a)^(1/3)/(-a*d^3+b*c^3)/(d^3*x^3+c^3)^2+3/2*c^2*d^4*x^2*(b*x^3+a)^(1/3)/(-a*d^3+b*c^3)/(d^3
*x^3+c^3)^2+5/3*b*c^4*d^2*(b*x^3+a)^(1/3)/(-a*d^3+b*c^3)^2/(d^3*x^3+c^3)-1/6*c*d^2*(-6*a*d^3+b*c^3)*(b*x^3+a)^
(1/3)/(-a*d^3+b*c^3)^2/(d^3*x^3+c^3)+1/6*d^4*(-4*a*d^3+9*b*c^3)*x^2*(b*x^3+a)^(1/3)/c/(-a*d^3+b*c^3)^2/(d^3*x^
3+c^3)+1/3*d^4*(2*a*d^3+3*b*c^3)*x^2*(b*x^3+a)^(1/3)/c/(-a*d^3+b*c^3)^2/(d^3*x^3+c^3)+x*(1+b*x^3/a)^(2/3)*Appe
llF1(1/3,2/3,3,4/3,-b*x^3/a,-d^3*x^3/c^3)/c^3/(b*x^3+a)^(2/3)-7/4*d^3*x^4*(1+b*x^3/a)^(2/3)*AppellF1(4/3,2/3,3
,7/3,-b*x^3/a,-d^3*x^3/c^3)/c^6/(b*x^3+a)^(2/3)+1/7*d^6*x^7*(1+b*x^3/a)^(2/3)*AppellF1(7/3,2/3,3,10/3,-b*x^3/a
,-d^3*x^3/c^3)/c^9/(b*x^3+a)^(2/3)-5/9*b^2*c^4*d*ln(d^3*x^3+c^3)/(-a*d^3+b*c^3)^(8/3)+1/18*b*c*d*(-6*a*d^3+b*c
^3)*ln(d^3*x^3+c^3)/(-a*d^3+b*c^3)^(8/3)-1/9*a*d^4*(-a*d^3+6*b*c^3)*ln(d^3*x^3+c^3)/c^2/(-a*d^3+b*c^3)^(8/3)-1
/18*d*(2*a^2*d^6-6*a*b*c^3*d^3+9*b^2*c^6)*ln(d^3*x^3+c^3)/c^2/(-a*d^3+b*c^3)^(8/3)+1/3*a*d^4*(-a*d^3+6*b*c^3)*
ln((-a*d^3+b*c^3)^(1/3)*x/c-(b*x^3+a)^(1/3))/c^2/(-a*d^3+b*c^3)^(8/3)+1/6*d*(2*a^2*d^6-6*a*b*c^3*d^3+9*b^2*c^6
)*ln((-a*d^3+b*c^3)^(1/3)*x/c-(b*x^3+a)^(1/3))/c^2/(-a*d^3+b*c^3)^(8/3)+5/3*b^2*c^4*d*ln((-a*d^3+b*c^3)^(1/3)+
d*(b*x^3+a)^(1/3))/(-a*d^3+b*c^3)^(8/3)-1/6*b*c*d*(-6*a*d^3+b*c^3)*ln((-a*d^3+b*c^3)^(1/3)+d*(b*x^3+a)^(1/3))/
(-a*d^3+b*c^3)^(8/3)+2/9*a*d^4*(-a*d^3+6*b*c^3)*arctan(1/3*(1+2*(-a*d^3+b*c^3)^(1/3)*x/c/(b*x^3+a)^(1/3))*3^(1
/2))/c^2/(-a*d^3+b*c^3)^(8/3)*3^(1/2)+1/9*d*(2*a^2*d^6-6*a*b*c^3*d^3+9*b^2*c^6)*arctan(1/3*(1+2*(-a*d^3+b*c^3)
^(1/3)*x/c/(b*x^3+a)^(1/3))*3^(1/2))/c^2/(-a*d^3+b*c^3)^(8/3)*3^(1/2)-10/9*b^2*c^4*d*arctan(1/3*(1-2*d*(b*x^3+
a)^(1/3)/(-a*d^3+b*c^3)^(1/3))*3^(1/2))/(-a*d^3+b*c^3)^(8/3)*3^(1/2)+1/9*b*c*d*(-6*a*d^3+b*c^3)*arctan(1/3*(1-
2*d*(b*x^3+a)^(1/3)/(-a*d^3+b*c^3)^(1/3))*3^(1/2))/(-a*d^3+b*c^3)^(8/3)*3^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 1.18, antiderivative size = 1357, normalized size of antiderivative = 1.00, number of steps used = 30, number of rules used = 18, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.947, Rules used = {2181, 441, 440, 483, 593, 12, 503, 455, 44, 60, 631, 210, 31, 525, 524, 482, 457, 79} \begin {gather*} \frac {d^6 \left (\frac {b x^3}{a}+1\right )^{2/3} F_1\left (\frac {7}{3};\frac {2}{3},3;\frac {10}{3};-\frac {b x^3}{a},-\frac {d^3 x^3}{c^3}\right ) x^7}{7 c^9 \left (b x^3+a\right )^{2/3}}-\frac {7 d^3 \left (\frac {b x^3}{a}+1\right )^{2/3} F_1\left (\frac {4}{3};\frac {2}{3},3;\frac {7}{3};-\frac {b x^3}{a},-\frac {d^3 x^3}{c^3}\right ) x^4}{4 c^6 \left (b x^3+a\right )^{2/3}}+\frac {d^4 \left (3 b c^3+2 a d^3\right ) \sqrt [3]{b x^3+a} x^2}{3 c \left (b c^3-a d^3\right )^2 \left (c^3+d^3 x^3\right )}+\frac {d^4 \left (9 b c^3-4 a d^3\right ) \sqrt [3]{b x^3+a} x^2}{6 c \left (b c^3-a d^3\right )^2 \left (c^3+d^3 x^3\right )}+\frac {3 c^2 d^4 \sqrt [3]{b x^3+a} x^2}{2 \left (b c^3-a d^3\right ) \left (c^3+d^3 x^3\right )^2}+\frac {\left (\frac {b x^3}{a}+1\right )^{2/3} F_1\left (\frac {1}{3};\frac {2}{3},3;\frac {4}{3};-\frac {b x^3}{a},-\frac {d^3 x^3}{c^3}\right ) x}{c^3 \left (b x^3+a\right )^{2/3}}+\frac {2 a d^4 \left (6 b c^3-a d^3\right ) \text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{b c^3-a d^3} x}{c \sqrt [3]{b x^3+a}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} c^2 \left (b c^3-a d^3\right )^{8/3}}+\frac {d \left (9 b^2 c^6-6 a b d^3 c^3+2 a^2 d^6\right ) \text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{b c^3-a d^3} x}{c \sqrt [3]{b x^3+a}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} c^2 \left (b c^3-a d^3\right )^{8/3}}-\frac {10 b^2 c^4 d \text {ArcTan}\left (\frac {1-\frac {2 d \sqrt [3]{b x^3+a}}{\sqrt [3]{b c^3-a d^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} \left (b c^3-a d^3\right )^{8/3}}+\frac {b c d \left (b c^3-6 a d^3\right ) \text {ArcTan}\left (\frac {1-\frac {2 d \sqrt [3]{b x^3+a}}{\sqrt [3]{b c^3-a d^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} \left (b c^3-a d^3\right )^{8/3}}-\frac {a d^4 \left (6 b c^3-a d^3\right ) \log \left (c^3+d^3 x^3\right )}{9 c^2 \left (b c^3-a d^3\right )^{8/3}}-\frac {d \left (9 b^2 c^6-6 a b d^3 c^3+2 a^2 d^6\right ) \log \left (c^3+d^3 x^3\right )}{18 c^2 \left (b c^3-a d^3\right )^{8/3}}-\frac {5 b^2 c^4 d \log \left (c^3+d^3 x^3\right )}{9 \left (b c^3-a d^3\right )^{8/3}}+\frac {b c d \left (b c^3-6 a d^3\right ) \log \left (c^3+d^3 x^3\right )}{18 \left (b c^3-a d^3\right )^{8/3}}+\frac {a d^4 \left (6 b c^3-a d^3\right ) \log \left (\frac {\sqrt [3]{b c^3-a d^3} x}{c}-\sqrt [3]{b x^3+a}\right )}{3 c^2 \left (b c^3-a d^3\right )^{8/3}}+\frac {d \left (9 b^2 c^6-6 a b d^3 c^3+2 a^2 d^6\right ) \log \left (\frac {\sqrt [3]{b c^3-a d^3} x}{c}-\sqrt [3]{b x^3+a}\right )}{6 c^2 \left (b c^3-a d^3\right )^{8/3}}+\frac {5 b^2 c^4 d \log \left (\sqrt [3]{b x^3+a} d+\sqrt [3]{b c^3-a d^3}\right )}{3 \left (b c^3-a d^3\right )^{8/3}}-\frac {b c d \left (b c^3-6 a d^3\right ) \log \left (\sqrt [3]{b x^3+a} d+\sqrt [3]{b c^3-a d^3}\right )}{6 \left (b c^3-a d^3\right )^{8/3}}+\frac {5 b c^4 d^2 \sqrt [3]{b x^3+a}}{3 \left (b c^3-a d^3\right )^2 \left (c^3+d^3 x^3\right )}-\frac {c d^2 \left (b c^3-6 a d^3\right ) \sqrt [3]{b x^3+a}}{6 \left (b c^3-a d^3\right )^2 \left (c^3+d^3 x^3\right )}+\frac {3 c^4 d^2 \sqrt [3]{b x^3+a}}{2 \left (b c^3-a d^3\right ) \left (c^3+d^3 x^3\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*c^4*d^2*(a + b*x^3)^(1/3))/(2*(b*c^3 - a*d^3)*(c^3 + d^3*x^3)^2) + (3*c^2*d^4*x^2*(a + b*x^3)^(1/3))/(2*(b*
c^3 - a*d^3)*(c^3 + d^3*x^3)^2) + (5*b*c^4*d^2*(a + b*x^3)^(1/3))/(3*(b*c^3 - a*d^3)^2*(c^3 + d^3*x^3)) - (c*d
^2*(b*c^3 - 6*a*d^3)*(a + b*x^3)^(1/3))/(6*(b*c^3 - a*d^3)^2*(c^3 + d^3*x^3)) + (d^4*(9*b*c^3 - 4*a*d^3)*x^2*(
a + b*x^3)^(1/3))/(6*c*(b*c^3 - a*d^3)^2*(c^3 + d^3*x^3)) + (d^4*(3*b*c^3 + 2*a*d^3)*x^2*(a + b*x^3)^(1/3))/(3
*c*(b*c^3 - a*d^3)^2*(c^3 + d^3*x^3)) + (x*(1 + (b*x^3)/a)^(2/3)*AppellF1[1/3, 2/3, 3, 4/3, -((b*x^3)/a), -((d
^3*x^3)/c^3)])/(c^3*(a + b*x^3)^(2/3)) - (7*d^3*x^4*(1 + (b*x^3)/a)^(2/3)*AppellF1[4/3, 2/3, 3, 7/3, -((b*x^3)
/a), -((d^3*x^3)/c^3)])/(4*c^6*(a + b*x^3)^(2/3)) + (d^6*x^7*(1 + (b*x^3)/a)^(2/3)*AppellF1[7/3, 2/3, 3, 10/3,
-((b*x^3)/a), -((d^3*x^3)/c^3)])/(7*c^9*(a + b*x^3)^(2/3)) + (2*a*d^4*(6*b*c^3 - a*d^3)*ArcTan[(1 + (2*(b*c^3
- a*d^3)^(1/3)*x)/(c*(a + b*x^3)^(1/3)))/Sqrt[3]])/(3*Sqrt[3]*c^2*(b*c^3 - a*d^3)^(8/3)) + (d*(9*b^2*c^6 - 6*
a*b*c^3*d^3 + 2*a^2*d^6)*ArcTan[(1 + (2*(b*c^3 - a*d^3)^(1/3)*x)/(c*(a + b*x^3)^(1/3)))/Sqrt[3]])/(3*Sqrt[3]*c
^2*(b*c^3 - a*d^3)^(8/3)) - (10*b^2*c^4*d*ArcTan[(1 - (2*d*(a + b*x^3)^(1/3))/(b*c^3 - a*d^3)^(1/3))/Sqrt[3]])
/(3*Sqrt[3]*(b*c^3 - a*d^3)^(8/3)) + (b*c*d*(b*c^3 - 6*a*d^3)*ArcTan[(1 - (2*d*(a + b*x^3)^(1/3))/(b*c^3 - a*d
^3)^(1/3))/Sqrt[3]])/(3*Sqrt[3]*(b*c^3 - a*d^3)^(8/3)) - (5*b^2*c^4*d*Log[c^3 + d^3*x^3])/(9*(b*c^3 - a*d^3)^(
8/3)) + (b*c*d*(b*c^3 - 6*a*d^3)*Log[c^3 + d^3*x^3])/(18*(b*c^3 - a*d^3)^(8/3)) - (a*d^4*(6*b*c^3 - a*d^3)*Log
[c^3 + d^3*x^3])/(9*c^2*(b*c^3 - a*d^3)^(8/3)) - (d*(9*b^2*c^6 - 6*a*b*c^3*d^3 + 2*a^2*d^6)*Log[c^3 + d^3*x^3]
)/(18*c^2*(b*c^3 - a*d^3)^(8/3)) + (a*d^4*(6*b*c^3 - a*d^3)*Log[((b*c^3 - a*d^3)^(1/3)*x)/c - (a + b*x^3)^(1/3
)])/(3*c^2*(b*c^3 - a*d^3)^(8/3)) + (d*(9*b^2*c^6 - 6*a*b*c^3*d^3 + 2*a^2*d^6)*Log[((b*c^3 - a*d^3)^(1/3)*x)/c
- (a + b*x^3)^(1/3)])/(6*c^2*(b*c^3 - a*d^3)^(8/3)) + (5*b^2*c^4*d*Log[(b*c^3 - a*d^3)^(1/3) + d*(a + b*x^3)^
(1/3)])/(3*(b*c^3 - a*d^3)^(8/3)) - (b*c*d*(b*c^3 - 6*a*d^3)*Log[(b*c^3 - a*d^3)^(1/3) + d*(a + b*x^3)^(1/3)])
/(6*(b*c^3 - a*d^3)^(8/3))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 31

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

Rule 44

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

Rule 60

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-(b*c - a*d)/b, 3]}, Simp[-
Log[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (Dist[3/(2*b*q), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x
)^(1/3)], x] + Dist[3/(2*b*q^2), Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x]
&& NegQ[(b*c - a*d)/b]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 210

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

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 441

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^F
racPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rule 457

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]]

Rule 482

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[e^(n - 1
)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(n*(b*c - a*d)*(p + 1))), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 483

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

Rule 503

Int[(x_)/(((a_) + (b_.)*(x_)^3)^(2/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Si
mp[-ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q^2), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/
(2*c*q^2), x] + Simp[Log[c + d*x^3]/(6*c*q^2), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 593

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*g*n*(b*c - a*d)*(p +
1))), x] + Dist[1/(a*n*(b*c - a*d)*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f)
*(m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, e, f, g, m, q}, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 631

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 2181

Int[(Px_.)*((c_) + (d_.)*(x_))^(q_)*((a_) + (b_.)*(x_)^3)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c^3 + d^3*x
^3)^q*(a + b*x^3)^p, Px/(c^2 - c*d*x + d^2*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p}, x] && PolyQ[Px, x] && ILtQ
[q, 0] && RationalQ[p] && EqQ[Denominator[p], 3]

Rubi steps

\begin {align*} \int \frac {1}{(c+d x)^3 \left (a+b x^3\right )^{2/3}} \, dx &=\int \frac {1}{(c+d x)^3 \left (a+b x^3\right )^{2/3}} \, dx\\ \end {align*}

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Mathematica [F]
time = 20.27, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(c+d x)^3 \left (a+b x^3\right )^{2/3}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (d x +c \right )^{3} \left (b \,x^{3}+a \right )^{\frac {2}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
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(1/(d*x+c)^3/(b*x^3+a)^(2/3),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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