3.1.41 \(\int \frac {1}{(c+d x)^2 (a+b x^3)^{2/3}} \, dx\) [41]
Optimal. Leaf size=760 \[ \frac {c^2 d^2 \sqrt [3]{a+b x^3}}{\left (b c^3-a d^3\right ) \left (c^3+d^3 x^3\right )}+\frac {d^4 x^2 \sqrt [3]{a+b x^3}}{\left (b c^3-a d^3\right ) \left (c^3+d^3 x^3\right )}+\frac {x \left (1+\frac {b x^3}{a}\right )^{2/3} F_1\left (\frac {1}{3};\frac {2}{3},2;\frac {4}{3};-\frac {b x^3}{a},-\frac {d^3 x^3}{c^3}\right )}{c^2 \left (a+b x^3\right )^{2/3}}-\frac {d^3 x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} F_1\left (\frac {4}{3};\frac {2}{3},2;\frac {7}{3};-\frac {b x^3}{a},-\frac {d^3 x^3}{c^3}\right )}{2 c^5 \left (a+b x^3\right )^{2/3}}+\frac {2 a d^4 \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b c^3-a d^3} x}{c \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} c \left (b c^3-a d^3\right )^{5/3}}+\frac {2 d \left (3 b c^3-a d^3\right ) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b c^3-a d^3} x}{c \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} c \left (b c^3-a d^3\right )^{5/3}}-\frac {2 b c^2 d \tan ^{-1}\left (\frac {1-\frac {2 d \sqrt [3]{a+b x^3}}{\sqrt [3]{b c^3-a d^3}}}{\sqrt {3}}\right )}{\sqrt {3} \left (b c^3-a d^3\right )^{5/3}}-\frac {b c^2 d \log \left (c^3+d^3 x^3\right )}{3 \left (b c^3-a d^3\right )^{5/3}}-\frac {a d^4 \log \left (c^3+d^3 x^3\right )}{9 c \left (b c^3-a d^3\right )^{5/3}}-\frac {d \left (3 b c^3-a d^3\right ) \log \left (c^3+d^3 x^3\right )}{9 c \left (b c^3-a d^3\right )^{5/3}}+\frac {a d^4 \log \left (\frac {\sqrt [3]{b c^3-a d^3} x}{c}-\sqrt [3]{a+b x^3}\right )}{3 c \left (b c^3-a d^3\right )^{5/3}}+\frac {d \left (3 b c^3-a d^3\right ) \log \left (\frac {\sqrt [3]{b c^3-a d^3} x}{c}-\sqrt [3]{a+b x^3}\right )}{3 c \left (b c^3-a d^3\right )^{5/3}}+\frac {b c^2 d \log \left (\sqrt [3]{b c^3-a d^3}+d \sqrt [3]{a+b x^3}\right )}{\left (b c^3-a d^3\right )^{5/3}} \]
[Out]
c^2*d^2*(b*x^3+a)^(1/3)/(-a*d^3+b*c^3)/(d^3*x^3+c^3)+d^4*x^2*(b*x^3+a)^(1/3)/(-a*d^3+b*c^3)/(d^3*x^3+c^3)+x*(1
+b*x^3/a)^(2/3)*AppellF1(1/3,2/3,2,4/3,-b*x^3/a,-d^3*x^3/c^3)/c^2/(b*x^3+a)^(2/3)-1/2*d^3*x^4*(1+b*x^3/a)^(2/3
)*AppellF1(4/3,2/3,2,7/3,-b*x^3/a,-d^3*x^3/c^3)/c^5/(b*x^3+a)^(2/3)-1/3*b*c^2*d*ln(d^3*x^3+c^3)/(-a*d^3+b*c^3)
^(5/3)-1/9*a*d^4*ln(d^3*x^3+c^3)/c/(-a*d^3+b*c^3)^(5/3)-1/9*d*(-a*d^3+3*b*c^3)*ln(d^3*x^3+c^3)/c/(-a*d^3+b*c^3
)^(5/3)+1/3*a*d^4*ln((-a*d^3+b*c^3)^(1/3)*x/c-(b*x^3+a)^(1/3))/c/(-a*d^3+b*c^3)^(5/3)+1/3*d*(-a*d^3+3*b*c^3)*l
n((-a*d^3+b*c^3)^(1/3)*x/c-(b*x^3+a)^(1/3))/c/(-a*d^3+b*c^3)^(5/3)+b*c^2*d*ln((-a*d^3+b*c^3)^(1/3)+d*(b*x^3+a)
^(1/3))/(-a*d^3+b*c^3)^(5/3)+2/9*a*d^4*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/(-
a*d^3+b*c^3)^(5/3)*3^(1/2)+2/9*d*(-a*d^3+3*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/(-a*d^3+b*c^3)^(5/3)*3^(1/2)-2/3*b*c^2*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)^(5/3)*3^(1/2)
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Rubi [A]
time = 0.52, antiderivative size = 760, normalized size of antiderivative = 1.00, number of steps
used = 18, number of rules used = 15, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.790, Rules used = {2181, 441,
440, 483, 12, 503, 455, 44, 60, 631, 210, 31, 525, 524, 482} \begin {gather*} -\frac {d^3 x^4 \left (\frac {b x^3}{a}+1\right )^{2/3} F_1\left (\frac {4}{3};\frac {2}{3},2;\frac {7}{3};-\frac {b x^3}{a},-\frac {d^3 x^3}{c^3}\right )}{2 c^5 \left (a+b x^3\right )^{2/3}}+\frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} F_1\left (\frac {1}{3};\frac {2}{3},2;\frac {4}{3};-\frac {b x^3}{a},-\frac {d^3 x^3}{c^3}\right )}{c^2 \left (a+b x^3\right )^{2/3}}+\frac {2 d \left (3 b c^3-a d^3\right ) \text {ArcTan}\left (\frac {\frac {2 x \sqrt [3]{b c^3-a d^3}}{c \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} c \left (b c^3-a d^3\right )^{5/3}}+\frac {2 a d^4 \text {ArcTan}\left (\frac {\frac {2 x \sqrt [3]{b c^3-a d^3}}{c \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} c \left (b c^3-a d^3\right )^{5/3}}-\frac {2 b c^2 d \text {ArcTan}\left (\frac {1-\frac {2 d \sqrt [3]{a+b x^3}}{\sqrt [3]{b c^3-a d^3}}}{\sqrt {3}}\right )}{\sqrt {3} \left (b c^3-a d^3\right )^{5/3}}-\frac {d \left (3 b c^3-a d^3\right ) \log \left (c^3+d^3 x^3\right )}{9 c \left (b c^3-a d^3\right )^{5/3}}+\frac {d \left (3 b c^3-a d^3\right ) \log \left (\frac {x \sqrt [3]{b c^3-a d^3}}{c}-\sqrt [3]{a+b x^3}\right )}{3 c \left (b c^3-a d^3\right )^{5/3}}-\frac {a d^4 \log \left (c^3+d^3 x^3\right )}{9 c \left (b c^3-a d^3\right )^{5/3}}+\frac {a d^4 \log \left (\frac {x \sqrt [3]{b c^3-a d^3}}{c}-\sqrt [3]{a+b x^3}\right )}{3 c \left (b c^3-a d^3\right )^{5/3}}+\frac {d^4 x^2 \sqrt [3]{a+b x^3}}{\left (c^3+d^3 x^3\right ) \left (b c^3-a d^3\right )}-\frac {b c^2 d \log \left (c^3+d^3 x^3\right )}{3 \left (b c^3-a d^3\right )^{5/3}}+\frac {b c^2 d \log \left (\sqrt [3]{b c^3-a d^3}+d \sqrt [3]{a+b x^3}\right )}{\left (b c^3-a d^3\right )^{5/3}}+\frac {c^2 d^2 \sqrt [3]{a+b x^3}}{\left (c^3+d^3 x^3\right ) \left (b c^3-a d^3\right )} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[1/((c + d*x)^2*(a + b*x^3)^(2/3)),x]
[Out]
(c^2*d^2*(a + b*x^3)^(1/3))/((b*c^3 - a*d^3)*(c^3 + d^3*x^3)) + (d^4*x^2*(a + b*x^3)^(1/3))/((b*c^3 - a*d^3)*(
c^3 + d^3*x^3)) + (x*(1 + (b*x^3)/a)^(2/3)*AppellF1[1/3, 2/3, 2, 4/3, -((b*x^3)/a), -((d^3*x^3)/c^3)])/(c^2*(a
+ b*x^3)^(2/3)) - (d^3*x^4*(1 + (b*x^3)/a)^(2/3)*AppellF1[4/3, 2/3, 2, 7/3, -((b*x^3)/a), -((d^3*x^3)/c^3)])/
(2*c^5*(a + b*x^3)^(2/3)) + (2*a*d^4*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*(b*c^3 - a*d^3)^(5/3)) + (2*d*(3*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*(b*c^3 - a*d^3)^(5/3)) - (2*b*c^2*d*ArcTan[(1 - (2*d*(a + b*x^3)^(1/3))/(
b*c^3 - a*d^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*(b*c^3 - a*d^3)^(5/3)) - (b*c^2*d*Log[c^3 + d^3*x^3])/(3*(b*c^3 - a*d
^3)^(5/3)) - (a*d^4*Log[c^3 + d^3*x^3])/(9*c*(b*c^3 - a*d^3)^(5/3)) - (d*(3*b*c^3 - a*d^3)*Log[c^3 + d^3*x^3])
/(9*c*(b*c^3 - a*d^3)^(5/3)) + (a*d^4*Log[((b*c^3 - a*d^3)^(1/3)*x)/c - (a + b*x^3)^(1/3)])/(3*c*(b*c^3 - a*d^
3)^(5/3)) + (d*(3*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*(b*c^3 - a*d^3)^(5
/3)) + (b*c^2*d*Log[(b*c^3 - a*d^3)^(1/3) + d*(a + b*x^3)^(1/3)])/(b*c^3 - a*d^3)^(5/3)
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 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 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 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)^2 \left (a+b x^3\right )^{2/3}} \, dx &=\int \frac {1}{(c+d x)^2 \left (a+b x^3\right )^{2/3}} \, dx\\ \end {align*}
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Mathematica [F]
time = 20.22, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(c+d x)^2 \left (a+b x^3\right )^{2/3}} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
Integrate[1/((c + d*x)^2*(a + b*x^3)^(2/3)),x]
[Out]
Integrate[1/((c + d*x)^2*(a + b*x^3)^(2/3)), x]
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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (d x +c \right )^{2} \left (b \,x^{3}+a \right )^{\frac {2}{3}}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(d*x+c)^2/(b*x^3+a)^(2/3),x)
[Out]
int(1/(d*x+c)^2/(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)^2/(b*x^3+a)^(2/3),x, algorithm="maxima")
[Out]
integrate(1/((b*x^3 + a)^(2/3)*(d*x + c)^2), 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)^2/(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 )^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(d*x+c)**2/(b*x**3+a)**(2/3),x)
[Out]
Integral(1/((a + b*x**3)**(2/3)*(c + d*x)**2), 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)^2/(b*x^3+a)^(2/3),x, algorithm="giac")
[Out]
integrate(1/((b*x^3 + a)^(2/3)*(d*x + c)^2), 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 )}^2} \,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)^2),x)
[Out]
int(1/((a + b*x^3)^(2/3)*(c + d*x)^2), x)
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