\(\int \frac {1}{\sqrt [3]{x} \sqrt {c+d x} (4 c+d x)} \, dx\) [463]

Optimal result

Integrand size = 24, antiderivative size = 199 \[ \int \frac {1}{\sqrt [3]{x} \sqrt {c+d x} (4 c+d x)} \, dx=-\frac {\arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{d} \sqrt [3]{x}\right )}{\sqrt {c+d x}}\right )}{2^{2/3} \sqrt {3} c^{5/6} d^{2/3}}+\frac {\arctan \left (\frac {\sqrt {c+d x}}{\sqrt {3} \sqrt {c}}\right )}{2^{2/3} \sqrt {3} c^{5/6} d^{2/3}}-\frac {\text {arctanh}\left (\frac {\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{d} \sqrt [3]{x}\right )}{\sqrt {c+d x}}\right )}{2^{2/3} c^{5/6} d^{2/3}}+\frac {\text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{3\ 2^{2/3} c^{5/6} d^{2/3}} \]

[Out]

-1/2*arctanh(c^(1/6)*(c^(1/3)-2^(1/3)*d^(1/3)*x^(1/3))/(d*x+c)^(1/2))*2^(1/3)/c^(5/6)/d^(2/3)+1/6*arctanh((d*x
+c)^(1/2)/c^(1/2))*2^(1/3)/c^(5/6)/d^(2/3)-1/6*arctan(c^(1/6)*(c^(1/3)+2^(1/3)*d^(1/3)*x^(1/3))*3^(1/2)/(d*x+c
)^(1/2))*2^(1/3)/c^(5/6)/d^(2/3)*3^(1/2)+1/6*arctan(1/3*(d*x+c)^(1/2)*3^(1/2)/c^(1/2))*2^(1/3)/c^(5/6)/d^(2/3)
*3^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {129, 497} \[ \int \frac {1}{\sqrt [3]{x} \sqrt {c+d x} (4 c+d x)} \, dx=-\frac {\arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{d} \sqrt [3]{x}\right )}{\sqrt {c+d x}}\right )}{2^{2/3} \sqrt {3} c^{5/6} d^{2/3}}+\frac {\arctan \left (\frac {\sqrt {c+d x}}{\sqrt {3} \sqrt {c}}\right )}{2^{2/3} \sqrt {3} c^{5/6} d^{2/3}}-\frac {\text {arctanh}\left (\frac {\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{d} \sqrt [3]{x}\right )}{\sqrt {c+d x}}\right )}{2^{2/3} c^{5/6} d^{2/3}}+\frac {\text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{3\ 2^{2/3} c^{5/6} d^{2/3}} \]

[In]

Int[1/(x^(1/3)*Sqrt[c + d*x]*(4*c + d*x)),x]

[Out]

-(ArcTan[(Sqrt[3]*c^(1/6)*(c^(1/3) + 2^(1/3)*d^(1/3)*x^(1/3)))/Sqrt[c + d*x]]/(2^(2/3)*Sqrt[3]*c^(5/6)*d^(2/3)
)) + ArcTan[Sqrt[c + d*x]/(Sqrt[3]*Sqrt[c])]/(2^(2/3)*Sqrt[3]*c^(5/6)*d^(2/3)) - ArcTanh[(c^(1/6)*(c^(1/3) - 2
^(1/3)*d^(1/3)*x^(1/3)))/Sqrt[c + d*x]]/(2^(2/3)*c^(5/6)*d^(2/3)) + ArcTanh[Sqrt[c + d*x]/Sqrt[c]]/(3*2^(2/3)*
c^(5/6)*d^(2/3))

Rule 129

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + b*(x^k/e))^m*(c + d*(x^k/e))^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

Rule 497

Int[(x_)/(((a_) + (b_.)*(x_)^3)*Sqrt[(c_) + (d_.)*(x_)^3]), x_Symbol] :> With[{q = Rt[d/c, 3]}, Simp[q*(ArcTan
h[Sqrt[c + d*x^3]/Rt[c, 2]]/(9*2^(2/3)*b*Rt[c, 2])), x] + (-Simp[q*(ArcTanh[Rt[c, 2]*((1 - 2^(1/3)*q*x)/Sqrt[c
+ d*x^3])]/(3*2^(2/3)*b*Rt[c, 2])), x] + Simp[q*(ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Rt[c, 2])]/(3*2^(2/3)*Sqrt[3
]*b*Rt[c, 2])), x] - Simp[q*(ArcTan[Sqrt[3]*Rt[c, 2]*((1 + 2^(1/3)*q*x)/Sqrt[c + d*x^3])]/(3*2^(2/3)*Sqrt[3]*b
*Rt[c, 2])), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[4*b*c - a*d, 0] && PosQ[c]

Rubi steps \begin{align*} \text {integral}& = 3 \text {Subst}\left (\int \frac {x}{\sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx,x,\sqrt [3]{x}\right ) \\ & = -\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{d} \sqrt [3]{x}\right )}{\sqrt {c+d x}}\right )}{2^{2/3} \sqrt {3} c^{5/6} d^{2/3}}+\frac {\tan ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {3} \sqrt {c}}\right )}{2^{2/3} \sqrt {3} c^{5/6} d^{2/3}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{d} \sqrt [3]{x}\right )}{\sqrt {c+d x}}\right )}{2^{2/3} c^{5/6} d^{2/3}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{3\ 2^{2/3} c^{5/6} d^{2/3}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 10.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.31 \[ \int \frac {1}{\sqrt [3]{x} \sqrt {c+d x} (4 c+d x)} \, dx=\frac {3 x^{2/3} \sqrt {\frac {c+d x}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x}{c},-\frac {d x}{4 c}\right )}{8 c \sqrt {c+d x}} \]

[In]

Integrate[1/(x^(1/3)*Sqrt[c + d*x]*(4*c + d*x)),x]

[Out]

(3*x^(2/3)*Sqrt[(c + d*x)/c]*AppellF1[2/3, 1/2, 1, 5/3, -((d*x)/c), -1/4*(d*x)/c])/(8*c*Sqrt[c + d*x])

Maple [F]

\[\int \frac {1}{x^{\frac {1}{3}} \left (d x +4 c \right ) \sqrt {d x +c}}d x\]

[In]

int(1/x^(1/3)/(d*x+4*c)/(d*x+c)^(1/2),x)

[Out]

int(1/x^(1/3)/(d*x+4*c)/(d*x+c)^(1/2),x)

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [3]{x} \sqrt {c+d x} (4 c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(1/x^(1/3)/(d*x+4*c)/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {1}{\sqrt [3]{x} \sqrt {c+d x} (4 c+d x)} \, dx=\int \frac {1}{\sqrt [3]{x} \sqrt {c + d x} \left (4 c + d x\right )}\, dx \]

[In]

integrate(1/x**(1/3)/(d*x+4*c)/(d*x+c)**(1/2),x)

[Out]

Integral(1/(x**(1/3)*sqrt(c + d*x)*(4*c + d*x)), x)

Maxima [F]

\[ \int \frac {1}{\sqrt [3]{x} \sqrt {c+d x} (4 c+d x)} \, dx=\int { \frac {1}{{\left (d x + 4 \, c\right )} \sqrt {d x + c} x^{\frac {1}{3}}} \,d x } \]

[In]

integrate(1/x^(1/3)/(d*x+4*c)/(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((d*x + 4*c)*sqrt(d*x + c)*x^(1/3)), x)

Giac [F]

\[ \int \frac {1}{\sqrt [3]{x} \sqrt {c+d x} (4 c+d x)} \, dx=\int { \frac {1}{{\left (d x + 4 \, c\right )} \sqrt {d x + c} x^{\frac {1}{3}}} \,d x } \]

[In]

integrate(1/x^(1/3)/(d*x+4*c)/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate(1/((d*x + 4*c)*sqrt(d*x + c)*x^(1/3)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [3]{x} \sqrt {c+d x} (4 c+d x)} \, dx=\int \frac {1}{x^{1/3}\,\left (4\,c+d\,x\right )\,\sqrt {c+d\,x}} \,d x \]

[In]

int(1/(x^(1/3)*(4*c + d*x)*(c + d*x)^(1/2)),x)

[Out]

int(1/(x^(1/3)*(4*c + d*x)*(c + d*x)^(1/2)), x)