3.3.62 \(\int \frac {\sqrt {c+d x^3}}{x (4 c+d x^3)} \, dx\) [262]

Optimal. Leaf size=65 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{2 \sqrt {3} \sqrt {c}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{6 \sqrt {c}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {457, 85, 65, 214, 209} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{2 \sqrt {3} \sqrt {c}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{6 \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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 85

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

Rule 209

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

Rule 214

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

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]]

Rubi steps

\begin {align*} \int \frac {\sqrt {c+d x^3}}{x \left (4 c+d x^3\right )} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {\sqrt {c+d x}}{x (4 c+d x)} \, dx,x,x^3\right )\\ &=\frac {1}{12} \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^3\right )+\frac {1}{4} d \text {Subst}\left (\int \frac {1}{\sqrt {c+d x} (4 c+d x)} \, dx,x,x^3\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \frac {1}{3 c+x^2} \, dx,x,\sqrt {c+d x^3}\right )+\frac {\text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^3}\right )}{6 d}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{2 \sqrt {3} \sqrt {c}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{6 \sqrt {c}}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 59, normalized size = 0.91 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )-\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{6 \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.52, size = 468, normalized size = 7.20

method	result	size
default	\(-\frac {d \left (\frac {2 \sqrt {d \,x^{3}+c}}{3 d}+\frac {i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (d \,\textit {\_Z}^{3}+4 c \right )}{\sum }\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {\frac {i d \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 )}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {d \left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i d \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 )}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}\, d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}+2 \underline {\hspace {1.25 ex}}\alpha ^{2} d^{2}-\left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha d -\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 {2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha ^{2} d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha +i \sqrt {3}\, c d -3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -3 c d}{6 d c}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{d \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 )}}\right )}{2 \sqrt {d \,x^{3}+c}}\right )}{3 d^{3}}\right )}{4 c}+\frac {\frac {2 \sqrt {d \,x^{3}+c}}{3}-\frac {2 \arctanh \left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right ) \sqrt {c}}{3}}{4 c}\)	\(468\)
elliptic	\(\text {Expression too large to display}\)	\(1502\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*d/c*(2/3*(d*x^3+c)^(1/2)/d+1/3*I/d^3*2^(1/2)*sum((-c*d^2)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-c*d^2)^(1
/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)*(d*(x-1/d*(-c*d^2)^(1/3))/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/
3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)
*(I*(-c*d^2)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-c*d^2)^(2/3)+2*_alpha^2*d^2-(-c*d^2)^(1/3)*_alpha*d-(-c*d^2)^(
2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/
3))^(1/2),1/6/d*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-c*d^2
)^(2/3)*_alpha-3*c*d)/c,(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1
/2)),_alpha=RootOf(_Z^3*d+4*c)))+1/4/c*(2/3*(d*x^3+c)^(1/2)-2/3*arctanh((d*x^3+c)^(1/2)/c^(1/2))*c^(1/2))

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

[Out]

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

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Fricas [A]
time = 2.40, size = 147, normalized size = 2.26 \begin {gather*} \left [\frac {2 \, \sqrt {3} \sqrt {c} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right ) + \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right )}{12 \, c}, -\frac {\sqrt {3} \sqrt {-c} \log \left (\frac {d x^{3} - 2 \, \sqrt {3} \sqrt {d x^{3} + c} \sqrt {-c} - 2 \, c}{d x^{3} + 4 \, c}\right ) - 2 \, \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right )}{12 \, c}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(2*sqrt(3)*sqrt(c)*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c)) + sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + c)*
sqrt(c) + 2*c)/x^3))/c, -1/12*(sqrt(3)*sqrt(-c)*log((d*x^3 - 2*sqrt(3)*sqrt(d*x^3 + c)*sqrt(-c) - 2*c)/(d*x^3
+ 4*c)) - 2*sqrt(-c)*arctan(sqrt(d*x^3 + c)*sqrt(-c)/c))/c]

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Sympy [A]
time = 3.00, size = 66, normalized size = 1.02 \begin {gather*} \frac {2 \left (\frac {d \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{\sqrt {- c}} \right )}}{12 \sqrt {- c}} + \frac {\sqrt {3} d \operatorname {atan}{\left (\frac {\sqrt {3} \sqrt {c + d x^{3}}}{3 \sqrt {c}} \right )}}{12 \sqrt {c}}\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(d*atan(sqrt(c + d*x**3)/sqrt(-c))/(12*sqrt(-c)) + sqrt(3)*d*atan(sqrt(3)*sqrt(c + d*x**3)/(3*sqrt(c)))/(12*
sqrt(c)))/d

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Giac [A]
time = 0.71, size = 50, normalized size = 0.77 \begin {gather*} \frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right )}{6 \, \sqrt {c}} + \frac {\arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{6 \, \sqrt {-c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(3)*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c))/sqrt(c) + 1/6*arctan(sqrt(d*x^3 + c)/sqrt(-c))/sqrt(-c
)

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Mupad [B]
time = 4.66, size = 93, normalized size = 1.43 \begin {gather*} \frac {\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 )}{12\,\sqrt {c}}+\frac {\sqrt {3}\,\ln \left (\frac {\sqrt {3}\,d\,x^3-2\,\sqrt {3}\,c+\sqrt {c}\,\sqrt {d\,x^3+c}\,6{}\mathrm {i}}{d\,x^3+4\,c}\right )\,1{}\mathrm {i}}{12\,\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((((c + d*x^3)^(1/2) - c^(1/2))^3*((c + d*x^3)^(1/2) + c^(1/2)))/x^6)/(12*c^(1/2)) + (3^(1/2)*log((c^(1/2)*
(c + d*x^3)^(1/2)*6i - 2*3^(1/2)*c + 3^(1/2)*d*x^3)/(4*c + d*x^3))*1i)/(12*c^(1/2))

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