3.276 \(\int \frac{x}{\sqrt{c+d x^3} (4 c+d x^3)} \, dx\)

Optimal. Leaf size=206 \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt{c+d x^3}}\right )}{3\ 2^{2/3} \sqrt{3} c^{5/6} d^{2/3}}+\frac{\tan ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{3} \sqrt{c}}\right )}{3\ 2^{2/3} \sqrt{3} c^{5/6} d^{2/3}}-\frac{\tanh ^{-1}\left (\frac{\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt{c+d x^3}}\right )}{3\ 2^{2/3} c^{5/6} d^{2/3}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{9\ 2^{2/3} c^{5/6} d^{2/3}} \]

[Out]

-ArcTan[(Sqrt[3]*c^(1/6)*(c^(1/3) + 2^(1/3)*d^(1/3)*x))/Sqrt[c + d*x^3]]/(3*2^(2/3)*Sqrt[3]*c^(5/6)*d^(2/3)) +
ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqrt[c])]/(3*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))/Sqrt[c + d*x^3]]/(3*2^(2/3)*c^(5/6)*d^(2/3)) + ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]]/(9*2^(2/3)*
c^(5/6)*d^(2/3))

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Rubi [A]  time = 0.0324488, antiderivative size = 206, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.042, Rules used = {484} \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt{c+d x^3}}\right )}{3\ 2^{2/3} \sqrt{3} c^{5/6} d^{2/3}}+\frac{\tan ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{3} \sqrt{c}}\right )}{3\ 2^{2/3} \sqrt{3} c^{5/6} d^{2/3}}-\frac{\tanh ^{-1}\left (\frac{\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt{c+d x^3}}\right )}{3\ 2^{2/3} c^{5/6} d^{2/3}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{9\ 2^{2/3} c^{5/6} d^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTan[(Sqrt[3]*c^(1/6)*(c^(1/3) + 2^(1/3)*d^(1/3)*x))/Sqrt[c + d*x^3]]/(3*2^(2/3)*Sqrt[3]*c^(5/6)*d^(2/3)) +
ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqrt[c])]/(3*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))/Sqrt[c + d*x^3]]/(3*2^(2/3)*c^(5/6)*d^(2/3)) + ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]]/(9*2^(2/3)*
c^(5/6)*d^(2/3))

Rule 484

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*} \int \frac{x}{\sqrt{c+d x^3} \left (4 c+d x^3\right )} \, dx &=-\frac{\tan ^{-1}\left (\frac{\sqrt{3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt{c+d x^3}}\right )}{3\ 2^{2/3} \sqrt{3} c^{5/6} d^{2/3}}+\frac{\tan ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{3} \sqrt{c}}\right )}{3\ 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} x\right )}{\sqrt{c+d x^3}}\right )}{3\ 2^{2/3} c^{5/6} d^{2/3}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{9\ 2^{2/3} c^{5/6} d^{2/3}}\\ \end{align*}

Mathematica [C]  time = 0.0260982, size = 67, normalized size = 0.33 \[ \frac{x^2 \sqrt{\frac{c+d x^3}{c}} F_1\left (\frac{2}{3};\frac{1}{2},1;\frac{5}{3};-\frac{d x^3}{c},-\frac{d x^3}{4 c}\right )}{8 c \sqrt{c+d x^3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [C]  time = 0.007, size = 416, normalized size = 2. \begin{align*}{\frac{-{\frac{i}{9}}\sqrt{2}}{{d}^{3}c}\sum _{{\it \_alpha}={\it RootOf} \left ({{\it \_Z}}^{3}d+4\,c \right ) }{\frac{1}{{\it \_alpha}}\sqrt [3]{-{d}^{2}c}\sqrt{{{\frac{i}{2}}d \left ( 2\,x+{\frac{1}{d} \left ( -i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c} \right ) } \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}}\sqrt{{d \left ( x-{\frac{1}{d}\sqrt [3]{-{d}^{2}c}} \right ) \left ( -3\,\sqrt [3]{-{d}^{2}c}+i\sqrt{3}\sqrt [3]{-{d}^{2}c} \right ) ^{-1}}}\sqrt{{-{\frac{i}{2}}d \left ( 2\,x+{\frac{1}{d} \left ( i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c} \right ) } \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}} \left ( i\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,\sqrt{3}d-i\sqrt{3} \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}}+2\,{{\it \_alpha}}^{2}{d}^{2}-\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,d- \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}} \right ){\it EllipticPi} \left ({\frac{\sqrt{3}}{3}\sqrt{{i\sqrt{3}d \left ( x+{\frac{1}{2\,d}\sqrt [3]{-{d}^{2}c}}-{\frac{{\frac{i}{2}}\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c}} \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}}},{\frac{1}{6\,cd} \left ( 2\,i\sqrt [3]{-{d}^{2}c}\sqrt{3}{{\it \_alpha}}^{2}d-i \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}}\sqrt{3}{\it \_alpha}+i\sqrt{3}cd-3\, \left ( -{d}^{2}c \right ) ^{2/3}{\it \_alpha}-3\,cd \right ) },\sqrt{{\frac{i\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c} \left ( -{\frac{3}{2\,d}\sqrt [3]{-{d}^{2}c}}+{\frac{{\frac{i}{2}}\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c}} \right ) ^{-1}}} \right ){\frac{1}{\sqrt{d{x}^{3}+c}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*I/d^3/c*2^(1/2)*sum(1/_alpha*(-d^2*c)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))
/(-d^2*c)^(1/3))^(1/2)*(d*(x-1/d*(-d^2*c)^(1/3))/(-3*(-d^2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d
*(2*x+1/d*(I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_
alpha*3^(1/2)*d-I*3^(1/2)*(-d^2*c)^(2/3)+2*_alpha^2*d^2-(-d^2*c)^(1/3)*_alpha*d-(-d^2*c)^(2/3))*EllipticPi(1/3
*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),1/6/d*(2*I
*(-d^2*c)^(1/3)*3^(1/2)*_alpha^2*d-I*(-d^2*c)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*d
)/c,(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)),_alpha=RootOf(_
Z^3*d+4*c))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (d x^{3} + 4 \, c\right )} \sqrt{d x^{3} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 7.15857, size = 5685, normalized size = 27.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*sqrt(3)*(1/432)^(1/6)*(-1/(c^5*d^4))^(1/6)*arctan(-1/3*(3*(sqrt(3)*sqrt(1/3)*c^3*d^2*x*sqrt(-1/(c^5*d^4))
+ 2*sqrt(3)*(1/432)^(1/6)*c*d*x^2*(-1/(c^5*d^4))^(1/6) + 24*sqrt(3)*(1/432)^(5/6)*(c^4*d^4*x^3 + 4*c^5*d^3)*(-
1/(c^5*d^4))^(5/6))*sqrt(d*x^3 + c) + (2*sqrt(3)*(1/2)^(1/3)*(c^2*d^2*x^3 + c^3*d)*(-1/(c^5*d^4))^(1/3) + sqrt
(3)*(d*x^4 + c*x) + 3*(sqrt(3)*sqrt(1/3)*c^3*d^2*x*sqrt(-1/(c^5*d^4)) + 2*sqrt(3)*(1/432)^(1/6)*c*d*x^2*(-1/(c
^5*d^4))^(1/6) - 24*sqrt(3)*(1/432)^(5/6)*(c^4*d^4*x^3 - 2*c^5*d^3)*(-1/(c^5*d^4))^(5/6))*sqrt(d*x^3 + c))*sqr
t((d^3*x^9 + 60*c*d^2*x^6 - 32*c^3 - 24*(1/2)^(2/3)*(c^4*d^5*x^7 + 5*c^5*d^4*x^4 + 4*c^6*d^3*x)*(-1/(c^5*d^4))
^(2/3) + 12*(1/2)^(1/3)*(c^2*d^4*x^8 - 7*c^3*d^3*x^5 - 8*c^4*d^2*x^2)*(-1/(c^5*d^4))^(1/3) + 12*(648*(1/432)^(
5/6)*c^5*d^5*x^5*(-1/(c^5*d^4))^(5/6) - sqrt(1/3)*(c^3*d^4*x^6 - 16*c^4*d^3*x^3 - 8*c^5*d^2)*sqrt(-1/(c^5*d^4)
) - (1/432)^(1/6)*(c*d^3*x^7 + 2*c^2*d^2*x^4 - 8*c^3*d*x)*(-1/(c^5*d^4))^(1/6))*sqrt(d*x^3 + c))/(d^3*x^9 + 12
*c*d^2*x^6 + 48*c^2*d*x^3 + 64*c^3)))/(d*x^4 + c*x)) + 1/9*sqrt(3)*(1/432)^(1/6)*(-1/(c^5*d^4))^(1/6)*arctan(-
1/3*(3*(sqrt(3)*sqrt(1/3)*c^3*d^2*x*sqrt(-1/(c^5*d^4)) + 2*sqrt(3)*(1/432)^(1/6)*c*d*x^2*(-1/(c^5*d^4))^(1/6)
+ 24*sqrt(3)*(1/432)^(5/6)*(c^4*d^4*x^3 + 4*c^5*d^3)*(-1/(c^5*d^4))^(5/6))*sqrt(d*x^3 + c) - (2*sqrt(3)*(1/2)^
(1/3)*(c^2*d^2*x^3 + c^3*d)*(-1/(c^5*d^4))^(1/3) + sqrt(3)*(d*x^4 + c*x) - 3*(sqrt(3)*sqrt(1/3)*c^3*d^2*x*sqrt
(-1/(c^5*d^4)) + 2*sqrt(3)*(1/432)^(1/6)*c*d*x^2*(-1/(c^5*d^4))^(1/6) - 24*sqrt(3)*(1/432)^(5/6)*(c^4*d^4*x^3
- 2*c^5*d^3)*(-1/(c^5*d^4))^(5/6))*sqrt(d*x^3 + c))*sqrt((d^3*x^9 + 60*c*d^2*x^6 - 32*c^3 - 24*(1/2)^(2/3)*(c^
4*d^5*x^7 + 5*c^5*d^4*x^4 + 4*c^6*d^3*x)*(-1/(c^5*d^4))^(2/3) + 12*(1/2)^(1/3)*(c^2*d^4*x^8 - 7*c^3*d^3*x^5 -
8*c^4*d^2*x^2)*(-1/(c^5*d^4))^(1/3) - 12*(648*(1/432)^(5/6)*c^5*d^5*x^5*(-1/(c^5*d^4))^(5/6) - sqrt(1/3)*(c^3*
d^4*x^6 - 16*c^4*d^3*x^3 - 8*c^5*d^2)*sqrt(-1/(c^5*d^4)) - (1/432)^(1/6)*(c*d^3*x^7 + 2*c^2*d^2*x^4 - 8*c^3*d*
x)*(-1/(c^5*d^4))^(1/6))*sqrt(d*x^3 + c))/(d^3*x^9 + 12*c*d^2*x^6 + 48*c^2*d*x^3 + 64*c^3)))/(d*x^4 + c*x)) -
1/36*(1/432)^(1/6)*(-1/(c^5*d^4))^(1/6)*log((d^3*x^9 + 60*c*d^2*x^6 - 32*c^3 - 24*(1/2)^(2/3)*(c^4*d^5*x^7 + 5
*c^5*d^4*x^4 + 4*c^6*d^3*x)*(-1/(c^5*d^4))^(2/3) + 12*(1/2)^(1/3)*(c^2*d^4*x^8 - 7*c^3*d^3*x^5 - 8*c^4*d^2*x^2
)*(-1/(c^5*d^4))^(1/3) + 12*(648*(1/432)^(5/6)*c^5*d^5*x^5*(-1/(c^5*d^4))^(5/6) - sqrt(1/3)*(c^3*d^4*x^6 - 16*
c^4*d^3*x^3 - 8*c^5*d^2)*sqrt(-1/(c^5*d^4)) - (1/432)^(1/6)*(c*d^3*x^7 + 2*c^2*d^2*x^4 - 8*c^3*d*x)*(-1/(c^5*d
^4))^(1/6))*sqrt(d*x^3 + c))/(d^3*x^9 + 12*c*d^2*x^6 + 48*c^2*d*x^3 + 64*c^3)) + 1/36*(1/432)^(1/6)*(-1/(c^5*d
^4))^(1/6)*log((d^3*x^9 + 60*c*d^2*x^6 - 32*c^3 - 24*(1/2)^(2/3)*(c^4*d^5*x^7 + 5*c^5*d^4*x^4 + 4*c^6*d^3*x)*(
-1/(c^5*d^4))^(2/3) + 12*(1/2)^(1/3)*(c^2*d^4*x^8 - 7*c^3*d^3*x^5 - 8*c^4*d^2*x^2)*(-1/(c^5*d^4))^(1/3) - 12*(
648*(1/432)^(5/6)*c^5*d^5*x^5*(-1/(c^5*d^4))^(5/6) - sqrt(1/3)*(c^3*d^4*x^6 - 16*c^4*d^3*x^3 - 8*c^5*d^2)*sqrt
(-1/(c^5*d^4)) - (1/432)^(1/6)*(c*d^3*x^7 + 2*c^2*d^2*x^4 - 8*c^3*d*x)*(-1/(c^5*d^4))^(1/6))*sqrt(d*x^3 + c))/
(d^3*x^9 + 12*c*d^2*x^6 + 48*c^2*d*x^3 + 64*c^3)) + 1/18*(1/432)^(1/6)*(-1/(c^5*d^4))^(1/6)*log((d^3*x^9 - 66*
c*d^2*x^6 - 72*c^2*d*x^3 - 32*c^3 + 48*(1/2)^(2/3)*(c^4*d^5*x^7 - c^5*d^4*x^4 - 2*c^6*d^3*x)*(-1/(c^5*d^4))^(2
/3) + 12*(1/2)^(1/3)*(c^2*d^4*x^8 - 7*c^3*d^3*x^5 - 8*c^4*d^2*x^2)*(-1/(c^5*d^4))^(1/3) + 6*(1296*(1/432)^(5/6
)*c^5*d^5*x^5*(-1/(c^5*d^4))^(5/6) + sqrt(1/3)*(5*c^3*d^4*x^6 - 20*c^4*d^3*x^3 - 16*c^5*d^2)*sqrt(-1/(c^5*d^4)
) + 2*(1/432)^(1/6)*(c*d^3*x^7 - 16*c^2*d^2*x^4 - 8*c^3*d*x)*(-1/(c^5*d^4))^(1/6))*sqrt(d*x^3 + c))/(d^3*x^9 +
12*c*d^2*x^6 + 48*c^2*d*x^3 + 64*c^3)) - 1/18*(1/432)^(1/6)*(-1/(c^5*d^4))^(1/6)*log((d^3*x^9 - 66*c*d^2*x^6
- 72*c^2*d*x^3 - 32*c^3 + 48*(1/2)^(2/3)*(c^4*d^5*x^7 - c^5*d^4*x^4 - 2*c^6*d^3*x)*(-1/(c^5*d^4))^(2/3) + 12*(
1/2)^(1/3)*(c^2*d^4*x^8 - 7*c^3*d^3*x^5 - 8*c^4*d^2*x^2)*(-1/(c^5*d^4))^(1/3) - 6*(1296*(1/432)^(5/6)*c^5*d^5*
x^5*(-1/(c^5*d^4))^(5/6) + sqrt(1/3)*(5*c^3*d^4*x^6 - 20*c^4*d^3*x^3 - 16*c^5*d^2)*sqrt(-1/(c^5*d^4)) + 2*(1/4
32)^(1/6)*(c*d^3*x^7 - 16*c^2*d^2*x^4 - 8*c^3*d*x)*(-1/(c^5*d^4))^(1/6))*sqrt(d*x^3 + c))/(d^3*x^9 + 12*c*d^2*
x^6 + 48*c^2*d*x^3 + 64*c^3))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{c + d x^{3}} \left (4 c + d x^{3}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\left (d x^{3} + 4 \, c\right )} \sqrt{d x^{3} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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