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

Optimal. Leaf size=246 \[ \frac {\sqrt [3]{2} \sqrt {2+\sqrt {3}} \left (c+2^{2/3} d x\right ) \sqrt {\frac {c^2-2^{2/3} c d x+2 \sqrt [3]{2} d^2 x^2}{\left (\left (1+\sqrt {3}\right ) c+2^{2/3} d x\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) c+2^{2/3} d x}{\left (1+\sqrt {3}\right ) c+2^{2/3} d x}\right )|-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} d^2 \sqrt {\frac {c \left (c+2^{2/3} d x\right )}{\left (\left (1+\sqrt {3}\right ) c+2^{2/3} d x\right )^2}} \sqrt {c^3+4 d^3 x^3}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {3} \sqrt {c} (c+2 d x)}{\sqrt {c^3+4 d^3 x^3}}\right )}{3 \sqrt {3} \sqrt {c} d^2} \]

[Out]

-2/9*arctan((2*d*x+c)*3^(1/2)*c^(1/2)/(4*d^3*x^3+c^3)^(1/2))/d^2*3^(1/2)/c^(1/2)+1/9*2^(1/3)*(c+2^(2/3)*d*x)*E
llipticF((2^(2/3)*d*x+c*(1-3^(1/2)))/(2^(2/3)*d*x+c*(1+3^(1/2))),I*3^(1/2)+2*I)*(1/2*6^(1/2)+1/2*2^(1/2))*((c^
2-2^(2/3)*c*d*x+2*2^(1/3)*d^2*x^2)/(2^(2/3)*d*x+c*(1+3^(1/2)))^2)^(1/2)*3^(3/4)/d^2/(4*d^3*x^3+c^3)^(1/2)/(c*(
c+2^(2/3)*d*x)/(2^(2/3)*d*x+c*(1+3^(1/2)))^2)^(1/2)

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Rubi [A]  time = 0.27, antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2139, 218, 2137, 203} \[ \frac {\sqrt [3]{2} \sqrt {2+\sqrt {3}} \left (c+2^{2/3} d x\right ) \sqrt {\frac {c^2-2^{2/3} c d x+2 \sqrt [3]{2} d^2 x^2}{\left (\left (1+\sqrt {3}\right ) c+2^{2/3} d x\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) c+2^{2/3} d x}{\left (1+\sqrt {3}\right ) c+2^{2/3} d x}\right )|-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} d^2 \sqrt {\frac {c \left (c+2^{2/3} d x\right )}{\left (\left (1+\sqrt {3}\right ) c+2^{2/3} d x\right )^2}} \sqrt {c^3+4 d^3 x^3}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {3} \sqrt {c} (c+2 d x)}{\sqrt {c^3+4 d^3 x^3}}\right )}{3 \sqrt {3} \sqrt {c} d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcTan[(Sqrt[3]*Sqrt[c]*(c + 2*d*x))/Sqrt[c^3 + 4*d^3*x^3]])/(3*Sqrt[3]*Sqrt[c]*d^2) + (2^(1/3)*Sqrt[2 + S
qrt[3]]*(c + 2^(2/3)*d*x)*Sqrt[(c^2 - 2^(2/3)*c*d*x + 2*2^(1/3)*d^2*x^2)/((1 + Sqrt[3])*c + 2^(2/3)*d*x)^2]*El
lipticF[ArcSin[((1 - Sqrt[3])*c + 2^(2/3)*d*x)/((1 + Sqrt[3])*c + 2^(2/3)*d*x)], -7 - 4*Sqrt[3]])/(3*3^(1/4)*d
^2*Sqrt[(c*(c + 2^(2/3)*d*x))/((1 + Sqrt[3])*c + 2^(2/3)*d*x)^2]*Sqrt[c^3 + 4*d^3*x^3])

Rule 203

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

Rule 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 2137

Int[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Dist[(2*e)/d, Subst[Int[
1/(1 + 3*a*x^2), x], x, (1 + (2*d*x)/c)/Sqrt[a + b*x^3]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[d*e - c*f,
0] && EqQ[b*c^3 - 4*a*d^3, 0] && EqQ[2*d*e + c*f, 0]

Rule 2139

Int[((e_.) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Dist[(2*d*e + c*f)/(3*c
*d), Int[1/Sqrt[a + b*x^3], x], x] + Dist[(d*e - c*f)/(3*c*d), Int[(c - 2*d*x)/((c + d*x)*Sqrt[a + b*x^3]), x]
, x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] && (EqQ[b*c^3 - 4*a*d^3, 0] || EqQ[b*c^3 + 8*a*d^3,
0]) && NeQ[2*d*e + c*f, 0]

Rubi steps

\begin {align*} \int \frac {x}{(c+d x) \sqrt {c^3+4 d^3 x^3}} \, dx &=\frac {\int \frac {1}{\sqrt {c^3+4 d^3 x^3}} \, dx}{3 d}-\frac {\int \frac {c-2 d x}{(c+d x) \sqrt {c^3+4 d^3 x^3}} \, dx}{3 d}\\ &=\frac {\sqrt [3]{2} \sqrt {2+\sqrt {3}} \left (c+2^{2/3} d x\right ) \sqrt {\frac {c^2-2^{2/3} c d x+2 \sqrt [3]{2} d^2 x^2}{\left (\left (1+\sqrt {3}\right ) c+2^{2/3} d x\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) c+2^{2/3} d x}{\left (1+\sqrt {3}\right ) c+2^{2/3} d x}\right )|-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} d^2 \sqrt {\frac {c \left (c+2^{2/3} d x\right )}{\left (\left (1+\sqrt {3}\right ) c+2^{2/3} d x\right )^2}} \sqrt {c^3+4 d^3 x^3}}-\frac {(2 c) \operatorname {Subst}\left (\int \frac {1}{1+3 c^3 x^2} \, dx,x,\frac {1+\frac {2 d x}{c}}{\sqrt {c^3+4 d^3 x^3}}\right )}{3 d^2}\\ &=-\frac {2 \tan ^{-1}\left (\frac {\sqrt {3} \sqrt {c} (c+2 d x)}{\sqrt {c^3+4 d^3 x^3}}\right )}{3 \sqrt {3} \sqrt {c} d^2}+\frac {\sqrt [3]{2} \sqrt {2+\sqrt {3}} \left (c+2^{2/3} d x\right ) \sqrt {\frac {c^2-2^{2/3} c d x+2 \sqrt [3]{2} d^2 x^2}{\left (\left (1+\sqrt {3}\right ) c+2^{2/3} d x\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) c+2^{2/3} d x}{\left (1+\sqrt {3}\right ) c+2^{2/3} d x}\right )|-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} d^2 \sqrt {\frac {c \left (c+2^{2/3} d x\right )}{\left (\left (1+\sqrt {3}\right ) c+2^{2/3} d x\right )^2}} \sqrt {c^3+4 d^3 x^3}}\\ \end {align*}

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Mathematica [C]  time = 1.07, size = 372, normalized size = 1.51 \[ \frac {\sqrt [6]{2} \sqrt {\frac {\sqrt [3]{2} c+2 d x}{\left (1+\sqrt [3]{-1}\right ) c}} \left (-\sqrt {\frac {\sqrt [3]{-2} c-2 (-1)^{2/3} d x}{\left (1+\sqrt [3]{-1}\right ) c}} \left (\sqrt [3]{-1} \left (2+\sqrt [3]{-2}\right ) c-2 \left (\sqrt [3]{-1}+2^{2/3}\right ) d x\right ) F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {\sqrt [3]{2} c+2 (-1)^{2/3} d x}{\left (1+\sqrt [3]{-1}\right ) c}}}{\sqrt [6]{2}}\right )|\sqrt [3]{-1}\right )+\frac {\sqrt [3]{-1} 2^{2/3} \left (1+\sqrt [3]{-1}\right ) c \sqrt {\frac {\sqrt [3]{2} c+2 (-1)^{2/3} d x}{\left (1+\sqrt [3]{-1}\right ) c}} \sqrt {\frac {4 d^2 x^2}{c^2}-\frac {2 \sqrt [3]{2} d x}{c}+2^{2/3}} \Pi \left (\frac {i \sqrt [3]{2} \sqrt {3}}{2+\sqrt [3]{-2}};\sin ^{-1}\left (\frac {\sqrt {\frac {\sqrt [3]{2} c+2 (-1)^{2/3} d x}{\left (1+\sqrt [3]{-1}\right ) c}}}{\sqrt [6]{2}}\right )|\sqrt [3]{-1}\right )}{\sqrt {3}}\right )}{\left (2+\sqrt [3]{-2}\right ) d^2 \sqrt {\frac {\sqrt [3]{2} c+2 (-1)^{2/3} d x}{\left (1+\sqrt [3]{-1}\right ) c}} \sqrt {c^3+4 d^3 x^3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2^(1/6)*Sqrt[(2^(1/3)*c + 2*d*x)/((1 + (-1)^(1/3))*c)]*(-(Sqrt[((-2)^(1/3)*c - 2*(-1)^(2/3)*d*x)/((1 + (-1)^(
1/3))*c)]*((-1)^(1/3)*(2 + (-2)^(1/3))*c - 2*((-1)^(1/3) + 2^(2/3))*d*x)*EllipticF[ArcSin[Sqrt[(2^(1/3)*c + 2*
(-1)^(2/3)*d*x)/((1 + (-1)^(1/3))*c)]/2^(1/6)], (-1)^(1/3)]) + ((-1)^(1/3)*2^(2/3)*(1 + (-1)^(1/3))*c*Sqrt[(2^
(1/3)*c + 2*(-1)^(2/3)*d*x)/((1 + (-1)^(1/3))*c)]*Sqrt[2^(2/3) - (2*2^(1/3)*d*x)/c + (4*d^2*x^2)/c^2]*Elliptic
Pi[(I*2^(1/3)*Sqrt[3])/(2 + (-2)^(1/3)), ArcSin[Sqrt[(2^(1/3)*c + 2*(-1)^(2/3)*d*x)/((1 + (-1)^(1/3))*c)]/2^(1
/6)], (-1)^(1/3)])/Sqrt[3]))/((2 + (-2)^(1/3))*d^2*Sqrt[(2^(1/3)*c + 2*(-1)^(2/3)*d*x)/((1 + (-1)^(1/3))*c)]*S
qrt[c^3 + 4*d^3*x^3])

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fricas [F]  time = 1.11, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {4 \, d^{3} x^{3} + c^{3}} x}{4 \, d^{4} x^{4} + 4 \, c d^{3} x^{3} + c^{3} d x + c^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {4 \, d^{3} x^{3} + c^{3}} {\left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 892, normalized size = 3.63 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*((1/4*2^(1/3)-1/4*I*3^(1/2)*2^(1/3))*c/d-(1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d)*((x-(1/4*2^(1/3)+1/4*I*3
^(1/2)*2^(1/3))*c/d)/((1/4*2^(1/3)-1/4*I*3^(1/2)*2^(1/3))*c/d-(1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d))^(1/2)*
((x+1/2*2^(1/3)*c/d)/((1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d+1/2*2^(1/3)*c/d))^(1/2)*((x-(1/4*2^(1/3)-1/4*I*3
^(1/2)*2^(1/3))*c/d)/((1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d-(1/4*2^(1/3)-1/4*I*3^(1/2)*2^(1/3))*c/d))^(1/2)/
(4*d^3*x^3+c^3)^(1/2)*EllipticF(((x-(1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d)/((1/4*2^(1/3)-1/4*I*3^(1/2)*2^(1/
3))*c/d-(1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d))^(1/2),(((1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d-(1/4*2^(1/3)
-1/4*I*3^(1/2)*2^(1/3))*c/d)/((1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d+1/2*2^(1/3)*c/d))^(1/2))-2*c/d^2*((1/4*2
^(1/3)-1/4*I*3^(1/2)*2^(1/3))*c/d-(1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d)*((x-(1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1
/3))*c/d)/((1/4*2^(1/3)-1/4*I*3^(1/2)*2^(1/3))*c/d-(1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d))^(1/2)*((x+1/2*2^(
1/3)*c/d)/((1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d+1/2*2^(1/3)*c/d))^(1/2)*((x-(1/4*2^(1/3)-1/4*I*3^(1/2)*2^(1
/3))*c/d)/((1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d-(1/4*2^(1/3)-1/4*I*3^(1/2)*2^(1/3))*c/d))^(1/2)/(4*d^3*x^3+
c^3)^(1/2)/((1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d+c/d)*EllipticPi(((x-(1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/
d)/((1/4*2^(1/3)-1/4*I*3^(1/2)*2^(1/3))*c/d-(1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d))^(1/2),((1/4*2^(1/3)+1/4*
I*3^(1/2)*2^(1/3))*c/d-(1/4*2^(1/3)-1/4*I*3^(1/2)*2^(1/3))*c/d)/((1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d+c/d),
(((1/4*2^(1/3)+1/4*I*3^(1/2)*2^(1/3))*c/d-(1/4*2^(1/3)-1/4*I*3^(1/2)*2^(1/3))*c/d)/((1/4*2^(1/3)+1/4*I*3^(1/2)
*2^(1/3))*c/d+1/2*2^(1/3)*c/d))^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {4 \, d^{3} x^{3} + c^{3}} {\left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x}{\sqrt {c^3+4\,d^3\,x^3}\,\left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\left (c + d x\right ) \sqrt {c^{3} + 4 d^{3} x^{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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