3.91 \(\int \frac {e+f x}{(c+d x) \sqrt {c^3-8 d^3 x^3}} \, dx\)

Optimal. Leaf size=221 \[ -\frac {2 (d e-c f) \tanh ^{-1}\left (\frac {(c-2 d x)^2}{3 \sqrt {c} \sqrt {c^3-8 d^3 x^3}}\right )}{9 c^{3/2} d^2}-\frac {\sqrt {2+\sqrt {3}} (c-2 d x) \sqrt {\frac {c^2+2 c d x+4 d^2 x^2}{\left (\left (1+\sqrt {3}\right ) c-2 d x\right )^2}} (c f+2 d e) F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) c-2 d x}{\left (1+\sqrt {3}\right ) c-2 d x}\right )|-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} c d^2 \sqrt {\frac {c (c-2 d x)}{\left (\left (1+\sqrt {3}\right ) c-2 d x\right )^2}} \sqrt {c^3-8 d^3 x^3}} \]

[Out]

-2/9*(-c*f+d*e)*arctanh(1/3*(-2*d*x+c)^2/c^(1/2)/(-8*d^3*x^3+c^3)^(1/2))/c^(3/2)/d^2-1/9*(c*f+2*d*e)*(-2*d*x+c
)*EllipticF((-2*d*x+c*(1-3^(1/2)))/(-2*d*x+c*(1+3^(1/2))),I*3^(1/2)+2*I)*(1/2*6^(1/2)+1/2*2^(1/2))*((4*d^2*x^2
+2*c*d*x+c^2)/(-2*d*x+c*(1+3^(1/2)))^2)^(1/2)*3^(3/4)/c/d^2/(-8*d^3*x^3+c^3)^(1/2)/(c*(-2*d*x+c)/(-2*d*x+c*(1+
3^(1/2)))^2)^(1/2)

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Rubi [A]  time = 0.28, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2139, 218, 2138, 206} \[ -\frac {2 (d e-c f) \tanh ^{-1}\left (\frac {(c-2 d x)^2}{3 \sqrt {c} \sqrt {c^3-8 d^3 x^3}}\right )}{9 c^{3/2} d^2}-\frac {\sqrt {2+\sqrt {3}} (c-2 d x) \sqrt {\frac {c^2+2 c d x+4 d^2 x^2}{\left (\left (1+\sqrt {3}\right ) c-2 d x\right )^2}} (c f+2 d e) F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) c-2 d x}{\left (1+\sqrt {3}\right ) c-2 d x}\right )|-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} c d^2 \sqrt {\frac {c (c-2 d x)}{\left (\left (1+\sqrt {3}\right ) c-2 d x\right )^2}} \sqrt {c^3-8 d^3 x^3}} \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)/((c + d*x)*Sqrt[c^3 - 8*d^3*x^3]),x]

[Out]

(-2*(d*e - c*f)*ArcTanh[(c - 2*d*x)^2/(3*Sqrt[c]*Sqrt[c^3 - 8*d^3*x^3])])/(9*c^(3/2)*d^2) - (Sqrt[2 + Sqrt[3]]
*(2*d*e + c*f)*(c - 2*d*x)*Sqrt[(c^2 + 2*c*d*x + 4*d^2*x^2)/((1 + Sqrt[3])*c - 2*d*x)^2]*EllipticF[ArcSin[((1
- Sqrt[3])*c - 2*d*x)/((1 + Sqrt[3])*c - 2*d*x)], -7 - 4*Sqrt[3]])/(3*3^(1/4)*c*d^2*Sqrt[(c*(c - 2*d*x))/((1 +
Sqrt[3])*c - 2*d*x)^2]*Sqrt[c^3 - 8*d^3*x^3])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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 2138

Int[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Dist[(-2*e)/d, Subst[Int
[1/(9 - a*x^2), x], x, (1 + (f*x)/e)^2/Sqrt[a + b*x^3]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[d*e - c*f,
0] && EqQ[b*c^3 + 8*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 {e+f x}{(c+d x) \sqrt {c^3-8 d^3 x^3}} \, dx &=\frac {(d e-c f) \int \frac {c-2 d x}{(c+d x) \sqrt {c^3-8 d^3 x^3}} \, dx}{3 c d}+\frac {(2 d e+c f) \int \frac {1}{\sqrt {c^3-8 d^3 x^3}} \, dx}{3 c d}\\ &=-\frac {\sqrt {2+\sqrt {3}} (2 d e+c f) (c-2 d x) \sqrt {\frac {c^2+2 c d x+4 d^2 x^2}{\left (\left (1+\sqrt {3}\right ) c-2 d x\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) c-2 d x}{\left (1+\sqrt {3}\right ) c-2 d x}\right )|-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} c d^2 \sqrt {\frac {c (c-2 d x)}{\left (\left (1+\sqrt {3}\right ) c-2 d x\right )^2}} \sqrt {c^3-8 d^3 x^3}}-\frac {(2 (d e-c f)) \operatorname {Subst}\left (\int \frac {1}{9-c^3 x^2} \, dx,x,\frac {\left (1-\frac {2 d x}{c}\right )^2}{\sqrt {c^3-8 d^3 x^3}}\right )}{3 d^2}\\ &=-\frac {2 (d e-c f) \tanh ^{-1}\left (\frac {(c-2 d x)^2}{3 \sqrt {c} \sqrt {c^3-8 d^3 x^3}}\right )}{9 c^{3/2} d^2}-\frac {\sqrt {2+\sqrt {3}} (2 d e+c f) (c-2 d x) \sqrt {\frac {c^2+2 c d x+4 d^2 x^2}{\left (\left (1+\sqrt {3}\right ) c-2 d x\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) c-2 d x}{\left (1+\sqrt {3}\right ) c-2 d x}\right )|-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} c d^2 \sqrt {\frac {c (c-2 d x)}{\left (\left (1+\sqrt {3}\right ) c-2 d x\right )^2}} \sqrt {c^3-8 d^3 x^3}}\\ \end {align*}

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Mathematica [C]  time = 1.18, size = 384, normalized size = 1.74 \[ -\frac {i \sqrt {\frac {c-2 d x}{\left (1+\sqrt [3]{-1}\right ) c}} \left (4 \sqrt {2} \sqrt {\frac {i c+\sqrt {3} d x+i d x}{-\sqrt {3} c+3 i c}} \sqrt {\frac {c^2+2 c d x+4 d^2 x^2}{c^2}} (d e-c f) \Pi \left (\frac {2 \sqrt {3}}{3 i+\sqrt {3}};\sin ^{-1}\left (\sqrt {2} \sqrt {\frac {i c+\sqrt {3} d x+i d x}{3 i c-\sqrt {3} c}}\right )|\frac {1}{2} \left (1+i \sqrt {3}\right )\right )+f \sqrt {\frac {\left (\sqrt {3}-i\right ) c+2 \left (\sqrt {3}+i\right ) d x}{\left (\sqrt {3}-3 i\right ) c}} \left (\left (\sqrt {3}-3 i\right ) c-2 \left (\sqrt {3}+3 i\right ) d x\right ) F\left (\sin ^{-1}\left (\sqrt {2} \sqrt {\frac {i c+\sqrt {3} d x+i d x}{3 i c-\sqrt {3} c}}\right )|\frac {1}{2} \left (1+i \sqrt {3}\right )\right )\right )}{2 \left (\sqrt [3]{-1}-2\right ) d^2 \sqrt {\frac {c-2 (-1)^{2/3} d x}{\left (1+\sqrt [3]{-1}\right ) c}} \sqrt {c^3-8 d^3 x^3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(e + f*x)/((c + d*x)*Sqrt[c^3 - 8*d^3*x^3]),x]

[Out]

((-1/2*I)*Sqrt[(c - 2*d*x)/((1 + (-1)^(1/3))*c)]*(f*Sqrt[((-I + Sqrt[3])*c + 2*(I + Sqrt[3])*d*x)/((-3*I + Sqr
t[3])*c)]*((-3*I + Sqrt[3])*c - 2*(3*I + Sqrt[3])*d*x)*EllipticF[ArcSin[Sqrt[2]*Sqrt[(I*c + I*d*x + Sqrt[3]*d*
x)/((3*I)*c - Sqrt[3]*c)]], (1 + I*Sqrt[3])/2] + 4*Sqrt[2]*(d*e - c*f)*Sqrt[(I*c + I*d*x + Sqrt[3]*d*x)/((3*I)
*c - Sqrt[3]*c)]*Sqrt[(c^2 + 2*c*d*x + 4*d^2*x^2)/c^2]*EllipticPi[(2*Sqrt[3])/(3*I + Sqrt[3]), ArcSin[Sqrt[2]*
Sqrt[(I*c + I*d*x + Sqrt[3]*d*x)/((3*I)*c - Sqrt[3]*c)]], (1 + I*Sqrt[3])/2]))/((-2 + (-1)^(1/3))*d^2*Sqrt[(c
- 2*(-1)^(2/3)*d*x)/((1 + (-1)^(1/3))*c)]*Sqrt[c^3 - 8*d^3*x^3])

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fricas [F]  time = 1.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-8 \, d^{3} x^{3} + c^{3}} {\left (f x + e\right )}}{8 \, d^{4} x^{4} + 8 \, c d^{3} x^{3} - c^{3} d x - c^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)/(d*x+c)/(-8*d^3*x^3+c^3)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-8*d^3*x^3 + c^3)*(f*x + e)/(8*d^4*x^4 + 8*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 \[ \int \frac {f x + e}{\sqrt {-8 \, d^{3} x^{3} + c^{3}} {\left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)/(d*x+c)/(-8*d^3*x^3+c^3)^(1/2),x, algorithm="giac")

[Out]

integrate((f*x + e)/(sqrt(-8*d^3*x^3 + c^3)*(d*x + c)), x)

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maple [B]  time = 0.01, size = 661, normalized size = 2.99 \[ \frac {2 \left (\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) c}{2 d}-\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}\right ) \sqrt {\frac {x -\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}}{\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) c}{2 d}-\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}}}\, \sqrt {\frac {x -\frac {c}{2 d}}{\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}-\frac {c}{2 d}}}\, \sqrt {\frac {x -\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) c}{2 d}}{\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}-\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) c}{2 d}}}\, f \EllipticF \left (\sqrt {\frac {x -\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}}{\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) c}{2 d}-\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}}}, \sqrt {\frac {\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}-\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) c}{2 d}}{\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}-\frac {c}{2 d}}}\right )}{\sqrt {-8 d^{3} x^{3}+c^{3}}\, d}+\frac {2 \left (-c f +d e \right ) \left (\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) c}{2 d}-\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}\right ) \sqrt {\frac {x -\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}}{\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) c}{2 d}-\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}}}\, \sqrt {\frac {x -\frac {c}{2 d}}{\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}-\frac {c}{2 d}}}\, \sqrt {\frac {x -\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) c}{2 d}}{\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}-\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) c}{2 d}}}\, \EllipticPi \left (\sqrt {\frac {x -\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}}{\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) c}{2 d}-\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}}}, \frac {\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}-\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) c}{2 d}}{\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}+\frac {c}{d}}, \sqrt {\frac {\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}-\frac {\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) c}{2 d}}{\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}-\frac {c}{2 d}}}\right )}{\sqrt {-8 d^{3} x^{3}+c^{3}}\, \left (\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) c}{2 d}+\frac {c}{d}\right ) d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)/(d*x+c)/(-8*d^3*x^3+c^3)^(1/2),x)

[Out]

2*f/d*(1/2*(-1/2+1/2*I*3^(1/2))*c/d-1/2*(-1/2-1/2*I*3^(1/2))*c/d)*((x-1/2*(-1/2-1/2*I*3^(1/2))*c/d)/(1/2*(-1/2
+1/2*I*3^(1/2))*c/d-1/2*(-1/2-1/2*I*3^(1/2))*c/d))^(1/2)*((x-1/2*c/d)/(1/2*(-1/2-1/2*I*3^(1/2))*c/d-1/2*c/d))^
(1/2)*((x-1/2*(-1/2+1/2*I*3^(1/2))*c/d)/(1/2*(-1/2-1/2*I*3^(1/2))*c/d-1/2*(-1/2+1/2*I*3^(1/2))*c/d))^(1/2)/(-8
*d^3*x^3+c^3)^(1/2)*EllipticF(((x-1/2*(-1/2-1/2*I*3^(1/2))*c/d)/(1/2*(-1/2+1/2*I*3^(1/2))*c/d-1/2*(-1/2-1/2*I*
3^(1/2))*c/d))^(1/2),((1/2*(-1/2-1/2*I*3^(1/2))*c/d-1/2*(-1/2+1/2*I*3^(1/2))*c/d)/(1/2*(-1/2-1/2*I*3^(1/2))*c/
d-1/2*c/d))^(1/2))+2*(-c*f+d*e)/d^2*(1/2*(-1/2+1/2*I*3^(1/2))*c/d-1/2*(-1/2-1/2*I*3^(1/2))*c/d)*((x-1/2*(-1/2-
1/2*I*3^(1/2))*c/d)/(1/2*(-1/2+1/2*I*3^(1/2))*c/d-1/2*(-1/2-1/2*I*3^(1/2))*c/d))^(1/2)*((x-1/2*c/d)/(1/2*(-1/2
-1/2*I*3^(1/2))*c/d-1/2*c/d))^(1/2)*((x-1/2*(-1/2+1/2*I*3^(1/2))*c/d)/(1/2*(-1/2-1/2*I*3^(1/2))*c/d-1/2*(-1/2+
1/2*I*3^(1/2))*c/d))^(1/2)/(-8*d^3*x^3+c^3)^(1/2)/(1/2*(-1/2-1/2*I*3^(1/2))*c/d+c/d)*EllipticPi(((x-1/2*(-1/2-
1/2*I*3^(1/2))*c/d)/(1/2*(-1/2+1/2*I*3^(1/2))*c/d-1/2*(-1/2-1/2*I*3^(1/2))*c/d))^(1/2),(1/2*(-1/2-1/2*I*3^(1/2
))*c/d-1/2*(-1/2+1/2*I*3^(1/2))*c/d)/(1/2*(-1/2-1/2*I*3^(1/2))*c/d+c/d),((1/2*(-1/2-1/2*I*3^(1/2))*c/d-1/2*(-1
/2+1/2*I*3^(1/2))*c/d)/(1/2*(-1/2-1/2*I*3^(1/2))*c/d-1/2*c/d))^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f x + e}{\sqrt {-8 \, d^{3} x^{3} + c^{3}} {\left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)/(d*x+c)/(-8*d^3*x^3+c^3)^(1/2),x, algorithm="maxima")

[Out]

integrate((f*x + e)/(sqrt(-8*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 \[ \int \frac {e+f\,x}{\sqrt {c^3-8\,d^3\,x^3}\,\left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e + f*x)/((c^3 - 8*d^3*x^3)^(1/2)*(c + d*x)),x)

[Out]

int((e + f*x)/((c^3 - 8*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 \[ \int \frac {e + f x}{\sqrt {- \left (- c + 2 d x\right ) \left (c^{2} + 2 c d x + 4 d^{2} x^{2}\right )} \left (c + d x\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)/(d*x+c)/(-8*d**3*x**3+c**3)**(1/2),x)

[Out]

Integral((e + f*x)/(sqrt(-(-c + 2*d*x)*(c**2 + 2*c*d*x + 4*d**2*x**2))*(c + d*x)), x)

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