3.1.1 \(\int \frac {1}{(c+d x) \sqrt [3]{-c^3+d^3 x^3}} \, dx\)

Optimal. Leaf size=139 \[ -\frac {3 \log \left (2^{2/3} d \sqrt [3]{d^3 x^3-c^3}+d (c-d x)\right )}{4 \sqrt [3]{2} c d}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {1-\frac {\sqrt [3]{2} (c-d x)}{\sqrt [3]{d^3 x^3-c^3}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} c d}+\frac {\log \left ((c-d x) (c+d x)^2\right )}{4 \sqrt [3]{2} c d} \]

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Rubi [A]  time = 0.07, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2148} \begin {gather*} -\frac {3 \log \left (2^{2/3} d \sqrt [3]{d^3 x^3-c^3}+d (c-d x)\right )}{4 \sqrt [3]{2} c d}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {1-\frac {\sqrt [3]{2} (c-d x)}{\sqrt [3]{d^3 x^3-c^3}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} c d}+\frac {\log \left ((c-d x) (c+d x)^2\right )}{4 \sqrt [3]{2} c d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3]*ArcTan[(1 - (2^(1/3)*(c - d*x))/(-c^3 + d^3*x^3)^(1/3))/Sqrt[3]])/(2*2^(1/3)*c*d) + Log[(c - d*x)*(c
+ d*x)^2]/(4*2^(1/3)*c*d) - (3*Log[d*(c - d*x) + 2^(2/3)*d*(-c^3 + d^3*x^3)^(1/3)])/(4*2^(1/3)*c*d)

Rule 2148

Int[1/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^3)^(1/3)), x_Symbol] :> Simp[(Sqrt[3]*ArcTan[(1 - (2^(1/3)*Rt[b,
3]*(c - d*x))/(d*(a + b*x^3)^(1/3)))/Sqrt[3]])/(2^(4/3)*Rt[b, 3]*c), x] + (Simp[Log[(c + d*x)^2*(c - d*x)]/(2
^(7/3)*Rt[b, 3]*c), x] - Simp[(3*Log[Rt[b, 3]*(c - d*x) + 2^(2/3)*d*(a + b*x^3)^(1/3)])/(2^(7/3)*Rt[b, 3]*c),
x]) /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^3 + a*d^3, 0]

Rubi steps

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

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Mathematica [F]  time = 0.08, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(c+d x) \sqrt [3]{-c^3+d^3 x^3}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [C]  time = 2.89, size = 385, normalized size = 2.77 \begin {gather*} \frac {(-1)^{5/6} \sqrt {3} \tanh ^{-1}\left (\frac {\frac {i \sqrt [3]{d^3 x^3-c^3}}{\sqrt {3}}+\frac {\sqrt {3} c+i c}{2^{2/3} \sqrt {3}}+\frac {\left (-\sqrt {3} d-i d\right ) x}{2^{2/3} \sqrt {3}}}{\sqrt [3]{d^3 x^3-c^3}}\right )}{2 \sqrt [3]{2} c d}+\frac {\sqrt [3]{-\frac {1}{2}} \log \left (i \sqrt {3} c^{3/2} \sqrt {d}-c^{3/2} \sqrt {d}+2\ 2^{2/3} \sqrt {c} \sqrt {d} \sqrt [3]{d^3 x^3-c^3}+\sqrt {c} d^{3/2} \left (x-i \sqrt {3} x\right )\right )}{2 c d}-\frac {\sqrt [3]{-\frac {1}{2}} \log \left (4 \sqrt [3]{2} c d \left (d^3 x^3-c^3\right )^{2/3}-i \sqrt {3} c^3 d-c^3 d+2 i \sqrt {3} c^2 d^2 x+2 c^2 d^2 x+\left (2 (-2)^{2/3} c d^2 x-2 (-2)^{2/3} c^2 d\right ) \sqrt [3]{d^3 x^3-c^3}-i \sqrt {3} c d^3 x^2-c d^3 x^2\right )}{4 c d} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((c + d*x)*(-c^3 + d^3*x^3)^(1/3)),x]

[Out]

((-1)^(5/6)*Sqrt[3]*ArcTanh[((I*c + Sqrt[3]*c)/(2^(2/3)*Sqrt[3]) + (((-I)*d - Sqrt[3]*d)*x)/(2^(2/3)*Sqrt[3])
+ (I*(-c^3 + d^3*x^3)^(1/3))/Sqrt[3])/(-c^3 + d^3*x^3)^(1/3)])/(2*2^(1/3)*c*d) + ((-1/2)^(1/3)*Log[-(c^(3/2)*S
qrt[d]) + I*Sqrt[3]*c^(3/2)*Sqrt[d] + Sqrt[c]*d^(3/2)*(x - I*Sqrt[3]*x) + 2*2^(2/3)*Sqrt[c]*Sqrt[d]*(-c^3 + d^
3*x^3)^(1/3)])/(2*c*d) - ((-1/2)^(1/3)*Log[-(c^3*d) - I*Sqrt[3]*c^3*d + 2*c^2*d^2*x + (2*I)*Sqrt[3]*c^2*d^2*x
- c*d^3*x^2 - I*Sqrt[3]*c*d^3*x^2 + (-2*(-2)^(2/3)*c^2*d + 2*(-2)^(2/3)*c*d^2*x)*(-c^3 + d^3*x^3)^(1/3) + 4*2^
(1/3)*c*d*(-c^3 + d^3*x^3)^(2/3)])/(4*c*d)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (d^{3} x^{3} - c^{3}\right )}^{\frac {1}{3}} {\left (d x + c\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (d x +c \right ) \left (d^{3} x^{3}-c^{3}\right )^{\frac {1}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (d^{3} x^{3} - c^{3}\right )}^{\frac {1}{3}} {\left (d x + c\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (d^3\,x^3-c^3\right )}^{1/3}\,\left (c+d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [3]{\left (- c + d x\right ) \left (c^{2} + c d x + d^{2} x^{2}\right )} \left (c + d x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(((-c + d*x)*(c**2 + c*d*x + d**2*x**2))**(1/3)*(c + d*x)), x)

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