3.1.19 \(\int \frac {1}{(c+d x) \sqrt [3]{2 c^3+d^3 x^3}} \, dx\) [19]

Optimal. Leaf size=186 \[ \frac {\tan ^{-1}\left (\frac {1+\frac {2 d x}{\sqrt [3]{2 c^3+d^3 x^3}}}{\sqrt {3}}\right )}{2 \sqrt {3} c d}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 (2 c+d x)}{\sqrt [3]{2 c^3+d^3 x^3}}}{\sqrt {3}}\right )}{2 c d}-\frac {\log (c+d x)}{2 c d}-\frac {\log \left (-d x+\sqrt [3]{2 c^3+d^3 x^3}\right )}{4 c d}+\frac {3 \log \left (d (2 c+d x)-d \sqrt [3]{2 c^3+d^3 x^3}\right )}{4 c d} \]

[Out]

-1/2*ln(d*x+c)/c/d-1/4*ln(-d*x+(d^3*x^3+2*c^3)^(1/3))/c/d+3/4*ln(d*(d*x+2*c)-d*(d^3*x^3+2*c^3)^(1/3))/c/d+1/6*
arctan(1/3*(1+2*d*x/(d^3*x^3+2*c^3)^(1/3))*3^(1/2))/c/d*3^(1/2)-1/2*arctan(1/3*(1+2*(d*x+2*c)/(d^3*x^3+2*c^3)^
(1/3))*3^(1/2))*3^(1/2)/c/d

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Rubi [A]
time = 0.13, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2175, 245, 2176} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\frac {2 d x}{\sqrt [3]{2 c^3+d^3 x^3}}+1}{\sqrt {3}}\right )}{2 \sqrt {3} c d}-\frac {\sqrt {3} \text {ArcTan}\left (\frac {\frac {2 (2 c+d x)}{\sqrt [3]{2 c^3+d^3 x^3}}+1}{\sqrt {3}}\right )}{2 c d}-\frac {\log \left (\sqrt [3]{2 c^3+d^3 x^3}-d x\right )}{4 c d}+\frac {3 \log \left (d (2 c+d x)-d \sqrt [3]{2 c^3+d^3 x^3}\right )}{4 c d}-\frac {\log (c+d x)}{2 c d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((c + d*x)*(2*c^3 + d^3*x^3)^(1/3)),x]

[Out]

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

Rule 245

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rule 2175

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

Rule 2176

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

Rubi steps

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

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Mathematica [F]
time = 7.22, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(c+d x) \sqrt [3]{2 c^3+d^3 x^3}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[1/((c + d*x)*(2*c^3 + d^3*x^3)^(1/3)),x]

[Out]

Integrate[1/((c + d*x)*(2*c^3 + d^3*x^3)^(1/3)), x]

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (d x +c \right ) \left (d^{3} x^{3}+2 c^{3}\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [F(-1)] Timed out
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+2*c^3)^(1/3),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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