3.26.19 \(\int \frac {1}{(a+b x) \sqrt [3]{c+d x} \sqrt [3]{b c+a d+2 b d x}} \, dx\)

Optimal. Leaf size=178 \[ -\frac {\log (a+b x)}{2 b^{2/3} (b c-a d)^{2/3}}+\frac {3 \log \left (\frac {b^{2/3} (c+d x)^{2/3}}{\sqrt [3]{b c-a d}}-\sqrt [3]{a d+b c+2 b d x}\right )}{4 b^{2/3} (b c-a d)^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {2 b^{2/3} (c+d x)^{2/3}}{\sqrt {3} \sqrt [3]{b c-a d} \sqrt [3]{a d+b c+2 b d x}}+\frac {1}{\sqrt {3}}\right )}{2 b^{2/3} (b c-a d)^{2/3}} \]

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Rubi [A]  time = 0.07, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {123} \begin {gather*} -\frac {\log (a+b x)}{2 b^{2/3} (b c-a d)^{2/3}}+\frac {3 \log \left (\frac {b^{2/3} (c+d x)^{2/3}}{\sqrt [3]{b c-a d}}-\sqrt [3]{a d+b c+2 b d x}\right )}{4 b^{2/3} (b c-a d)^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {2 b^{2/3} (c+d x)^{2/3}}{\sqrt {3} \sqrt [3]{b c-a d} \sqrt [3]{a d+b c+2 b d x}}+\frac {1}{\sqrt {3}}\right )}{2 b^{2/3} (b c-a d)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 123

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

Rubi steps

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

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Mathematica [C]  time = 0.16, size = 140, normalized size = 0.79 \begin {gather*} -\frac {3 \sqrt [3]{\frac {b c-a d}{2 d (a+b x)}+1} \sqrt [3]{\frac {b c-a d}{d (a+b x)}+1} F_1\left (\frac {2}{3};\frac {1}{3},\frac {1}{3};\frac {5}{3};-\frac {b c-a d}{d (a+b x)},\frac {a d-b c}{2 d (a+b x)}\right )}{2 b \sqrt [3]{c+d x} \sqrt [3]{a d+b (c+2 d x)}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-3*(1 + (b*c - a*d)/(2*d*(a + b*x)))^(1/3)*(1 + (b*c - a*d)/(d*(a + b*x)))^(1/3)*AppellF1[2/3, 1/3, 1/3, 5/3,
-((b*c - a*d)/(d*(a + b*x))), (-(b*c) + a*d)/(2*d*(a + b*x))])/(2*b*(c + d*x)^(1/3)*(a*d + b*(c + 2*d*x))^(1/
3))

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IntegrateAlgebraic [C]  time = 1.47, size = 439, normalized size = 2.47 \begin {gather*} \frac {i \left (\sqrt {3}+i\right ) \log \left (2 \sqrt [3]{b c-a d} \sqrt [3]{a d+2 b (c+d x)-b c}+b^{2/3} \left ((c+d x)^{2/3}+i \sqrt {3} (c+d x)^{2/3}\right )\right )}{4 b^{2/3} (b c-a d)^{2/3}}+\frac {\left (1-i \sqrt {3}\right ) \log \left (b^{2/3} \left (-(c+d x)^{2/3} \sqrt [3]{b c-a d} \sqrt [3]{a d+2 b (c+d x)-b c}-i \sqrt {3} (c+d x)^{2/3} \sqrt [3]{b c-a d} \sqrt [3]{a d+2 b (c+d x)-b c}\right )+2 (b c-a d)^{2/3} (a d+2 b (c+d x)-b c)^{2/3}+b^{4/3} \left (-(c+d x)^{4/3}+i \sqrt {3} (c+d x)^{4/3}\right )\right )}{8 b^{2/3} (b c-a d)^{2/3}}-\frac {i \left (\sqrt {3}-3 i\right ) \tanh ^{-1}\left (\frac {i \sqrt [3]{b c-a d} \sqrt [3]{a d+2 b (c+d x)-b c}+\left (\sqrt {3}-i\right ) b^{2/3} (c+d x)^{2/3}}{\sqrt {3} \sqrt [3]{b c-a d} \sqrt [3]{a d+2 b (c+d x)-b c}}\right )}{4 b^{2/3} (b c-a d)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((a + b*x)*(c + d*x)^(1/3)*(b*c + a*d + 2*b*d*x)^(1/3)),x]

[Out]

((-1/4*I)*(-3*I + Sqrt[3])*ArcTanh[((-I + Sqrt[3])*b^(2/3)*(c + d*x)^(2/3) + I*(b*c - a*d)^(1/3)*(-(b*c) + a*d
+ 2*b*(c + d*x))^(1/3))/(Sqrt[3]*(b*c - a*d)^(1/3)*(-(b*c) + a*d + 2*b*(c + d*x))^(1/3))])/(b^(2/3)*(b*c - a*
d)^(2/3)) + ((I/4)*(I + Sqrt[3])*Log[b^(2/3)*((c + d*x)^(2/3) + I*Sqrt[3]*(c + d*x)^(2/3)) + 2*(b*c - a*d)^(1/
3)*(-(b*c) + a*d + 2*b*(c + d*x))^(1/3)])/(b^(2/3)*(b*c - a*d)^(2/3)) + ((1 - I*Sqrt[3])*Log[2*(b*c - a*d)^(2/
3)*(-(b*c) + a*d + 2*b*(c + d*x))^(2/3) + b^(4/3)*(-(c + d*x)^(4/3) + I*Sqrt[3]*(c + d*x)^(4/3)) + b^(2/3)*(-(
(b*c - a*d)^(1/3)*(c + d*x)^(2/3)*(-(b*c) + a*d + 2*b*(c + d*x))^(1/3)) - I*Sqrt[3]*(b*c - a*d)^(1/3)*(c + d*x
)^(2/3)*(-(b*c) + a*d + 2*b*(c + d*x))^(1/3))])/(8*b^(2/3)*(b*c - a*d)^(2/3))

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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/(b*x+a)/(d*x+c)^(1/3)/(2*b*d*x+a*d+b*c)^(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 (2 \, b d x + b c + a d\right )}^{\frac {1}{3}} {\left (b x + a\right )} {\left (d x + c\right )}^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.24, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (b x +a \right ) \left (d x +c \right )^{\frac {1}{3}} \left (2 b d x +a d +b c \right )^{\frac {1}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\left (a+b\,x\right )\,{\left (c+d\,x\right )}^{1/3}\,{\left (a\,d+b\,c+2\,b\,d\,x\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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