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

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

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {55, 617, 204, 31} \begin {gather*} -\frac {\log (a+b x)}{2 b^{2/3} \sqrt [3]{b c-a d}}+\frac {3 \log \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{2 b^{2/3} \sqrt [3]{b c-a d}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}}+1}{\sqrt {3}}\right )}{b^{2/3} \sqrt [3]{b c-a d}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)*(c + d*x)^(1/3)),x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

Rule 204

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

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 106, normalized size = 0.76 \begin {gather*} \frac {3 \log \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}}+1}{\sqrt {3}}\right )-\log (a+b x)}{2 b^{2/3} \sqrt [3]{b c-a d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.23, size = 191, normalized size = 1.37 \begin {gather*} -\frac {\log \left (\sqrt [3]{a d-b c}+\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{b^{2/3} \sqrt [3]{a d-b c}}+\frac {\log \left (-\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{a d-b c}+(a d-b c)^{2/3}+b^{2/3} (c+d x)^{2/3}\right )}{2 b^{2/3} \sqrt [3]{a d-b c}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{a d-b c}}\right )}{b^{2/3} \sqrt [3]{a d-b c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 1.27, size = 570, normalized size = 4.10 \begin {gather*} \left [\frac {\sqrt {3} {\left (b^{2} c - a b d\right )} \sqrt {-\frac {{\left (b^{3} c - a b^{2} d\right )}^{\frac {1}{3}}}{b c - a d}} \log \left (\frac {2 \, b^{2} d x + 3 \, b^{2} c - a b d - \sqrt {3} {\left ({\left (b^{3} c - a b^{2} d\right )}^{\frac {1}{3}} {\left (b c - a d\right )} + {\left (b^{2} c - a b d\right )} {\left (d x + c\right )}^{\frac {1}{3}} - 2 \, {\left (b^{3} c - a b^{2} d\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {2}{3}}\right )} \sqrt {-\frac {{\left (b^{3} c - a b^{2} d\right )}^{\frac {1}{3}}}{b c - a d}} - 3 \, {\left (b^{3} c - a b^{2} d\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{b x + a}\right ) - {\left (b^{3} c - a b^{2} d\right )}^{\frac {2}{3}} \log \left ({\left (d x + c\right )}^{\frac {2}{3}} b^{2} + {\left (b^{3} c - a b^{2} d\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {1}{3}} b + {\left (b^{3} c - a b^{2} d\right )}^{\frac {2}{3}}\right ) + 2 \, {\left (b^{3} c - a b^{2} d\right )}^{\frac {2}{3}} \log \left ({\left (d x + c\right )}^{\frac {1}{3}} b - {\left (b^{3} c - a b^{2} d\right )}^{\frac {1}{3}}\right )}{2 \, {\left (b^{3} c - a b^{2} d\right )}}, \frac {2 \, \sqrt {3} {\left (b^{2} c - a b d\right )} \sqrt {\frac {{\left (b^{3} c - a b^{2} d\right )}^{\frac {1}{3}}}{b c - a d}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (d x + c\right )}^{\frac {1}{3}} b + {\left (b^{3} c - a b^{2} d\right )}^{\frac {1}{3}}\right )} \sqrt {\frac {{\left (b^{3} c - a b^{2} d\right )}^{\frac {1}{3}}}{b c - a d}}}{3 \, b}\right ) - {\left (b^{3} c - a b^{2} d\right )}^{\frac {2}{3}} \log \left ({\left (d x + c\right )}^{\frac {2}{3}} b^{2} + {\left (b^{3} c - a b^{2} d\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {1}{3}} b + {\left (b^{3} c - a b^{2} d\right )}^{\frac {2}{3}}\right ) + 2 \, {\left (b^{3} c - a b^{2} d\right )}^{\frac {2}{3}} \log \left ({\left (d x + c\right )}^{\frac {1}{3}} b - {\left (b^{3} c - a b^{2} d\right )}^{\frac {1}{3}}\right )}{2 \, {\left (b^{3} c - a b^{2} d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(3)*(b^2*c - a*b*d)*sqrt(-(b^3*c - a*b^2*d)^(1/3)/(b*c - a*d))*log((2*b^2*d*x + 3*b^2*c - a*b*d - sq
rt(3)*((b^3*c - a*b^2*d)^(1/3)*(b*c - a*d) + (b^2*c - a*b*d)*(d*x + c)^(1/3) - 2*(b^3*c - a*b^2*d)^(2/3)*(d*x
+ c)^(2/3))*sqrt(-(b^3*c - a*b^2*d)^(1/3)/(b*c - a*d)) - 3*(b^3*c - a*b^2*d)^(2/3)*(d*x + c)^(1/3))/(b*x + a))
- (b^3*c - a*b^2*d)^(2/3)*log((d*x + c)^(2/3)*b^2 + (b^3*c - a*b^2*d)^(1/3)*(d*x + c)^(1/3)*b + (b^3*c - a*b^
2*d)^(2/3)) + 2*(b^3*c - a*b^2*d)^(2/3)*log((d*x + c)^(1/3)*b - (b^3*c - a*b^2*d)^(1/3)))/(b^3*c - a*b^2*d), 1
/2*(2*sqrt(3)*(b^2*c - a*b*d)*sqrt((b^3*c - a*b^2*d)^(1/3)/(b*c - a*d))*arctan(1/3*sqrt(3)*(2*(d*x + c)^(1/3)*
b + (b^3*c - a*b^2*d)^(1/3))*sqrt((b^3*c - a*b^2*d)^(1/3)/(b*c - a*d))/b) - (b^3*c - a*b^2*d)^(2/3)*log((d*x +
c)^(2/3)*b^2 + (b^3*c - a*b^2*d)^(1/3)*(d*x + c)^(1/3)*b + (b^3*c - a*b^2*d)^(2/3)) + 2*(b^3*c - a*b^2*d)^(2/
3)*log((d*x + c)^(1/3)*b - (b^3*c - a*b^2*d)^(1/3)))/(b^3*c - a*b^2*d)]

________________________________________________________________________________________

giac [A]  time = 1.10, size = 196, normalized size = 1.41 \begin {gather*} \frac {3 \, {\left (b^{3} c - a b^{2} d\right )}^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (d x + c\right )}^{\frac {1}{3}} + \left (\frac {b c - a d}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {b c - a d}{b}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} b^{3} c - \sqrt {3} a b^{2} d} - \frac {\log \left ({\left (d x + c\right )}^{\frac {2}{3}} + {\left (d x + c\right )}^{\frac {1}{3}} \left (\frac {b c - a d}{b}\right )^{\frac {1}{3}} + \left (\frac {b c - a d}{b}\right )^{\frac {2}{3}}\right )}{2 \, {\left (b^{3} c - a b^{2} d\right )}^{\frac {1}{3}}} + \frac {\left (\frac {b c - a d}{b}\right )^{\frac {2}{3}} \log \left ({\left | {\left (d x + c\right )}^{\frac {1}{3}} - \left (\frac {b c - a d}{b}\right )^{\frac {1}{3}} \right |}\right )}{b c - a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*(b^3*c - a*b^2*d)^(2/3)*arctan(1/3*sqrt(3)*(2*(d*x + c)^(1/3) + ((b*c - a*d)/b)^(1/3))/((b*c - a*d)/b)^(1/3)
)/(sqrt(3)*b^3*c - sqrt(3)*a*b^2*d) - 1/2*log((d*x + c)^(2/3) + (d*x + c)^(1/3)*((b*c - a*d)/b)^(1/3) + ((b*c
- a*d)/b)^(2/3))/(b^3*c - a*b^2*d)^(1/3) + ((b*c - a*d)/b)^(2/3)*log(abs((d*x + c)^(1/3) - ((b*c - a*d)/b)^(1/
3)))/(b*c - a*d)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 161, normalized size = 1.16 \begin {gather*} \frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (d x +c \right )^{\frac {1}{3}}}{\left (\frac {a d -b c}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{\left (\frac {a d -b c}{b}\right )^{\frac {1}{3}} b}-\frac {\ln \left (\left (d x +c \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{b}\right )^{\frac {1}{3}}\right )}{\left (\frac {a d -b c}{b}\right )^{\frac {1}{3}} b}+\frac {\ln \left (\left (d x +c \right )^{\frac {2}{3}}-\left (\frac {a d -b c}{b}\right )^{\frac {1}{3}} \left (d x +c \right )^{\frac {1}{3}}+\left (\frac {a d -b c}{b}\right )^{\frac {2}{3}}\right )}{2 \left (\frac {a d -b c}{b}\right )^{\frac {1}{3}} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/b/((a*d-b*c)/b)^(1/3)*ln((d*x+c)^(1/3)+((a*d-b*c)/b)^(1/3))+1/2/b/((a*d-b*c)/b)^(1/3)*ln((d*x+c)^(2/3)-((a*
d-b*c)/b)^(1/3)*(d*x+c)^(1/3)+((a*d-b*c)/b)^(2/3))+3^(1/2)/b/((a*d-b*c)/b)^(1/3)*arctan(1/3*3^(1/2)*(2/((a*d-b
*c)/b)^(1/3)*(d*x+c)^(1/3)-1))

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c positive or negative?

________________________________________________________________________________________

mupad [B]  time = 0.21, size = 204, normalized size = 1.47 \begin {gather*} \frac {\ln \left (9\,b\,{\left (c+d\,x\right )}^{1/3}-\frac {9\,b^3\,c-9\,a\,b^2\,d}{b^{4/3}\,{\left (b\,c-a\,d\right )}^{2/3}}\right )}{b^{2/3}\,{\left (b\,c-a\,d\right )}^{1/3}}+\frac {\ln \left (9\,b\,{\left (c+d\,x\right )}^{1/3}-\frac {{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (9\,b^3\,c-9\,a\,b^2\,d\right )}{4\,b^{4/3}\,{\left (b\,c-a\,d\right )}^{2/3}}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,b^{2/3}\,{\left (b\,c-a\,d\right )}^{1/3}}-\frac {\ln \left (9\,b\,{\left (c+d\,x\right )}^{1/3}-\frac {{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (9\,b^3\,c-9\,a\,b^2\,d\right )}{4\,b^{4/3}\,{\left (b\,c-a\,d\right )}^{2/3}}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2\,b^{2/3}\,{\left (b\,c-a\,d\right )}^{1/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)*(c + d*x)^(1/3)),x)

[Out]

log(9*b*(c + d*x)^(1/3) - (9*b^3*c - 9*a*b^2*d)/(b^(4/3)*(b*c - a*d)^(2/3)))/(b^(2/3)*(b*c - a*d)^(1/3)) + (lo
g(9*b*(c + d*x)^(1/3) - ((3^(1/2)*1i - 1)^2*(9*b^3*c - 9*a*b^2*d))/(4*b^(4/3)*(b*c - a*d)^(2/3)))*(3^(1/2)*1i
- 1))/(2*b^(2/3)*(b*c - a*d)^(1/3)) - (log(9*b*(c + d*x)^(1/3) - ((3^(1/2)*1i + 1)^2*(9*b^3*c - 9*a*b^2*d))/(4
*b^(4/3)*(b*c - a*d)^(2/3)))*(3^(1/2)*1i + 1))/(2*b^(2/3)*(b*c - a*d)^(1/3))

________________________________________________________________________________________

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}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________