3.1.19 \(\int \sqrt [3]{b \tan (c+d x)} \, dx\) [19]

3.1.19.1 Optimal result

Integrand size = 12, antiderivative size = 131 \[ \int \sqrt [3]{b \tan (c+d x)} \, dx=-\frac {\sqrt {3} \sqrt [3]{b} \arctan \left (\frac {b^{2/3}-2 (b \tan (c+d x))^{2/3}}{\sqrt {3} b^{2/3}}\right )}{2 d}-\frac {\sqrt [3]{b} \log \left (b^{2/3}+(b \tan (c+d x))^{2/3}\right )}{2 d}+\frac {\sqrt [3]{b} \log \left (b^{4/3}-b^{2/3} (b \tan (c+d x))^{2/3}+(b \tan (c+d x))^{4/3}\right )}{4 d} \]

output

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

3.1.19.2 Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.81 \[ \int \sqrt [3]{b \tan (c+d x)} \, dx=-\frac {\left (\log \left (1+\sqrt [3]{\tan ^2(c+d x)}\right )-\sqrt [3]{-1} \log \left (1-\sqrt [3]{-1} \sqrt [3]{\tan ^2(c+d x)}\right )+(-1)^{2/3} \log \left (1+(-1)^{2/3} \sqrt [3]{\tan ^2(c+d x)}\right )\right ) (b \tan (c+d x))^{4/3}}{2 b d \tan ^2(c+d x)^{2/3}} \]

input

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

output

-1/2*((Log[1 + (Tan[c + d*x]^2)^(1/3)] - (-1)^(1/3)*Log[1 - (-1)^(1/3)*(Ta 
n[c + d*x]^2)^(1/3)] + (-1)^(2/3)*Log[1 + (-1)^(2/3)*(Tan[c + d*x]^2)^(1/3 
)])*(b*Tan[c + d*x])^(4/3))/(b*d*(Tan[c + d*x]^2)^(2/3))
 

3.1.19.3 Rubi [A] (warning: unable to verify)

Time = 0.32 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.85, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {3042, 3957, 266, 807, 821, 16, 1142, 25, 1082, 217, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sqrt [3]{b \tan (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt [3]{b \tan (c+d x)}dx\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {b \int \frac {\sqrt [3]{b \tan (c+d x)}}{\tan ^2(c+d x) b^2+b^2}d(b \tan (c+d x))}{d}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {3 b \int \frac {b^3 \tan ^3(c+d x)}{b^6 \tan ^6(c+d x)+b^2}d\sqrt [3]{b \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {3 b \int \frac {b^2 \tan ^2(c+d x)}{b^3 \tan ^3(c+d x)+b^2}d\left (b^2 \tan ^2(c+d x)\right )}{2 d}\)

\(\Big \downarrow \) 821

\(\displaystyle \frac {3 b \left (\frac {\int \frac {b^2 \tan ^2(c+d x)+b^{2/3}}{b^2 \tan ^2(c+d x)-b^{5/3} \tan (c+d x)+b^{4/3}}d\left (b^2 \tan ^2(c+d x)\right )}{3 b^{2/3}}-\frac {\int \frac {1}{b^2 \tan ^2(c+d x)+b^{2/3}}d\left (b^2 \tan ^2(c+d x)\right )}{3 b^{2/3}}\right )}{2 d}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {3 b \left (\frac {\int \frac {b^2 \tan ^2(c+d x)+b^{2/3}}{b^2 \tan ^2(c+d x)-b^{5/3} \tan (c+d x)+b^{4/3}}d\left (b^2 \tan ^2(c+d x)\right )}{3 b^{2/3}}-\frac {\log \left (b^{2/3}+b^2 \tan ^2(c+d x)\right )}{3 b^{2/3}}\right )}{2 d}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {3 b \left (\frac {\frac {3}{2} b^{2/3} \int \frac {1}{b^2 \tan ^2(c+d x)-b^{5/3} \tan (c+d x)+b^{4/3}}d\left (b^2 \tan ^2(c+d x)\right )+\frac {1}{2} \int -\frac {b^{2/3}-2 b^2 \tan ^2(c+d x)}{b^2 \tan ^2(c+d x)-b^{5/3} \tan (c+d x)+b^{4/3}}d\left (b^2 \tan ^2(c+d x)\right )}{3 b^{2/3}}-\frac {\log \left (b^{2/3}+b^2 \tan ^2(c+d x)\right )}{3 b^{2/3}}\right )}{2 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {3 b \left (\frac {\frac {3}{2} b^{2/3} \int \frac {1}{b^2 \tan ^2(c+d x)-b^{5/3} \tan (c+d x)+b^{4/3}}d\left (b^2 \tan ^2(c+d x)\right )-\frac {1}{2} \int \frac {b^{2/3}-2 b^2 \tan ^2(c+d x)}{b^2 \tan ^2(c+d x)-b^{5/3} \tan (c+d x)+b^{4/3}}d\left (b^2 \tan ^2(c+d x)\right )}{3 b^{2/3}}-\frac {\log \left (b^{2/3}+b^2 \tan ^2(c+d x)\right )}{3 b^{2/3}}\right )}{2 d}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {3 b \left (\frac {3 \int \frac {1}{2 \sqrt [3]{b} \tan (c+d x)-4}d\left (1-2 \sqrt [3]{b} \tan (c+d x)\right )-\frac {1}{2} \int \frac {b^{2/3}-2 b^2 \tan ^2(c+d x)}{b^2 \tan ^2(c+d x)-b^{5/3} \tan (c+d x)+b^{4/3}}d\left (b^2 \tan ^2(c+d x)\right )}{3 b^{2/3}}-\frac {\log \left (b^{2/3}+b^2 \tan ^2(c+d x)\right )}{3 b^{2/3}}\right )}{2 d}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {3 b \left (\frac {-\frac {1}{2} \int \frac {b^{2/3}-2 b^2 \tan ^2(c+d x)}{b^2 \tan ^2(c+d x)-b^{5/3} \tan (c+d x)+b^{4/3}}d\left (b^2 \tan ^2(c+d x)\right )-\sqrt {3} \arctan \left (\frac {1-2 \sqrt [3]{b} \tan (c+d x)}{\sqrt {3}}\right )}{3 b^{2/3}}-\frac {\log \left (b^{2/3}+b^2 \tan ^2(c+d x)\right )}{3 b^{2/3}}\right )}{2 d}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {3 b \left (\frac {\frac {1}{2} \log \left (-b^{5/3} \tan (c+d x)+b^{4/3}+b^2 \tan ^2(c+d x)\right )-\sqrt {3} \arctan \left (\frac {1-2 \sqrt [3]{b} \tan (c+d x)}{\sqrt {3}}\right )}{3 b^{2/3}}-\frac {\log \left (b^{2/3}+b^2 \tan ^2(c+d x)\right )}{3 b^{2/3}}\right )}{2 d}\)

input

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

output

(3*b*(-1/3*Log[b^(2/3) + b^2*Tan[c + d*x]^2]/b^(2/3) + (-(Sqrt[3]*ArcTan[( 
1 - 2*b^(1/3)*Tan[c + d*x])/Sqrt[3]]) + Log[b^(4/3) - b^(5/3)*Tan[c + d*x] 
 + b^2*Tan[c + d*x]^2]/2)/(3*b^(2/3))))/(2*d)
 

3.1.19.3.1 Defintions of rubi rules used

rule 16

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

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 217

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

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 821

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

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[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])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

3.1.19.4 Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(\frac {3 b \left (-\frac {\ln \left (\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}+\left (b^{2}\right )^{\frac {1}{3}}\right )}{6 \left (b^{2}\right )^{\frac {1}{3}}}+\frac {\ln \left (\left (b \tan \left (d x +c \right )\right )^{\frac {4}{3}}-\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}} \left (b^{2}\right )^{\frac {1}{3}}+\left (b^{2}\right )^{\frac {2}{3}}\right )}{12 \left (b^{2}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{\left (b^{2}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{6 \left (b^{2}\right )^{\frac {1}{3}}}\right )}{d}\)	\(108\)
default	\(\frac {3 b \left (-\frac {\ln \left (\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}+\left (b^{2}\right )^{\frac {1}{3}}\right )}{6 \left (b^{2}\right )^{\frac {1}{3}}}+\frac {\ln \left (\left (b \tan \left (d x +c \right )\right )^{\frac {4}{3}}-\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}} \left (b^{2}\right )^{\frac {1}{3}}+\left (b^{2}\right )^{\frac {2}{3}}\right )}{12 \left (b^{2}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{\left (b^{2}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{6 \left (b^{2}\right )^{\frac {1}{3}}}\right )}{d}\)	\(108\)

input

int((b*tan(d*x+c))^(1/3),x,method=_RETURNVERBOSE)
 

output

3/d*b*(-1/6/(b^2)^(1/3)*ln((b*tan(d*x+c))^(2/3)+(b^2)^(1/3))+1/12/(b^2)^(1 
/3)*ln((b*tan(d*x+c))^(4/3)-(b*tan(d*x+c))^(2/3)*(b^2)^(1/3)+(b^2)^(2/3))+ 
1/6*3^(1/2)/(b^2)^(1/3)*arctan(1/3*3^(1/2)*(2*(b*tan(d*x+c))^(2/3)/(b^2)^( 
1/3)-1)))
 

3.1.19.5 Fricas [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.95 \[ \int \sqrt [3]{b \tan (c+d x)} \, dx=\frac {2 \, \sqrt {3} \left (-b\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} \left (-b\right )^{\frac {1}{3}} + \sqrt {3} b}{3 \, b}\right ) - \left (-b\right )^{\frac {1}{3}} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b \tan \left (d x + c\right ) - \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} \left (-b\right )^{\frac {2}{3}} - \left (-b\right )^{\frac {1}{3}} b\right ) + 2 \, \left (-b\right )^{\frac {1}{3}} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + \left (-b\right )^{\frac {2}{3}}\right )}{4 \, d} \]

input

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

output

1/4*(2*sqrt(3)*(-b)^(1/3)*arctan(1/3*(2*sqrt(3)*(b*tan(d*x + c))^(2/3)*(-b 
)^(1/3) + sqrt(3)*b)/b) - (-b)^(1/3)*log((b*tan(d*x + c))^(1/3)*b*tan(d*x 
+ c) - (b*tan(d*x + c))^(2/3)*(-b)^(2/3) - (-b)^(1/3)*b) + 2*(-b)^(1/3)*lo 
g((b*tan(d*x + c))^(2/3) + (-b)^(2/3)))/d
 

3.1.19.6 Sympy [F]

\[ \int \sqrt [3]{b \tan (c+d x)} \, dx=\int \sqrt [3]{b \tan {\left (c + d x \right )}}\, dx \]

input

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

output

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

3.1.19.7 Maxima [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.75 \[ \int \sqrt [3]{b \tan (c+d x)} \, dx=\frac {2 \, \sqrt {3} b^{\frac {4}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} - b^{\frac {2}{3}}\right )}}{3 \, b^{\frac {2}{3}}}\right ) + b^{\frac {4}{3}} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac {4}{3}} - \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} b^{\frac {2}{3}} + b^{\frac {4}{3}}\right ) - 2 \, b^{\frac {4}{3}} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + b^{\frac {2}{3}}\right )}{4 \, b d} \]

input

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

output

1/4*(2*sqrt(3)*b^(4/3)*arctan(1/3*sqrt(3)*(2*(b*tan(d*x + c))^(2/3) - b^(2 
/3))/b^(2/3)) + b^(4/3)*log((b*tan(d*x + c))^(4/3) - (b*tan(d*x + c))^(2/3 
)*b^(2/3) + b^(4/3)) - 2*b^(4/3)*log((b*tan(d*x + c))^(2/3) + b^(2/3)))/(b 
*d)
 

3.1.19.8 Giac [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.97 \[ \int \sqrt [3]{b \tan (c+d x)} \, dx=\frac {1}{4} \, b {\left (\frac {2 \, \sqrt {3} {\left | b \right |}^{\frac {4}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} - {\left | b \right |}^{\frac {2}{3}}\right )}}{3 \, {\left | b \right |}^{\frac {2}{3}}}\right )}{b^{2} d} + \frac {{\left | b \right |}^{\frac {4}{3}} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b \tan \left (d x + c\right ) - \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} {\left | b \right |}^{\frac {2}{3}} + {\left | b \right |}^{\frac {4}{3}}\right )}{b^{2} d} - \frac {2 \, {\left | b \right |}^{\frac {4}{3}} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + {\left | b \right |}^{\frac {2}{3}}\right )}{b^{2} d}\right )} \]

input

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

output

1/4*b*(2*sqrt(3)*abs(b)^(4/3)*arctan(1/3*sqrt(3)*(2*(b*tan(d*x + c))^(2/3) 
 - abs(b)^(2/3))/abs(b)^(2/3))/(b^2*d) + abs(b)^(4/3)*log((b*tan(d*x + c)) 
^(1/3)*b*tan(d*x + c) - (b*tan(d*x + c))^(2/3)*abs(b)^(2/3) + abs(b)^(4/3) 
)/(b^2*d) - 2*abs(b)^(4/3)*log((b*tan(d*x + c))^(2/3) + abs(b)^(2/3))/(b^2 
*d))
 

3.1.19.9 Mupad [B] (verification not implemented)

Time = 3.39 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.11 \[ \int \sqrt [3]{b \tan (c+d x)} \, dx=\frac {{\left (-b\right )}^{1/3}\,\ln \left (81\,{\left (-b\right )}^{16/3}\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{2/3}+81\,b^6\right )}{2\,d}-\frac {{\left (-b\right )}^{1/3}\,\ln \left (\frac {81\,b^6}{d^4}-\frac {81\,{\left (-b\right )}^{16/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{2/3}}{d^4}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,d}+\frac {{\left (-b\right )}^{1/3}\,\ln \left (\frac {81\,b^6}{d^4}+\frac {162\,{\left (-b\right )}^{16/3}\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{2/3}}{d^4}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )}{d} \]

input

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

output

((-b)^(1/3)*log(81*(-b)^(16/3)*(b*tan(c + d*x))^(2/3) + 81*b^6))/(2*d) - ( 
(-b)^(1/3)*log((81*b^6)/d^4 - (81*(-b)^(16/3)*((3^(1/2)*1i)/2 + 1/2)*(b*ta 
n(c + d*x))^(2/3))/d^4)*((3^(1/2)*1i)/2 + 1/2))/(2*d) + ((-b)^(1/3)*log((8 
1*b^6)/d^4 + (162*(-b)^(16/3)*((3^(1/2)*1i)/4 - 1/4)*(b*tan(c + d*x))^(2/3 
))/d^4)*((3^(1/2)*1i)/4 - 1/4))/d