\(\int \frac {1}{\sqrt [4]{2+3 x} (4+3 x)} \, dx\) [296]

Optimal result

Integrand size = 17, antiderivative size = 85 \[ \int \frac {1}{\sqrt [4]{2+3 x} (4+3 x)} \, dx=-\frac {1}{3} \sqrt [4]{2} \arctan \left (\frac {\sqrt {2}-\sqrt {2+3 x}}{2^{3/4} \sqrt [4]{2+3 x}}\right )-\frac {1}{3} \sqrt [4]{2} \text {arctanh}\left (\frac {2^{3/4} \sqrt [4]{2+3 x}}{\sqrt {2}+\sqrt {2+3 x}}\right ) \] Output:

-1/3*2^(1/4)*arctan(1/2*(2^(1/2)-(2+3*x)^(1/2))*2^(1/4)/(2+3*x)^(1/4))-1/3 
*2^(1/4)*arctanh(2^(3/4)*(2+3*x)^(1/4)/(2^(1/2)+(2+3*x)^(1/2)))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\sqrt [4]{2+3 x} (4+3 x)} \, dx=\frac {1}{3} \sqrt [4]{2} \left (\arctan \left (\frac {-\sqrt {2}+\sqrt {2+3 x}}{2^{3/4} \sqrt [4]{2+3 x}}\right )-\text {arctanh}\left (\frac {2 \sqrt [4]{4+6 x}}{2+\sqrt {4+6 x}}\right )\right ) \] Input:

Integrate[1/((2 + 3*x)^(1/4)*(4 + 3*x)),x]
 

Output:

(2^(1/4)*(ArcTan[(-Sqrt[2] + Sqrt[2 + 3*x])/(2^(3/4)*(2 + 3*x)^(1/4))] - A 
rcTanh[(2*(4 + 6*x)^(1/4))/(2 + Sqrt[4 + 6*x])]))/3
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.72, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {73, 826, 1476, 1082, 217, 1479, 25, 27, 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.

\(\displaystyle \int \frac {1}{\sqrt [4]{3 x+2} (3 x+4)} \, dx\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {4}{3} \int \frac {\sqrt {3 x+2}}{3 x+4}d\sqrt [4]{3 x+2}\)

\(\Big \downarrow \) 826

\(\displaystyle \frac {4}{3} \left (\frac {1}{2} \int \frac {\sqrt {3 x+2}+\sqrt {2}}{3 x+4}d\sqrt [4]{3 x+2}-\frac {1}{2} \int \frac {\sqrt {2}-\sqrt {3 x+2}}{3 x+4}d\sqrt [4]{3 x+2}\right )\)

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[1/((2 + 3*x)^(1/4)*(4 + 3*x)),x]
 

Output:

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

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, 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 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ 
a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] 
 && AtomQ[SplitProduct[SumBaseQ, b]]))
 

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 1476

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

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

Maple [A] (verified)

Time = 0.95 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(\frac {2^{\frac {1}{4}} \left (\ln \left (\frac {\sqrt {2+3 x}-2^{\frac {3}{4}} \left (2+3 x \right )^{\frac {1}{4}}+\sqrt {2}}{\sqrt {2+3 x}+2^{\frac {3}{4}} \left (2+3 x \right )^{\frac {1}{4}}+\sqrt {2}}\right )+2 \arctan \left (2^{\frac {1}{4}} \left (2+3 x \right )^{\frac {1}{4}}+1\right )+2 \arctan \left (2^{\frac {1}{4}} \left (2+3 x \right )^{\frac {1}{4}}-1\right )\right )}{6}\)	\(88\)
default	\(\frac {2^{\frac {1}{4}} \left (\ln \left (\frac {\sqrt {2+3 x}-2^{\frac {3}{4}} \left (2+3 x \right )^{\frac {1}{4}}+\sqrt {2}}{\sqrt {2+3 x}+2^{\frac {3}{4}} \left (2+3 x \right )^{\frac {1}{4}}+\sqrt {2}}\right )+2 \arctan \left (2^{\frac {1}{4}} \left (2+3 x \right )^{\frac {1}{4}}+1\right )+2 \arctan \left (2^{\frac {1}{4}} \left (2+3 x \right )^{\frac {1}{4}}-1\right )\right )}{6}\)	\(88\)
pseudoelliptic	\(\frac {2^{\frac {1}{4}} \left (\ln \left (\frac {\sqrt {2+3 x}-2^{\frac {3}{4}} \left (2+3 x \right )^{\frac {1}{4}}+\sqrt {2}}{\sqrt {2+3 x}+2^{\frac {3}{4}} \left (2+3 x \right )^{\frac {1}{4}}+\sqrt {2}}\right )+2 \arctan \left (2^{\frac {1}{4}} \left (2+3 x \right )^{\frac {1}{4}}+1\right )+2 \arctan \left (2^{\frac {1}{4}} \left (2+3 x \right )^{\frac {1}{4}}-1\right )\right )}{6}\)	\(88\)
trager	\(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+8\right )^{2}\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+8\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}+8\right )^{2} \left (2+3 x \right )^{\frac {3}{4}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+8\right )^{2} \sqrt {2+3 x}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+8\right )^{2}\right ) \left (2+3 x \right )^{\frac {1}{4}}-6 x}{3 x +4}\right )}{6}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+8\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+8\right )^{3} \left (2+3 x \right )^{\frac {3}{4}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+8\right )^{2} \sqrt {2+3 x}+4 \operatorname {RootOf}\left (\textit {\_Z}^{4}+8\right ) \left (2+3 x \right )^{\frac {1}{4}}-6 x}{3 x +4}\right )}{6}\)	\(169\)

Input:

int(1/(2+3*x)^(1/4)/(3*x+4),x,method=_RETURNVERBOSE)
 

Output:

1/6*2^(1/4)*(ln(((2+3*x)^(1/2)-2^(3/4)*(2+3*x)^(1/4)+2^(1/2))/((2+3*x)^(1/ 
2)+2^(3/4)*(2+3*x)^(1/4)+2^(1/2)))+2*arctan(2^(1/4)*(2+3*x)^(1/4)+1)+2*arc 
tan(2^(1/4)*(2+3*x)^(1/4)-1))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.26 \[ \int \frac {1}{\sqrt [4]{2+3 x} (4+3 x)} \, dx=\frac {1}{12} \cdot 8^{\frac {3}{4}} \arctan \left (\frac {1}{4} \cdot 8^{\frac {3}{4}} {\left (3 \, x + 2\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{12} \cdot 8^{\frac {3}{4}} \arctan \left (\frac {1}{4} \cdot 8^{\frac {3}{4}} {\left (3 \, x + 2\right )}^{\frac {1}{4}} - 1\right ) - \frac {1}{24} \cdot 8^{\frac {3}{4}} \log \left (2 \, \sqrt {2} + 2 \cdot 8^{\frac {1}{4}} {\left (3 \, x + 2\right )}^{\frac {1}{4}} + 2 \, \sqrt {3 \, x + 2}\right ) + \frac {1}{24} \cdot 8^{\frac {3}{4}} \log \left (2 \, \sqrt {2} - 2 \cdot 8^{\frac {1}{4}} {\left (3 \, x + 2\right )}^{\frac {1}{4}} + 2 \, \sqrt {3 \, x + 2}\right ) \] Input:

integrate(1/(2+3*x)^(1/4)/(4+3*x),x, algorithm="fricas")
 

Output:

1/12*8^(3/4)*arctan(1/4*8^(3/4)*(3*x + 2)^(1/4) + 1) + 1/12*8^(3/4)*arctan 
(1/4*8^(3/4)*(3*x + 2)^(1/4) - 1) - 1/24*8^(3/4)*log(2*sqrt(2) + 2*8^(1/4) 
*(3*x + 2)^(1/4) + 2*sqrt(3*x + 2)) + 1/24*8^(3/4)*log(2*sqrt(2) - 2*8^(1/ 
4)*(3*x + 2)^(1/4) + 2*sqrt(3*x + 2))
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.80 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.72 \[ \int \frac {1}{\sqrt [4]{2+3 x} (4+3 x)} \, dx=- \frac {2^{\frac {3}{4}} e^{- \frac {3 i \pi }{4}} \log {\left (- \frac {2^{\frac {3}{4}} \cdot \sqrt [4]{3} \sqrt [4]{x + \frac {2}{3}} e^{\frac {i \pi }{4}}}{2} + 1 \right )} \Gamma \left (\frac {3}{4}\right )}{8 \Gamma \left (\frac {7}{4}\right )} - \frac {2^{\frac {3}{4}} i e^{- \frac {3 i \pi }{4}} \log {\left (- \frac {2^{\frac {3}{4}} \cdot \sqrt [4]{3} \sqrt [4]{x + \frac {2}{3}} e^{\frac {3 i \pi }{4}}}{2} + 1 \right )} \Gamma \left (\frac {3}{4}\right )}{8 \Gamma \left (\frac {7}{4}\right )} + \frac {2^{\frac {3}{4}} e^{- \frac {3 i \pi }{4}} \log {\left (- \frac {2^{\frac {3}{4}} \cdot \sqrt [4]{3} \sqrt [4]{x + \frac {2}{3}} e^{\frac {5 i \pi }{4}}}{2} + 1 \right )} \Gamma \left (\frac {3}{4}\right )}{8 \Gamma \left (\frac {7}{4}\right )} + \frac {2^{\frac {3}{4}} i e^{- \frac {3 i \pi }{4}} \log {\left (- \frac {2^{\frac {3}{4}} \cdot \sqrt [4]{3} \sqrt [4]{x + \frac {2}{3}} e^{\frac {7 i \pi }{4}}}{2} + 1 \right )} \Gamma \left (\frac {3}{4}\right )}{8 \Gamma \left (\frac {7}{4}\right )} \] Input:

integrate(1/(2+3*x)**(1/4)/(4+3*x),x)
 

Output:

-2**(3/4)*exp(-3*I*pi/4)*log(-2**(3/4)*3**(1/4)*(x + 2/3)**(1/4)*exp_polar 
(I*pi/4)/2 + 1)*gamma(3/4)/(8*gamma(7/4)) - 2**(3/4)*I*exp(-3*I*pi/4)*log( 
-2**(3/4)*3**(1/4)*(x + 2/3)**(1/4)*exp_polar(3*I*pi/4)/2 + 1)*gamma(3/4)/ 
(8*gamma(7/4)) + 2**(3/4)*exp(-3*I*pi/4)*log(-2**(3/4)*3**(1/4)*(x + 2/3)* 
*(1/4)*exp_polar(5*I*pi/4)/2 + 1)*gamma(3/4)/(8*gamma(7/4)) + 2**(3/4)*I*e 
xp(-3*I*pi/4)*log(-2**(3/4)*3**(1/4)*(x + 2/3)**(1/4)*exp_polar(7*I*pi/4)/ 
2 + 1)*gamma(3/4)/(8*gamma(7/4))
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.25 \[ \int \frac {1}{\sqrt [4]{2+3 x} (4+3 x)} \, dx=\frac {1}{3} \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} + 2 \, {\left (3 \, x + 2\right )}^{\frac {1}{4}}\right )}\right ) + \frac {1}{3} \cdot 2^{\frac {1}{4}} \arctan \left (-\frac {1}{2} \cdot 2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} - 2 \, {\left (3 \, x + 2\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{6} \cdot 2^{\frac {1}{4}} \log \left (2^{\frac {3}{4}} {\left (3 \, x + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {3 \, x + 2}\right ) + \frac {1}{6} \cdot 2^{\frac {1}{4}} \log \left (-2^{\frac {3}{4}} {\left (3 \, x + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {3 \, x + 2}\right ) \] Input:

integrate(1/(2+3*x)^(1/4)/(4+3*x),x, algorithm="maxima")
 

Output:

1/3*2^(1/4)*arctan(1/2*2^(1/4)*(2^(3/4) + 2*(3*x + 2)^(1/4))) + 1/3*2^(1/4 
)*arctan(-1/2*2^(1/4)*(2^(3/4) - 2*(3*x + 2)^(1/4))) - 1/6*2^(1/4)*log(2^( 
3/4)*(3*x + 2)^(1/4) + sqrt(2) + sqrt(3*x + 2)) + 1/6*2^(1/4)*log(-2^(3/4) 
*(3*x + 2)^(1/4) + sqrt(2) + sqrt(3*x + 2))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.25 \[ \int \frac {1}{\sqrt [4]{2+3 x} (4+3 x)} \, dx=\frac {1}{3} \cdot 2^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} + 2 \, {\left (3 \, x + 2\right )}^{\frac {1}{4}}\right )}\right ) + \frac {1}{3} \cdot 2^{\frac {1}{4}} \arctan \left (-\frac {1}{2} \cdot 2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} - 2 \, {\left (3 \, x + 2\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{6} \cdot 2^{\frac {1}{4}} \log \left (2^{\frac {3}{4}} {\left (3 \, x + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {3 \, x + 2}\right ) + \frac {1}{6} \cdot 2^{\frac {1}{4}} \log \left (-2^{\frac {3}{4}} {\left (3 \, x + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {3 \, x + 2}\right ) \] Input:

integrate(1/(2+3*x)^(1/4)/(4+3*x),x, algorithm="giac")
 

Output:

1/3*2^(1/4)*arctan(1/2*2^(1/4)*(2^(3/4) + 2*(3*x + 2)^(1/4))) + 1/3*2^(1/4 
)*arctan(-1/2*2^(1/4)*(2^(3/4) - 2*(3*x + 2)^(1/4))) - 1/6*2^(1/4)*log(2^( 
3/4)*(3*x + 2)^(1/4) + sqrt(2) + sqrt(3*x + 2)) + 1/6*2^(1/4)*log(-2^(3/4) 
*(3*x + 2)^(1/4) + sqrt(2) + sqrt(3*x + 2))
 

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.53 \[ \int \frac {1}{\sqrt [4]{2+3 x} (4+3 x)} \, dx=2^{1/4}\,\mathrm {atan}\left (2^{1/4}\,{\left (3\,x+2\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{3}-\frac {1}{3}{}\mathrm {i}\right )+2^{1/4}\,\mathrm {atan}\left (2^{1/4}\,{\left (3\,x+2\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{3}+\frac {1}{3}{}\mathrm {i}\right ) \] Input:

int(1/((3*x + 2)^(1/4)*(3*x + 4)),x)
 

Output:

2^(1/4)*atan(2^(1/4)*(3*x + 2)^(1/4)*(1/2 - 1i/2))*(1/3 - 1i/3) + 2^(1/4)* 
atan(2^(1/4)*(3*x + 2)^(1/4)*(1/2 + 1i/2))*(1/3 + 1i/3)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.32 \[ \int \frac {1}{\sqrt [4]{2+3 x} (4+3 x)} \, dx=\frac {2^{\frac {1}{4}} \left (2 \mathit {atan} \left (\frac {\left (2 \left (3 x +2\right )^{\frac {1}{4}}-\sqrt {2}\, 2^{\frac {1}{4}}\right ) 2^{\frac {3}{4}}}{2 \sqrt {2}}\right )+2 \mathit {atan} \left (\frac {\left (2 \left (3 x +2\right )^{\frac {1}{4}}+\sqrt {2}\, 2^{\frac {1}{4}}\right ) 2^{\frac {3}{4}}}{2 \sqrt {2}}\right )+\mathrm {log}\left (-\left (3 x +2\right )^{\frac {1}{4}} \sqrt {2}\, 2^{\frac {1}{4}}+\sqrt {3 x +2}+\sqrt {2}\right )-\mathrm {log}\left (\left (3 x +2\right )^{\frac {1}{4}} \sqrt {2}\, 2^{\frac {1}{4}}+\sqrt {3 x +2}+\sqrt {2}\right )\right )}{6} \] Input:

int(1/(2+3*x)^(1/4)/(4+3*x),x)
 

Output:

(2**(1/4)*(2*atan((2*(3*x + 2)**(1/4) - sqrt(2)*2**(1/4))/(sqrt(2)*2**(1/4 
))) + 2*atan((2*(3*x + 2)**(1/4) + sqrt(2)*2**(1/4))/(sqrt(2)*2**(1/4))) + 
 log( - (3*x + 2)**(1/4)*sqrt(2)*2**(1/4) + sqrt(3*x + 2) + sqrt(2)) - log 
((3*x + 2)**(1/4)*sqrt(2)*2**(1/4) + sqrt(3*x + 2) + sqrt(2))))/6