\(\int \frac {(2+x)^{4/3} (3+5 x)}{4+7 x+2 x^2} \, dx\) [854]

Optimal result

Integrand size = 25, antiderivative size = 387 \[ \int \frac {(2+x)^{4/3} (3+5 x)}{4+7 x+2 x^2} \, dx=-\frac {3}{136} \left (153-31 \sqrt {17}\right ) \sqrt [3]{2+x}-\frac {3}{136} \left (153+31 \sqrt {17}\right ) \sqrt [3]{2+x}+\frac {3}{272} \left (85-23 \sqrt {17}\right ) (2+x)^{4/3}+\frac {3}{272} \left (85+23 \sqrt {17}\right ) (2+x)^{4/3}-\frac {\sqrt {\frac {3}{17}} \sqrt [3]{61741+14939 \sqrt {17}} \arctan \left (\frac {1-\frac {2\ 2^{2/3} \sqrt [3]{2+x}}{\sqrt [3]{-1+\sqrt {17}}}}{\sqrt {3}}\right )}{4\ 2^{2/3}}+\frac {\sqrt {\frac {3}{17}} \sqrt [3]{3951424-956096 \sqrt {17}} \arctan \left (\frac {1+\frac {2\ 2^{2/3} \sqrt [3]{2+x}}{\sqrt [3]{1+\sqrt {17}}}}{\sqrt {3}}\right )}{16\ 2^{2/3}}+\frac {\sqrt [3]{3951424-956096 \sqrt {17}} \log \left (7-\sqrt {17}+4 x\right )}{32\ 2^{2/3} \sqrt {17}}-\frac {\sqrt [3]{61741+14939 \sqrt {17}} \log \left (7+\sqrt {17}+4 x\right )}{8\ 2^{2/3} \sqrt {17}}-\frac {3 \sqrt [3]{3951424-956096 \sqrt {17}} \log \left (\sqrt [3]{2 \left (1+\sqrt {17}\right )}-2 \sqrt [3]{2+x}\right )}{32\ 2^{2/3} \sqrt {17}}+\frac {3 \sqrt [3]{61741+14939 \sqrt {17}} \log \left (\sqrt [3]{2 \left (-1+\sqrt {17}\right )}+2 \sqrt [3]{2+x}\right )}{8\ 2^{2/3} \sqrt {17}} \] Output:

-3/136*(153-31*17^(1/2))*(2+x)^(1/3)-3/136*(153+31*17^(1/2))*(2+x)^(1/3)+3 
/272*(85-23*17^(1/2))*(2+x)^(4/3)+3/272*(85+23*17^(1/2))*(2+x)^(4/3)-1/136 
*51^(1/2)*(61741+14939*17^(1/2))^(1/3)*arctan(1/3*(1-2*2^(2/3)*(2+x)^(1/3) 
/(-1+17^(1/2))^(1/3))*3^(1/2))*2^(1/3)+1/544*51^(1/2)*(3951424-956096*17^( 
1/2))^(1/3)*arctan(1/3*(1+2*2^(2/3)*(2+x)^(1/3)/(1+17^(1/2))^(1/3))*3^(1/2 
))*2^(1/3)+1/1088*(3951424-956096*17^(1/2))^(1/3)*ln(7-17^(1/2)+4*x)*2^(1/ 
3)*17^(1/2)-1/272*(61741+14939*17^(1/2))^(1/3)*ln(7+17^(1/2)+4*x)*2^(1/3)* 
17^(1/2)-3/1088*(3951424-956096*17^(1/2))^(1/3)*ln((2+2*17^(1/2))^(1/3)-2* 
(2+x)^(1/3))*2^(1/3)*17^(1/2)+3/272*(61741+14939*17^(1/2))^(1/3)*ln((-2+2* 
17^(1/2))^(1/3)+2*(2+x)^(1/3))*2^(1/3)*17^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.06 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.24 \[ \int \frac {(2+x)^{4/3} (3+5 x)}{4+7 x+2 x^2} \, dx=\frac {3}{8} \sqrt [3]{2+x} (-18+5 (2+x))+\frac {1}{4} \text {RootSum}\left [-2-\text {$\#$1}^3+2 \text {$\#$1}^6\&,\frac {-18 \log \left (\sqrt [3]{2+x}-\text {$\#$1}\right )+11 \log \left (\sqrt [3]{2+x}-\text {$\#$1}\right ) \text {$\#$1}^3}{-\text {$\#$1}^2+4 \text {$\#$1}^5}\&\right ] \] Input:

Integrate[((2 + x)^(4/3)*(3 + 5*x))/(4 + 7*x + 2*x^2),x]
 

Output:

(3*(2 + x)^(1/3)*(-18 + 5*(2 + x)))/8 + RootSum[-2 - #1^3 + 2*#1^6 & , (-1 
8*Log[(2 + x)^(1/3) - #1] + 11*Log[(2 + x)^(1/3) - #1]*#1^3)/(-#1^2 + 4*#1 
^5) & ]/4
 

Rubi [A] (warning: unable to verify)

Time = 0.99 (sec) , antiderivative size = 314, normalized size of antiderivative = 0.81, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1196, 25, 1196, 25, 1200, 2009}

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.

\(\displaystyle \int \frac {(x+2)^{4/3} (5 x+3)}{2 x^2+7 x+4} \, dx\)

\(\Big \downarrow \) 1196

\(\displaystyle \frac {1}{2} \int -\frac {\sqrt [3]{x+2} (9 x+8)}{2 x^2+7 x+4}dx+\frac {15}{8} (x+2)^{4/3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {15}{8} (x+2)^{4/3}-\frac {1}{2} \int \frac {\sqrt [3]{x+2} (9 x+8)}{2 x^2+7 x+4}dx\)

\(\Big \downarrow \) 1196

\(\displaystyle \frac {1}{2} \left (-\frac {1}{2} \int -\frac {11 x+4}{(x+2)^{2/3} \left (2 x^2+7 x+4\right )}dx-\frac {27 \sqrt [3]{x+2}}{2}\right )+\frac {15}{8} (x+2)^{4/3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {11 x+4}{(x+2)^{2/3} \left (2 x^2+7 x+4\right )}dx-\frac {27 \sqrt [3]{x+2}}{2}\right )+\frac {15}{8} (x+2)^{4/3}\)

\(\Big \downarrow \) 1200

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \left (\frac {11-\frac {61}{\sqrt {17}}}{\left (4 x-\sqrt {17}+7\right ) (x+2)^{2/3}}+\frac {11+\frac {61}{\sqrt {17}}}{\left (4 x+\sqrt {17}+7\right ) (x+2)^{2/3}}\right )dx-\frac {27 \sqrt [3]{x+2}}{2}\right )+\frac {15}{8} (x+2)^{4/3}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\frac {\sqrt {\frac {3}{17}} \sqrt [3]{61741+14939 \sqrt {17}} \arctan \left (\frac {1-\frac {2\ 2^{2/3} \sqrt [3]{x+2}}{\sqrt [3]{\sqrt {17}-1}}}{\sqrt {3}}\right )}{2^{2/3}}+\frac {\sqrt {\frac {3}{17}} \sqrt [3]{61741-14939 \sqrt {17}} \arctan \left (\frac {\frac {2\ 2^{2/3} \sqrt [3]{x+2}}{\sqrt [3]{1+\sqrt {17}}}+1}{\sqrt {3}}\right )}{2^{2/3}}+\frac {\sqrt [3]{61741 \sqrt {17}-253963} \log \left (4 x-\sqrt {17}+7\right )}{2\ 34^{2/3}}-\frac {\sqrt [3]{253963+61741 \sqrt {17}} \log \left (4 x+\sqrt {17}+7\right )}{2\ 34^{2/3}}-\frac {3 \sqrt [3]{61741 \sqrt {17}-253963} \log \left (\sqrt [3]{2 \left (1+\sqrt {17}\right )}-2 \sqrt [3]{x+2}\right )}{2\ 34^{2/3}}+\frac {3 \sqrt [3]{253963+61741 \sqrt {17}} \log \left (2 \sqrt [3]{x+2}+\sqrt [3]{2 \left (\sqrt {17}-1\right )}\right )}{2\ 34^{2/3}}\right )-\frac {27 \sqrt [3]{x+2}}{2}\right )+\frac {15}{8} (x+2)^{4/3}\)

Input:

Int[((2 + x)^(4/3)*(3 + 5*x))/(4 + 7*x + 2*x^2),x]
 

Output:

(15*(2 + x)^(4/3))/8 + ((-27*(2 + x)^(1/3))/2 + (-((Sqrt[3/17]*(61741 + 14 
939*Sqrt[17])^(1/3)*ArcTan[(1 - (2*2^(2/3)*(2 + x)^(1/3))/(-1 + Sqrt[17])^ 
(1/3))/Sqrt[3]])/2^(2/3)) + (Sqrt[3/17]*(61741 - 14939*Sqrt[17])^(1/3)*Arc 
Tan[(1 + (2*2^(2/3)*(2 + x)^(1/3))/(1 + Sqrt[17])^(1/3))/Sqrt[3]])/2^(2/3) 
 + ((-253963 + 61741*Sqrt[17])^(1/3)*Log[7 - Sqrt[17] + 4*x])/(2*34^(2/3)) 
 - ((253963 + 61741*Sqrt[17])^(1/3)*Log[7 + Sqrt[17] + 4*x])/(2*34^(2/3)) 
- (3*(-253963 + 61741*Sqrt[17])^(1/3)*Log[(2*(1 + Sqrt[17]))^(1/3) - 2*(2 
+ x)^(1/3)])/(2*34^(2/3)) + (3*(253963 + 61741*Sqrt[17])^(1/3)*Log[(2*(-1 
+ Sqrt[17]))^(1/3) + 2*(2 + x)^(1/3)])/(2*34^(2/3)))/2)/2
 

Defintions of rubi rules used

rule 25

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

rule 1196

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + 
(c_.)*(x_)^2), x_Symbol] :> Simp[g*((d + e*x)^m/(c*m)), x] + Simp[1/c   Int 
[(d + e*x)^(m - 1)*(Simp[c*d*f - a*e*g + (g*c*d - b*e*g + c*e*f)*x, x]/(a + 
 b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && FractionQ[m] & 
& GtQ[m, 0]
 

rule 1200

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* 
(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* 
x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In 
tegersQ[n]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 41.76 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.17

method	result	size
derivativedivides	\(\frac {15 \left (2+x \right )^{\frac {4}{3}}}{8}-\frac {27 \left (2+x \right )^{\frac {1}{3}}}{4}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (2 \textit {\_Z}^{6}-\textit {\_Z}^{3}-2\right )}{\sum }\frac {\left (11 \textit {\_R}^{3}-18\right ) \ln \left (\left (2+x \right )^{\frac {1}{3}}-\textit {\_R} \right )}{4 \textit {\_R}^{5}-\textit {\_R}^{2}}\right )}{4}\)	\(65\)
default	\(\frac {15 \left (2+x \right )^{\frac {4}{3}}}{8}-\frac {27 \left (2+x \right )^{\frac {1}{3}}}{4}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (2 \textit {\_Z}^{6}-\textit {\_Z}^{3}-2\right )}{\sum }\frac {\left (11 \textit {\_R}^{3}-18\right ) \ln \left (\left (2+x \right )^{\frac {1}{3}}-\textit {\_R} \right )}{4 \textit {\_R}^{5}-\textit {\_R}^{2}}\right )}{4}\)	\(65\)
trager	\(\text {Expression too large to display}\)	\(8056\)
risch	\(\text {Expression too large to display}\)	\(13962\)

Input:

int((2+x)^(4/3)*(5*x+3)/(2*x^2+7*x+4),x,method=_RETURNVERBOSE)
 

Output:

15/8*(2+x)^(4/3)-27/4*(2+x)^(1/3)+1/4*sum((11*_R^3-18)/(4*_R^5-_R^2)*ln((2 
+x)^(1/3)-_R),_R=RootOf(2*_Z^6-_Z^3-2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 298, normalized size of antiderivative = 0.77 \[ \int \frac {(2+x)^{4/3} (3+5 x)}{4+7 x+2 x^2} \, dx =\text {Too large to display} \] Input:

integrate((2+x)^(4/3)*(3+5*x)/(2*x^2+7*x+4),x, algorithm="fricas")
 

Output:

-1/8*(61741/1156*sqrt(17) + 14939/68)^(1/3)*(sqrt(-3) + 1)*log((31*sqrt(17 
)*(sqrt(-3) + 1) - 153*sqrt(-3) - 153)*(61741/1156*sqrt(17) + 14939/68)^(1 
/3) + 416*(x + 2)^(1/3)) + 1/8*(61741/1156*sqrt(17) + 14939/68)^(1/3)*(sqr 
t(-3) - 1)*log(-(31*sqrt(17)*(sqrt(-3) - 1) - 153*sqrt(-3) + 153)*(61741/1 
156*sqrt(17) + 14939/68)^(1/3) + 416*(x + 2)^(1/3)) - 1/8*(-61741/1156*sqr 
t(17) + 14939/68)^(1/3)*(sqrt(-3) + 1)*log(-(31*sqrt(17)*(sqrt(-3) + 1) + 
153*sqrt(-3) + 153)*(-61741/1156*sqrt(17) + 14939/68)^(1/3) + 416*(x + 2)^ 
(1/3)) + 1/8*(-61741/1156*sqrt(17) + 14939/68)^(1/3)*(sqrt(-3) - 1)*log((3 
1*sqrt(17)*(sqrt(-3) - 1) + 153*sqrt(-3) - 153)*(-61741/1156*sqrt(17) + 14 
939/68)^(1/3) + 416*(x + 2)^(1/3)) + 3/8*(5*x - 8)*(x + 2)^(1/3) + 1/4*(61 
741/1156*sqrt(17) + 14939/68)^(1/3)*log(-(61741/1156*sqrt(17) + 14939/68)^ 
(1/3)*(31*sqrt(17) - 153) + 208*(x + 2)^(1/3)) + 1/4*(-61741/1156*sqrt(17) 
 + 14939/68)^(1/3)*log((31*sqrt(17) + 153)*(-61741/1156*sqrt(17) + 14939/6 
8)^(1/3) + 208*(x + 2)^(1/3))
 

Sympy [F]

\[ \int \frac {(2+x)^{4/3} (3+5 x)}{4+7 x+2 x^2} \, dx=\int \frac {\left (x + 2\right )^{\frac {4}{3}} \cdot \left (5 x + 3\right )}{2 x^{2} + 7 x + 4}\, dx \] Input:

integrate((2+x)**(4/3)*(3+5*x)/(2*x**2+7*x+4),x)
 

Output:

Integral((x + 2)**(4/3)*(5*x + 3)/(2*x**2 + 7*x + 4), x)
 

Maxima [F]

\[ \int \frac {(2+x)^{4/3} (3+5 x)}{4+7 x+2 x^2} \, dx=\int { \frac {{\left (5 \, x + 3\right )} {\left (x + 2\right )}^{\frac {4}{3}}}{2 \, x^{2} + 7 \, x + 4} \,d x } \] Input:

integrate((2+x)^(4/3)*(3+5*x)/(2*x^2+7*x+4),x, algorithm="maxima")
 

Output:

integrate((5*x + 3)*(x + 2)^(4/3)/(2*x^2 + 7*x + 4), x)
 

Giac [F]

\[ \int \frac {(2+x)^{4/3} (3+5 x)}{4+7 x+2 x^2} \, dx=\int { \frac {{\left (5 \, x + 3\right )} {\left (x + 2\right )}^{\frac {4}{3}}}{2 \, x^{2} + 7 \, x + 4} \,d x } \] Input:

integrate((2+x)^(4/3)*(3+5*x)/(2*x^2+7*x+4),x, algorithm="giac")
 

Output:

integrate((5*x + 3)*(x + 2)^(4/3)/(2*x^2 + 7*x + 4), x)
 

Mupad [B] (verification not implemented)

Time = 11.75 (sec) , antiderivative size = 567, normalized size of antiderivative = 1.47 \[ \int \frac {(2+x)^{4/3} (3+5 x)}{4+7 x+2 x^2} \, dx=\text {Too large to display} \] Input:

int(((5*x + 3)*(x + 2)^(4/3))/(7*x + 2*x^2 + 4),x)
 

Output:

(15*(x + 2)^(4/3))/8 - (27*(x + 2)^(1/3))/4 + (34^(1/3)*log((9126351*(x + 
2)^(1/3))/2048 + (34^(1/3)*(253963 - 61741*17^(1/2))^(1/3)*((34^(2/3)*(253 
963 - 61741*17^(1/2))^(2/3)*((351135*(x + 2)^(1/3))/32 + (4131*34^(1/3)*(2 
53963 - 61741*17^(1/2))^(1/3))/256))/18496 + 96669531/4096))/136)*(253963 
- 61741*17^(1/2))^(1/3))/136 - (34^(1/3)*log((9126351*(x + 2)^(1/3))/2048 
- (34^(1/3)*(61741*17^(1/2) - 253963)^(1/3)*((34^(2/3)*(61741*17^(1/2) - 2 
53963)^(2/3)*((351135*(x + 2)^(1/3))/32 - (4131*34^(1/3)*(61741*17^(1/2) - 
 253963)^(1/3))/256))/18496 + 96669531/4096))/136)*(61741*17^(1/2) - 25396 
3)^(1/3))/136 + (34^(1/3)*log((9126351*(x + 2)^(1/3))/2048 + (34^(1/3)*(61 
741*17^(1/2) + 253963)^(1/3)*((34^(2/3)*(61741*17^(1/2) + 253963)^(2/3)*(( 
351135*(x + 2)^(1/3))/32 + (4131*34^(1/3)*(61741*17^(1/2) + 253963)^(1/3)) 
/256))/18496 + 96669531/4096))/136)*(61741*17^(1/2) + 253963)^(1/3))/136 + 
 (34^(1/3)*log((9126351*(x + 2)^(1/3))/2048 + (34^(1/3)*(3^(1/2)*1i - 1)*( 
61741*17^(1/2) + 253963)^(1/3)*((34^(2/3)*(3^(1/2)*1i - 1)^2*(61741*17^(1/ 
2) + 253963)^(2/3)*((351135*(x + 2)^(1/3))/32 + (4131*34^(1/3)*(3^(1/2)*1i 
 - 1)*(61741*17^(1/2) + 253963)^(1/3))/512))/73984 + 96669531/4096))/272)* 
(3^(1/2)*1i - 1)*(61741*17^(1/2) + 253963)^(1/3))/272 - (34^(1/3)*log((912 
6351*(x + 2)^(1/3))/2048 - (34^(1/3)*(3^(1/2)*1i + 1)*(253963 - 61741*17^( 
1/2))^(1/3)*((34^(2/3)*(3^(1/2)*1i + 1)^2*(253963 - 61741*17^(1/2))^(2/3)* 
((351135*(x + 2)^(1/3))/32 - (4131*34^(1/3)*(3^(1/2)*1i + 1)*(253963 - ...
 

Reduce [F]

\[ \int \frac {(2+x)^{4/3} (3+5 x)}{4+7 x+2 x^2} \, dx=\int \frac {\left (x +2\right )^{\frac {4}{3}} \left (5 x +3\right )}{2 x^{2}+7 x +4}d x \] Input:

int((2+x)^(4/3)*(3+5*x)/(2*x^2+7*x+4),x)
 

Output:

int((2+x)^(4/3)*(3+5*x)/(2*x^2+7*x+4),x)