\(\int \frac {(3+5 x)^{3/2}}{(1-2 x)^{3/2} (2+3 x)} \, dx\) [1138]

Optimal result

Integrand size = 26, antiderivative size = 86 \[ \int \frac {(3+5 x)^{3/2}}{(1-2 x)^{3/2} (2+3 x)} \, dx=\frac {11 \sqrt {3+5 x}}{7 \sqrt {1-2 x}}-\frac {5}{3} \sqrt {\frac {5}{2}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )-\frac {2 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{21 \sqrt {7}} \] Output:

11/7*(3+5*x)^(1/2)/(1-2*x)^(1/2)-5/6*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*1 
0^(1/2)-2/147*7^(1/2)*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))
 

Mathematica [A] (verified)

Time = 1.27 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.86 \[ \int \frac {(3+5 x)^{3/2}}{(1-2 x)^{3/2} (2+3 x)} \, dx=\frac {1}{147} \left (\frac {231 \sqrt {3+5 x}}{\sqrt {1-2 x}}+2 \sqrt {7} \arctan \left (\frac {\sqrt {2 \left (34+\sqrt {1155}\right )} \sqrt {3+5 x}}{-\sqrt {11}+\sqrt {5-10 x}}\right )+245 \sqrt {10} \arctan \left (\frac {\sqrt {6+10 x}}{\sqrt {11}-\sqrt {5-10 x}}\right )+2 \sqrt {7} \arctan \left (\frac {\sqrt {6+10 x}}{\sqrt {34+\sqrt {1155}} \left (-\sqrt {11}+\sqrt {5-10 x}\right )}\right )\right ) \] Input:

Integrate[(3 + 5*x)^(3/2)/((1 - 2*x)^(3/2)*(2 + 3*x)),x]
 

Output:

((231*Sqrt[3 + 5*x])/Sqrt[1 - 2*x] + 2*Sqrt[7]*ArcTan[(Sqrt[2*(34 + Sqrt[1 
155])]*Sqrt[3 + 5*x])/(-Sqrt[11] + Sqrt[5 - 10*x])] + 245*Sqrt[10]*ArcTan[ 
Sqrt[6 + 10*x]/(Sqrt[11] - Sqrt[5 - 10*x])] + 2*Sqrt[7]*ArcTan[Sqrt[6 + 10 
*x]/(Sqrt[34 + Sqrt[1155]]*(-Sqrt[11] + Sqrt[5 - 10*x]))])/147
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {109, 27, 175, 64, 104, 217, 223}

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.

\(\displaystyle \int \frac {(5 x+3)^{3/2}}{(1-2 x)^{3/2} (3 x+2)} \, dx\)

\(\Big \downarrow \) 109

\(\displaystyle \frac {11 \sqrt {5 x+3}}{7 \sqrt {1-2 x}}-\frac {1}{7} \int \frac {175 x+116}{2 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {11 \sqrt {5 x+3}}{7 \sqrt {1-2 x}}-\frac {1}{14} \int \frac {175 x+116}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx\)

\(\Big \downarrow \) 175

\(\displaystyle \frac {1}{14} \left (\frac {2}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {175}{3} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx\right )+\frac {11 \sqrt {5 x+3}}{7 \sqrt {1-2 x}}\)

\(\Big \downarrow \) 64

\(\displaystyle \frac {1}{14} \left (\frac {2}{3} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {70}{3} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )+\frac {11 \sqrt {5 x+3}}{7 \sqrt {1-2 x}}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {1}{14} \left (\frac {4}{3} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {70}{3} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}\right )+\frac {11 \sqrt {5 x+3}}{7 \sqrt {1-2 x}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{14} \left (-\frac {70}{3} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {4 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{3 \sqrt {7}}\right )+\frac {11 \sqrt {5 x+3}}{7 \sqrt {1-2 x}}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {1}{14} \left (-\frac {35}{3} \sqrt {10} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {4 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{3 \sqrt {7}}\right )+\frac {11 \sqrt {5 x+3}}{7 \sqrt {1-2 x}}\)

Input:

Int[(3 + 5*x)^(3/2)/((1 - 2*x)^(3/2)*(2 + 3*x)),x]
 

Output:

(11*Sqrt[3 + 5*x])/(7*Sqrt[1 - 2*x]) + ((-35*Sqrt[10]*ArcSin[Sqrt[2/11]*Sq 
rt[3 + 5*x]])/3 - (4*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(3*Sqr 
t[7]))/14
 

Defintions of rubi rules used

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 64

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp 
[2/b   Subst[Int[1/Sqrt[c - a*(d/b) + d*(x^2/b)], x], x, Sqrt[a + b*x]], x] 
 /; FreeQ[{a, b, c, d}, x] && GtQ[c - a*(d/b), 0] && ( !GtQ[a - c*(b/d), 0] 
 || PosQ[b])
 

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 109

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f 
*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) 
 Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) 
+ c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) 
 + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || 
IntegersQ[m, n + p] || IntegersQ[p, m + n])
 

rule 175

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ 
)))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b   Int[(c + d*x)^n*(e + f*x)^p, x] 
, x] + Simp[(b*g - a*h)/b   Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] 
 /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, 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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(123\) vs. \(2(60)=120\).

Time = 0.26 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.44

method	result	size
default	\(\frac {\left (490 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -8 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x -245 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+4 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+924 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {3+5 x}}{588 \sqrt {1-2 x}\, \sqrt {-10 x^{2}-x +3}}\)	\(124\)

Input:

int((3+5*x)^(3/2)/(1-2*x)^(3/2)/(2+3*x),x,method=_RETURNVERBOSE)
 

Output:

1/588*(490*10^(1/2)*arcsin(20/11*x+1/11)*x-8*7^(1/2)*arctan(1/14*(37*x+20) 
*7^(1/2)/(-10*x^2-x+3)^(1/2))*x-245*10^(1/2)*arcsin(20/11*x+1/11)+4*7^(1/2 
)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+924*(-10*x^2-x+3)^(1/ 
2))*(3+5*x)^(1/2)/(1-2*x)^(1/2)/(-10*x^2-x+3)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.35 \[ \int \frac {(3+5 x)^{3/2}}{(1-2 x)^{3/2} (2+3 x)} \, dx=-\frac {2 \, \sqrt {7} {\left (2 \, x - 1\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 245 \, \sqrt {\frac {5}{2}} {\left (2 \, x - 1\right )} \arctan \left (\frac {\sqrt {\frac {5}{2}} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{10 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 462 \, \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{294 \, {\left (2 \, x - 1\right )}} \] Input:

integrate((3+5*x)^(3/2)/(1-2*x)^(3/2)/(2+3*x),x, algorithm="fricas")
 

Output:

-1/294*(2*sqrt(7)*(2*x - 1)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)* 
sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 245*sqrt(5/2)*(2*x - 1)*arctan(1/10*sqr 
t(5/2)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 462*sqr 
t(5*x + 3)*sqrt(-2*x + 1))/(2*x - 1)
 

Sympy [F]

\[ \int \frac {(3+5 x)^{3/2}}{(1-2 x)^{3/2} (2+3 x)} \, dx=\int \frac {\left (5 x + 3\right )^{\frac {3}{2}}}{\left (1 - 2 x\right )^{\frac {3}{2}} \cdot \left (3 x + 2\right )}\, dx \] Input:

integrate((3+5*x)**(3/2)/(1-2*x)**(3/2)/(2+3*x),x)
 

Output:

Integral((5*x + 3)**(3/2)/((1 - 2*x)**(3/2)*(3*x + 2)), x)
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.80 \[ \int \frac {(3+5 x)^{3/2}}{(1-2 x)^{3/2} (2+3 x)} \, dx=-\frac {5}{12} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {1}{147} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {55 \, x}{7 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {33}{7 \, \sqrt {-10 \, x^{2} - x + 3}} \] Input:

integrate((3+5*x)^(3/2)/(1-2*x)^(3/2)/(2+3*x),x, algorithm="maxima")
 

Output:

-5/12*sqrt(10)*arcsin(20/11*x + 1/11) + 1/147*sqrt(7)*arcsin(37/11*x/abs(3 
*x + 2) + 20/11/abs(3*x + 2)) + 55/7*x/sqrt(-10*x^2 - x + 3) + 33/7/sqrt(- 
10*x^2 - x + 3)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 167 vs. \(2 (60) = 120\).

Time = 0.18 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.94 \[ \int \frac {(3+5 x)^{3/2}}{(1-2 x)^{3/2} (2+3 x)} \, dx=\frac {1}{1470} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {5}{12} \, \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {11 \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{35 \, {\left (2 \, x - 1\right )}} \] Input:

integrate((3+5*x)^(3/2)/(1-2*x)^(3/2)/(2+3*x),x, algorithm="giac")
 

Output:

1/1470*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((sq 
rt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5 
) - sqrt(22)))) - 5/12*sqrt(10)*(pi + 2*arctan(-1/4*sqrt(5*x + 3)*((sqrt(2 
)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - 
sqrt(22)))) - 11/35*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(3+5 x)^{3/2}}{(1-2 x)^{3/2} (2+3 x)} \, dx=\int \frac {{\left (5\,x+3\right )}^{3/2}}{{\left (1-2\,x\right )}^{3/2}\,\left (3\,x+2\right )} \,d x \] Input:

int((5*x + 3)^(3/2)/((1 - 2*x)^(3/2)*(3*x + 2)),x)
                                                                                    
                                                                                    
 

Output:

int((5*x + 3)^(3/2)/((1 - 2*x)^(3/2)*(3*x + 2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.42 \[ \int \frac {(3+5 x)^{3/2}}{(1-2 x)^{3/2} (2+3 x)} \, dx=\frac {245 \sqrt {-2 x +1}\, \sqrt {10}\, \mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )+4 \sqrt {-2 x +1}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )-4 \sqrt {-2 x +1}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )+462 \sqrt {5 x +3}}{294 \sqrt {-2 x +1}} \] Input:

int((3+5*x)^(3/2)/(1-2*x)^(3/2)/(2+3*x),x)
 

Output:

(245*sqrt( - 2*x + 1)*sqrt(10)*asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11)) + 
 4*sqrt( - 2*x + 1)*sqrt(7)*atan((sqrt(33) - sqrt(35)*tan(asin((sqrt( - 2* 
x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2)) - 4*sqrt( - 2*x + 1)*sqrt(7)*atan(( 
sqrt(33) + sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt 
(2)) + 462*sqrt(5*x + 3))/(294*sqrt( - 2*x + 1))