\(\int \frac {1}{(1-2 x)^{3/2} (2+3 x)^2 \sqrt {3+5 x}} \, dx\) [1161]

Optimal result

Integrand size = 26, antiderivative size = 86 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^2 \sqrt {3+5 x}} \, dx=-\frac {58 \sqrt {3+5 x}}{539 \sqrt {1-2 x}}+\frac {3 \sqrt {3+5 x}}{7 \sqrt {1-2 x} (2+3 x)}-\frac {123 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{49 \sqrt {7}} \] Output:

-58/539*(3+5*x)^(1/2)/(1-2*x)^(1/2)+3/7*(3+5*x)^(1/2)/(1-2*x)^(1/2)/(2+3*x 
)-123/343*7^(1/2)*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.91 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^2 \sqrt {3+5 x}} \, dx=\frac {7 (115-174 x) \sqrt {3+5 x}-1353 \sqrt {7-14 x} (2+3 x) \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{3773 \sqrt {1-2 x} (2+3 x)} \] Input:

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

Output:

(7*(115 - 174*x)*Sqrt[3 + 5*x] - 1353*Sqrt[7 - 14*x]*(2 + 3*x)*ArcTan[Sqrt 
[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(3773*Sqrt[1 - 2*x]*(2 + 3*x))
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {114, 27, 169, 27, 104, 217}

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.

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

\(\Big \downarrow \) 114

\(\displaystyle \frac {1}{7} \int \frac {1-60 x}{2 (1-2 x)^{3/2} (3 x+2) \sqrt {5 x+3}}dx+\frac {3 \sqrt {5 x+3}}{7 \sqrt {1-2 x} (3 x+2)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{14} \int \frac {1-60 x}{(1-2 x)^{3/2} (3 x+2) \sqrt {5 x+3}}dx+\frac {3 \sqrt {5 x+3}}{7 \sqrt {1-2 x} (3 x+2)}\)

\(\Big \downarrow \) 169

\(\displaystyle \frac {1}{14} \left (-\frac {2}{77} \int -\frac {1353}{2 \sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {116 \sqrt {5 x+3}}{77 \sqrt {1-2 x}}\right )+\frac {3 \sqrt {5 x+3}}{7 \sqrt {1-2 x} (3 x+2)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{14} \left (\frac {123}{7} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {116 \sqrt {5 x+3}}{77 \sqrt {1-2 x}}\right )+\frac {3 \sqrt {5 x+3}}{7 \sqrt {1-2 x} (3 x+2)}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {1}{14} \left (\frac {246}{7} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {116 \sqrt {5 x+3}}{77 \sqrt {1-2 x}}\right )+\frac {3 \sqrt {5 x+3}}{7 \sqrt {1-2 x} (3 x+2)}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{14} \left (-\frac {246 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{7 \sqrt {7}}-\frac {116 \sqrt {5 x+3}}{77 \sqrt {1-2 x}}\right )+\frac {3 \sqrt {5 x+3}}{7 \sqrt {1-2 x} (3 x+2)}\)

Input:

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

Output:

(3*Sqrt[3 + 5*x])/(7*Sqrt[1 - 2*x]*(2 + 3*x)) + ((-116*Sqrt[3 + 5*x])/(77* 
Sqrt[1 - 2*x]) - (246*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(7*Sq 
rt[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 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 114

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e 
 - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) 
 - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || 
 IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
 

rule 169

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S 
imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n 
*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n, 2*p]
 

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])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(153\) vs. \(2(65)=130\).

Time = 0.26 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.79

method	result	size
default	\(-\frac {\left (8118 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}+1353 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x -2706 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+2436 x \sqrt {-10 x^{2}-x +3}-1610 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {3+5 x}}{7546 \left (2+3 x \right ) \sqrt {1-2 x}\, \sqrt {-10 x^{2}-x +3}}\)	\(154\)

Input:

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

Output:

-1/7546*(8118*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x 
^2+1353*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x-2706* 
7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+2436*x*(-10*x^2 
-x+3)^(1/2)-1610*(-10*x^2-x+3)^(1/2))*(3+5*x)^(1/2)/(2+3*x)/(1-2*x)^(1/2)/ 
(-10*x^2-x+3)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^2 \sqrt {3+5 x}} \, dx=-\frac {1353 \, \sqrt {7} {\left (6 \, x^{2} + x - 2\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (174 \, x - 115\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{7546 \, {\left (6 \, x^{2} + x - 2\right )}} \] Input:

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

Output:

-1/7546*(1353*sqrt(7)*(6*x^2 + x - 2)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt 
(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 14*(174*x - 115)*sqrt(5*x + 3 
)*sqrt(-2*x + 1))/(6*x^2 + x - 2)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (65) = 130\).

Time = 0.19 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.55 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^2 \sqrt {3+5 x}} \, dx=\frac {123}{6860} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {8 \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{2695 \, {\left (2 \, x - 1\right )}} + \frac {198 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{49 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}} \] Input:

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

Output:

123/6860*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*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)))) - 8/2695*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1) 
 + 198/49*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4 
*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))/(((sqrt(2)*sqrt(-10*x 
 + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) 
 - sqrt(22)))^2 + 280)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.27 \[ \int \frac {1}{(1-2 x)^{3/2} (2+3 x)^2 \sqrt {3+5 x}} \, dx=\frac {4059 \sqrt {-2 x +1}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x +2706 \sqrt {-2 x +1}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )-4059 \sqrt {-2 x +1}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x -2706 \sqrt {-2 x +1}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )-1218 \sqrt {5 x +3}\, x +805 \sqrt {5 x +3}}{3773 \sqrt {-2 x +1}\, \left (3 x +2\right )} \] Input:

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

Output:

(4059*sqrt( - 2*x + 1)*sqrt(7)*atan((sqrt(33) - sqrt(35)*tan(asin((sqrt( - 
 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x + 2706*sqrt( - 2*x + 1)*sqrt(7 
)*atan((sqrt(33) - sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/ 
2))/sqrt(2)) - 4059*sqrt( - 2*x + 1)*sqrt(7)*atan((sqrt(33) + sqrt(35)*tan 
(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x - 2706*sqrt( - 2 
*x + 1)*sqrt(7)*atan((sqrt(33) + sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt( 
5))/sqrt(11))/2))/sqrt(2)) - 1218*sqrt(5*x + 3)*x + 805*sqrt(5*x + 3))/(37 
73*sqrt( - 2*x + 1)*(3*x + 2))