3.26.74 \(\int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx\) [2574]

3.26.74.1 Optimal result

Integrand size = 26, antiderivative size = 67 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx=\frac {49}{22 \sqrt {1-2 x} (3+5 x)^{3/2}}-\frac {3679 \sqrt {1-2 x}}{3630 (3+5 x)^{3/2}}-\frac {4091 \sqrt {1-2 x}}{19965 \sqrt {3+5 x}} \]

output

49/22/(3+5*x)^(3/2)/(1-2*x)^(1/2)-3679/3630*(1-2*x)^(1/2)/(3+5*x)^(3/2)-40 
91/19965*(1-2*x)^(1/2)/(3+5*x)^(1/2)
 

3.26.74.2 Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.48 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx=\frac {2 \left (1196+4456 x+4091 x^2\right )}{3993 \sqrt {1-2 x} (3+5 x)^{3/2}} \]

input

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

output

(2*(1196 + 4456*x + 4091*x^2))/(3993*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))
 

3.26.74.3 Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {100, 27, 87, 48}

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.

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

\(\Big \downarrow \) 100

\(\displaystyle \frac {49}{22 \sqrt {1-2 x} (5 x+3)^{3/2}}-\frac {1}{22} \int -\frac {617-198 x}{2 \sqrt {1-2 x} (5 x+3)^{5/2}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{44} \int \frac {617-198 x}{\sqrt {1-2 x} (5 x+3)^{5/2}}dx+\frac {49}{22 \sqrt {1-2 x} (5 x+3)^{3/2}}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {1}{44} \left (\frac {8182}{165} \int \frac {1}{\sqrt {1-2 x} (5 x+3)^{3/2}}dx-\frac {7358 \sqrt {1-2 x}}{165 (5 x+3)^{3/2}}\right )+\frac {49}{22 \sqrt {1-2 x} (5 x+3)^{3/2}}\)

\(\Big \downarrow \) 48

\(\displaystyle \frac {1}{44} \left (-\frac {16364 \sqrt {1-2 x}}{1815 \sqrt {5 x+3}}-\frac {7358 \sqrt {1-2 x}}{165 (5 x+3)^{3/2}}\right )+\frac {49}{22 \sqrt {1-2 x} (5 x+3)^{3/2}}\)

input

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

output

49/(22*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) + ((-7358*Sqrt[1 - 2*x])/(165*(3 + 5 
*x)^(3/2)) - (16364*Sqrt[1 - 2*x])/(1815*Sqrt[3 + 5*x]))/44
 

3.26.74.3.1 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 48

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp 
[(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ 
a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 100

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

3.26.74.4 Maple [A] (verified)

Time = 1.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.40

method	result	size
gosper	\(\frac {\frac {8182}{3993} x^{2}+\frac {8912}{3993} x +\frac {2392}{3993}}{\left (3+5 x \right )^{\frac {3}{2}} \sqrt {1-2 x}}\)	\(27\)
default	\(-\frac {2 \sqrt {1-2 x}\, \left (4091 x^{2}+4456 x +1196\right )}{3993 \left (3+5 x \right )^{\frac {3}{2}} \left (-1+2 x \right )}\)	\(34\)

input

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

output

2/3993/(1-2*x)^(1/2)/(3+5*x)^(3/2)*(4091*x^2+4456*x+1196)
 

3.26.74.5 Fricas [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.64 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx=-\frac {2 \, {\left (4091 \, x^{2} + 4456 \, x + 1196\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{3993 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \]

input

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

output

-2/3993*(4091*x^2 + 4456*x + 1196)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(50*x^3 + 
35*x^2 - 12*x - 9)
 

3.26.74.6 Sympy [F]

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

input

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

output

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

3.26.74.7 Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.96 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx=\frac {8182 \, x}{19965 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {20014}{99825 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {2}{825 \, {\left (5 \, \sqrt {-10 \, x^{2} - x + 3} x + 3 \, \sqrt {-10 \, x^{2} - x + 3}\right )}} \]

input

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

output

8182/19965*x/sqrt(-10*x^2 - x + 3) + 20014/99825/sqrt(-10*x^2 - x + 3) - 2 
/825/(5*sqrt(-10*x^2 - x + 3)*x + 3*sqrt(-10*x^2 - x + 3))
 

3.26.74.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (49) = 98\).

Time = 0.33 (sec) , antiderivative size = 148, normalized size of antiderivative = 2.21 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx=-\frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{1597200 \, {\left (5 \, x + 3\right )}^{\frac {3}{2}}} - \frac {139 \, \sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{133100 \, \sqrt {5 \, x + 3}} - \frac {98 \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{6655 \, {\left (2 \, x - 1\right )}} + \frac {\sqrt {10} {\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (\frac {417 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} + 4\right )}}{99825 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \]

input

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

output

-1/1597200*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) 
 - 139/133100*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) 
- 98/6655*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1) + 1/99825*sqrt(1 
0)*(5*x + 3)^(3/2)*(417*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) + 
 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3
 

3.26.74.9 Mupad [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.76 \[ \int \frac {(2+3 x)^2}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx=\frac {\sqrt {5\,x+3}\,\left (\frac {8182\,x^2}{99825}+\frac {8912\,x}{99825}+\frac {2392}{99825}\right )}{\frac {6\,x\,\sqrt {1-2\,x}}{5}+\frac {9\,\sqrt {1-2\,x}}{25}+x^2\,\sqrt {1-2\,x}} \]

input

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

output

((5*x + 3)^(1/2)*((8912*x)/99825 + (8182*x^2)/99825 + 2392/99825))/((6*x*( 
1 - 2*x)^(1/2))/5 + (9*(1 - 2*x)^(1/2))/25 + x^2*(1 - 2*x)^(1/2))