\(\int \frac {\sqrt {1-2 x}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx\) [910]

Optimal result

Integrand size = 26, antiderivative size = 93 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\frac {\sqrt {1-2 x} \sqrt {3+5 x}}{2 (2+3 x)^2}+\frac {103 \sqrt {1-2 x} \sqrt {3+5 x}}{28 (2+3 x)}-\frac {1177 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{28 \sqrt {7}} \] Output:

1/2*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x)^2+103*(1-2*x)^(1/2)*(3+5*x)^(1/2)/ 
(56+84*x)-1177/196*7^(1/2)*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))
 

Mathematica [A] (verified)

Time = 1.66 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.45 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\frac {1}{196} \left (\frac {7 \sqrt {1-2 x} \sqrt {3+5 x} (220+309 x)}{(2+3 x)^2}+1177 \sqrt {7} \arctan \left (\frac {\sqrt {2 \left (34+\sqrt {1155}\right )} \sqrt {3+5 x}}{-\sqrt {11}+\sqrt {5-10 x}}\right )+1177 \sqrt {7} \arctan \left (\frac {\sqrt {6+10 x}}{\sqrt {34+\sqrt {1155}} \left (-\sqrt {11}+\sqrt {5-10 x}\right )}\right )\right ) \] Input:

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

Output:

((7*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(220 + 309*x))/(2 + 3*x)^2 + 1177*Sqrt[7]* 
ArcTan[(Sqrt[2*(34 + Sqrt[1155])]*Sqrt[3 + 5*x])/(-Sqrt[11] + Sqrt[5 - 10* 
x])] + 1177*Sqrt[7]*ArcTan[Sqrt[6 + 10*x]/(Sqrt[34 + Sqrt[1155]]*(-Sqrt[11 
] + Sqrt[5 - 10*x]))])/196
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {107, 105, 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.

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

\(\Big \downarrow \) 107

\(\displaystyle \frac {107}{28} \int \frac {\sqrt {1-2 x}}{(3 x+2)^2 \sqrt {5 x+3}}dx+\frac {3 \sqrt {5 x+3} (1-2 x)^{3/2}}{14 (3 x+2)^2}\)

\(\Big \downarrow \) 105

\(\displaystyle \frac {107}{28} \left (\frac {11}{2} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx+\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{3 x+2}\right )+\frac {3 \sqrt {5 x+3} (1-2 x)^{3/2}}{14 (3 x+2)^2}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {107}{28} \left (11 \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}+\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{3 x+2}\right )+\frac {3 \sqrt {5 x+3} (1-2 x)^{3/2}}{14 (3 x+2)^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {107}{28} \left (\frac {\sqrt {1-2 x} \sqrt {5 x+3}}{3 x+2}-\frac {11 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{\sqrt {7}}\right )+\frac {3 \sqrt {5 x+3} (1-2 x)^{3/2}}{14 (3 x+2)^2}\)

Input:

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

Output:

(3*(1 - 2*x)^(3/2)*Sqrt[3 + 5*x])/(14*(2 + 3*x)^2) + (107*((Sqrt[1 - 2*x]* 
Sqrt[3 + 5*x])/(2 + 3*x) - (11*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x] 
)])/Sqrt[7]))/28
 

Defintions of rubi rules used

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 105

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

rule 107

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[(a*d*f*(m + 1) + b*c*f*(n + 
 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x 
] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
 

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 [A] (verified)

Time = 0.34 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.28

method	result	size
risch	\(-\frac {\left (-1+2 x \right ) \sqrt {3+5 x}\, \left (220+309 x \right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{28 \left (2+3 x \right )^{2} \sqrt {-\left (-1+2 x \right ) \left (3+5 x \right )}\, \sqrt {1-2 x}}+\frac {1177 \sqrt {7}\, \arctan \left (\frac {9 \left (\frac {20}{3}+\frac {37 x}{3}\right ) \sqrt {7}}{14 \sqrt {-90 \left (\frac {2}{3}+x \right )^{2}+67+111 x}}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{392 \sqrt {1-2 x}\, \sqrt {3+5 x}}\)	\(119\)
default	\(\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (10593 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}+14124 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +4708 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+4326 x \sqrt {-10 x^{2}-x +3}+3080 \sqrt {-10 x^{2}-x +3}\right )}{392 \sqrt {-10 x^{2}-x +3}\, \left (2+3 x \right )^{2}}\)	\(154\)

Input:

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

Output:

-1/28*(-1+2*x)*(3+5*x)^(1/2)*(220+309*x)/(2+3*x)^2/(-(-1+2*x)*(3+5*x))^(1/ 
2)*((1-2*x)*(3+5*x))^(1/2)/(1-2*x)^(1/2)+1177/392*7^(1/2)*arctan(9/14*(20/ 
3+37/3*x)*7^(1/2)/(-90*(2/3+x)^2+67+111*x)^(1/2))*((1-2*x)*(3+5*x))^(1/2)/ 
(1-2*x)^(1/2)/(3+5*x)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=-\frac {1177 \, \sqrt {7} {\left (9 \, x^{2} + 12 \, x + 4\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (309 \, x + 220\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{392 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \] Input:

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

Output:

-1/392*(1177*sqrt(7)*(9*x^2 + 12*x + 4)*arctan(1/14*sqrt(7)*(37*x + 20)*sq 
rt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 14*(309*x + 220)*sqrt(5*x + 
 3)*sqrt(-2*x + 1))/(9*x^2 + 12*x + 4)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\frac {1177}{392} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {\sqrt {-10 \, x^{2} - x + 3}}{2 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} + \frac {103 \, \sqrt {-10 \, x^{2} - x + 3}}{28 \, {\left (3 \, x + 2\right )}} \] Input:

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

Output:

1177/392*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 1/2*s 
qrt(-10*x^2 - x + 3)/(9*x^2 + 12*x + 4) + 103/28*sqrt(-10*x^2 - x + 3)/(3* 
x + 2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (72) = 144\).

Time = 0.18 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.76 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\frac {11}{3920} \, \sqrt {5} {\left (107 \, \sqrt {70} \sqrt {2} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {280 \, \sqrt {2} {\left (173 \, {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} + \frac {29960 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} - \frac {119840 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{{\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{2}}\right )} \] Input:

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

Output:

11/3920*sqrt(5)*(107*sqrt(70)*sqrt(2)*(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)*s 
qrt(-10*x + 5) - sqrt(22)))) + 280*sqrt(2)*(173*((sqrt(2)*sqrt(-10*x + 5) 
- sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqr 
t(22)))^3 + 29960*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 119 
840*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))/(((sqrt(2)*sqrt(-1 
0*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 
 5) - sqrt(22)))^2 + 280)^2)
 

Mupad [B] (verification not implemented)

Time = 8.28 (sec) , antiderivative size = 1037, normalized size of antiderivative = 11.15 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\text {Too large to display} \] Input:

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

Output:

((18323*((1 - 2*x)^(1/2) - 1)^5)/(875*(3^(1/2) - (5*x + 3)^(1/2))^5) - (36 
646*((1 - 2*x)^(1/2) - 1)^3)/(4375*(3^(1/2) - (5*x + 3)^(1/2))^3) - (2326* 
((1 - 2*x)^(1/2) - 1))/(4375*(3^(1/2) - (5*x + 3)^(1/2))) + (1163*((1 - 2* 
x)^(1/2) - 1)^7)/(140*(3^(1/2) - (5*x + 3)^(1/2))^7) + (10607*3^(1/2)*((1 
- 2*x)^(1/2) - 1)^2)/(4375*(3^(1/2) - (5*x + 3)^(1/2))^2) - (3646*3^(1/2)* 
((1 - 2*x)^(1/2) - 1)^4)/(875*(3^(1/2) - (5*x + 3)^(1/2))^4) + (10607*3^(1 
/2)*((1 - 2*x)^(1/2) - 1)^6)/(700*(3^(1/2) - (5*x + 3)^(1/2))^6))/((544*(( 
1 - 2*x)^(1/2) - 1)^2)/(625*(3^(1/2) - (5*x + 3)^(1/2))^2) - (1764*((1 - 2 
*x)^(1/2) - 1)^4)/(625*(3^(1/2) - (5*x + 3)^(1/2))^4) + (136*((1 - 2*x)^(1 
/2) - 1)^6)/(25*(3^(1/2) - (5*x + 3)^(1/2))^6) + ((1 - 2*x)^(1/2) - 1)^8/( 
3^(1/2) - (5*x + 3)^(1/2))^8 - (96*3^(1/2)*((1 - 2*x)^(1/2) - 1)^3)/(625*( 
3^(1/2) - (5*x + 3)^(1/2))^3) + (48*3^(1/2)*((1 - 2*x)^(1/2) - 1)^5)/(125* 
(3^(1/2) - (5*x + 3)^(1/2))^5) + (12*3^(1/2)*((1 - 2*x)^(1/2) - 1)^7)/(5*( 
3^(1/2) - (5*x + 3)^(1/2))^7) - (96*3^(1/2)*((1 - 2*x)^(1/2) - 1))/(625*(3 
^(1/2) - (5*x + 3)^(1/2))) + 16/625) - (1177*7^(1/2)*atan(((1177*7^(1/2)*( 
(7062*3^(1/2))/875 + (3531*((1 - 2*x)^(1/2) - 1))/(875*(3^(1/2) - (5*x + 3 
)^(1/2))) - (7^(1/2)*((212*((1 - 2*x)^(1/2) - 1)^2)/(25*(3^(1/2) - (5*x + 
3)^(1/2))^2) + (888*3^(1/2)*((1 - 2*x)^(1/2) - 1))/(125*(3^(1/2) - (5*x + 
3)^(1/2))) - 536/125)*1177i)/392 - (3531*3^(1/2)*((1 - 2*x)^(1/2) - 1)^2)/ 
(175*(3^(1/2) - (5*x + 3)^(1/2))^2)))/392 + (1177*7^(1/2)*((7062*3^(1/2...
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.71 \[ \int \frac {\sqrt {1-2 x}}{(2+3 x)^3 \sqrt {3+5 x}} \, dx=\frac {10593 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x^{2}+14124 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x +4708 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )-10593 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x^{2}-14124 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x -4708 \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )+2163 \sqrt {5 x +3}\, \sqrt {-2 x +1}\, x +1540 \sqrt {5 x +3}\, \sqrt {-2 x +1}}{1764 x^{2}+2352 x +784} \] Input:

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

Output:

(10593*sqrt(7)*atan((sqrt(33) - sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5 
))/sqrt(11))/2))/sqrt(2))*x**2 + 14124*sqrt(7)*atan((sqrt(33) - sqrt(35)*t 
an(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x + 4708*sqrt(7) 
*atan((sqrt(33) - sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2 
))/sqrt(2)) - 10593*sqrt(7)*atan((sqrt(33) + sqrt(35)*tan(asin((sqrt( - 2* 
x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x**2 - 14124*sqrt(7)*atan((sqrt(33) 
 + sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x - 
 4708*sqrt(7)*atan((sqrt(33) + sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5) 
)/sqrt(11))/2))/sqrt(2)) + 2163*sqrt(5*x + 3)*sqrt( - 2*x + 1)*x + 1540*sq 
rt(5*x + 3)*sqrt( - 2*x + 1))/(196*(9*x**2 + 12*x + 4))