Integrand size = 26, antiderivative size = 79 \[ \int \frac {\sqrt {3+5 x}}{(1-2 x)^{5/2} (2+3 x)} \, dx=\frac {2 \sqrt {3+5 x}}{21 (1-2 x)^{3/2}}+\frac {128 \sqrt {3+5 x}}{1617 \sqrt {1-2 x}}+\frac {6 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{49 \sqrt {7}} \] Output:
2/21*(3+5*x)^(1/2)/(1-2*x)^(3/2)+128/1617*(3+5*x)^(1/2)/(1-2*x)^(1/2)+6/34 3*7^(1/2)*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))
Time = 0.11 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {3+5 x}}{(1-2 x)^{5/2} (2+3 x)} \, dx=-\frac {2 \sqrt {3+5 x} (-141+128 x)}{1617 (1-2 x)^{3/2}}+\frac {6 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{49 \sqrt {7}} \] Input:
Integrate[Sqrt[3 + 5*x]/((1 - 2*x)^(5/2)*(2 + 3*x)),x]
Output:
(-2*Sqrt[3 + 5*x]*(-141 + 128*x))/(1617*(1 - 2*x)^(3/2)) + (6*ArcTan[Sqrt[ 1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(49*Sqrt[7])
Time = 0.19 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.06, 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.
\(\displaystyle \int \frac {\sqrt {5 x+3}}{(1-2 x)^{5/2} (3 x+2)} \, dx\) |
\(\Big \downarrow \) 107 |
\(\displaystyle \frac {3}{7} \int \frac {\sqrt {5 x+3}}{(1-2 x)^{3/2} (3 x+2)}dx+\frac {4 (5 x+3)^{3/2}}{231 (1-2 x)^{3/2}}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {3}{7} \left (\frac {2 \sqrt {5 x+3}}{7 \sqrt {1-2 x}}-\frac {1}{7} \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx\right )+\frac {4 (5 x+3)^{3/2}}{231 (1-2 x)^{3/2}}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {3}{7} \left (\frac {2 \sqrt {5 x+3}}{7 \sqrt {1-2 x}}-\frac {2}{7} \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}\right )+\frac {4 (5 x+3)^{3/2}}{231 (1-2 x)^{3/2}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {3}{7} \left (\frac {2 \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{7 \sqrt {7}}+\frac {2 \sqrt {5 x+3}}{7 \sqrt {1-2 x}}\right )+\frac {4 (5 x+3)^{3/2}}{231 (1-2 x)^{3/2}}\) |
Input:
Int[Sqrt[3 + 5*x]/((1 - 2*x)^(5/2)*(2 + 3*x)),x]
Output:
(4*(3 + 5*x)^(3/2))/(231*(1 - 2*x)^(3/2)) + (3*((2*Sqrt[3 + 5*x])/(7*Sqrt[ 1 - 2*x]) + (2*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(7*Sqrt[7])) )/7
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]
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]
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])
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])
Leaf count of result is larger than twice the leaf count of optimal. \(146\) vs. \(2(58)=116\).
Time = 0.28 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.86
method | result | size |
default | \(-\frac {\left (396 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}-396 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +99 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+1792 x \sqrt {-10 x^{2}-x +3}-1974 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {3+5 x}}{11319 \left (1-2 x \right )^{\frac {3}{2}} \sqrt {-10 x^{2}-x +3}}\) | \(147\) |
Input:
int((3+5*x)^(1/2)/(1-2*x)^(5/2)/(2+3*x),x,method=_RETURNVERBOSE)
Output:
-1/11319*(396*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x ^2-396*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+99*7^( 1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+1792*x*(-10*x^2-x+ 3)^(1/2)-1974*(-10*x^2-x+3)^(1/2))*(3+5*x)^(1/2)/(1-2*x)^(3/2)/(-10*x^2-x+ 3)^(1/2)
Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {3+5 x}}{(1-2 x)^{5/2} (2+3 x)} \, dx=\frac {99 \, \sqrt {7} {\left (4 \, x^{2} - 4 \, x + 1\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (128 \, x - 141\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{11319 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \] Input:
integrate((3+5*x)^(1/2)/(1-2*x)^(5/2)/(2+3*x),x, algorithm="fricas")
Output:
1/11319*(99*sqrt(7)*(4*x^2 - 4*x + 1)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt (5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 14*(128*x - 141)*sqrt(5*x + 3 )*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1)
\[ \int \frac {\sqrt {3+5 x}}{(1-2 x)^{5/2} (2+3 x)} \, dx=\int \frac {\sqrt {5 x + 3}}{\left (1 - 2 x\right )^{\frac {5}{2}} \cdot \left (3 x + 2\right )}\, dx \] Input:
integrate((3+5*x)**(1/2)/(1-2*x)**(5/2)/(2+3*x),x)
Output:
Integral(sqrt(5*x + 3)/((1 - 2*x)**(5/2)*(3*x + 2)), x)
Time = 0.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.10 \[ \int \frac {\sqrt {3+5 x}}{(1-2 x)^{5/2} (2+3 x)} \, dx=-\frac {3}{343} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {640 \, x}{1617 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {1}{1617 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {55 \, x}{21 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} + \frac {11}{7 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \] Input:
integrate((3+5*x)^(1/2)/(1-2*x)^(5/2)/(2+3*x),x, algorithm="maxima")
Output:
-3/343*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 640/161 7*x/sqrt(-10*x^2 - x + 3) - 1/1617/sqrt(-10*x^2 - x + 3) + 55/21*x/(-10*x^ 2 - x + 3)^(3/2) + 11/7/(-10*x^2 - x + 3)^(3/2)
Time = 0.16 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.43 \[ \int \frac {\sqrt {3+5 x}}{(1-2 x)^{5/2} (2+3 x)} \, dx=-\frac {3}{3430} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {2 \, {\left (128 \, \sqrt {5} {\left (5 \, x + 3\right )} - 1089 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{40425 \, {\left (2 \, x - 1\right )}^{2}} \] Input:
integrate((3+5*x)^(1/2)/(1-2*x)^(5/2)/(2+3*x),x, algorithm="giac")
Output:
-3/3430*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((s qrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) - 2/40425*(128*sqrt(5)*(5*x + 3) - 1089*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2
Timed out. \[ \int \frac {\sqrt {3+5 x}}{(1-2 x)^{5/2} (2+3 x)} \, dx=\int \frac {\sqrt {5\,x+3}}{{\left (1-2\,x\right )}^{5/2}\,\left (3\,x+2\right )} \,d x \] Input:
int((5*x + 3)^(1/2)/((1 - 2*x)^(5/2)*(3*x + 2)),x)
Output:
int((5*x + 3)^(1/2)/((1 - 2*x)^(5/2)*(3*x + 2)), x)
Time = 0.19 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.47 \[ \int \frac {\sqrt {3+5 x}}{(1-2 x)^{5/2} (2+3 x)} \, dx=\frac {-\frac {12 \sqrt {-2 x +1}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x}{343}+\frac {6 \sqrt {-2 x +1}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )}{343}+\frac {12 \sqrt {-2 x +1}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x}{343}-\frac {6 \sqrt {-2 x +1}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )}{343}+\frac {256 \sqrt {5 x +3}\, x}{1617}-\frac {94 \sqrt {5 x +3}}{539}}{\sqrt {-2 x +1}\, \left (2 x -1\right )} \] Input:
int((3+5*x)^(1/2)/(1-2*x)^(5/2)/(2+3*x),x)
Output:
(2*( - 198*sqrt( - 2*x + 1)*sqrt(7)*atan((sqrt(33) - sqrt(35)*tan(asin((sq rt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x + 99*sqrt( - 2*x + 1)*sqr t(7)*atan((sqrt(33) - sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11 ))/2))/sqrt(2)) + 198*sqrt( - 2*x + 1)*sqrt(7)*atan((sqrt(33) + sqrt(35)*t an(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x - 99*sqrt( - 2 *x + 1)*sqrt(7)*atan((sqrt(33) + sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt( 5))/sqrt(11))/2))/sqrt(2)) + 896*sqrt(5*x + 3)*x - 987*sqrt(5*x + 3)))/(11 319*sqrt( - 2*x + 1)*(2*x - 1))