Integrand size = 26, antiderivative size = 75 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)^{5/2}} \, dx=-\frac {2 (1-2 x)^{3/2}}{3 (3+5 x)^{3/2}}+\frac {14 \sqrt {1-2 x}}{\sqrt {3+5 x}}-14 \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right ) \] Output:
-2/3*(1-2*x)^(3/2)/(3+5*x)^(3/2)+14*(1-2*x)^(1/2)/(3+5*x)^(1/2)-14*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 = 60, normalized size of antiderivative = 0.80 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)^{5/2}} \, dx=\frac {2 \sqrt {1-2 x} (62+107 x)}{3 (3+5 x)^{3/2}}-14 \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right ) \] Input:
Integrate[(1 - 2*x)^(3/2)/((2 + 3*x)*(3 + 5*x)^(5/2)),x]
Output:
(2*Sqrt[1 - 2*x]*(62 + 107*x))/(3*(3 + 5*x)^(3/2)) - 14*Sqrt[7]*ArcTan[Sqr t[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])]
Time = 0.19 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {105, 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.
\(\displaystyle \int \frac {(1-2 x)^{3/2}}{(3 x+2) (5 x+3)^{5/2}} \, dx\) |
\(\Big \downarrow \) 105 |
\(\displaystyle -7 \int \frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)^{3/2}}dx-\frac {2 (1-2 x)^{3/2}}{3 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 105 |
\(\displaystyle -7 \left (-7 \int \frac {1}{\sqrt {1-2 x} (3 x+2) \sqrt {5 x+3}}dx-\frac {2 \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{3/2}}{3 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle -7 \left (-14 \int \frac {1}{-\frac {1-2 x}{5 x+3}-7}d\frac {\sqrt {1-2 x}}{\sqrt {5 x+3}}-\frac {2 \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{3/2}}{3 (5 x+3)^{3/2}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -7 \left (2 \sqrt {7} \arctan \left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )-\frac {2 \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )-\frac {2 (1-2 x)^{3/2}}{3 (5 x+3)^{3/2}}\) |
Input:
Int[(1 - 2*x)^(3/2)/((2 + 3*x)*(3 + 5*x)^(5/2)),x]
Output:
(-2*(1 - 2*x)^(3/2))/(3*(3 + 5*x)^(3/2)) - 7*((-2*Sqrt[1 - 2*x])/Sqrt[3 + 5*x] + 2*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])
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_)^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 = 0.20 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.96
method | result | size |
default | \(\frac {\left (525 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}+630 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +189 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+214 x \sqrt {-10 x^{2}-x +3}+124 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{3 \sqrt {-10 x^{2}-x +3}\, \left (3+5 x \right )^{\frac {3}{2}}}\) | \(147\) |
Input:
int((1-2*x)^(3/2)/(2+3*x)/(3+5*x)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/3*(525*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2+63 0*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+189*7^(1/2) *arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+214*x*(-10*x^2-x+3)^(1 /2)+124*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(3/ 2)
Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.15 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)^{5/2}} \, dx=-\frac {21 \, \sqrt {7} {\left (25 \, x^{2} + 30 \, x + 9\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 2 \, {\left (107 \, x + 62\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{3 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \] Input:
integrate((1-2*x)^(3/2)/(2+3*x)/(3+5*x)^(5/2),x, algorithm="fricas")
Output:
-1/3*(21*sqrt(7)*(25*x^2 + 30*x + 9)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt( 5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 2*(107*x + 62)*sqrt(5*x + 3)*s qrt(-2*x + 1))/(25*x^2 + 30*x + 9)
\[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)^{5/2}} \, dx=\int \frac {\left (1 - 2 x\right )^{\frac {3}{2}}}{\left (3 x + 2\right ) \left (5 x + 3\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((1-2*x)**(3/2)/(2+3*x)/(3+5*x)**(5/2),x)
Output:
Integral((1 - 2*x)**(3/2)/((3*x + 2)*(5*x + 3)**(5/2)), x)
Time = 0.11 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.39 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)^{5/2}} \, dx=7 \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) - \frac {428 \, x}{15 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {8 \, x^{2}}{15 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} + \frac {1118}{75 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {488 \, x}{75 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {254}{75 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \] Input:
integrate((1-2*x)^(3/2)/(2+3*x)/(3+5*x)^(5/2),x, algorithm="maxima")
Output:
7*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) - 428/15*x/sqr t(-10*x^2 - x + 3) + 8/15*x^2/(-10*x^2 - x + 3)^(3/2) + 1118/75/sqrt(-10*x ^2 - x + 3) + 488/75*x/(-10*x^2 - x + 3)^(3/2) - 254/75/(-10*x^2 - x + 3)^ (3/2)
Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (58) = 116\).
Time = 0.17 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.53 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)^{5/2}} \, dx=\frac {7}{10} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {1}{1200} \, \sqrt {10} {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} - \frac {840 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} + \frac {3360 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )} \] Input:
integrate((1-2*x)^(3/2)/(2+3*x)/(3+5*x)^(5/2),x, algorithm="giac")
Output:
7/10*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)))) - 1/1200*sqrt(10)*(((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqr t(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^3 - 840 *(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 3360*sqrt(5*x + 3)/( sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))
Timed out. \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)^{5/2}} \, dx=\int \frac {{\left (1-2\,x\right )}^{3/2}}{\left (3\,x+2\right )\,{\left (5\,x+3\right )}^{5/2}} \,d x \] Input:
int((1 - 2*x)^(3/2)/((3*x + 2)*(5*x + 3)^(5/2)),x)
Output:
int((1 - 2*x)^(3/2)/((3*x + 2)*(5*x + 3)^(5/2)), x)
Time = 0.24 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.60 \[ \int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)^{5/2}} \, dx=\frac {70 \sqrt {5 x +3}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x +42 \sqrt {5 x +3}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}-\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )-70 \sqrt {5 x +3}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right ) x -42 \sqrt {5 x +3}\, \sqrt {7}\, \mathit {atan} \left (\frac {\sqrt {33}+\sqrt {35}\, \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-2 x +1}\, \sqrt {5}}{\sqrt {11}}\right )}{2}\right )}{\sqrt {2}}\right )+\frac {214 \sqrt {-2 x +1}\, x}{3}+\frac {124 \sqrt {-2 x +1}}{3}}{\sqrt {5 x +3}\, \left (5 x +3\right )} \] Input:
int((1-2*x)^(3/2)/(2+3*x)/(3+5*x)^(5/2),x)
Output:
(2*(105*sqrt(5*x + 3)*sqrt(7)*atan((sqrt(33) - sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x + 63*sqrt(5*x + 3)*sqrt(7)*atan ((sqrt(33) - sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sq rt(2)) - 105*sqrt(5*x + 3)*sqrt(7)*atan((sqrt(33) + sqrt(35)*tan(asin((sqr t( - 2*x + 1)*sqrt(5))/sqrt(11))/2))/sqrt(2))*x - 63*sqrt(5*x + 3)*sqrt(7) *atan((sqrt(33) + sqrt(35)*tan(asin((sqrt( - 2*x + 1)*sqrt(5))/sqrt(11))/2 ))/sqrt(2)) + 107*sqrt( - 2*x + 1)*x + 62*sqrt( - 2*x + 1)))/(3*sqrt(5*x + 3)*(5*x + 3))