Integrand size = 19, antiderivative size = 93 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx=\frac {825}{32} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {25}{8} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {(3+5 x)^{5/2}}{\sqrt {1-2 x}}-\frac {1815}{32} \sqrt {\frac {5}{2}} \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right ) \]
-1815/64*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+(3+5*x)^(5/2)/(1-2*x )^(1/2)+25/8*(3+5*x)^(3/2)*(1-2*x)^(1/2)+825/32*(1-2*x)^(1/2)*(3+5*x)^(1/2 )
Time = 0.15 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.69 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx=\frac {1}{64} \left (-\frac {2 \sqrt {3+5 x} \left (-1413+790 x+200 x^2\right )}{\sqrt {1-2 x}}-1815 \sqrt {10} \arctan \left (\frac {\sqrt {\frac {6}{5}+2 x}}{\sqrt {1-2 x}}\right )\right ) \]
((-2*Sqrt[3 + 5*x]*(-1413 + 790*x + 200*x^2))/Sqrt[1 - 2*x] - 1815*Sqrt[10 ]*ArcTan[Sqrt[6/5 + 2*x]/Sqrt[1 - 2*x]])/64
Time = 0.18 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {57, 60, 60, 64, 223}
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.
\(\displaystyle \int \frac {(5 x+3)^{5/2}}{(1-2 x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {(5 x+3)^{5/2}}{\sqrt {1-2 x}}-\frac {25}{2} \int \frac {(5 x+3)^{3/2}}{\sqrt {1-2 x}}dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(5 x+3)^{5/2}}{\sqrt {1-2 x}}-\frac {25}{2} \left (\frac {33}{8} \int \frac {\sqrt {5 x+3}}{\sqrt {1-2 x}}dx-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(5 x+3)^{5/2}}{\sqrt {1-2 x}}-\frac {25}{2} \left (\frac {33}{8} \left (\frac {11}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {5 x+3}}dx-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )\) |
\(\Big \downarrow \) 64 |
\(\displaystyle \frac {(5 x+3)^{5/2}}{\sqrt {1-2 x}}-\frac {25}{2} \left (\frac {33}{8} \left (\frac {11}{10} \int \frac {1}{\sqrt {\frac {11}{5}-\frac {2}{5} (5 x+3)}}d\sqrt {5 x+3}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {(5 x+3)^{5/2}}{\sqrt {1-2 x}}-\frac {25}{2} \left (\frac {33}{8} \left (\frac {11 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2 \sqrt {10}}-\frac {1}{2} \sqrt {1-2 x} \sqrt {5 x+3}\right )-\frac {1}{4} \sqrt {1-2 x} (5 x+3)^{3/2}\right )\) |
(3 + 5*x)^(5/2)/Sqrt[1 - 2*x] - (25*(-1/4*(Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) + (33*(-1/2*(Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + (11*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(2*Sqrt[10])))/8))/2
3.26.43.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c , d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp [2/b Subst[Int[1/Sqrt[c - a*(d/b) + d*(x^2/b)], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[c - a*(d/b), 0] && ( !GtQ[a - c*(b/d), 0] || PosQ[b])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
\[\int \frac {\left (3+5 x \right )^{\frac {5}{2}}}{\left (1-2 x \right )^{\frac {3}{2}}}d x\]
Time = 0.23 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.94 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx=\frac {1815 \, \sqrt {5} \sqrt {2} {\left (2 \, x - 1\right )} \arctan \left (\frac {\sqrt {5} \sqrt {2} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 4 \, {\left (200 \, x^{2} + 790 \, x - 1413\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{128 \, {\left (2 \, x - 1\right )}} \]
1/128*(1815*sqrt(5)*sqrt(2)*(2*x - 1)*arctan(1/20*sqrt(5)*sqrt(2)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 4*(200*x^2 + 790*x - 1 413)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(2*x - 1)
Result contains complex when optimal does not.
Time = 4.78 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.99 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx=\begin {cases} \frac {125 i \left (x + \frac {3}{5}\right )^{\frac {5}{2}}}{4 \sqrt {10 x - 5}} + \frac {1375 i \left (x + \frac {3}{5}\right )^{\frac {3}{2}}}{16 \sqrt {10 x - 5}} - \frac {9075 i \sqrt {x + \frac {3}{5}}}{32 \sqrt {10 x - 5}} + \frac {1815 \sqrt {10} i \operatorname {acosh}{\left (\frac {\sqrt {110} \sqrt {x + \frac {3}{5}}}{11} \right )}}{64} & \text {for}\: \left |{x + \frac {3}{5}}\right | > \frac {11}{10} \\- \frac {1815 \sqrt {10} \operatorname {asin}{\left (\frac {\sqrt {110} \sqrt {x + \frac {3}{5}}}{11} \right )}}{64} - \frac {125 \left (x + \frac {3}{5}\right )^{\frac {5}{2}}}{4 \sqrt {5 - 10 x}} - \frac {1375 \left (x + \frac {3}{5}\right )^{\frac {3}{2}}}{16 \sqrt {5 - 10 x}} + \frac {9075 \sqrt {x + \frac {3}{5}}}{32 \sqrt {5 - 10 x}} & \text {otherwise} \end {cases} \]
Piecewise((125*I*(x + 3/5)**(5/2)/(4*sqrt(10*x - 5)) + 1375*I*(x + 3/5)**( 3/2)/(16*sqrt(10*x - 5)) - 9075*I*sqrt(x + 3/5)/(32*sqrt(10*x - 5)) + 1815 *sqrt(10)*I*acosh(sqrt(110)*sqrt(x + 3/5)/11)/64, Abs(x + 3/5) > 11/10), ( -1815*sqrt(10)*asin(sqrt(110)*sqrt(x + 3/5)/11)/64 - 125*(x + 3/5)**(5/2)/ (4*sqrt(5 - 10*x)) - 1375*(x + 3/5)**(3/2)/(16*sqrt(5 - 10*x)) + 9075*sqrt (x + 3/5)/(32*sqrt(5 - 10*x)), True))
Time = 0.31 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.81 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx=-\frac {125 \, x^{3}}{4 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {2275 \, x^{2}}{16 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {1815}{128} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) + \frac {4695 \, x}{32 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {4239}{32 \, \sqrt {-10 \, x^{2} - x + 3}} \]
-125/4*x^3/sqrt(-10*x^2 - x + 3) - 2275/16*x^2/sqrt(-10*x^2 - x + 3) + 181 5/128*sqrt(10)*arcsin(-20/11*x - 1/11) + 4695/32*x/sqrt(-10*x^2 - x + 3) + 4239/32/sqrt(-10*x^2 - x + 3)
Time = 0.31 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.76 \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx=-\frac {1815}{64} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (2 \, {\left (4 \, \sqrt {5} {\left (5 \, x + 3\right )} + 55 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 1815 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{160 \, {\left (2 \, x - 1\right )}} \]
-1815/64*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/160*(2*(4*sqrt(5 )*(5*x + 3) + 55*sqrt(5))*(5*x + 3) - 1815*sqrt(5))*sqrt(5*x + 3)*sqrt(-10 *x + 5)/(2*x - 1)
Timed out. \[ \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx=\int \frac {{\left (5\,x+3\right )}^{5/2}}{{\left (1-2\,x\right )}^{3/2}} \,d x \]