Integrand size = 22, antiderivative size = 61 \[ \int \frac {3+5 x}{(1-2 x)^{3/2} (2+3 x)^2} \, dx=\frac {22}{49 \sqrt {1-2 x}}+\frac {\sqrt {1-2 x}}{49 (2+3 x)}-\frac {64 \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{49 \sqrt {21}} \] Output:
22/49/(1-2*x)^(1/2)+(1-2*x)^(1/2)/(98+147*x)-64/1029*21^(1/2)*arctanh(1/7* 21^(1/2)*(1-2*x)^(1/2))
Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int \frac {3+5 x}{(1-2 x)^{3/2} (2+3 x)^2} \, dx=\frac {45+64 x}{49 \sqrt {1-2 x} (2+3 x)}-\frac {64 \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{49 \sqrt {21}} \] Input:
Integrate[(3 + 5*x)/((1 - 2*x)^(3/2)*(2 + 3*x)^2),x]
Output:
(45 + 64*x)/(49*Sqrt[1 - 2*x]*(2 + 3*x)) - (64*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(49*Sqrt[21])
Time = 0.19 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.11, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {87, 61, 73, 219}
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.
\(\displaystyle \int \frac {5 x+3}{(1-2 x)^{3/2} (3 x+2)^2} \, dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {32}{21} \int \frac {1}{(1-2 x)^{3/2} (3 x+2)}dx+\frac {1}{21 \sqrt {1-2 x} (3 x+2)}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {32}{21} \left (\frac {3}{7} \int \frac {1}{\sqrt {1-2 x} (3 x+2)}dx+\frac {2}{7 \sqrt {1-2 x}}\right )+\frac {1}{21 \sqrt {1-2 x} (3 x+2)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {32}{21} \left (\frac {2}{7 \sqrt {1-2 x}}-\frac {3}{7} \int \frac {1}{\frac {7}{2}-\frac {3}{2} (1-2 x)}d\sqrt {1-2 x}\right )+\frac {1}{21 \sqrt {1-2 x} (3 x+2)}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {32}{21} \left (\frac {2}{7 \sqrt {1-2 x}}-\frac {2}{7} \sqrt {\frac {3}{7}} \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )\right )+\frac {1}{21 \sqrt {1-2 x} (3 x+2)}\) |
Input:
Int[(3 + 5*x)/((1 - 2*x)^(3/2)*(2 + 3*x)^2),x]
Output:
1/(21*Sqrt[1 - 2*x]*(2 + 3*x)) + (32*(2/(7*Sqrt[1 - 2*x]) - (2*Sqrt[3/7]*A rcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/7))/21
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] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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]))))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Time = 0.21 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.67
method | result | size |
risch | \(\frac {45+64 x}{49 \left (2+3 x \right ) \sqrt {1-2 x}}-\frac {64 \sqrt {21}\, \operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right )}{1029}\) | \(41\) |
derivativedivides | \(\frac {22}{49 \sqrt {1-2 x}}-\frac {2 \sqrt {1-2 x}}{147 \left (-\frac {4}{3}-2 x \right )}-\frac {64 \sqrt {21}\, \operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right )}{1029}\) | \(45\) |
default | \(\frac {22}{49 \sqrt {1-2 x}}-\frac {2 \sqrt {1-2 x}}{147 \left (-\frac {4}{3}-2 x \right )}-\frac {64 \sqrt {21}\, \operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right )}{1029}\) | \(45\) |
pseudoelliptic | \(-\frac {192 \left (\sqrt {21}\, \sqrt {1-2 x}\, \left (\frac {2}{3}+x \right ) \operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right )-7 x -\frac {315}{64}\right )}{\sqrt {1-2 x}\, \left (2058+3087 x \right )}\) | \(49\) |
trager | \(-\frac {\left (45+64 x \right ) \sqrt {1-2 x}}{49 \left (6 x^{2}+x -2\right )}+\frac {32 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{1029}\) | \(70\) |
Input:
int((3+5*x)/(1-2*x)^(3/2)/(2+3*x)^2,x,method=_RETURNVERBOSE)
Output:
1/49*(45+64*x)/(2+3*x)/(1-2*x)^(1/2)-64/1029*21^(1/2)*arctanh(1/7*21^(1/2) *(1-2*x)^(1/2))
Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.07 \[ \int \frac {3+5 x}{(1-2 x)^{3/2} (2+3 x)^2} \, dx=\frac {32 \, \sqrt {21} {\left (6 \, x^{2} + x - 2\right )} \log \left (\frac {3 \, x + \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) - 21 \, {\left (64 \, x + 45\right )} \sqrt {-2 \, x + 1}}{1029 \, {\left (6 \, x^{2} + x - 2\right )}} \] Input:
integrate((3+5*x)/(1-2*x)^(3/2)/(2+3*x)^2,x, algorithm="fricas")
Output:
1/1029*(32*sqrt(21)*(6*x^2 + x - 2)*log((3*x + sqrt(21)*sqrt(-2*x + 1) - 5 )/(3*x + 2)) - 21*(64*x + 45)*sqrt(-2*x + 1))/(6*x^2 + x - 2)
Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (49) = 98\).
Time = 34.76 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.84 \[ \int \frac {3+5 x}{(1-2 x)^{3/2} (2+3 x)^2} \, dx=\frac {11 \sqrt {21} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {21}}{3} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {21}}{3} \right )}\right )}{343} + \frac {4 \left (\begin {cases} \frac {\sqrt {21} \left (- \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1\right )}\right )}{147} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {21}}{3} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {21}}{3} \end {cases}\right )}{7} + \frac {22}{49 \sqrt {1 - 2 x}} \] Input:
integrate((3+5*x)/(1-2*x)**(3/2)/(2+3*x)**2,x)
Output:
11*sqrt(21)*(log(sqrt(1 - 2*x) - sqrt(21)/3) - log(sqrt(1 - 2*x) + sqrt(21 )/3))/343 + 4*Piecewise((sqrt(21)*(-log(sqrt(21)*sqrt(1 - 2*x)/7 - 1)/4 + log(sqrt(21)*sqrt(1 - 2*x)/7 + 1)/4 - 1/(4*(sqrt(21)*sqrt(1 - 2*x)/7 + 1)) - 1/(4*(sqrt(21)*sqrt(1 - 2*x)/7 - 1)))/147, (sqrt(1 - 2*x) > -sqrt(21)/3 ) & (sqrt(1 - 2*x) < sqrt(21)/3)))/7 + 22/(49*sqrt(1 - 2*x))
Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.07 \[ \int \frac {3+5 x}{(1-2 x)^{3/2} (2+3 x)^2} \, dx=\frac {32}{1029} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {2 \, {\left (64 \, x + 45\right )}}{49 \, {\left (3 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 7 \, \sqrt {-2 \, x + 1}\right )}} \] Input:
integrate((3+5*x)/(1-2*x)^(3/2)/(2+3*x)^2,x, algorithm="maxima")
Output:
32/1029*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2* x + 1))) - 2/49*(64*x + 45)/(3*(-2*x + 1)^(3/2) - 7*sqrt(-2*x + 1))
Time = 0.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.11 \[ \int \frac {3+5 x}{(1-2 x)^{3/2} (2+3 x)^2} \, dx=\frac {32}{1029} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {2 \, {\left (64 \, x + 45\right )}}{49 \, {\left (3 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 7 \, \sqrt {-2 \, x + 1}\right )}} \] Input:
integrate((3+5*x)/(1-2*x)^(3/2)/(2+3*x)^2,x, algorithm="giac")
Output:
32/1029*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3 *sqrt(-2*x + 1))) - 2/49*(64*x + 45)/(3*(-2*x + 1)^(3/2) - 7*sqrt(-2*x + 1 ))
Time = 1.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.75 \[ \int \frac {3+5 x}{(1-2 x)^{3/2} (2+3 x)^2} \, dx=\frac {\frac {128\,x}{147}+\frac {30}{49}}{\frac {7\,\sqrt {1-2\,x}}{3}-{\left (1-2\,x\right )}^{3/2}}-\frac {64\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{1029} \] Input:
int((5*x + 3)/((1 - 2*x)^(3/2)*(3*x + 2)^2),x)
Output:
((128*x)/147 + 30/49)/((7*(1 - 2*x)^(1/2))/3 - (1 - 2*x)^(3/2)) - (64*21^( 1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/1029
Time = 0.15 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.90 \[ \int \frac {3+5 x}{(1-2 x)^{3/2} (2+3 x)^2} \, dx=\frac {96 \sqrt {-2 x +1}\, \sqrt {21}\, \mathrm {log}\left (3 \sqrt {-2 x +1}-\sqrt {21}\right ) x +64 \sqrt {-2 x +1}\, \sqrt {21}\, \mathrm {log}\left (3 \sqrt {-2 x +1}-\sqrt {21}\right )-96 \sqrt {-2 x +1}\, \sqrt {21}\, \mathrm {log}\left (3 \sqrt {-2 x +1}+\sqrt {21}\right ) x -64 \sqrt {-2 x +1}\, \sqrt {21}\, \mathrm {log}\left (3 \sqrt {-2 x +1}+\sqrt {21}\right )+1344 x +945}{1029 \sqrt {-2 x +1}\, \left (3 x +2\right )} \] Input:
int((3+5*x)/(1-2*x)^(3/2)/(2+3*x)^2,x)
Output:
(96*sqrt( - 2*x + 1)*sqrt(21)*log(3*sqrt( - 2*x + 1) - sqrt(21))*x + 64*sq rt( - 2*x + 1)*sqrt(21)*log(3*sqrt( - 2*x + 1) - sqrt(21)) - 96*sqrt( - 2* x + 1)*sqrt(21)*log(3*sqrt( - 2*x + 1) + sqrt(21))*x - 64*sqrt( - 2*x + 1) *sqrt(21)*log(3*sqrt( - 2*x + 1) + sqrt(21)) + 1344*x + 945)/(1029*sqrt( - 2*x + 1)*(3*x + 2))