\(\int \frac {3+5 x}{(1-2 x)^{3/2} (2+3 x)^2} \, dx\) [2078]

Optimal result

Integrand size = 22, antiderivative size = 61 \[ \int \frac {3+5 x}{(1-2 x)^{3/2} (2+3 x)^2} \, dx=\frac {64}{147 \sqrt {1-2 x}}+\frac {1}{21 \sqrt {1-2 x} (2+3 x)}-\frac {64 \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{49 \sqrt {21}} \]

[Out]

-64/1029*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+64/147/(1-2*x)^(1/2)+1/21/(2+3*x)/(1-2*x)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {79, 53, 65, 212} \[ \int \frac {3+5 x}{(1-2 x)^{3/2} (2+3 x)^2} \, dx=-\frac {64 \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{49 \sqrt {21}}+\frac {64}{147 \sqrt {1-2 x}}+\frac {1}{21 \sqrt {1-2 x} (3 x+2)} \]

[In]

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

[Out]

64/(147*Sqrt[1 - 2*x]) + 1/(21*Sqrt[1 - 2*x]*(2 + 3*x)) - (64*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(49*Sqrt[21])

Rule 53

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] - Dist[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] && NeQ[b*c - a*d, 0] && 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]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(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] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{21 \sqrt {1-2 x} (2+3 x)}+\frac {32}{21} \int \frac {1}{(1-2 x)^{3/2} (2+3 x)} \, dx \\ & = \frac {64}{147 \sqrt {1-2 x}}+\frac {1}{21 \sqrt {1-2 x} (2+3 x)}+\frac {32}{49} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx \\ & = \frac {64}{147 \sqrt {1-2 x}}+\frac {1}{21 \sqrt {1-2 x} (2+3 x)}-\frac {32}{49} \text {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right ) \\ & = \frac {64}{147 \sqrt {1-2 x}}+\frac {1}{21 \sqrt {1-2 x} (2+3 x)}-\frac {64 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{49 \sqrt {21}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int \frac {3+5 x}{(1-2 x)^{3/2} (2+3 x)^2} \, dx=\frac {45+64 x}{49 \sqrt {1-2 x} (2+3 x)}-\frac {64 \text {arctanh}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{49 \sqrt {21}} \]

[In]

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

[Out]

(45 + 64*x)/(49*Sqrt[1 - 2*x]*(2 + 3*x)) - (64*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(49*Sqrt[21])

Maple [A] (verified)

Time = 1.05 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.67

method	result	size
risch	\(\frac {64 x +45}{49 \left (2+3 x \right ) \sqrt {1-2 x}}-\frac {64 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{1029}\)	\(41\)
derivativedivides	\(-\frac {2 \sqrt {1-2 x}}{147 \left (-\frac {4}{3}-2 x \right )}-\frac {64 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{1029}+\frac {22}{49 \sqrt {1-2 x}}\)	\(45\)
default	\(-\frac {2 \sqrt {1-2 x}}{147 \left (-\frac {4}{3}-2 x \right )}-\frac {64 \,\operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right ) \sqrt {21}}{1029}+\frac {22}{49 \sqrt {1-2 x}}\)	\(45\)
pseudoelliptic	\(-\frac {192 \left (\sqrt {1-2 x}\, \sqrt {21}\, \left (\frac {2}{3}+x \right ) \operatorname {arctanh}\left (\frac {\sqrt {21}\, \sqrt {1-2 x}}{7}\right )-7 x -\frac {315}{64}\right )}{\sqrt {1-2 x}\, \left (2058+3087 x \right )}\)	\(49\)
trager	\(-\frac {\left (64 x +45\right ) \sqrt {1-2 x}}{49 \left (6 x^{2}+x -2\right )}-\frac {32 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) \ln \left (\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right ) x +21 \sqrt {1-2 x}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-21\right )}{2+3 x}\right )}{1029}\)	\(70\)

[In]

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

[Out]

1/49*(64*x+45)/(2+3*x)/(1-2*x)^(1/2)-64/1029*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.07 \[ \int \frac {3+5 x}{(1-2 x)^{3/2} (2+3 x)^2} \, dx=\frac {32 \, \sqrt {21} {\left (6 \, x^{2} + x - 2\right )} \log \left (\frac {3 \, x + \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) - 21 \, {\left (64 \, x + 45\right )} \sqrt {-2 \, x + 1}}{1029 \, {\left (6 \, x^{2} + x - 2\right )}} \]

[In]

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

[Out]

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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (53) = 106\).

Time = 32.90 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.84 \[ \int \frac {3+5 x}{(1-2 x)^{3/2} (2+3 x)^2} \, dx=\frac {11 \sqrt {21} \left (\log {\left (\sqrt {1 - 2 x} - \frac {\sqrt {21}}{3} \right )} - \log {\left (\sqrt {1 - 2 x} + \frac {\sqrt {21}}{3} \right )}\right )}{343} + \frac {4 \left (\begin {cases} \frac {\sqrt {21} \left (- \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} - 1\right )}\right )}{147} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {21}}{3} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {21}}{3} \end {cases}\right )}{7} + \frac {22}{49 \sqrt {1 - 2 x}} \]

[In]

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

[Out]

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))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.07 \[ \int \frac {3+5 x}{(1-2 x)^{3/2} (2+3 x)^2} \, dx=\frac {32}{1029} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {2 \, {\left (64 \, x + 45\right )}}{49 \, {\left (3 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 7 \, \sqrt {-2 \, x + 1}\right )}} \]

[In]

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

[Out]

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))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.11 \[ \int \frac {3+5 x}{(1-2 x)^{3/2} (2+3 x)^2} \, dx=\frac {32}{1029} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {2 \, {\left (64 \, x + 45\right )}}{49 \, {\left (3 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 7 \, \sqrt {-2 \, x + 1}\right )}} \]

[In]

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

[Out]

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))

Mupad [B] (verification not implemented)

Time = 1.41 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.75 \[ \int \frac {3+5 x}{(1-2 x)^{3/2} (2+3 x)^2} \, dx=\frac {\frac {128\,x}{147}+\frac {30}{49}}{\frac {7\,\sqrt {1-2\,x}}{3}-{\left (1-2\,x\right )}^{3/2}}-\frac {64\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{1029} \]

[In]

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

[Out]

((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