3.18.15 \(\int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^3} \, dx\)

Optimal. Leaf size=121 \[ \frac {995 \sqrt {1-2 x}}{22 (5 x+3)}-\frac {15 \sqrt {1-2 x}}{2 (5 x+3)^2}+\frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)^2}+624 \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {6665}{11} \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {99, 151, 156, 63, 206} \begin {gather*} \frac {995 \sqrt {1-2 x}}{22 (5 x+3)}-\frac {15 \sqrt {1-2 x}}{2 (5 x+3)^2}+\frac {\sqrt {1-2 x}}{(3 x+2) (5 x+3)^2}+624 \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {6665}{11} \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*Sqrt[1 - 2*x])/(2*(3 + 5*x)^2) + Sqrt[1 - 2*x]/((2 + 3*x)*(3 + 5*x)^2) + (995*Sqrt[1 - 2*x])/(22*(3 + 5*x
)) + 624*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] - (6665*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/11

Rule 63

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 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(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] + Dist[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*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin {align*} \int \frac {\sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)^3} \, dx &=\frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^2}-\int \frac {-18+25 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)^3} \, dx\\ &=-\frac {15 \sqrt {1-2 x}}{2 (3+5 x)^2}+\frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^2}+\frac {1}{22} \int \frac {-1298+1485 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)^2} \, dx\\ &=-\frac {15 \sqrt {1-2 x}}{2 (3+5 x)^2}+\frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^2}+\frac {995 \sqrt {1-2 x}}{22 (3+5 x)}-\frac {1}{242} \int \frac {-53614+32835 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx\\ &=-\frac {15 \sqrt {1-2 x}}{2 (3+5 x)^2}+\frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^2}+\frac {995 \sqrt {1-2 x}}{22 (3+5 x)}-936 \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx+\frac {33325}{22} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {15 \sqrt {1-2 x}}{2 (3+5 x)^2}+\frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^2}+\frac {995 \sqrt {1-2 x}}{22 (3+5 x)}+936 \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-\frac {33325}{22} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {15 \sqrt {1-2 x}}{2 (3+5 x)^2}+\frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^2}+\frac {995 \sqrt {1-2 x}}{22 (3+5 x)}+624 \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {6665}{11} \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 94, normalized size = 0.78 \begin {gather*} \frac {\sqrt {1-2 x} \left (14925 x^2+18410 x+5662\right )}{22 (3 x+2) (5 x+3)^2}+624 \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {6665}{11} \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*(5662 + 18410*x + 14925*x^2))/(22*(2 + 3*x)*(3 + 5*x)^2) + 624*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt
[1 - 2*x]] - (6665*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/11

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.32, size = 117, normalized size = 0.97 \begin {gather*} \frac {-14925 (1-2 x)^{5/2}+66670 (1-2 x)^{3/2}-74393 \sqrt {1-2 x}}{11 (3 (1-2 x)-7) (5 (1-2 x)-11)^2}+624 \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {6665}{11} \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-74393*Sqrt[1 - 2*x] + 66670*(1 - 2*x)^(3/2) - 14925*(1 - 2*x)^(5/2))/(11*(-7 + 3*(1 - 2*x))*(-11 + 5*(1 - 2*
x))^2) + 624*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] - (6665*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/
11

________________________________________________________________________________________

fricas [A]  time = 0.91, size = 142, normalized size = 1.17 \begin {gather*} \frac {46655 \, \sqrt {11} \sqrt {5} {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 75504 \, \sqrt {7} \sqrt {3} {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) + 77 \, {\left (14925 \, x^{2} + 18410 \, x + 5662\right )} \sqrt {-2 \, x + 1}}{1694 \, {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1694*(46655*sqrt(11)*sqrt(5)*(75*x^3 + 140*x^2 + 87*x + 18)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/
(5*x + 3)) + 75504*sqrt(7)*sqrt(3)*(75*x^3 + 140*x^2 + 87*x + 18)*log(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x +
5)/(3*x + 2)) + 77*(14925*x^2 + 18410*x + 5662)*sqrt(-2*x + 1))/(75*x^3 + 140*x^2 + 87*x + 18)

________________________________________________________________________________________

giac [A]  time = 1.21, size = 123, normalized size = 1.02 \begin {gather*} \frac {6665}{242} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {312}{7} \, \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 {9 \, \sqrt {-2 \, x + 1}}{3 \, x + 2} - \frac {5 \, {\left (665 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 1441 \, \sqrt {-2 \, x + 1}\right )}}{44 \, {\left (5 \, x + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6665/242*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 312/7*sqrt(21)
*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 9*sqrt(-2*x + 1)/(3*x + 2) - 5/4
4*(665*(-2*x + 1)^(3/2) - 1441*sqrt(-2*x + 1))/(5*x + 3)^2

________________________________________________________________________________________

maple [A]  time = 0.01, size = 82, normalized size = 0.68 \begin {gather*} \frac {624 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{7}-\frac {6665 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{121}+\frac {-\frac {3325 \left (-2 x +1\right )^{\frac {3}{2}}}{11}+655 \sqrt {-2 x +1}}{\left (-10 x -6\right )^{2}}-\frac {6 \sqrt {-2 x +1}}{-2 x -\frac {4}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

250*(-133/110*(-2*x+1)^(3/2)+131/50*(-2*x+1)^(1/2))/(-10*x-6)^2-6665/121*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))
*55^(1/2)-6*(-2*x+1)^(1/2)/(-2*x-4/3)+624/7*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)

________________________________________________________________________________________

maxima [A]  time = 1.15, size = 128, normalized size = 1.06 \begin {gather*} \frac {6665}{242} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {312}{7} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {14925 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 66670 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 74393 \, \sqrt {-2 \, x + 1}}{11 \, {\left (75 \, {\left (2 \, x - 1\right )}^{3} + 505 \, {\left (2 \, x - 1\right )}^{2} + 2266 \, x - 286\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6665/242*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 312/7*sqrt(21)*log(-(sqr
t(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 1/11*(14925*(-2*x + 1)^(5/2) - 66670*(-2*x + 1)^(3/
2) + 74393*sqrt(-2*x + 1))/(75*(2*x - 1)^3 + 505*(2*x - 1)^2 + 2266*x - 286)

________________________________________________________________________________________

mupad [B]  time = 0.10, size = 89, normalized size = 0.74 \begin {gather*} \frac {624\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{7}-\frac {6665\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{121}+\frac {\frac {6763\,\sqrt {1-2\,x}}{75}-\frac {13334\,{\left (1-2\,x\right )}^{3/2}}{165}+\frac {199\,{\left (1-2\,x\right )}^{5/2}}{11}}{\frac {2266\,x}{75}+\frac {101\,{\left (2\,x-1\right )}^2}{15}+{\left (2\,x-1\right )}^3-\frac {286}{75}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(624*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/7 - (6665*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/12
1 + ((6763*(1 - 2*x)^(1/2))/75 - (13334*(1 - 2*x)^(3/2))/165 + (199*(1 - 2*x)^(5/2))/11)/((2266*x)/75 + (101*(
2*x - 1)^2)/15 + (2*x - 1)^3 - 286/75)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________