3.1847 \(\int \frac {\sqrt {1-2 x}}{(2+3 x)^4 (3+5 x)^2} \, dx\)

Optimal. Leaf size=158 \[ -\frac {48645 \sqrt {1-2 x}}{98 (5 x+3)}+\frac {7261 \sqrt {1-2 x}}{147 (3 x+2) (5 x+3)}+\frac {139 \sqrt {1-2 x}}{42 (3 x+2)^2 (5 x+3)}+\frac {\sqrt {1-2 x}}{3 (3 x+2)^3 (5 x+3)}-\frac {335579}{49} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+6650 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

[Out]

-335579/343*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+6650/11*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)
-48645/98*(1-2*x)^(1/2)/(3+5*x)+1/3*(1-2*x)^(1/2)/(2+3*x)^3/(3+5*x)+139/42*(1-2*x)^(1/2)/(2+3*x)^2/(3+5*x)+726
1/147*(1-2*x)^(1/2)/(2+3*x)/(3+5*x)

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Rubi [A]  time = 0.06, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 9, 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} \[ -\frac {48645 \sqrt {1-2 x}}{98 (5 x+3)}+\frac {7261 \sqrt {1-2 x}}{147 (3 x+2) (5 x+3)}+\frac {139 \sqrt {1-2 x}}{42 (3 x+2)^2 (5 x+3)}+\frac {\sqrt {1-2 x}}{3 (3 x+2)^3 (5 x+3)}-\frac {335579}{49} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+6650 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - 2*x]/((2 + 3*x)^4*(3 + 5*x)^2),x]

[Out]

(-48645*Sqrt[1 - 2*x])/(98*(3 + 5*x)) + Sqrt[1 - 2*x]/(3*(2 + 3*x)^3*(3 + 5*x)) + (139*Sqrt[1 - 2*x])/(42*(2 +
3*x)^2*(3 + 5*x)) + (7261*Sqrt[1 - 2*x])/(147*(2 + 3*x)*(3 + 5*x)) - (335579*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt
[1 - 2*x]])/49 + 6650*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]

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)^4 (3+5 x)^2} \, dx &=\frac {\sqrt {1-2 x}}{3 (2+3 x)^3 (3+5 x)}-\frac {1}{3} \int \frac {-23+35 x}{\sqrt {1-2 x} (2+3 x)^3 (3+5 x)^2} \, dx\\ &=\frac {\sqrt {1-2 x}}{3 (2+3 x)^3 (3+5 x)}+\frac {139 \sqrt {1-2 x}}{42 (2+3 x)^2 (3+5 x)}-\frac {1}{42} \int \frac {-2524+3475 x}{\sqrt {1-2 x} (2+3 x)^2 (3+5 x)^2} \, dx\\ &=\frac {\sqrt {1-2 x}}{3 (2+3 x)^3 (3+5 x)}+\frac {139 \sqrt {1-2 x}}{42 (2+3 x)^2 (3+5 x)}+\frac {7261 \sqrt {1-2 x}}{147 (2+3 x) (3+5 x)}-\frac {1}{294} \int \frac {-190359+217830 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)^2} \, dx\\ &=-\frac {48645 \sqrt {1-2 x}}{98 (3+5 x)}+\frac {\sqrt {1-2 x}}{3 (2+3 x)^3 (3+5 x)}+\frac {139 \sqrt {1-2 x}}{42 (2+3 x)^2 (3+5 x)}+\frac {7261 \sqrt {1-2 x}}{147 (2+3 x) (3+5 x)}+\frac {\int \frac {-7863537+4815855 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx}{3234}\\ &=-\frac {48645 \sqrt {1-2 x}}{98 (3+5 x)}+\frac {\sqrt {1-2 x}}{3 (2+3 x)^3 (3+5 x)}+\frac {139 \sqrt {1-2 x}}{42 (2+3 x)^2 (3+5 x)}+\frac {7261 \sqrt {1-2 x}}{147 (2+3 x) (3+5 x)}+\frac {1006737}{98} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx-16625 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {48645 \sqrt {1-2 x}}{98 (3+5 x)}+\frac {\sqrt {1-2 x}}{3 (2+3 x)^3 (3+5 x)}+\frac {139 \sqrt {1-2 x}}{42 (2+3 x)^2 (3+5 x)}+\frac {7261 \sqrt {1-2 x}}{147 (2+3 x) (3+5 x)}-\frac {1006737}{98} \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )+16625 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {48645 \sqrt {1-2 x}}{98 (3+5 x)}+\frac {\sqrt {1-2 x}}{3 (2+3 x)^3 (3+5 x)}+\frac {139 \sqrt {1-2 x}}{42 (2+3 x)^2 (3+5 x)}+\frac {7261 \sqrt {1-2 x}}{147 (2+3 x) (3+5 x)}-\frac {335579}{49} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+6650 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 99, normalized size = 0.63 \[ -\frac {\sqrt {1-2 x} \left (1313415 x^3+2583264 x^2+1692159 x+369116\right )}{98 (3 x+2)^3 (5 x+3)}-\frac {335579}{49} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+6650 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - 2*x]/((2 + 3*x)^4*(3 + 5*x)^2),x]

[Out]

-1/98*(Sqrt[1 - 2*x]*(369116 + 1692159*x + 2583264*x^2 + 1313415*x^3))/((2 + 3*x)^3*(3 + 5*x)) - (335579*Sqrt[
3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/49 + 6650*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]

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fricas [A]  time = 0.81, size = 162, normalized size = 1.03 \[ \frac {2280950 \, \sqrt {11} \sqrt {5} {\left (135 \, x^{4} + 351 \, x^{3} + 342 \, x^{2} + 148 \, x + 24\right )} \log \left (-\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} - 5 \, x + 8}{5 \, x + 3}\right ) + 3691369 \, \sqrt {7} \sqrt {3} {\left (135 \, x^{4} + 351 \, x^{3} + 342 \, x^{2} + 148 \, x + 24\right )} \log \left (\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} + 3 \, x - 5}{3 \, x + 2}\right ) - 77 \, {\left (1313415 \, x^{3} + 2583264 \, x^{2} + 1692159 \, x + 369116\right )} \sqrt {-2 \, x + 1}}{7546 \, {\left (135 \, x^{4} + 351 \, x^{3} + 342 \, x^{2} + 148 \, x + 24\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(1/2)/(2+3*x)^4/(3+5*x)^2,x, algorithm="fricas")

[Out]

1/7546*(2280950*sqrt(11)*sqrt(5)*(135*x^4 + 351*x^3 + 342*x^2 + 148*x + 24)*log(-(sqrt(11)*sqrt(5)*sqrt(-2*x +
1) - 5*x + 8)/(5*x + 3)) + 3691369*sqrt(7)*sqrt(3)*(135*x^4 + 351*x^3 + 342*x^2 + 148*x + 24)*log((sqrt(7)*sq
rt(3)*sqrt(-2*x + 1) + 3*x - 5)/(3*x + 2)) - 77*(1313415*x^3 + 2583264*x^2 + 1692159*x + 369116)*sqrt(-2*x + 1
))/(135*x^4 + 351*x^3 + 342*x^2 + 148*x + 24)

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giac [A]  time = 1.29, size = 139, normalized size = 0.88 \[ -\frac {3325}{11} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {335579}{686} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {125 \, \sqrt {-2 \, x + 1}}{5 \, x + 3} - \frac {3 \, {\left (65511 \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - 308644 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 363629 \, \sqrt {-2 \, x + 1}\right )}}{392 \, {\left (3 \, x + 2\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(1/2)/(2+3*x)^4/(3+5*x)^2,x, algorithm="giac")

[Out]

-3325/11*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 335579/686*sqr
t(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 125*sqrt(-2*x + 1)/(5*x + 3
) - 3/392*(65511*(2*x - 1)^2*sqrt(-2*x + 1) - 308644*(-2*x + 1)^(3/2) + 363629*sqrt(-2*x + 1))/(3*x + 2)^3

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maple [A]  time = 0.02, size = 91, normalized size = 0.58 \[ -\frac {335579 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{343}+\frac {6650 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{11}+\frac {50 \sqrt {-2 x +1}}{-2 x -\frac {6}{5}}+\frac {\frac {196533 \left (-2 x +1\right )^{\frac {5}{2}}}{49}-\frac {132276 \left (-2 x +1\right )^{\frac {3}{2}}}{7}+22263 \sqrt {-2 x +1}}{\left (-6 x -4\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x+1)^(1/2)/(3*x+2)^4/(5*x+3)^2,x)

[Out]

50*(-2*x+1)^(1/2)/(-2*x-6/5)+6650/11*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)+324*(7279/588*(-2*x+1)^(5/
2)-11023/189*(-2*x+1)^(3/2)+7421/108*(-2*x+1)^(1/2))/(-6*x-4)^3-335579/343*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2)
)*21^(1/2)

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maxima [A]  time = 1.21, size = 146, normalized size = 0.92 \[ -\frac {3325}{11} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {335579}{686} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {1313415 \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - 9106773 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + 21041937 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 16201507 \, \sqrt {-2 \, x + 1}}{49 \, {\left (135 \, {\left (2 \, x - 1\right )}^{4} + 1242 \, {\left (2 \, x - 1\right )}^{3} + 4284 \, {\left (2 \, x - 1\right )}^{2} + 13132 \, x - 2793\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(1/2)/(2+3*x)^4/(3+5*x)^2,x, algorithm="maxima")

[Out]

-3325/11*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 335579/686*sqrt(21)*log(
-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 1/49*(1313415*(-2*x + 1)^(7/2) - 9106773*(-2*x
+ 1)^(5/2) + 21041937*(-2*x + 1)^(3/2) - 16201507*sqrt(-2*x + 1))/(135*(2*x - 1)^4 + 1242*(2*x - 1)^3 + 4284*
(2*x - 1)^2 + 13132*x - 2793)

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mupad [B]  time = 1.22, size = 108, normalized size = 0.68 \[ \frac {6650\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{11}-\frac {335579\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{343}-\frac {\frac {330643\,\sqrt {1-2\,x}}{135}-\frac {111333\,{\left (1-2\,x\right )}^{3/2}}{35}+\frac {3035591\,{\left (1-2\,x\right )}^{5/2}}{2205}-\frac {9729\,{\left (1-2\,x\right )}^{7/2}}{49}}{\frac {13132\,x}{135}+\frac {476\,{\left (2\,x-1\right )}^2}{15}+\frac {46\,{\left (2\,x-1\right )}^3}{5}+{\left (2\,x-1\right )}^4-\frac {931}{45}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - 2*x)^(1/2)/((3*x + 2)^4*(5*x + 3)^2),x)

[Out]

(6650*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/11 - (335579*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7)
)/343 - ((330643*(1 - 2*x)^(1/2))/135 - (111333*(1 - 2*x)^(3/2))/35 + (3035591*(1 - 2*x)^(5/2))/2205 - (9729*(
1 - 2*x)^(7/2))/49)/((13132*x)/135 + (476*(2*x - 1)^2)/15 + (46*(2*x - 1)^3)/5 + (2*x - 1)^4 - 931/45)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)**(1/2)/(2+3*x)**4/(3+5*x)**2,x)

[Out]

Timed out

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