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]

(-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]]

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Rubi [A]  time = 0.0597056, antiderivative size = 158, normalized size of antiderivative = 1., 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 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 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 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*}

Mathematica [A]  time = 0.118317, 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]

-(Sqrt[1 - 2*x]*(369116 + 1692159*x + 2583264*x^2 + 1313415*x^3))/(98*(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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Maple [A]  time = 0.013, size = 91, normalized size = 0.6 \begin{align*} 324\,{\frac{1}{ \left ( -6\,x-4 \right ) ^{3}} \left ({\frac{7279\, \left ( 1-2\,x \right ) ^{5/2}}{588}}-{\frac{11023\, \left ( 1-2\,x \right ) ^{3/2}}{189}}+{\frac{7421\,\sqrt{1-2\,x}}{108}} \right ) }-{\frac{335579\,\sqrt{21}}{343}{\it Artanh} \left ({\frac{\sqrt{21}}{7}\sqrt{1-2\,x}} \right ) }+50\,{\frac{\sqrt{1-2\,x}}{-2\,x-6/5}}+{\frac{6650\,\sqrt{55}}{11}{\it Artanh} \left ({\frac{\sqrt{55}}{11}\sqrt{1-2\,x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.6016, size = 197, normalized size = 1.25 \begin{align*} -\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 )}} \end{align*}

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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Fricas [A]  time = 1.65951, size = 509, normalized size = 3.22 \begin{align*} \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 )}} \end{align*}

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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Sympy [A]  time = 140.568, size = 665, normalized size = 4.21 \begin{align*} \text{result too large to display} \end{align*}

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]

-6060*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, (x <= 1/2) & (x > -2/3))) + 1632*P
iecewise((sqrt(21)*(3*log(sqrt(21)*sqrt(1 - 2*x)/7 - 1)/16 - 3*log(sqrt(21)*sqrt(1 - 2*x)/7 + 1)/16 + 3/(16*(s
qrt(21)*sqrt(1 - 2*x)/7 + 1)) + 1/(16*(sqrt(21)*sqrt(1 - 2*x)/7 + 1)**2) + 3/(16*(sqrt(21)*sqrt(1 - 2*x)/7 - 1
)) - 1/(16*(sqrt(21)*sqrt(1 - 2*x)/7 - 1)**2))/1029, (x <= 1/2) & (x > -2/3))) - 336*Piecewise((sqrt(21)*(-5*l
og(sqrt(21)*sqrt(1 - 2*x)/7 - 1)/32 + 5*log(sqrt(21)*sqrt(1 - 2*x)/7 + 1)/32 - 5/(32*(sqrt(21)*sqrt(1 - 2*x)/7
+ 1)) - 1/(16*(sqrt(21)*sqrt(1 - 2*x)/7 + 1)**2) - 1/(48*(sqrt(21)*sqrt(1 - 2*x)/7 + 1)**3) - 5/(32*(sqrt(21)
*sqrt(1 - 2*x)/7 - 1)) + 1/(16*(sqrt(21)*sqrt(1 - 2*x)/7 - 1)**2) - 1/(48*(sqrt(21)*sqrt(1 - 2*x)/7 - 1)**3))/
7203, (x <= 1/2) & (x > -2/3))) - 5500*Piecewise((sqrt(55)*(-log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/4 + log(sqrt(5
5)*sqrt(1 - 2*x)/11 + 1)/4 - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)) - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)))/60
5, (x <= 1/2) & (x > -3/5))) + 20100*Piecewise((-sqrt(21)*acoth(sqrt(21)*sqrt(1 - 2*x)/7)/21, 2*x - 1 < -7/3),
(-sqrt(21)*atanh(sqrt(21)*sqrt(1 - 2*x)/7)/21, 2*x - 1 > -7/3)) - 33500*Piecewise((-sqrt(55)*acoth(sqrt(55)*s
qrt(1 - 2*x)/11)/55, 2*x - 1 < -11/5), (-sqrt(55)*atanh(sqrt(55)*sqrt(1 - 2*x)/11)/55, 2*x - 1 > -11/5))

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Giac [A]  time = 1.98556, size = 188, normalized size = 1.19 \begin{align*} -\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}} \end{align*}

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