3.2066 \(\int \frac{1}{\sqrt{1-2 x} (2+3 x)^2 (3+5 x)^3} \, dx\)

Optimal. Leaf size=126 \[ \frac{33465 \sqrt{1-2 x}}{1694 (5 x+3)}-\frac{505 \sqrt{1-2 x}}{154 (5 x+3)^2}+\frac{3 \sqrt{1-2 x}}{7 (3 x+2) (5 x+3)^2}+\frac{1908}{7} \sqrt{\frac{3}{7}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )-\frac{32025}{121} \sqrt{\frac{5}{11}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

[Out]

(-505*Sqrt[1 - 2*x])/(154*(3 + 5*x)^2) + (3*Sqrt[1 - 2*x])/(7*(2 + 3*x)*(3 + 5*x)^2) + (33465*Sqrt[1 - 2*x])/(
1694*(3 + 5*x)) + (1908*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/7 - (32025*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*S
qrt[1 - 2*x]])/121

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Rubi [A]  time = 0.0482673, antiderivative size = 126, normalized size of antiderivative = 1., 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 = {103, 151, 156, 63, 206} \[ \frac{33465 \sqrt{1-2 x}}{1694 (5 x+3)}-\frac{505 \sqrt{1-2 x}}{154 (5 x+3)^2}+\frac{3 \sqrt{1-2 x}}{7 (3 x+2) (5 x+3)^2}+\frac{1908}{7} \sqrt{\frac{3}{7}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )-\frac{32025}{121} \sqrt{\frac{5}{11}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-505*Sqrt[1 - 2*x])/(154*(3 + 5*x)^2) + (3*Sqrt[1 - 2*x])/(7*(2 + 3*x)*(3 + 5*x)^2) + (33465*Sqrt[1 - 2*x])/(
1694*(3 + 5*x)) + (1908*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/7 - (32025*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*S
qrt[1 - 2*x]])/121

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(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*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

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{1}{\sqrt{1-2 x} (2+3 x)^2 (3+5 x)^3} \, dx &=\frac{3 \sqrt{1-2 x}}{7 (2+3 x) (3+5 x)^2}+\frac{1}{7} \int \frac{56-75 x}{\sqrt{1-2 x} (2+3 x) (3+5 x)^3} \, dx\\ &=-\frac{505 \sqrt{1-2 x}}{154 (3+5 x)^2}+\frac{3 \sqrt{1-2 x}}{7 (2+3 x) (3+5 x)^2}-\frac{1}{154} \int \frac{3966-4545 x}{\sqrt{1-2 x} (2+3 x) (3+5 x)^2} \, dx\\ &=-\frac{505 \sqrt{1-2 x}}{154 (3+5 x)^2}+\frac{3 \sqrt{1-2 x}}{7 (2+3 x) (3+5 x)^2}+\frac{33465 \sqrt{1-2 x}}{1694 (3+5 x)}+\frac{\int \frac{163938-100395 x}{\sqrt{1-2 x} (2+3 x) (3+5 x)} \, dx}{1694}\\ &=-\frac{505 \sqrt{1-2 x}}{154 (3+5 x)^2}+\frac{3 \sqrt{1-2 x}}{7 (2+3 x) (3+5 x)^2}+\frac{33465 \sqrt{1-2 x}}{1694 (3+5 x)}-\frac{2862}{7} \int \frac{1}{\sqrt{1-2 x} (2+3 x)} \, dx+\frac{160125}{242} \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=-\frac{505 \sqrt{1-2 x}}{154 (3+5 x)^2}+\frac{3 \sqrt{1-2 x}}{7 (2+3 x) (3+5 x)^2}+\frac{33465 \sqrt{1-2 x}}{1694 (3+5 x)}+\frac{2862}{7} \operatorname{Subst}\left (\int \frac{1}{\frac{7}{2}-\frac{3 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )-\frac{160125}{242} \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=-\frac{505 \sqrt{1-2 x}}{154 (3+5 x)^2}+\frac{3 \sqrt{1-2 x}}{7 (2+3 x) (3+5 x)^2}+\frac{33465 \sqrt{1-2 x}}{1694 (3+5 x)}+\frac{1908}{7} \sqrt{\frac{3}{7}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )-\frac{32025}{121} \sqrt{\frac{5}{11}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0849703, size = 95, normalized size = 0.75 \[ \frac{\frac{11 \sqrt{1-2 x} \left (501975 x^2+619170 x+190406\right )}{(3 x+2) (5 x+3)^2}-448350 \sqrt{55} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{18634}+\frac{1908}{7} \sqrt{\frac{3}{7}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1908*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/7 + ((11*Sqrt[1 - 2*x]*(190406 + 619170*x + 501975*x^2))/((2
+ 3*x)*(3 + 5*x)^2) - 448350*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/18634

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Maple [A]  time = 0.011, size = 82, normalized size = 0.7 \begin{align*} -{\frac{18}{7}\sqrt{1-2\,x} \left ( -2\,x-{\frac{4}{3}} \right ) ^{-1}}+{\frac{1908\,\sqrt{21}}{49}{\it Artanh} \left ({\frac{\sqrt{21}}{7}\sqrt{1-2\,x}} \right ) }+1250\,{\frac{1}{ \left ( -10\,x-6 \right ) ^{2}} \left ( -{\frac{129\, \left ( 1-2\,x \right ) ^{3/2}}{1210}}+{\frac{127\,\sqrt{1-2\,x}}{550}} \right ) }-{\frac{32025\,\sqrt{55}}{1331}{\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+3*x)^2/(3+5*x)^3/(1-2*x)^(1/2),x)

[Out]

-18/7*(1-2*x)^(1/2)/(-2*x-4/3)+1908/49*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+1250*(-129/1210*(1-2*x)^(3
/2)+127/550*(1-2*x)^(1/2))/(-10*x-6)^2-32025/1331*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 3.34388, size = 173, normalized size = 1.37 \begin{align*} \frac{32025}{2662} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) - \frac{954}{49} \, \sqrt{21} \log \left (-\frac{\sqrt{21} - 3 \, \sqrt{-2 \, x + 1}}{\sqrt{21} + 3 \, \sqrt{-2 \, x + 1}}\right ) + \frac{501975 \,{\left (-2 \, x + 1\right )}^{\frac{5}{2}} - 2242290 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + 2501939 \, \sqrt{-2 \, x + 1}}{847 \,{\left (75 \,{\left (2 \, x - 1\right )}^{3} + 505 \,{\left (2 \, x - 1\right )}^{2} + 2266 \, x - 286\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32025/2662*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 954/49*sqrt(21)*log(-(
sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 1/847*(501975*(-2*x + 1)^(5/2) - 2242290*(-2*x +
1)^(3/2) + 2501939*sqrt(-2*x + 1))/(75*(2*x - 1)^3 + 505*(2*x - 1)^2 + 2266*x - 286)

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Fricas [A]  time = 1.66477, size = 441, normalized size = 3.5 \begin{align*} \frac{1569225 \, \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 ) + 2539548 \, \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 (501975 \, x^{2} + 619170 \, x + 190406\right )} \sqrt{-2 \, x + 1}}{130438 \,{\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/130438*(1569225*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)) + 2539548*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*(501975*x^2 + 619170*x + 190406)*sqrt(-2*x + 1))/(75*x^3 + 140*x^2 + 87*x + 18)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Giac [A]  time = 2.59971, size = 166, normalized size = 1.32 \begin{align*} \frac{32025}{2662} \, \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{954}{49} \, \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{27 \, \sqrt{-2 \, x + 1}}{7 \,{\left (3 \, x + 2\right )}} - \frac{25 \,{\left (645 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - 1397 \, \sqrt{-2 \, x + 1}\right )}}{484 \,{\left (5 \, x + 3\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32025/2662*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 954/49*sqrt(
21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 27/7*sqrt(-2*x + 1)/(3*x + 2)
- 25/484*(645*(-2*x + 1)^(3/2) - 1397*sqrt(-2*x + 1))/(5*x + 3)^2