3.2183 \(\int \frac {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^2} \, dx\)

Optimal. Leaf size=80 \[ \frac {7 (3 x+2)^2}{33 (1-2 x)^{3/2} (5 x+3)}-\frac {2 (17112 x+10309)}{19965 \sqrt {1-2 x} (5 x+3)}-\frac {208 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{6655 \sqrt {55}} \]

[Out]

7/33*(2+3*x)^2/(1-2*x)^(3/2)/(3+5*x)-208/366025*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-2/19965*(10309+1
7112*x)/(3+5*x)/(1-2*x)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {98, 144, 63, 206} \[ \frac {7 (3 x+2)^2}{33 (1-2 x)^{3/2} (5 x+3)}-\frac {2 (17112 x+10309)}{19965 \sqrt {1-2 x} (5 x+3)}-\frac {208 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{6655 \sqrt {55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*(2 + 3*x)^2)/(33*(1 - 2*x)^(3/2)*(3 + 5*x)) - (2*(10309 + 17112*x))/(19965*Sqrt[1 - 2*x]*(3 + 5*x)) - (208*
ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(6655*Sqrt[55])

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 98

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

Rule 144

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :>
Simp[((b^2*c*d*e*g*(n + 1) + a^2*c*d*f*h*(n + 1) + a*b*(d^2*e*g*(m + 1) + c^2*f*h*(m + 1) - c*d*(f*g + e*h)*(
m + n + 2)) + (a^2*d^2*f*h*(n + 1) - a*b*d^2*(f*g + e*h)*(n + 1) + b^2*(c^2*f*h*(m + 1) - c*d*(f*g + e*h)*(m +
1) + d^2*e*g*(m + n + 2)))*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b*d*(b*c - a*d)^2*(m + 1)*(n + 1)), x] -
Dist[(a^2*d^2*f*h*(2 + 3*n + n^2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h
*(2 + 3*m + m^2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(6 + m^2 + 5*n + n^2 + m*(2*n + 5))))/(b*d*(b
*c - a*d)^2*(m + 1)*(n + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, h
}, x] && LtQ[m, -1] && LtQ[n, -1]

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 {(2+3 x)^3}{(1-2 x)^{5/2} (3+5 x)^2} \, dx &=\frac {7 (2+3 x)^2}{33 (1-2 x)^{3/2} (3+5 x)}-\frac {1}{33} \int \frac {(2+3 x) (50+96 x)}{(1-2 x)^{3/2} (3+5 x)^2} \, dx\\ &=\frac {7 (2+3 x)^2}{33 (1-2 x)^{3/2} (3+5 x)}-\frac {2 (10309+17112 x)}{19965 \sqrt {1-2 x} (3+5 x)}+\frac {104 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{6655}\\ &=\frac {7 (2+3 x)^2}{33 (1-2 x)^{3/2} (3+5 x)}-\frac {2 (10309+17112 x)}{19965 \sqrt {1-2 x} (3+5 x)}-\frac {104 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{6655}\\ &=\frac {7 (2+3 x)^2}{33 (1-2 x)^{3/2} (3+5 x)}-\frac {2 (10309+17112 x)}{19965 \sqrt {1-2 x} (3+5 x)}-\frac {208 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{6655 \sqrt {55}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 70, normalized size = 0.88 \[ \frac {624 \sqrt {55} \sqrt {2 x-1} \tan ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )-\frac {55 \left (106563 x^2+57832 x-3678\right )}{10 x^2+x-3}}{1098075 \sqrt {1-2 x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-55*(-3678 + 57832*x + 106563*x^2))/(-3 + x + 10*x^2) + 624*Sqrt[55]*Sqrt[-1 + 2*x]*ArcTan[Sqrt[5/11]*Sqrt[-
1 + 2*x]])/(1098075*Sqrt[1 - 2*x])

________________________________________________________________________________________

fricas [A]  time = 0.62, size = 84, normalized size = 1.05 \[ \frac {312 \, \sqrt {55} {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (106563 \, x^{2} + 57832 \, x - 3678\right )} \sqrt {-2 \, x + 1}}{1098075 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1098075*(312*sqrt(55)*(20*x^3 - 8*x^2 - 7*x + 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(10
6563*x^2 + 57832*x - 3678)*sqrt(-2*x + 1))/(20*x^3 - 8*x^2 - 7*x + 3)

________________________________________________________________________________________

giac [A]  time = 1.30, size = 77, normalized size = 0.96 \[ \frac {104}{366025} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {49 \, {\left (87 \, x - 5\right )}}{3993 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} - \frac {\sqrt {-2 \, x + 1}}{6655 \, {\left (5 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

104/366025*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 49/3993*(87*
x - 5)/((2*x - 1)*sqrt(-2*x + 1)) - 1/6655*sqrt(-2*x + 1)/(5*x + 3)

________________________________________________________________________________________

maple [A]  time = 0.02, size = 54, normalized size = 0.68 \[ -\frac {208 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{366025}+\frac {343}{726 \left (-2 x +1\right )^{\frac {3}{2}}}-\frac {1421}{2662 \sqrt {-2 x +1}}+\frac {2 \sqrt {-2 x +1}}{33275 \left (-2 x -\frac {6}{5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

343/726/(-2*x+1)^(3/2)-1421/2662/(-2*x+1)^(1/2)+2/33275*(-2*x+1)^(1/2)/(-2*x-6/5)-208/366025*arctanh(1/11*55^(
1/2)*(-2*x+1)^(1/2))*55^(1/2)

________________________________________________________________________________________

maxima [A]  time = 1.08, size = 74, normalized size = 0.92 \[ \frac {104}{366025} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {106563 \, {\left (2 \, x - 1\right )}^{2} + 657580 \, x - 121275}{39930 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 11 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

104/366025*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/39930*(106563*(2*x -
1)^2 + 657580*x - 121275)/(5*(-2*x + 1)^(5/2) - 11*(-2*x + 1)^(3/2))

________________________________________________________________________________________

mupad [B]  time = 1.20, size = 55, normalized size = 0.69 \[ \frac {\frac {5978\,x}{1815}+\frac {35521\,{\left (2\,x-1\right )}^2}{66550}-\frac {147}{242}}{\frac {11\,{\left (1-2\,x\right )}^{3/2}}{5}-{\left (1-2\,x\right )}^{5/2}}-\frac {208\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{366025} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((5978*x)/1815 + (35521*(2*x - 1)^2)/66550 - 147/242)/((11*(1 - 2*x)^(3/2))/5 - (1 - 2*x)^(5/2)) - (208*55^(1/
2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/366025

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________