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

Optimal. Leaf size=112 \[ -\frac {2525}{3773 \sqrt {1-2 x}}+\frac {225}{98 \sqrt {1-2 x} (3 x+2)}+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2}+\frac {8025}{343} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {250}{11} \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {103, 151, 152, 156, 63, 206} \begin {gather*} -\frac {2525}{3773 \sqrt {1-2 x}}+\frac {225}{98 \sqrt {1-2 x} (3 x+2)}+\frac {3}{14 \sqrt {1-2 x} (3 x+2)^2}+\frac {8025}{343} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {250}{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[1/((1 - 2*x)^(3/2)*(2 + 3*x)^3*(3 + 5*x)),x]

[Out]

-2525/(3773*Sqrt[1 - 2*x]) + 3/(14*Sqrt[1 - 2*x]*(2 + 3*x)^2) + 225/(98*Sqrt[1 - 2*x]*(2 + 3*x)) + (8025*Sqrt[
3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/343 - (250*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 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 152

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] && IntegersQ[2*m, 2*n, 2*p]

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 {1}{(1-2 x)^{3/2} (2+3 x)^3 (3+5 x)} \, dx &=\frac {3}{14 \sqrt {1-2 x} (2+3 x)^2}+\frac {1}{14} \int \frac {25-75 x}{(1-2 x)^{3/2} (2+3 x)^2 (3+5 x)} \, dx\\ &=\frac {3}{14 \sqrt {1-2 x} (2+3 x)^2}+\frac {225}{98 \sqrt {1-2 x} (2+3 x)}+\frac {1}{98} \int \frac {425-3375 x}{(1-2 x)^{3/2} (2+3 x) (3+5 x)} \, dx\\ &=-\frac {2525}{3773 \sqrt {1-2 x}}+\frac {3}{14 \sqrt {1-2 x} (2+3 x)^2}+\frac {225}{98 \sqrt {1-2 x} (2+3 x)}-\frac {\int \frac {-\frac {63025}{2}+\frac {37875 x}{2}}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx}{3773}\\ &=-\frac {2525}{3773 \sqrt {1-2 x}}+\frac {3}{14 \sqrt {1-2 x} (2+3 x)^2}+\frac {225}{98 \sqrt {1-2 x} (2+3 x)}-\frac {24075}{686} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx+\frac {625}{11} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {2525}{3773 \sqrt {1-2 x}}+\frac {3}{14 \sqrt {1-2 x} (2+3 x)^2}+\frac {225}{98 \sqrt {1-2 x} (2+3 x)}+\frac {24075}{686} \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-\frac {625}{11} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {2525}{3773 \sqrt {1-2 x}}+\frac {3}{14 \sqrt {1-2 x} (2+3 x)^2}+\frac {225}{98 \sqrt {1-2 x} (2+3 x)}+\frac {8025}{343} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {250}{11} \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.05, size = 80, normalized size = 0.71 \begin {gather*} \frac {7 \left (24500 (3 x+2)^2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\frac {5}{11} (2 x-1)\right )+7425 x+5181\right )-176550 (3 x+2)^2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {3}{7}-\frac {6 x}{7}\right )}{7546 \sqrt {1-2 x} (3 x+2)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-176550*(2 + 3*x)^2*Hypergeometric2F1[-1/2, 1, 1/2, 3/7 - (6*x)/7] + 7*(5181 + 7425*x + 24500*(2 + 3*x)^2*Hyp
ergeometric2F1[-1/2, 1, 1/2, (-5*(-1 + 2*x))/11]))/(7546*Sqrt[1 - 2*x]*(2 + 3*x)^2)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.26, size = 101, normalized size = 0.90 \begin {gather*} \frac {-22725 (1-2 x)^2+54075 (1-2 x)+784}{3773 (3 (1-2 x)-7)^2 \sqrt {1-2 x}}+\frac {8025}{343} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {250}{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[1/((1 - 2*x)^(3/2)*(2 + 3*x)^3*(3 + 5*x)),x]

[Out]

(784 + 54075*(1 - 2*x) - 22725*(1 - 2*x)^2)/(3773*(-7 + 3*(1 - 2*x))^2*Sqrt[1 - 2*x]) + (8025*Sqrt[3/7]*ArcTan
h[Sqrt[3/7]*Sqrt[1 - 2*x]])/343 - (250*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/11

________________________________________________________________________________________

fricas [A]  time = 0.81, size = 142, normalized size = 1.27 \begin {gather*} \frac {600250 \, \sqrt {11} \sqrt {5} {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 971025 \, \sqrt {7} \sqrt {3} {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) + 77 \, {\left (45450 \, x^{2} + 8625 \, x - 16067\right )} \sqrt {-2 \, x + 1}}{581042 \, {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/581042*(600250*sqrt(11)*sqrt(5)*(18*x^3 + 15*x^2 - 4*x - 4)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/
(5*x + 3)) + 971025*sqrt(7)*sqrt(3)*(18*x^3 + 15*x^2 - 4*x - 4)*log(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x + 5
)/(3*x + 2)) + 77*(45450*x^2 + 8625*x - 16067)*sqrt(-2*x + 1))/(18*x^3 + 15*x^2 - 4*x - 4)

________________________________________________________________________________________

giac [A]  time = 1.25, size = 116, normalized size = 1.04 \begin {gather*} \frac {125}{121} \, \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 {8025}{4802} \, \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 {16}{3773 \, \sqrt {-2 \, x + 1}} - \frac {9 \, {\left (33 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 79 \, \sqrt {-2 \, x + 1}\right )}}{196 \, {\left (3 \, x + 2\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125/121*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 8025/4802*sqrt(
21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 16/3773/sqrt(-2*x + 1) - 9/19
6*(33*(-2*x + 1)^(3/2) - 79*sqrt(-2*x + 1))/(3*x + 2)^2

________________________________________________________________________________________

maple [A]  time = 0.02, size = 75, normalized size = 0.67 \begin {gather*} \frac {8025 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{2401}-\frac {250 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{121}+\frac {16}{3773 \sqrt {-2 x +1}}-\frac {486 \left (\frac {77 \left (-2 x +1\right )^{\frac {3}{2}}}{18}-\frac {553 \sqrt {-2 x +1}}{54}\right )}{343 \left (-6 x -4\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/3773/(-2*x+1)^(1/2)-250/121*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)-486/343*(77/18*(-2*x+1)^(3/2)-55
3/54*(-2*x+1)^(1/2))/(-6*x-4)^2+8025/2401*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)

________________________________________________________________________________________

maxima [A]  time = 1.15, size = 119, normalized size = 1.06 \begin {gather*} \frac {125}{121} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {8025}{4802} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {22725 \, {\left (2 \, x - 1\right )}^{2} + 108150 \, x - 54859}{3773 \, {\left (9 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 42 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 49 \, \sqrt {-2 \, x + 1}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125/121*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 8025/4802*sqrt(21)*log(-(
sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 1/3773*(22725*(2*x - 1)^2 + 108150*x - 54859)/(9
*(-2*x + 1)^(5/2) - 42*(-2*x + 1)^(3/2) + 49*sqrt(-2*x + 1))

________________________________________________________________________________________

mupad [B]  time = 0.10, size = 81, normalized size = 0.72 \begin {gather*} \frac {8025\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{2401}-\frac {250\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{121}-\frac {\frac {5150\,x}{1617}+\frac {2525\,{\left (2\,x-1\right )}^2}{3773}-\frac {7837}{4851}}{\frac {49\,\sqrt {1-2\,x}}{9}-\frac {14\,{\left (1-2\,x\right )}^{3/2}}{3}+{\left (1-2\,x\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8025*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/2401 - (250*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))
/121 - ((5150*x)/1617 + (2525*(2*x - 1)^2)/3773 - 7837/4851)/((49*(1 - 2*x)^(1/2))/9 - (14*(1 - 2*x)^(3/2))/3
+ (1 - 2*x)^(5/2))

________________________________________________________________________________________

sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: MellinTransformStripError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: MellinTransformStripError

________________________________________________________________________________________