3.2195 \(\int \frac {2+3 x}{(1-2 x)^{5/2} (3+5 x)^3} \, dx\)

Optimal. Leaf size=96 \[ \frac {365}{14641 \sqrt {1-2 x}}-\frac {73}{1210 (1-2 x)^{3/2} (5 x+3)}+\frac {73}{3993 (1-2 x)^{3/2}}-\frac {1}{110 (1-2 x)^{3/2} (5 x+3)^2}-\frac {365 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{14641} \]

[Out]

73/3993/(1-2*x)^(3/2)-1/110/(1-2*x)^(3/2)/(3+5*x)^2-73/1210/(1-2*x)^(3/2)/(3+5*x)-365/161051*arctanh(1/11*55^(
1/2)*(1-2*x)^(1/2))*55^(1/2)+365/14641/(1-2*x)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {78, 51, 63, 206} \[ \frac {365}{14641 \sqrt {1-2 x}}-\frac {73}{1210 (1-2 x)^{3/2} (5 x+3)}+\frac {73}{3993 (1-2 x)^{3/2}}-\frac {1}{110 (1-2 x)^{3/2} (5 x+3)^2}-\frac {365 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{14641} \]

Antiderivative was successfully verified.

[In]

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

[Out]

73/(3993*(1 - 2*x)^(3/2)) + 365/(14641*Sqrt[1 - 2*x]) - 1/(110*(1 - 2*x)^(3/2)*(3 + 5*x)^2) - 73/(1210*(1 - 2*
x)^(3/2)*(3 + 5*x)) - (365*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/14641

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, 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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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}{(1-2 x)^{5/2} (3+5 x)^3} \, dx &=-\frac {1}{110 (1-2 x)^{3/2} (3+5 x)^2}+\frac {73}{110} \int \frac {1}{(1-2 x)^{5/2} (3+5 x)^2} \, dx\\ &=-\frac {1}{110 (1-2 x)^{3/2} (3+5 x)^2}-\frac {73}{1210 (1-2 x)^{3/2} (3+5 x)}+\frac {73}{242} \int \frac {1}{(1-2 x)^{5/2} (3+5 x)} \, dx\\ &=\frac {73}{3993 (1-2 x)^{3/2}}-\frac {1}{110 (1-2 x)^{3/2} (3+5 x)^2}-\frac {73}{1210 (1-2 x)^{3/2} (3+5 x)}+\frac {365 \int \frac {1}{(1-2 x)^{3/2} (3+5 x)} \, dx}{2662}\\ &=\frac {73}{3993 (1-2 x)^{3/2}}+\frac {365}{14641 \sqrt {1-2 x}}-\frac {1}{110 (1-2 x)^{3/2} (3+5 x)^2}-\frac {73}{1210 (1-2 x)^{3/2} (3+5 x)}+\frac {1825 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{29282}\\ &=\frac {73}{3993 (1-2 x)^{3/2}}+\frac {365}{14641 \sqrt {1-2 x}}-\frac {1}{110 (1-2 x)^{3/2} (3+5 x)^2}-\frac {73}{1210 (1-2 x)^{3/2} (3+5 x)}-\frac {1825 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{29282}\\ &=\frac {73}{3993 (1-2 x)^{3/2}}+\frac {365}{14641 \sqrt {1-2 x}}-\frac {1}{110 (1-2 x)^{3/2} (3+5 x)^2}-\frac {73}{1210 (1-2 x)^{3/2} (3+5 x)}-\frac {365 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{14641}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 48, normalized size = 0.50 \[ -\frac {363-292 (5 x+3)^2 \, _2F_1\left (-\frac {3}{2},2;-\frac {1}{2};-\frac {5}{11} (2 x-1)\right )}{39930 (1-2 x)^{3/2} (5 x+3)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/39930*(363 - 292*(3 + 5*x)^2*Hypergeometric2F1[-3/2, 2, -1/2, (-5*(-1 + 2*x))/11])/((1 - 2*x)^(3/2)*(3 + 5*
x)^2)

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fricas [A]  time = 1.01, size = 105, normalized size = 1.09 \[ \frac {1095 \, \sqrt {11} \sqrt {5} {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) - 11 \, {\left (109500 \, x^{3} + 36500 \, x^{2} - 47961 \, x - 17466\right )} \sqrt {-2 \, x + 1}}{966306 \, {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/966306*(1095*sqrt(11)*sqrt(5)*(100*x^4 + 20*x^3 - 59*x^2 - 6*x + 9)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5
*x - 8)/(5*x + 3)) - 11*(109500*x^3 + 36500*x^2 - 47961*x - 17466)*sqrt(-2*x + 1))/(100*x^4 + 20*x^3 - 59*x^2
- 6*x + 9)

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giac [A]  time = 1.32, size = 89, normalized size = 0.93 \[ \frac {365}{322102} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {4 \, {\left (432 \, x - 293\right )}}{43923 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} + \frac {5 \, {\left (35 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 79 \, \sqrt {-2 \, x + 1}\right )}}{5324 \, {\left (5 \, x + 3\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

365/322102*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 4/43923*(432
*x - 293)/((2*x - 1)*sqrt(-2*x + 1)) + 5/5324*(35*(-2*x + 1)^(3/2) - 79*sqrt(-2*x + 1))/(5*x + 3)^2

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maple [A]  time = 0.02, size = 66, normalized size = 0.69 \[ -\frac {365 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{161051}+\frac {28}{3993 \left (-2 x +1\right )^{\frac {3}{2}}}+\frac {288}{14641 \sqrt {-2 x +1}}+\frac {\frac {175 \left (-2 x +1\right )^{\frac {3}{2}}}{1331}-\frac {395 \sqrt {-2 x +1}}{1331}}{\left (-10 x -6\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

28/3993/(-2*x+1)^(3/2)+288/14641/(-2*x+1)^(1/2)+500/14641*(77/20*(-2*x+1)^(3/2)-869/100*(-2*x+1)^(1/2))/(-10*x
-6)^2-365/161051*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.08, size = 92, normalized size = 0.96 \[ \frac {365}{322102} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {27375 \, {\left (2 \, x - 1\right )}^{3} + 100375 \, {\left (2 \, x - 1\right )}^{2} + 141328 \, x - 107932}{43923 \, {\left (25 \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - 110 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + 121 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

365/322102*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/43923*(27375*(2*x -
1)^3 + 100375*(2*x - 1)^2 + 141328*x - 107932)/(25*(-2*x + 1)^(7/2) - 110*(-2*x + 1)^(5/2) + 121*(-2*x + 1)^(3
/2))

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mupad [B]  time = 0.08, size = 72, normalized size = 0.75 \[ -\frac {365\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{161051}-\frac {\frac {1168\,x}{9075}+\frac {365\,{\left (2\,x-1\right )}^2}{3993}+\frac {365\,{\left (2\,x-1\right )}^3}{14641}-\frac {892}{9075}}{\frac {121\,{\left (1-2\,x\right )}^{3/2}}{25}-\frac {22\,{\left (1-2\,x\right )}^{5/2}}{5}+{\left (1-2\,x\right )}^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (365*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/161051 - ((1168*x)/9075 + (365*(2*x - 1)^2)/3993 + (365*
(2*x - 1)^3)/14641 - 892/9075)/((121*(1 - 2*x)^(3/2))/25 - (22*(1 - 2*x)^(5/2))/5 + (1 - 2*x)^(7/2))

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

[Out]

Timed out

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