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

Optimal. Leaf size=127 \[ \frac {7 (3 x+2)^4}{33 (1-2 x)^{3/2} (5 x+3)^2}-\frac {73 (3 x+2)^3}{3630 \sqrt {1-2 x} (5 x+3)^2}-\frac {317 (3 x+2)^2}{19965 \sqrt {1-2 x} (5 x+3)}-\frac {3 (544568-333311 x)}{732050 \sqrt {1-2 x}}-\frac {4693 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{366025 \sqrt {55}} \]

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Rubi [A]  time = 0.04, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {98, 149, 146, 63, 206} \begin {gather*} \frac {7 (3 x+2)^4}{33 (1-2 x)^{3/2} (5 x+3)^2}-\frac {73 (3 x+2)^3}{3630 \sqrt {1-2 x} (5 x+3)^2}-\frac {317 (3 x+2)^2}{19965 \sqrt {1-2 x} (5 x+3)}-\frac {3 (544568-333311 x)}{732050 \sqrt {1-2 x}}-\frac {4693 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{366025 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(544568 - 333311*x))/(732050*Sqrt[1 - 2*x]) - (73*(2 + 3*x)^3)/(3630*Sqrt[1 - 2*x]*(3 + 5*x)^2) + (7*(2 +
3*x)^4)/(33*(1 - 2*x)^(3/2)*(3 + 5*x)^2) - (317*(2 + 3*x)^2)/(19965*Sqrt[1 - 2*x]*(3 + 5*x)) - (4693*ArcTanh[S
qrt[5/11]*Sqrt[1 - 2*x]])/(366025*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 146

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n + 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(
b*c - a*d)*(m + 1)*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)), x] - Dist[
(a^2*d^2*f*h*(n + 1)*(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*(m +
1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d*(b*c - a*d)*(m +
1)*(m + n + 3)), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((Ge
Q[m, -2] && LtQ[m, -1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 0]

Rule 149

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*(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 - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[m]

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)^5}{(1-2 x)^{5/2} (3+5 x)^3} \, dx &=\frac {7 (2+3 x)^4}{33 (1-2 x)^{3/2} (3+5 x)^2}-\frac {1}{33} \int \frac {(2+3 x)^3 (106+201 x)}{(1-2 x)^{3/2} (3+5 x)^3} \, dx\\ &=-\frac {73 (2+3 x)^3}{3630 \sqrt {1-2 x} (3+5 x)^2}+\frac {7 (2+3 x)^4}{33 (1-2 x)^{3/2} (3+5 x)^2}-\frac {\int \frac {(2+3 x)^2 (7457+13485 x)}{(1-2 x)^{3/2} (3+5 x)^2} \, dx}{3630}\\ &=-\frac {73 (2+3 x)^3}{3630 \sqrt {1-2 x} (3+5 x)^2}+\frac {7 (2+3 x)^4}{33 (1-2 x)^{3/2} (3+5 x)^2}-\frac {317 (2+3 x)^2}{19965 \sqrt {1-2 x} (3+5 x)}-\frac {\int \frac {(2+3 x) (258630+454515 x)}{(1-2 x)^{3/2} (3+5 x)} \, dx}{199650}\\ &=-\frac {3 (544568-333311 x)}{732050 \sqrt {1-2 x}}-\frac {73 (2+3 x)^3}{3630 \sqrt {1-2 x} (3+5 x)^2}+\frac {7 (2+3 x)^4}{33 (1-2 x)^{3/2} (3+5 x)^2}-\frac {317 (2+3 x)^2}{19965 \sqrt {1-2 x} (3+5 x)}+\frac {4693 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{732050}\\ &=-\frac {3 (544568-333311 x)}{732050 \sqrt {1-2 x}}-\frac {73 (2+3 x)^3}{3630 \sqrt {1-2 x} (3+5 x)^2}+\frac {7 (2+3 x)^4}{33 (1-2 x)^{3/2} (3+5 x)^2}-\frac {317 (2+3 x)^2}{19965 \sqrt {1-2 x} (3+5 x)}-\frac {4693 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{732050}\\ &=-\frac {3 (544568-333311 x)}{732050 \sqrt {1-2 x}}-\frac {73 (2+3 x)^3}{3630 \sqrt {1-2 x} (3+5 x)^2}+\frac {7 (2+3 x)^4}{33 (1-2 x)^{3/2} (3+5 x)^2}-\frac {317 (2+3 x)^2}{19965 \sqrt {1-2 x} (3+5 x)}-\frac {4693 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{366025 \sqrt {55}}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 100, normalized size = 0.79 \begin {gather*} -\frac {-5327 (5 x+3)^2 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {5}{11} (1-2 x)\right )+5535 (2 x-1) (5 x+3)^2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {5}{11} (1-2 x)\right )+33 \left (7350750 x^4-17151750 x^3-21347475 x^2-741695 x+2582641\right )}{4991250 (1-2 x)^{3/2} (5 x+3)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4991250*(33*(2582641 - 741695*x - 21347475*x^2 - 17151750*x^3 + 7350750*x^4) - 5327*(3 + 5*x)^2*Hypergeomet
ric2F1[-3/2, 1, -1/2, (5*(1 - 2*x))/11] + 5535*(-1 + 2*x)*(3 + 5*x)^2*Hypergeometric2F1[-1/2, 1, 1/2, (5*(1 -
2*x))/11])/((1 - 2*x)^(3/2)*(3 + 5*x)^2)

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IntegrateAlgebraic [A]  time = 0.20, size = 88, normalized size = 0.69 \begin {gather*} \frac {-53366445 (1-2 x)^4-35296050 (1-2 x)^3+1045740476 (1-2 x)^2-1815756250 (1-2 x)[phone]}{4392300 (5 (1-2 x)-11)^2 (1-2 x)^{3/2}}-\frac {4693 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{366025 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(559252925 - 1815756250*(1 - 2*x) + 1045740476*(1 - 2*x)^2 - 35296050*(1 - 2*x)^3 - 53366445*(1 - 2*x)^4)/(439
2300*(-11 + 5*(1 - 2*x))^2*(1 - 2*x)^(3/2)) - (4693*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(366025*Sqrt[55])

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fricas [A]  time = 1.52, size = 104, normalized size = 0.82 \begin {gather*} \frac {14079 \, \sqrt {55} {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (106732890 \, x^{4} - 248761830 \, x^{3} - 309826828 \, x^{2} - 10907307 \, x + 37428168\right )} \sqrt {-2 \, x + 1}}{120788250 \, {\left (100 \, x^{4} + 20 \, x^{3} - 59 \, x^{2} - 6 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120788250*(14079*sqrt(55)*(100*x^4 + 20*x^3 - 59*x^2 - 6*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x
+ 3)) - 55*(106732890*x^4 - 248761830*x^3 - 309826828*x^2 - 10907307*x + 37428168)*sqrt(-2*x + 1))/(100*x^4 +
20*x^3 - 59*x^2 - 6*x + 9)

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giac [A]  time = 1.26, size = 98, normalized size = 0.77 \begin {gather*} \frac {4693}{40262750} \, \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 {243}{500} \, \sqrt {-2 \, x + 1} - \frac {2401 \, {\left (360 \, x - 103\right )}}{175692 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} + \frac {155 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 343 \, \sqrt {-2 \, x + 1}}{665500 \, {\left (5 \, x + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4693/40262750*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 243/500*s
qrt(-2*x + 1) - 2401/175692*(360*x - 103)/((2*x - 1)*sqrt(-2*x + 1)) + 1/665500*(155*(-2*x + 1)^(3/2) - 343*sq
rt(-2*x + 1))/(5*x + 3)^2

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maple [A]  time = 0.02, size = 75, normalized size = 0.59 \begin {gather*} -\frac {4693 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{20131375}-\frac {243 \sqrt {-2 x +1}}{500}+\frac {16807}{15972 \left (-2 x +1\right )^{\frac {3}{2}}}-\frac {36015}{14641 \sqrt {-2 x +1}}+\frac {\frac {31 \left (-2 x +1\right )^{\frac {3}{2}}}{33275}-\frac {343 \sqrt {-2 x +1}}{166375}}{\left (-10 x -6\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-243/500*(-2*x+1)^(1/2)+16807/15972/(-2*x+1)^(3/2)-36015/14641/(-2*x+1)^(1/2)+4/73205*(341/20*(-2*x+1)^(3/2)-3
773/100*(-2*x+1)^(1/2))/(-10*x-6)^2-4693/20131375*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.22, size = 101, normalized size = 0.80 \begin {gather*} \frac {4693}{40262750} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {243}{500} \, \sqrt {-2 \, x + 1} + \frac {1350542040 \, {\left (2 \, x - 1\right )}^{3} + 6520170349 \, {\left (2 \, x - 1\right )}^{2} + 18157562500 \, x - 6282516625}{21961500 \, {\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4693/40262750*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 243/500*sqrt(-2*x +
1) + 1/21961500*(1350542040*(2*x - 1)^3 + 6520170349*(2*x - 1)^2 + 18157562500*x - 6282516625)/(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 = 0.06, size = 80, normalized size = 0.63 \begin {gather*} \frac {\frac {12005\,x}{363}+\frac {592742759\,{\left (2\,x-1\right )}^2}{49912500}+\frac {22509034\,{\left (2\,x-1\right )}^3}{9150625}-\frac {415373}{36300}}{\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}}-\frac {243\,\sqrt {1-2\,x}}{500}-\frac {4693\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{20131375} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((12005*x)/363 + (592742759*(2*x - 1)^2)/49912500 + (22509034*(2*x - 1)^3)/9150625 - 415373/36300)/((121*(1 -
2*x)^(3/2))/25 - (22*(1 - 2*x)^(5/2))/5 + (1 - 2*x)^(7/2)) - (243*(1 - 2*x)^(1/2))/500 - (4693*55^(1/2)*atanh(
(55^(1/2)*(1 - 2*x)^(1/2))/11))/20131375

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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