3.2105 \(\int \frac {(2+3 x)^5}{(1-2 x)^{3/2} (3+5 x)} \, dx\)

Optimal. Leaf size=93 \[ -\frac {243}{560} (1-2 x)^{7/2}+\frac {5751 (1-2 x)^{5/2}}{1000}-\frac {17019}{500} (1-2 x)^{3/2}+\frac {806121 \sqrt {1-2 x}}{5000}+\frac {16807}{176 \sqrt {1-2 x}}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{6875 \sqrt {55}} \]

[Out]

-17019/500*(1-2*x)^(3/2)+5751/1000*(1-2*x)^(5/2)-243/560*(1-2*x)^(7/2)-2/378125*arctanh(1/11*55^(1/2)*(1-2*x)^
(1/2))*55^(1/2)+16807/176/(1-2*x)^(1/2)+806121/5000*(1-2*x)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {87, 43, 63, 206} \[ -\frac {243}{560} (1-2 x)^{7/2}+\frac {5751 (1-2 x)^{5/2}}{1000}-\frac {17019}{500} (1-2 x)^{3/2}+\frac {806121 \sqrt {1-2 x}}{5000}+\frac {16807}{176 \sqrt {1-2 x}}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{6875 \sqrt {55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

16807/(176*Sqrt[1 - 2*x]) + (806121*Sqrt[1 - 2*x])/5000 - (17019*(1 - 2*x)^(3/2))/500 + (5751*(1 - 2*x)^(5/2))
/1000 - (243*(1 - 2*x)^(7/2))/560 - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(6875*Sqrt[55])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

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)^{3/2} (3+5 x)} \, dx &=\int \left (\frac {16807}{176 (1-2 x)^{3/2}}-\frac {848277}{10000 \sqrt {1-2 x}}-\frac {107433 x}{1000 \sqrt {1-2 x}}-\frac {7857 x^2}{100 \sqrt {1-2 x}}-\frac {243 x^3}{10 \sqrt {1-2 x}}+\frac {1}{6875 \sqrt {1-2 x} (3+5 x)}\right ) \, dx\\ &=\frac {16807}{176 \sqrt {1-2 x}}+\frac {848277 \sqrt {1-2 x}}{10000}+\frac {\int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{6875}-\frac {243}{10} \int \frac {x^3}{\sqrt {1-2 x}} \, dx-\frac {7857}{100} \int \frac {x^2}{\sqrt {1-2 x}} \, dx-\frac {107433 \int \frac {x}{\sqrt {1-2 x}} \, dx}{1000}\\ &=\frac {16807}{176 \sqrt {1-2 x}}+\frac {848277 \sqrt {1-2 x}}{10000}-\frac {\operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{6875}-\frac {243}{10} \int \left (\frac {1}{8 \sqrt {1-2 x}}-\frac {3}{8} \sqrt {1-2 x}+\frac {3}{8} (1-2 x)^{3/2}-\frac {1}{8} (1-2 x)^{5/2}\right ) \, dx-\frac {7857}{100} \int \left (\frac {1}{4 \sqrt {1-2 x}}-\frac {1}{2} \sqrt {1-2 x}+\frac {1}{4} (1-2 x)^{3/2}\right ) \, dx-\frac {107433 \int \left (\frac {1}{2 \sqrt {1-2 x}}-\frac {1}{2} \sqrt {1-2 x}\right ) \, dx}{1000}\\ &=\frac {16807}{176 \sqrt {1-2 x}}+\frac {806121 \sqrt {1-2 x}}{5000}-\frac {17019}{500} (1-2 x)^{3/2}+\frac {5751 (1-2 x)^{5/2}}{1000}-\frac {243}{560} (1-2 x)^{7/2}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{6875 \sqrt {55}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 55, normalized size = 0.59 \[ \frac {14 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {5}{11} (1-2 x)\right )-33 \left (50625 x^4+234225 x^3+565500 x^2+1584705 x-1662482\right )}{240625 \sqrt {1-2 x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-33*(-1662482 + 1584705*x + 565500*x^2 + 234225*x^3 + 50625*x^4) + 14*Hypergeometric2F1[-1/2, 1, 1/2, (5*(1 -
 2*x))/11])/(240625*Sqrt[1 - 2*x])

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fricas [A]  time = 0.98, size = 74, normalized size = 0.80 \[ \frac {7 \, \sqrt {55} {\left (2 \, x - 1\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (334125 \, x^{4} + 1545885 \, x^{3} + 3732300 \, x^{2} + 10459053 \, x - 10972384\right )} \sqrt {-2 \, x + 1}}{2646875 \, {\left (2 \, x - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2646875*(7*sqrt(55)*(2*x - 1)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(334125*x^4 + 1545885*
x^3 + 3732300*x^2 + 10459053*x - 10972384)*sqrt(-2*x + 1))/(2*x - 1)

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giac [A]  time = 1.29, size = 99, normalized size = 1.06 \[ \frac {243}{560} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} + \frac {5751}{1000} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - \frac {17019}{500} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{378125} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {806121}{5000} \, \sqrt {-2 \, x + 1} + \frac {16807}{176 \, \sqrt {-2 \, x + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

243/560*(2*x - 1)^3*sqrt(-2*x + 1) + 5751/1000*(2*x - 1)^2*sqrt(-2*x + 1) - 17019/500*(-2*x + 1)^(3/2) + 1/378
125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 806121/5000*sqrt(-2
*x + 1) + 16807/176/sqrt(-2*x + 1)

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maple [A]  time = 0.01, size = 65, normalized size = 0.70 \[ -\frac {2 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{378125}-\frac {17019 \left (-2 x +1\right )^{\frac {3}{2}}}{500}+\frac {5751 \left (-2 x +1\right )^{\frac {5}{2}}}{1000}-\frac {243 \left (-2 x +1\right )^{\frac {7}{2}}}{560}+\frac {16807}{176 \sqrt {-2 x +1}}+\frac {806121 \sqrt {-2 x +1}}{5000} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-17019/500*(-2*x+1)^(3/2)+5751/1000*(-2*x+1)^(5/2)-243/560*(-2*x+1)^(7/2)-2/378125*arctanh(1/11*55^(1/2)*(-2*x
+1)^(1/2))*55^(1/2)+16807/176/(-2*x+1)^(1/2)+806121/5000*(-2*x+1)^(1/2)

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maxima [A]  time = 1.32, size = 82, normalized size = 0.88 \[ -\frac {243}{560} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} + \frac {5751}{1000} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - \frac {17019}{500} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{378125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {806121}{5000} \, \sqrt {-2 \, x + 1} + \frac {16807}{176 \, \sqrt {-2 \, x + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-243/560*(-2*x + 1)^(7/2) + 5751/1000*(-2*x + 1)^(5/2) - 17019/500*(-2*x + 1)^(3/2) + 1/378125*sqrt(55)*log(-(
sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 806121/5000*sqrt(-2*x + 1) + 16807/176/sqrt(-2*x
 + 1)

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mupad [B]  time = 1.22, size = 66, normalized size = 0.71 \[ \frac {16807}{176\,\sqrt {1-2\,x}}+\frac {806121\,\sqrt {1-2\,x}}{5000}-\frac {17019\,{\left (1-2\,x\right )}^{3/2}}{500}+\frac {5751\,{\left (1-2\,x\right )}^{5/2}}{1000}-\frac {243\,{\left (1-2\,x\right )}^{7/2}}{560}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,2{}\mathrm {i}}{378125} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*2i)/378125 + 16807/(176*(1 - 2*x)^(1/2)) + (806121*(1 - 2*x)^
(1/2))/5000 - (17019*(1 - 2*x)^(3/2))/500 + (5751*(1 - 2*x)^(5/2))/1000 - (243*(1 - 2*x)^(7/2))/560

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sympy [A]  time = 126.28, size = 126, normalized size = 1.35 \[ - \frac {243 \left (1 - 2 x\right )^{\frac {7}{2}}}{560} + \frac {5751 \left (1 - 2 x\right )^{\frac {5}{2}}}{1000} - \frac {17019 \left (1 - 2 x\right )^{\frac {3}{2}}}{500} + \frac {806121 \sqrt {1 - 2 x}}{5000} + \frac {2 \left (\begin {cases} - \frac {\sqrt {55} \operatorname {acoth}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: 2 x - 1 < - \frac {11}{5} \\- \frac {\sqrt {55} \operatorname {atanh}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: 2 x - 1 > - \frac {11}{5} \end {cases}\right )}{6875} + \frac {16807}{176 \sqrt {1 - 2 x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-243*(1 - 2*x)**(7/2)/560 + 5751*(1 - 2*x)**(5/2)/1000 - 17019*(1 - 2*x)**(3/2)/500 + 806121*sqrt(1 - 2*x)/500
0 + 2*Piecewise((-sqrt(55)*acoth(sqrt(55)*sqrt(1 - 2*x)/11)/55, 2*x - 1 < -11/5), (-sqrt(55)*atanh(sqrt(55)*sq
rt(1 - 2*x)/11)/55, 2*x - 1 > -11/5))/6875 + 16807/(176*sqrt(1 - 2*x))

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