3.21.32 \(\int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)} \, dx\)

Optimal. Leaf size=80 \[ \frac {27}{40} (1-2 x)^{3/2}-\frac {2889}{200} \sqrt {1-2 x}-\frac {33271}{968 \sqrt {1-2 x}}+\frac {2401}{264 (1-2 x)^{3/2}}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}} \]

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Rubi [A]  time = 0.04, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {87, 43, 63, 206} \begin {gather*} \frac {27}{40} (1-2 x)^{3/2}-\frac {2889}{200} \sqrt {1-2 x}-\frac {33271}{968 \sqrt {1-2 x}}+\frac {2401}{264 (1-2 x)^{3/2}}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x)^4/((1 - 2*x)^(5/2)*(3 + 5*x)),x]

[Out]

2401/(264*(1 - 2*x)^(3/2)) - 33271/(968*Sqrt[1 - 2*x]) - (2889*Sqrt[1 - 2*x])/200 + (27*(1 - 2*x)^(3/2))/40 -
(2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(3025*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)^4}{(1-2 x)^{5/2} (3+5 x)} \, dx &=\int \left (\frac {2401}{88 (1-2 x)^{5/2}}-\frac {33271}{968 (1-2 x)^{3/2}}+\frac {621}{50 \sqrt {1-2 x}}+\frac {81 x}{20 \sqrt {1-2 x}}+\frac {1}{3025 \sqrt {1-2 x} (3+5 x)}\right ) \, dx\\ &=\frac {2401}{264 (1-2 x)^{3/2}}-\frac {33271}{968 \sqrt {1-2 x}}-\frac {621}{50} \sqrt {1-2 x}+\frac {\int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{3025}+\frac {81}{20} \int \frac {x}{\sqrt {1-2 x}} \, dx\\ &=\frac {2401}{264 (1-2 x)^{3/2}}-\frac {33271}{968 \sqrt {1-2 x}}-\frac {621}{50} \sqrt {1-2 x}-\frac {\operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{3025}+\frac {81}{20} \int \left (\frac {1}{2 \sqrt {1-2 x}}-\frac {1}{2} \sqrt {1-2 x}\right ) \, dx\\ &=\frac {2401}{264 (1-2 x)^{3/2}}-\frac {33271}{968 \sqrt {1-2 x}}-\frac {2889}{200} \sqrt {1-2 x}+\frac {27}{40} (1-2 x)^{3/2}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 50, normalized size = 0.62 \begin {gather*} \frac {2 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {5}{11} (1-2 x)\right )-33 \left (3375 x^3+31050 x^2-76545 x+24404\right )}{20625 (1-2 x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 3*x)^4/((1 - 2*x)^(5/2)*(3 + 5*x)),x]

[Out]

(-33*(24404 - 76545*x + 31050*x^2 + 3375*x^3) + 2*Hypergeometric2F1[-3/2, 1, -1/2, (5*(1 - 2*x))/11])/(20625*(
1 - 2*x)^(3/2))

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IntegrateAlgebraic [A]  time = 0.08, size = 68, normalized size = 0.85 \begin {gather*} \frac {49005 (1-2 x)^3-1048707 (1-2 x)^2-2495325 (1-2 x)+660275}{72600 (1-2 x)^{3/2}}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(2 + 3*x)^4/((1 - 2*x)^(5/2)*(3 + 5*x)),x]

[Out]

(660275 - 2495325*(1 - 2*x) - 1048707*(1 - 2*x)^2 + 49005*(1 - 2*x)^3)/(72600*(1 - 2*x)^(3/2)) - (2*ArcTanh[Sq
rt[5/11]*Sqrt[1 - 2*x]])/(3025*Sqrt[55])

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fricas [A]  time = 1.15, size = 79, normalized size = 0.99 \begin {gather*} \frac {3 \, \sqrt {55} {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (49005 \, x^{3} + 450846 \, x^{2} - 1111431 \, x + 354344\right )} \sqrt {-2 \, x + 1}}{499125 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4/(1-2*x)^(5/2)/(3+5*x),x, algorithm="fricas")

[Out]

1/499125*(3*sqrt(55)*(4*x^2 - 4*x + 1)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(49005*x^3 + 45
0846*x^2 - 1111431*x + 354344)*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1)

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giac [A]  time = 1.65, size = 79, normalized size = 0.99 \begin {gather*} \frac {27}{40} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{166375} \, \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 {2889}{200} \, \sqrt {-2 \, x + 1} - \frac {343 \, {\left (291 \, x - 107\right )}}{1452 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4/(1-2*x)^(5/2)/(3+5*x),x, algorithm="giac")

[Out]

27/40*(-2*x + 1)^(3/2) + 1/166375*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*
x + 1))) - 2889/200*sqrt(-2*x + 1) - 343/1452*(291*x - 107)/((2*x - 1)*sqrt(-2*x + 1))

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maple [A]  time = 0.01, size = 56, normalized size = 0.70 \begin {gather*} -\frac {2 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{166375}+\frac {2401}{264 \left (-2 x +1\right )^{\frac {3}{2}}}+\frac {27 \left (-2 x +1\right )^{\frac {3}{2}}}{40}-\frac {33271}{968 \sqrt {-2 x +1}}-\frac {2889 \sqrt {-2 x +1}}{200} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x+2)^4/(-2*x+1)^(5/2)/(5*x+3),x)

[Out]

2401/264/(-2*x+1)^(3/2)+27/40*(-2*x+1)^(3/2)-2/166375*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)-33271/968
/(-2*x+1)^(1/2)-2889/200*(-2*x+1)^(1/2)

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maxima [A]  time = 1.24, size = 69, normalized size = 0.86 \begin {gather*} \frac {27}{40} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{166375} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {2889}{200} \, \sqrt {-2 \, x + 1} + \frac {343 \, {\left (291 \, x - 107\right )}}{1452 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4/(1-2*x)^(5/2)/(3+5*x),x, algorithm="maxima")

[Out]

27/40*(-2*x + 1)^(3/2) + 1/166375*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) -
2889/200*sqrt(-2*x + 1) + 343/1452*(291*x - 107)/(-2*x + 1)^(3/2)

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mupad [B]  time = 0.07, size = 52, normalized size = 0.65 \begin {gather*} \frac {\frac {33271\,x}{484}-\frac {36701}{1452}}{{\left (1-2\,x\right )}^{3/2}}-\frac {2889\,\sqrt {1-2\,x}}{200}+\frac {27\,{\left (1-2\,x\right )}^{3/2}}{40}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,2{}\mathrm {i}}{166375} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x + 2)^4/((1 - 2*x)^(5/2)*(5*x + 3)),x)

[Out]

((33271*x)/484 - 36701/1452)/(1 - 2*x)^(3/2) + (55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*2i)/166375 - (
2889*(1 - 2*x)^(1/2))/200 + (27*(1 - 2*x)^(3/2))/40

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sympy [A]  time = 81.34, size = 114, normalized size = 1.42 \begin {gather*} \frac {27 \left (1 - 2 x\right )^{\frac {3}{2}}}{40} - \frac {2889 \sqrt {1 - 2 x}}{200} + \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 )}{3025} - \frac {33271}{968 \sqrt {1 - 2 x}} + \frac {2401}{264 \left (1 - 2 x\right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)**4/(1-2*x)**(5/2)/(3+5*x),x)

[Out]

27*(1 - 2*x)**(3/2)/40 - 2889*sqrt(1 - 2*x)/200 + 2*Piecewise((-sqrt(55)*acoth(sqrt(55)*sqrt(1 - 2*x)/11)/55,
2*x - 1 < -11/5), (-sqrt(55)*atanh(sqrt(55)*sqrt(1 - 2*x)/11)/55, 2*x - 1 > -11/5))/3025 - 33271/(968*sqrt(1 -
2*x)) + 2401/(264*(1 - 2*x)**(3/2))

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