3.22.71 \(\int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)} \, dx\) [2171]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.03, 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 = {89, 45, 65, 212} \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 45

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 65

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 89

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 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 {\text {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*}

________________________________________________________________________________________

Mathematica [A]
time = 0.08, size = 56, normalized size = 0.70 \begin {gather*} \frac {-\frac {55 \left (354344-1111431 x+450846 x^2+49005 x^3\right )}{(1-2 x)^{3/2}}-6 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{499125} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-55*(354344 - 1111431*x + 450846*x^2 + 49005*x^3))/(1 - 2*x)^(3/2) - 6*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 -
2*x]])/499125

________________________________________________________________________________________

Maple [A]
time = 0.14, size = 56, normalized size = 0.70

method result size
derivativedivides \(\frac {2401}{264 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {27 \left (1-2 x \right )^{\frac {3}{2}}}{40}-\frac {2 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{166375}-\frac {33271}{968 \sqrt {1-2 x}}-\frac {2889 \sqrt {1-2 x}}{200}\) \(56\)
default \(\frac {2401}{264 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {27 \left (1-2 x \right )^{\frac {3}{2}}}{40}-\frac {2 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{166375}-\frac {33271}{968 \sqrt {1-2 x}}-\frac {2889 \sqrt {1-2 x}}{200}\) \(56\)
trager \(-\frac {\left (49005 x^{3}+450846 x^{2}-1111431 x +354344\right ) \sqrt {1-2 x}}{9075 \left (-1+2 x \right )^{2}}-\frac {\RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (-\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x -8 \RootOf \left (\textit {\_Z}^{2}-55\right )-55 \sqrt {1-2 x}}{3+5 x}\right )}{166375}\) \(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.54, 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)

________________________________________________________________________________________

Fricas [A]
time = 1.34, 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)

________________________________________________________________________________________

Sympy [A]
time = 41.63, size = 107, normalized size = 1.34 \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}\: x < - \frac {3}{5} \\- \frac {\sqrt {55} \operatorname {atanh}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: x > - \frac {3}{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,
x < -3/5), (-sqrt(55)*atanh(sqrt(55)*sqrt(1 - 2*x)/11)/55, x > -3/5))/3025 - 33271/(968*sqrt(1 - 2*x)) + 2401/
(264*(1 - 2*x)**(3/2))

________________________________________________________________________________________

Giac [A]
time = 1.25, 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))

________________________________________________________________________________________

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

________________________________________________________________________________________