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

Optimal. Leaf size=93 \[ -\frac {243}{400} (1-2 x)^{5/2}+\frac {1917}{200} (1-2 x)^{3/2}-\frac {51057}{500} \sqrt {1-2 x}-\frac {156065}{968 \sqrt {1-2 x}}+\frac {16807}{528 (1-2 x)^{3/2}}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15125 \sqrt {55}} \]

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Rubi [A]  time = 0.05, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 8, 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 {243}{400} (1-2 x)^{5/2}+\frac {1917}{200} (1-2 x)^{3/2}-\frac {51057}{500} \sqrt {1-2 x}-\frac {156065}{968 \sqrt {1-2 x}}+\frac {16807}{528 (1-2 x)^{3/2}}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15125 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

16807/(528*(1 - 2*x)^(3/2)) - 156065/(968*Sqrt[1 - 2*x]) - (51057*Sqrt[1 - 2*x])/500 + (1917*(1 - 2*x)^(3/2))/
200 - (243*(1 - 2*x)^(5/2))/400 - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(15125*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)^{5/2} (3+5 x)} \, dx &=\int \left (\frac {16807}{176 (1-2 x)^{5/2}}-\frac {156065}{968 (1-2 x)^{3/2}}+\frac {152793}{2000 \sqrt {1-2 x}}+\frac {1134 x}{25 \sqrt {1-2 x}}+\frac {243 x^2}{20 \sqrt {1-2 x}}+\frac {1}{15125 \sqrt {1-2 x} (3+5 x)}\right ) \, dx\\ &=\frac {16807}{528 (1-2 x)^{3/2}}-\frac {156065}{968 \sqrt {1-2 x}}-\frac {152793 \sqrt {1-2 x}}{2000}+\frac {\int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{15125}+\frac {243}{20} \int \frac {x^2}{\sqrt {1-2 x}} \, dx+\frac {1134}{25} \int \frac {x}{\sqrt {1-2 x}} \, dx\\ &=\frac {16807}{528 (1-2 x)^{3/2}}-\frac {156065}{968 \sqrt {1-2 x}}-\frac {152793 \sqrt {1-2 x}}{2000}-\frac {\operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{15125}+\frac {243}{20} \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 {1134}{25} \int \left (\frac {1}{2 \sqrt {1-2 x}}-\frac {1}{2} \sqrt {1-2 x}\right ) \, dx\\ &=\frac {16807}{528 (1-2 x)^{3/2}}-\frac {156065}{968 \sqrt {1-2 x}}-\frac {51057}{500} \sqrt {1-2 x}+\frac {1917}{200} (1-2 x)^{3/2}-\frac {243}{400} (1-2 x)^{5/2}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15125 \sqrt {55}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 55, normalized size = 0.59 \begin {gather*} \frac {2 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {5}{11} (1-2 x)\right )-33 \left (30375 x^4+178875 x^3+962550 x^2-2119545 x+695404\right )}{103125 (1-2 x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-33*(695404 - 2119545*x + 962550*x^2 + 178875*x^3 + 30375*x^4) + 2*Hypergeometric2F1[-3/2, 1, -1/2, (5*(1 - 2
*x))/11])/(103125*(1 - 2*x)^(3/2))

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IntegrateAlgebraic [A]  time = 0.08, size = 77, normalized size = 0.83 \begin {gather*} \frac {-441045 (1-2 x)^4+6958710 (1-2 x)^3-74134764 (1-2 x)^2-117048750 (1-2 x)[phone]}{726000 (1-2 x)^{3/2}}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15125 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(23109625 - 117048750*(1 - 2*x) - 74134764*(1 - 2*x)^2 + 6958710*(1 - 2*x)^3 - 441045*(1 - 2*x)^4)/(726000*(1
- 2*x)^(3/2)) - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(15125*Sqrt[55])

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fricas [A]  time = 1.35, size = 84, normalized size = 0.90 \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 (441045 \, x^{4} + 2597265 \, x^{3} + 13976226 \, x^{2} - 30775791 \, x + 10097264\right )} \sqrt {-2 \, x + 1}}{2495625 \, {\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)^5/(1-2*x)^(5/2)/(3+5*x),x, algorithm="fricas")

[Out]

1/2495625*(3*sqrt(55)*(4*x^2 - 4*x + 1)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(441045*x^4 +
2597265*x^3 + 13976226*x^2 - 30775791*x + 10097264)*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1)

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giac [A]  time = 1.81, size = 95, normalized size = 1.02 \begin {gather*} -\frac {243}{400} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {1917}{200} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{831875} \, \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 {51057}{500} \, \sqrt {-2 \, x + 1} - \frac {2401 \, {\left (780 \, x - 313\right )}}{5808 \, {\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)^5/(1-2*x)^(5/2)/(3+5*x),x, algorithm="giac")

[Out]

-243/400*(2*x - 1)^2*sqrt(-2*x + 1) + 1917/200*(-2*x + 1)^(3/2) + 1/831875*sqrt(55)*log(1/2*abs(-2*sqrt(55) +
10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 51057/500*sqrt(-2*x + 1) - 2401/5808*(780*x - 313)/((2*x -
1)*sqrt(-2*x + 1))

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maple [A]  time = 0.01, size = 65, normalized size = 0.70 \begin {gather*} -\frac {2 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{831875}+\frac {16807}{528 \left (-2 x +1\right )^{\frac {3}{2}}}+\frac {1917 \left (-2 x +1\right )^{\frac {3}{2}}}{200}-\frac {243 \left (-2 x +1\right )^{\frac {5}{2}}}{400}-\frac {156065}{968 \sqrt {-2 x +1}}-\frac {51057 \sqrt {-2 x +1}}{500} \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),x)

[Out]

16807/528/(-2*x+1)^(3/2)+1917/200*(-2*x+1)^(3/2)-243/400*(-2*x+1)^(5/2)-2/831875*arctanh(1/11*55^(1/2)*(-2*x+1
)^(1/2))*55^(1/2)-156065/968/(-2*x+1)^(1/2)-51057/500*(-2*x+1)^(1/2)

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maxima [A]  time = 1.27, size = 78, normalized size = 0.84 \begin {gather*} -\frac {243}{400} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {1917}{200} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{831875} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {51057}{500} \, \sqrt {-2 \, x + 1} + \frac {2401 \, {\left (780 \, x - 313\right )}}{5808 \, {\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)^5/(1-2*x)^(5/2)/(3+5*x),x, algorithm="maxima")

[Out]

-243/400*(-2*x + 1)^(5/2) + 1917/200*(-2*x + 1)^(3/2) + 1/831875*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(
sqrt(55) + 5*sqrt(-2*x + 1))) - 51057/500*sqrt(-2*x + 1) + 2401/5808*(780*x - 313)/(-2*x + 1)^(3/2)

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mupad [B]  time = 1.21, size = 61, normalized size = 0.66 \begin {gather*} \frac {\frac {156065\,x}{484}-\frac {751513}{5808}}{{\left (1-2\,x\right )}^{3/2}}-\frac {51057\,\sqrt {1-2\,x}}{500}+\frac {1917\,{\left (1-2\,x\right )}^{3/2}}{200}-\frac {243\,{\left (1-2\,x\right )}^{5/2}}{400}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,2{}\mathrm {i}}{831875} \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)),x)

[Out]

((156065*x)/484 - 751513/5808)/(1 - 2*x)^(3/2) + (55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*2i)/831875 -
(51057*(1 - 2*x)^(1/2))/500 + (1917*(1 - 2*x)^(3/2))/200 - (243*(1 - 2*x)^(5/2))/400

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sympy [A]  time = 106.93, size = 126, normalized size = 1.35 \begin {gather*} - \frac {243 \left (1 - 2 x\right )^{\frac {5}{2}}}{400} + \frac {1917 \left (1 - 2 x\right )^{\frac {3}{2}}}{200} - \frac {51057 \sqrt {1 - 2 x}}{500} + \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 )}{15125} - \frac {156065}{968 \sqrt {1 - 2 x}} + \frac {16807}{528 \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)**5/(1-2*x)**(5/2)/(3+5*x),x)

[Out]

-243*(1 - 2*x)**(5/2)/400 + 1917*(1 - 2*x)**(3/2)/200 - 51057*sqrt(1 - 2*x)/500 + 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 > -1
1/5))/15125 - 156065/(968*sqrt(1 - 2*x)) + 16807/(528*(1 - 2*x)**(3/2))

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