3.2037 \(\int \frac {(2+3 x)^4}{\sqrt {1-2 x} (3+5 x)} \, dx\)

Optimal. Leaf size=80 \[ \frac {81}{280} (1-2 x)^{7/2}-\frac {2889 (1-2 x)^{5/2}}{1000}+\frac {11457 (1-2 x)^{3/2}}{1000}-\frac {136419 \sqrt {1-2 x}}{5000}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{625 \sqrt {55}} \]

[Out]

11457/1000*(1-2*x)^(3/2)-2889/1000*(1-2*x)^(5/2)+81/280*(1-2*x)^(7/2)-2/34375*arctanh(1/11*55^(1/2)*(1-2*x)^(1
/2))*55^(1/2)-136419/5000*(1-2*x)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {88, 63, 206} \[ \frac {81}{280} (1-2 x)^{7/2}-\frac {2889 (1-2 x)^{5/2}}{1000}+\frac {11457 (1-2 x)^{3/2}}{1000}-\frac {136419 \sqrt {1-2 x}}{5000}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{625 \sqrt {55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-136419*Sqrt[1 - 2*x])/5000 + (11457*(1 - 2*x)^(3/2))/1000 - (2889*(1 - 2*x)^(5/2))/1000 + (81*(1 - 2*x)^(7/2
))/280 - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(625*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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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}{\sqrt {1-2 x} (3+5 x)} \, dx &=\int \left (\frac {136419}{5000 \sqrt {1-2 x}}-\frac {34371 \sqrt {1-2 x}}{1000}+\frac {2889}{200} (1-2 x)^{3/2}-\frac {81}{40} (1-2 x)^{5/2}+\frac {1}{625 \sqrt {1-2 x} (3+5 x)}\right ) \, dx\\ &=-\frac {136419 \sqrt {1-2 x}}{5000}+\frac {11457 (1-2 x)^{3/2}}{1000}-\frac {2889 (1-2 x)^{5/2}}{1000}+\frac {81}{280} (1-2 x)^{7/2}+\frac {1}{625} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {136419 \sqrt {1-2 x}}{5000}+\frac {11457 (1-2 x)^{3/2}}{1000}-\frac {2889 (1-2 x)^{5/2}}{1000}+\frac {81}{280} (1-2 x)^{7/2}-\frac {1}{625} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {136419 \sqrt {1-2 x}}{5000}+\frac {11457 (1-2 x)^{3/2}}{1000}-\frac {2889 (1-2 x)^{5/2}}{1000}+\frac {81}{280} (1-2 x)^{7/2}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{625 \sqrt {55}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 56, normalized size = 0.70 \[ -\frac {3 \sqrt {1-2 x} \left (3375 x^3+11790 x^2+19095 x+26872\right )}{4375}-\frac {2 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{625 \sqrt {55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[1 - 2*x]*(26872 + 19095*x + 11790*x^2 + 3375*x^3))/4375 - (2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(625*
Sqrt[55])

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fricas [A]  time = 0.80, size = 55, normalized size = 0.69 \[ -\frac {3}{4375} \, {\left (3375 \, x^{3} + 11790 \, x^{2} + 19095 \, x + 26872\right )} \sqrt {-2 \, x + 1} + \frac {1}{34375} \, \sqrt {55} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4375*(3375*x^3 + 11790*x^2 + 19095*x + 26872)*sqrt(-2*x + 1) + 1/34375*sqrt(55)*log((5*x + sqrt(55)*sqrt(-2
*x + 1) - 8)/(5*x + 3))

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giac [A]  time = 1.22, size = 90, normalized size = 1.12 \[ -\frac {81}{280} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} - \frac {2889}{1000} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {11457}{1000} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{34375} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {136419}{5000} \, \sqrt {-2 \, x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-81/280*(2*x - 1)^3*sqrt(-2*x + 1) - 2889/1000*(2*x - 1)^2*sqrt(-2*x + 1) + 11457/1000*(-2*x + 1)^(3/2) + 1/34
375*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 136419/5000*sqrt(-2
*x + 1)

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maple [A]  time = 0.01, size = 56, normalized size = 0.70 \[ -\frac {2 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{34375}+\frac {11457 \left (-2 x +1\right )^{\frac {3}{2}}}{1000}-\frac {2889 \left (-2 x +1\right )^{\frac {5}{2}}}{1000}+\frac {81 \left (-2 x +1\right )^{\frac {7}{2}}}{280}-\frac {136419 \sqrt {-2 x +1}}{5000} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11457/1000*(-2*x+1)^(3/2)-2889/1000*(-2*x+1)^(5/2)+81/280*(-2*x+1)^(7/2)-2/34375*arctanh(1/11*55^(1/2)*(-2*x+1
)^(1/2))*55^(1/2)-136419/5000*(-2*x+1)^(1/2)

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maxima [A]  time = 1.33, size = 73, normalized size = 0.91 \[ \frac {81}{280} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - \frac {2889}{1000} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {11457}{1000} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{34375} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {136419}{5000} \, \sqrt {-2 \, x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

81/280*(-2*x + 1)^(7/2) - 2889/1000*(-2*x + 1)^(5/2) + 11457/1000*(-2*x + 1)^(3/2) + 1/34375*sqrt(55)*log(-(sq
rt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 136419/5000*sqrt(-2*x + 1)

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mupad [B]  time = 1.19, size = 57, normalized size = 0.71 \[ \frac {11457\,{\left (1-2\,x\right )}^{3/2}}{1000}-\frac {136419\,\sqrt {1-2\,x}}{5000}-\frac {2889\,{\left (1-2\,x\right )}^{5/2}}{1000}+\frac {81\,{\left (1-2\,x\right )}^{7/2}}{280}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,2{}\mathrm {i}}{34375} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*2i)/34375 - (136419*(1 - 2*x)^(1/2))/5000 + (11457*(1 - 2*x)^
(3/2))/1000 - (2889*(1 - 2*x)^(5/2))/1000 + (81*(1 - 2*x)^(7/2))/280

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sympy [A]  time = 55.93, size = 114, normalized size = 1.42 \[ \frac {81 \left (1 - 2 x\right )^{\frac {7}{2}}}{280} - \frac {2889 \left (1 - 2 x\right )^{\frac {5}{2}}}{1000} + \frac {11457 \left (1 - 2 x\right )^{\frac {3}{2}}}{1000} - \frac {136419 \sqrt {1 - 2 x}}{5000} + \frac {2 \left (\begin {cases} - \frac {\sqrt {55} \operatorname {acoth}{\left (\frac {\sqrt {55}}{5 \sqrt {1 - 2 x}} \right )}}{55} & \text {for}\: \frac {1}{1 - 2 x} > \frac {5}{11} \\- \frac {\sqrt {55} \operatorname {atanh}{\left (\frac {\sqrt {55}}{5 \sqrt {1 - 2 x}} \right )}}{55} & \text {for}\: \frac {1}{1 - 2 x} < \frac {5}{11} \end {cases}\right )}{625} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

81*(1 - 2*x)**(7/2)/280 - 2889*(1 - 2*x)**(5/2)/1000 + 11457*(1 - 2*x)**(3/2)/1000 - 136419*sqrt(1 - 2*x)/5000
+ 2*Piecewise((-sqrt(55)*acoth(sqrt(55)/(5*sqrt(1 - 2*x)))/55, 1/(1 - 2*x) > 5/11), (-sqrt(55)*atanh(sqrt(55)
/(5*sqrt(1 - 2*x)))/55, 1/(1 - 2*x) < 5/11))/625

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