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

Optimal. Leaf size=133 \[ -\frac {\sqrt {1-2 x} (3 x+2)^5}{5 (5 x+3)}+\frac {11}{75} \sqrt {1-2 x} (3 x+2)^4+\frac {64 \sqrt {1-2 x} (3 x+2)^3}{2625}-\frac {172 \sqrt {1-2 x} (3 x+2)^2}{3125}-\frac {4 \sqrt {1-2 x} (3625 x+10998)}{15625}-\frac {328 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15625 \sqrt {55}} \]

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Rubi [A]  time = 0.05, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {97, 153, 147, 63, 206} \begin {gather*} -\frac {\sqrt {1-2 x} (3 x+2)^5}{5 (5 x+3)}+\frac {11}{75} \sqrt {1-2 x} (3 x+2)^4+\frac {64 \sqrt {1-2 x} (3 x+2)^3}{2625}-\frac {172 \sqrt {1-2 x} (3 x+2)^2}{3125}-\frac {4 \sqrt {1-2 x} (3625 x+10998)}{15625}-\frac {328 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15625 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-172*Sqrt[1 - 2*x]*(2 + 3*x)^2)/3125 + (64*Sqrt[1 - 2*x]*(2 + 3*x)^3)/2625 + (11*Sqrt[1 - 2*x]*(2 + 3*x)^4)/7
5 - (Sqrt[1 - 2*x]*(2 + 3*x)^5)/(5*(3 + 5*x)) - (4*Sqrt[1 - 2*x]*(10998 + 3625*x))/15625 - (328*ArcTanh[Sqrt[5
/11]*Sqrt[1 - 2*x]])/(15625*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 97

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p)/(b*(m + 1)), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 153

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]

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 {\sqrt {1-2 x} (2+3 x)^5}{(3+5 x)^2} \, dx &=-\frac {\sqrt {1-2 x} (2+3 x)^5}{5 (3+5 x)}+\frac {1}{5} \int \frac {(13-33 x) (2+3 x)^4}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {11}{75} \sqrt {1-2 x} (2+3 x)^4-\frac {\sqrt {1-2 x} (2+3 x)^5}{5 (3+5 x)}-\frac {1}{225} \int \frac {(2+3 x)^3 (-180+192 x)}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {64 \sqrt {1-2 x} (2+3 x)^3}{2625}+\frac {11}{75} \sqrt {1-2 x} (2+3 x)^4-\frac {\sqrt {1-2 x} (2+3 x)^5}{5 (3+5 x)}+\frac {\int \frac {(2+3 x)^2 (8568+10836 x)}{\sqrt {1-2 x} (3+5 x)} \, dx}{7875}\\ &=-\frac {172 \sqrt {1-2 x} (2+3 x)^2}{3125}+\frac {64 \sqrt {1-2 x} (2+3 x)^3}{2625}+\frac {11}{75} \sqrt {1-2 x} (2+3 x)^4-\frac {\sqrt {1-2 x} (2+3 x)^5}{5 (3+5 x)}-\frac {\int \frac {(-558432-913500 x) (2+3 x)}{\sqrt {1-2 x} (3+5 x)} \, dx}{196875}\\ &=-\frac {172 \sqrt {1-2 x} (2+3 x)^2}{3125}+\frac {64 \sqrt {1-2 x} (2+3 x)^3}{2625}+\frac {11}{75} \sqrt {1-2 x} (2+3 x)^4-\frac {\sqrt {1-2 x} (2+3 x)^5}{5 (3+5 x)}-\frac {4 \sqrt {1-2 x} (10998+3625 x)}{15625}+\frac {164 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{15625}\\ &=-\frac {172 \sqrt {1-2 x} (2+3 x)^2}{3125}+\frac {64 \sqrt {1-2 x} (2+3 x)^3}{2625}+\frac {11}{75} \sqrt {1-2 x} (2+3 x)^4-\frac {\sqrt {1-2 x} (2+3 x)^5}{5 (3+5 x)}-\frac {4 \sqrt {1-2 x} (10998+3625 x)}{15625}-\frac {164 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{15625}\\ &=-\frac {172 \sqrt {1-2 x} (2+3 x)^2}{3125}+\frac {64 \sqrt {1-2 x} (2+3 x)^3}{2625}+\frac {11}{75} \sqrt {1-2 x} (2+3 x)^4-\frac {\sqrt {1-2 x} (2+3 x)^5}{5 (3+5 x)}-\frac {4 \sqrt {1-2 x} (10998+3625 x)}{15625}-\frac {328 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15625 \sqrt {55}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 73, normalized size = 0.55 \begin {gather*} \frac {\frac {55 \sqrt {1-2 x} \left (1181250 x^5+3864375 x^4+4760100 x^3+2225760 x^2-1133340 x-862072\right )}{5 x+3}-2296 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{6015625} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((55*Sqrt[1 - 2*x]*(-862072 - 1133340*x + 2225760*x^2 + 4760100*x^3 + 3864375*x^4 + 1181250*x^5))/(3 + 5*x) -
2296*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/6015625

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IntegrateAlgebraic [A]  time = 0.12, size = 97, normalized size = 0.73 \begin {gather*} \frac {\left (590625 (1-2 x)^5-6817500 (1-2 x)^4+30883950 (1-2 x)^3-66556140 (1-2 x)^2+55710585 (1-2 x)-18368\right ) \sqrt {1-2 x}}{875000 (5 (1-2 x)-11)}-\frac {328 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15625 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-18368 + 55710585*(1 - 2*x) - 66556140*(1 - 2*x)^2 + 30883950*(1 - 2*x)^3 - 6817500*(1 - 2*x)^4 + 590625*(1
- 2*x)^5)*Sqrt[1 - 2*x])/(875000*(-11 + 5*(1 - 2*x))) - (328*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(15625*Sqrt[55
])

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fricas [A]  time = 1.75, size = 79, normalized size = 0.59 \begin {gather*} \frac {1148 \, \sqrt {55} {\left (5 \, x + 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (1181250 \, x^{5} + 3864375 \, x^{4} + 4760100 \, x^{3} + 2225760 \, x^{2} - 1133340 \, x - 862072\right )} \sqrt {-2 \, x + 1}}{6015625 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6015625*(1148*sqrt(55)*(5*x + 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(1181250*x^5 + 3864
375*x^4 + 4760100*x^3 + 2225760*x^2 - 1133340*x - 862072)*sqrt(-2*x + 1))/(5*x + 3)

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giac [A]  time = 1.24, size = 122, normalized size = 0.92 \begin {gather*} \frac {27}{200} \, {\left (2 \, x - 1\right )}^{4} \sqrt {-2 \, x + 1} + \frac {8829}{7000} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} + \frac {107109}{25000} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - \frac {144681}{25000} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {164}{859375} \, \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 {6}{3125} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{15625 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/200*(2*x - 1)^4*sqrt(-2*x + 1) + 8829/7000*(2*x - 1)^3*sqrt(-2*x + 1) + 107109/25000*(2*x - 1)^2*sqrt(-2*x
+ 1) - 144681/25000*(-2*x + 1)^(3/2) + 164/859375*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(
55) + 5*sqrt(-2*x + 1))) + 6/3125*sqrt(-2*x + 1) - 1/15625*sqrt(-2*x + 1)/(5*x + 3)

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maple [A]  time = 0.01, size = 81, normalized size = 0.61 \begin {gather*} -\frac {328 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{859375}+\frac {27 \left (-2 x +1\right )^{\frac {9}{2}}}{200}-\frac {8829 \left (-2 x +1\right )^{\frac {7}{2}}}{7000}+\frac {107109 \left (-2 x +1\right )^{\frac {5}{2}}}{25000}-\frac {144681 \left (-2 x +1\right )^{\frac {3}{2}}}{25000}+\frac {6 \sqrt {-2 x +1}}{3125}+\frac {2 \sqrt {-2 x +1}}{78125 \left (-2 x -\frac {6}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/200*(-2*x+1)^(9/2)-8829/7000*(-2*x+1)^(7/2)+107109/25000*(-2*x+1)^(5/2)-144681/25000*(-2*x+1)^(3/2)+6/3125*
(-2*x+1)^(1/2)+2/78125*(-2*x+1)^(1/2)/(-6/5-2*x)-328/859375*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.10, size = 98, normalized size = 0.74 \begin {gather*} \frac {27}{200} \, {\left (-2 \, x + 1\right )}^{\frac {9}{2}} - \frac {8829}{7000} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} + \frac {107109}{25000} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - \frac {144681}{25000} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {164}{859375} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {6}{3125} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{15625 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/200*(-2*x + 1)^(9/2) - 8829/7000*(-2*x + 1)^(7/2) + 107109/25000*(-2*x + 1)^(5/2) - 144681/25000*(-2*x + 1)
^(3/2) + 164/859375*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 6/3125*sqrt(-
2*x + 1) - 1/15625*sqrt(-2*x + 1)/(5*x + 3)

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mupad [B]  time = 0.06, size = 82, normalized size = 0.62 \begin {gather*} \frac {6\,\sqrt {1-2\,x}}{3125}-\frac {2\,\sqrt {1-2\,x}}{78125\,\left (2\,x+\frac {6}{5}\right )}-\frac {144681\,{\left (1-2\,x\right )}^{3/2}}{25000}+\frac {107109\,{\left (1-2\,x\right )}^{5/2}}{25000}-\frac {8829\,{\left (1-2\,x\right )}^{7/2}}{7000}+\frac {27\,{\left (1-2\,x\right )}^{9/2}}{200}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,328{}\mathrm {i}}{859375} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*328i)/859375 - (2*(1 - 2*x)^(1/2))/(78125*(2*x + 6/5)) + (6*(
1 - 2*x)^(1/2))/3125 - (144681*(1 - 2*x)^(3/2))/25000 + (107109*(1 - 2*x)^(5/2))/25000 - (8829*(1 - 2*x)^(7/2)
)/7000 + (27*(1 - 2*x)^(9/2))/200

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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