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

Optimal. Leaf size=141 \[ -\frac {(1-2 x)^{5/2} (3 x+2)^4}{5 (5 x+3)}+\frac {39}{275} (1-2 x)^{5/2} (3 x+2)^3-\frac {32 (1-2 x)^{5/2} (3 x+2)^2}{4125}+\frac {254 (1-2 x)^{3/2}}{46875}-\frac {(1-2 x)^{5/2} (1110975 x+1347116)}{3609375}+\frac {2794 \sqrt {1-2 x}}{78125}-\frac {2794 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{78125} \]

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Rubi [A]  time = 0.05, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {97, 153, 147, 50, 63, 206} \begin {gather*} -\frac {(1-2 x)^{5/2} (3 x+2)^4}{5 (5 x+3)}+\frac {39}{275} (1-2 x)^{5/2} (3 x+2)^3-\frac {32 (1-2 x)^{5/2} (3 x+2)^2}{4125}+\frac {254 (1-2 x)^{3/2}}{46875}-\frac {(1-2 x)^{5/2} (1110975 x+1347116)}{3609375}+\frac {2794 \sqrt {1-2 x}}{78125}-\frac {2794 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{78125} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2794*Sqrt[1 - 2*x])/78125 + (254*(1 - 2*x)^(3/2))/46875 - (32*(1 - 2*x)^(5/2)*(2 + 3*x)^2)/4125 + (39*(1 - 2*
x)^(5/2)*(2 + 3*x)^3)/275 - ((1 - 2*x)^(5/2)*(2 + 3*x)^4)/(5*(3 + 5*x)) - ((1 - 2*x)^(5/2)*(1347116 + 1110975*
x))/3609375 - (2794*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/78125

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 {(1-2 x)^{5/2} (2+3 x)^4}{(3+5 x)^2} \, dx &=-\frac {(1-2 x)^{5/2} (2+3 x)^4}{5 (3+5 x)}+\frac {1}{5} \int \frac {(2-39 x) (1-2 x)^{3/2} (2+3 x)^3}{3+5 x} \, dx\\ &=\frac {39}{275} (1-2 x)^{5/2} (2+3 x)^3-\frac {(1-2 x)^{5/2} (2+3 x)^4}{5 (3+5 x)}-\frac {1}{275} \int \frac {(-337-96 x) (1-2 x)^{3/2} (2+3 x)^2}{3+5 x} \, dx\\ &=-\frac {32 (1-2 x)^{5/2} (2+3 x)^2}{4125}+\frac {39}{275} (1-2 x)^{5/2} (2+3 x)^3-\frac {(1-2 x)^{5/2} (2+3 x)^4}{5 (3+5 x)}+\frac {\int \frac {(1-2 x)^{3/2} (2+3 x) (29178+44439 x)}{3+5 x} \, dx}{12375}\\ &=-\frac {32 (1-2 x)^{5/2} (2+3 x)^2}{4125}+\frac {39}{275} (1-2 x)^{5/2} (2+3 x)^3-\frac {(1-2 x)^{5/2} (2+3 x)^4}{5 (3+5 x)}-\frac {(1-2 x)^{5/2} (1347116+1110975 x)}{3609375}+\frac {127 \int \frac {(1-2 x)^{3/2}}{3+5 x} \, dx}{3125}\\ &=\frac {254 (1-2 x)^{3/2}}{46875}-\frac {32 (1-2 x)^{5/2} (2+3 x)^2}{4125}+\frac {39}{275} (1-2 x)^{5/2} (2+3 x)^3-\frac {(1-2 x)^{5/2} (2+3 x)^4}{5 (3+5 x)}-\frac {(1-2 x)^{5/2} (1347116+1110975 x)}{3609375}+\frac {1397 \int \frac {\sqrt {1-2 x}}{3+5 x} \, dx}{15625}\\ &=\frac {2794 \sqrt {1-2 x}}{78125}+\frac {254 (1-2 x)^{3/2}}{46875}-\frac {32 (1-2 x)^{5/2} (2+3 x)^2}{4125}+\frac {39}{275} (1-2 x)^{5/2} (2+3 x)^3-\frac {(1-2 x)^{5/2} (2+3 x)^4}{5 (3+5 x)}-\frac {(1-2 x)^{5/2} (1347116+1110975 x)}{3609375}+\frac {15367 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{78125}\\ &=\frac {2794 \sqrt {1-2 x}}{78125}+\frac {254 (1-2 x)^{3/2}}{46875}-\frac {32 (1-2 x)^{5/2} (2+3 x)^2}{4125}+\frac {39}{275} (1-2 x)^{5/2} (2+3 x)^3-\frac {(1-2 x)^{5/2} (2+3 x)^4}{5 (3+5 x)}-\frac {(1-2 x)^{5/2} (1347116+1110975 x)}{3609375}-\frac {15367 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{78125}\\ &=\frac {2794 \sqrt {1-2 x}}{78125}+\frac {254 (1-2 x)^{3/2}}{46875}-\frac {32 (1-2 x)^{5/2} (2+3 x)^2}{4125}+\frac {39}{275} (1-2 x)^{5/2} (2+3 x)^3-\frac {(1-2 x)^{5/2} (2+3 x)^4}{5 (3+5 x)}-\frac {(1-2 x)^{5/2} (1347116+1110975 x)}{3609375}-\frac {2794 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{78125}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 78, normalized size = 0.55 \begin {gather*} \frac {\frac {5 \sqrt {1-2 x} \left (212625000 x^6+237037500 x^5-173598750 x^4-214071975 x^3+85482115 x^2+50081215 x-15982128\right )}{5 x+3}-645414 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{90234375} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((5*Sqrt[1 - 2*x]*(-15982128 + 50081215*x + 85482115*x^2 - 214071975*x^3 - 173598750*x^4 + 237037500*x^5 + 212
625000*x^6))/(3 + 5*x) - 645414*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/90234375

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IntegrateAlgebraic [A]  time = 0.13, size = 108, normalized size = 0.77 \begin {gather*} -\frac {\sqrt {1-2 x} \left (26578125 (1-2 x)^6-218728125 (1-2 x)^5+608169375 (1-2 x)^4-562886775 (1-2 x)^3-782320 (1-2 x)^2-8605520 (1-2 x)[phone]\right )}{72187500 (5 (1-2 x)-11)}-\frac {2794 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{78125} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/72187500*((28398216 - 8605520*(1 - 2*x) - 782320*(1 - 2*x)^2 - 562886775*(1 - 2*x)^3 + 608169375*(1 - 2*x)^
4 - 218728125*(1 - 2*x)^5 + 26578125*(1 - 2*x)^6)*Sqrt[1 - 2*x])/(-11 + 5*(1 - 2*x)) - (2794*Sqrt[11/5]*ArcTan
h[Sqrt[5/11]*Sqrt[1 - 2*x]])/78125

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fricas [A]  time = 1.65, size = 90, normalized size = 0.64 \begin {gather*} \frac {322707 \, \sqrt {11} \sqrt {5} {\left (5 \, x + 3\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 5 \, {\left (212625000 \, x^{6} + 237037500 \, x^{5} - 173598750 \, x^{4} - 214071975 \, x^{3} + 85482115 \, x^{2} + 50081215 \, x - 15982128\right )} \sqrt {-2 \, x + 1}}{90234375 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/90234375*(322707*sqrt(11)*sqrt(5)*(5*x + 3)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 5*(
212625000*x^6 + 237037500*x^5 - 173598750*x^4 - 214071975*x^3 + 85482115*x^2 + 50081215*x - 15982128)*sqrt(-2*
x + 1))/(5*x + 3)

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giac [A]  time = 0.98, size = 138, normalized size = 0.98 \begin {gather*} \frac {81}{1100} \, {\left (2 \, x - 1\right )}^{5} \sqrt {-2 \, x + 1} + \frac {111}{250} \, {\left (2 \, x - 1\right )}^{4} \sqrt {-2 \, x + 1} + \frac {12393}{17500} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} + \frac {24}{15625} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {52}{9375} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1397}{390625} \, \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 {2816}{78125} \, \sqrt {-2 \, x + 1} - \frac {121 \, \sqrt {-2 \, x + 1}}{78125 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

81/1100*(2*x - 1)^5*sqrt(-2*x + 1) + 111/250*(2*x - 1)^4*sqrt(-2*x + 1) + 12393/17500*(2*x - 1)^3*sqrt(-2*x +
1) + 24/15625*(2*x - 1)^2*sqrt(-2*x + 1) + 52/9375*(-2*x + 1)^(3/2) + 1397/390625*sqrt(55)*log(1/2*abs(-2*sqrt
(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 2816/78125*sqrt(-2*x + 1) - 121/78125*sqrt(-2*x + 1
)/(5*x + 3)

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maple [A]  time = 0.01, size = 90, normalized size = 0.64 \begin {gather*} -\frac {2794 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{390625}-\frac {81 \left (-2 x +1\right )^{\frac {11}{2}}}{1100}+\frac {111 \left (-2 x +1\right )^{\frac {9}{2}}}{250}-\frac {12393 \left (-2 x +1\right )^{\frac {7}{2}}}{17500}+\frac {24 \left (-2 x +1\right )^{\frac {5}{2}}}{15625}+\frac {52 \left (-2 x +1\right )^{\frac {3}{2}}}{9375}+\frac {2816 \sqrt {-2 x +1}}{78125}+\frac {242 \sqrt {-2 x +1}}{390625 \left (-2 x -\frac {6}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-81/1100*(-2*x+1)^(11/2)+111/250*(-2*x+1)^(9/2)-12393/17500*(-2*x+1)^(7/2)+24/15625*(-2*x+1)^(5/2)+52/9375*(-2
*x+1)^(3/2)+2816/78125*(-2*x+1)^(1/2)+242/390625*(-2*x+1)^(1/2)/(-2*x-6/5)-2794/390625*arctanh(1/11*55^(1/2)*(
-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.27, size = 107, normalized size = 0.76 \begin {gather*} -\frac {81}{1100} \, {\left (-2 \, x + 1\right )}^{\frac {11}{2}} + \frac {111}{250} \, {\left (-2 \, x + 1\right )}^{\frac {9}{2}} - \frac {12393}{17500} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} + \frac {24}{15625} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {52}{9375} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1397}{390625} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {2816}{78125} \, \sqrt {-2 \, x + 1} - \frac {121 \, \sqrt {-2 \, x + 1}}{78125 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-81/1100*(-2*x + 1)^(11/2) + 111/250*(-2*x + 1)^(9/2) - 12393/17500*(-2*x + 1)^(7/2) + 24/15625*(-2*x + 1)^(5/
2) + 52/9375*(-2*x + 1)^(3/2) + 1397/390625*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*
x + 1))) + 2816/78125*sqrt(-2*x + 1) - 121/78125*sqrt(-2*x + 1)/(5*x + 3)

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mupad [B]  time = 0.07, size = 91, normalized size = 0.65 \begin {gather*} \frac {2816\,\sqrt {1-2\,x}}{78125}-\frac {242\,\sqrt {1-2\,x}}{390625\,\left (2\,x+\frac {6}{5}\right )}+\frac {52\,{\left (1-2\,x\right )}^{3/2}}{9375}+\frac {24\,{\left (1-2\,x\right )}^{5/2}}{15625}-\frac {12393\,{\left (1-2\,x\right )}^{7/2}}{17500}+\frac {111\,{\left (1-2\,x\right )}^{9/2}}{250}-\frac {81\,{\left (1-2\,x\right )}^{11/2}}{1100}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,2794{}\mathrm {i}}{390625} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*2794i)/390625 - (242*(1 - 2*x)^(1/2))/(390625*(2*x + 6/5)) +
(2816*(1 - 2*x)^(1/2))/78125 + (52*(1 - 2*x)^(3/2))/9375 + (24*(1 - 2*x)^(5/2))/15625 - (12393*(1 - 2*x)^(7/2)
)/17500 + (111*(1 - 2*x)^(9/2))/250 - (81*(1 - 2*x)^(11/2))/1100

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

[Out]

Timed out

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