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

Optimal. Leaf size=128 \[ -\frac {(1-2 x)^{3/2} (3 x+2)^4}{5 (5 x+3)}+\frac {11}{75} (1-2 x)^{3/2} (3 x+2)^3-\frac {2}{875} (1-2 x)^{3/2} (3 x+2)^2-\frac {(1-2 x)^{3/2} (3663 x+5678)}{9375}+\frac {258 \sqrt {1-2 x}}{15625}-\frac {258 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15625} \]

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Rubi [A]  time = 0.04, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 7, 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)^{3/2} (3 x+2)^4}{5 (5 x+3)}+\frac {11}{75} (1-2 x)^{3/2} (3 x+2)^3-\frac {2}{875} (1-2 x)^{3/2} (3 x+2)^2-\frac {(1-2 x)^{3/2} (3663 x+5678)}{9375}+\frac {258 \sqrt {1-2 x}}{15625}-\frac {258 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15625} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(258*Sqrt[1 - 2*x])/15625 - (2*(1 - 2*x)^(3/2)*(2 + 3*x)^2)/875 + (11*(1 - 2*x)^(3/2)*(2 + 3*x)^3)/75 - ((1 -
2*x)^(3/2)*(2 + 3*x)^4)/(5*(3 + 5*x)) - ((1 - 2*x)^(3/2)*(5678 + 3663*x))/9375 - (258*Sqrt[11/5]*ArcTanh[Sqrt[
5/11]*Sqrt[1 - 2*x]])/15625

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)^{3/2} (2+3 x)^4}{(3+5 x)^2} \, dx &=-\frac {(1-2 x)^{3/2} (2+3 x)^4}{5 (3+5 x)}+\frac {1}{5} \int \frac {(6-33 x) \sqrt {1-2 x} (2+3 x)^3}{3+5 x} \, dx\\ &=\frac {11}{75} (1-2 x)^{3/2} (2+3 x)^3-\frac {(1-2 x)^{3/2} (2+3 x)^4}{5 (3+5 x)}-\frac {1}{225} \int \frac {(-243-18 x) \sqrt {1-2 x} (2+3 x)^2}{3+5 x} \, dx\\ &=-\frac {2}{875} (1-2 x)^{3/2} (2+3 x)^2+\frac {11}{75} (1-2 x)^{3/2} (2+3 x)^3-\frac {(1-2 x)^{3/2} (2+3 x)^4}{5 (3+5 x)}+\frac {\int \frac {\sqrt {1-2 x} (2+3 x) (17010+25641 x)}{3+5 x} \, dx}{7875}\\ &=-\frac {2}{875} (1-2 x)^{3/2} (2+3 x)^2+\frac {11}{75} (1-2 x)^{3/2} (2+3 x)^3-\frac {(1-2 x)^{3/2} (2+3 x)^4}{5 (3+5 x)}-\frac {(1-2 x)^{3/2} (5678+3663 x)}{9375}+\frac {129 \int \frac {\sqrt {1-2 x}}{3+5 x} \, dx}{3125}\\ &=\frac {258 \sqrt {1-2 x}}{15625}-\frac {2}{875} (1-2 x)^{3/2} (2+3 x)^2+\frac {11}{75} (1-2 x)^{3/2} (2+3 x)^3-\frac {(1-2 x)^{3/2} (2+3 x)^4}{5 (3+5 x)}-\frac {(1-2 x)^{3/2} (5678+3663 x)}{9375}+\frac {1419 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{15625}\\ &=\frac {258 \sqrt {1-2 x}}{15625}-\frac {2}{875} (1-2 x)^{3/2} (2+3 x)^2+\frac {11}{75} (1-2 x)^{3/2} (2+3 x)^3-\frac {(1-2 x)^{3/2} (2+3 x)^4}{5 (3+5 x)}-\frac {(1-2 x)^{3/2} (5678+3663 x)}{9375}-\frac {1419 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{15625}\\ &=\frac {258 \sqrt {1-2 x}}{15625}-\frac {2}{875} (1-2 x)^{3/2} (2+3 x)^2+\frac {11}{75} (1-2 x)^{3/2} (2+3 x)^3-\frac {(1-2 x)^{3/2} (2+3 x)^4}{5 (3+5 x)}-\frac {(1-2 x)^{3/2} (5678+3663 x)}{9375}-\frac {258 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15625}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 78, normalized size = 0.61 \begin {gather*} -\frac {5 \sqrt {1-2 x} \left (787500 x^5+1395000 x^4+157275 x^3-924335 x^2-143235 x+161312\right )+1806 \sqrt {55} (5 x+3) \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{546875 (5 x+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/546875*(5*Sqrt[1 - 2*x]*(161312 - 143235*x - 924335*x^2 + 157275*x^3 + 1395000*x^4 + 787500*x^5) + 1806*Sqr
t[55]*(3 + 5*x)*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(3 + 5*x)

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IntegrateAlgebraic [A]  time = 0.12, size = 99, normalized size = 0.77 \begin {gather*} -\frac {\sqrt {1-2 x} \left (196875 (1-2 x)^5-1681875 (1-2 x)^4+4916025 (1-2 x)^3-4776905 (1-2 x)^2-24080 (1-2 x)+79464\right )}{437500 (5 (1-2 x)-11)}-\frac {258 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15625} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/437500*((79464 - 24080*(1 - 2*x) - 4776905*(1 - 2*x)^2 + 4916025*(1 - 2*x)^3 - 1681875*(1 - 2*x)^4 + 196875
*(1 - 2*x)^5)*Sqrt[1 - 2*x])/(-11 + 5*(1 - 2*x)) - (258*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/15625

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fricas [A]  time = 1.43, size = 85, normalized size = 0.66 \begin {gather*} \frac {903 \, \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 (787500 \, x^{5} + 1395000 \, x^{4} + 157275 \, x^{3} - 924335 \, x^{2} - 143235 \, x + 161312\right )} \sqrt {-2 \, x + 1}}{546875 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/546875*(903*sqrt(11)*sqrt(5)*(5*x + 3)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) - 5*(78750
0*x^5 + 1395000*x^4 + 157275*x^3 - 924335*x^2 - 143235*x + 161312)*sqrt(-2*x + 1))/(5*x + 3)

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giac [A]  time = 1.03, size = 122, normalized size = 0.95 \begin {gather*} -\frac {9}{100} \, {\left (2 \, x - 1\right )}^{4} \sqrt {-2 \, x + 1} - \frac {999}{1750} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} - \frac {12393}{12500} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {8}{3125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {129}{78125} \, \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 {52}{3125} \, \sqrt {-2 \, x + 1} - \frac {11 \, \sqrt {-2 \, x + 1}}{15625 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/100*(2*x - 1)^4*sqrt(-2*x + 1) - 999/1750*(2*x - 1)^3*sqrt(-2*x + 1) - 12393/12500*(2*x - 1)^2*sqrt(-2*x +
1) + 8/3125*(-2*x + 1)^(3/2) + 129/78125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*s
qrt(-2*x + 1))) + 52/3125*sqrt(-2*x + 1) - 11/15625*sqrt(-2*x + 1)/(5*x + 3)

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maple [A]  time = 0.01, size = 81, normalized size = 0.63 \begin {gather*} -\frac {258 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{78125}-\frac {9 \left (-2 x +1\right )^{\frac {9}{2}}}{100}+\frac {999 \left (-2 x +1\right )^{\frac {7}{2}}}{1750}-\frac {12393 \left (-2 x +1\right )^{\frac {5}{2}}}{12500}+\frac {8 \left (-2 x +1\right )^{\frac {3}{2}}}{3125}+\frac {52 \sqrt {-2 x +1}}{3125}+\frac {22 \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((-2*x+1)^(3/2)*(3*x+2)^4/(5*x+3)^2,x)

[Out]

-9/100*(-2*x+1)^(9/2)+999/1750*(-2*x+1)^(7/2)-12393/12500*(-2*x+1)^(5/2)+8/3125*(-2*x+1)^(3/2)+52/3125*(-2*x+1
)^(1/2)+22/78125*(-2*x+1)^(1/2)/(-2*x-6/5)-258/78125*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.18, size = 98, normalized size = 0.77 \begin {gather*} -\frac {9}{100} \, {\left (-2 \, x + 1\right )}^{\frac {9}{2}} + \frac {999}{1750} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - \frac {12393}{12500} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {8}{3125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {129}{78125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {52}{3125} \, \sqrt {-2 \, x + 1} - \frac {11 \, \sqrt {-2 \, x + 1}}{15625 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/100*(-2*x + 1)^(9/2) + 999/1750*(-2*x + 1)^(7/2) - 12393/12500*(-2*x + 1)^(5/2) + 8/3125*(-2*x + 1)^(3/2) +
129/78125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 52/3125*sqrt(-2*x + 1)
- 11/15625*sqrt(-2*x + 1)/(5*x + 3)

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mupad [B]  time = 0.06, size = 82, normalized size = 0.64 \begin {gather*} \frac {52\,\sqrt {1-2\,x}}{3125}-\frac {22\,\sqrt {1-2\,x}}{78125\,\left (2\,x+\frac {6}{5}\right )}+\frac {8\,{\left (1-2\,x\right )}^{3/2}}{3125}-\frac {12393\,{\left (1-2\,x\right )}^{5/2}}{12500}+\frac {999\,{\left (1-2\,x\right )}^{7/2}}{1750}-\frac {9\,{\left (1-2\,x\right )}^{9/2}}{100}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,258{}\mathrm {i}}{78125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*258i)/78125 - (22*(1 - 2*x)^(1/2))/(78125*(2*x + 6/5)) + (52*
(1 - 2*x)^(1/2))/3125 + (8*(1 - 2*x)^(3/2))/3125 - (12393*(1 - 2*x)^(5/2))/12500 + (999*(1 - 2*x)^(7/2))/1750
- (9*(1 - 2*x)^(9/2))/100

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

[Out]

Timed out

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