3.1878 \(\int \frac {(1-2 x)^{3/2} (3+5 x)^2}{(2+3 x)^4} \, dx\)

Optimal. Leaf size=101 \[ \frac {209 (1-2 x)^{5/2}}{2646 (3 x+2)^2}-\frac {(1-2 x)^{5/2}}{189 (3 x+2)^3}-\frac {7559 (1-2 x)^{3/2}}{7938 (3 x+2)}-\frac {7559 \sqrt {1-2 x}}{3969}+\frac {7559 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{567 \sqrt {21}} \]

[Out]

-1/189*(1-2*x)^(5/2)/(2+3*x)^3+209/2646*(1-2*x)^(5/2)/(2+3*x)^2-7559/7938*(1-2*x)^(3/2)/(2+3*x)+7559/11907*arc
tanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-7559/3969*(1-2*x)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {89, 78, 47, 50, 63, 206} \[ \frac {209 (1-2 x)^{5/2}}{2646 (3 x+2)^2}-\frac {(1-2 x)^{5/2}}{189 (3 x+2)^3}-\frac {7559 (1-2 x)^{3/2}}{7938 (3 x+2)}-\frac {7559 \sqrt {1-2 x}}{3969}+\frac {7559 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{567 \sqrt {21}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7559*Sqrt[1 - 2*x])/3969 - (1 - 2*x)^(5/2)/(189*(2 + 3*x)^3) + (209*(1 - 2*x)^(5/2))/(2646*(2 + 3*x)^2) - (7
559*(1 - 2*x)^(3/2))/(7938*(2 + 3*x)) + (7559*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(567*Sqrt[21])

Rule 47

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

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 78

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

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 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 {(1-2 x)^{3/2} (3+5 x)^2}{(2+3 x)^4} \, dx &=-\frac {(1-2 x)^{5/2}}{189 (2+3 x)^3}+\frac {1}{189} \int \frac {(1-2 x)^{3/2} (841+1575 x)}{(2+3 x)^3} \, dx\\ &=-\frac {(1-2 x)^{5/2}}{189 (2+3 x)^3}+\frac {209 (1-2 x)^{5/2}}{2646 (2+3 x)^2}+\frac {7559 \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2} \, dx}{2646}\\ &=-\frac {(1-2 x)^{5/2}}{189 (2+3 x)^3}+\frac {209 (1-2 x)^{5/2}}{2646 (2+3 x)^2}-\frac {7559 (1-2 x)^{3/2}}{7938 (2+3 x)}-\frac {7559 \int \frac {\sqrt {1-2 x}}{2+3 x} \, dx}{2646}\\ &=-\frac {7559 \sqrt {1-2 x}}{3969}-\frac {(1-2 x)^{5/2}}{189 (2+3 x)^3}+\frac {209 (1-2 x)^{5/2}}{2646 (2+3 x)^2}-\frac {7559 (1-2 x)^{3/2}}{7938 (2+3 x)}-\frac {7559 \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx}{1134}\\ &=-\frac {7559 \sqrt {1-2 x}}{3969}-\frac {(1-2 x)^{5/2}}{189 (2+3 x)^3}+\frac {209 (1-2 x)^{5/2}}{2646 (2+3 x)^2}-\frac {7559 (1-2 x)^{3/2}}{7938 (2+3 x)}+\frac {7559 \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{1134}\\ &=-\frac {7559 \sqrt {1-2 x}}{3969}-\frac {(1-2 x)^{5/2}}{189 (2+3 x)^3}+\frac {209 (1-2 x)^{5/2}}{2646 (2+3 x)^2}-\frac {7559 (1-2 x)^{3/2}}{7938 (2+3 x)}+\frac {7559 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{567 \sqrt {21}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 54, normalized size = 0.53 \[ \frac {(1-2 x)^{5/2} \left (245 (627 x+404)-30236 (3 x+2)^3 \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {3}{7}-\frac {6 x}{7}\right )\right )}{648270 (3 x+2)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - 2*x)^(5/2)*(245*(404 + 627*x) - 30236*(2 + 3*x)^3*Hypergeometric2F1[2, 5/2, 7/2, 3/7 - (6*x)/7]))/(64827
0*(2 + 3*x)^3)

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fricas [A]  time = 1.02, size = 90, normalized size = 0.89 \[ \frac {7559 \, \sqrt {21} {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (\frac {3 \, x - \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) - 21 \, {\left (37800 \, x^{3} + 100809 \, x^{2} + 82493 \, x + 21424\right )} \sqrt {-2 \, x + 1}}{23814 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/23814*(7559*sqrt(21)*(27*x^3 + 54*x^2 + 36*x + 8)*log((3*x - sqrt(21)*sqrt(-2*x + 1) - 5)/(3*x + 2)) - 21*(3
7800*x^3 + 100809*x^2 + 82493*x + 21424)*sqrt(-2*x + 1))/(27*x^3 + 54*x^2 + 36*x + 8)

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giac [A]  time = 0.99, size = 93, normalized size = 0.92 \[ -\frac {7559}{23814} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {100}{81} \, \sqrt {-2 \, x + 1} - \frac {25209 \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - 114604 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 130291 \, \sqrt {-2 \, x + 1}}{4536 \, {\left (3 \, x + 2\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7559/23814*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 100/81*sqrt(
-2*x + 1) - 1/4536*(25209*(2*x - 1)^2*sqrt(-2*x + 1) - 114604*(-2*x + 1)^(3/2) + 130291*sqrt(-2*x + 1))/(3*x +
2)^3

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maple [A]  time = 0.01, size = 66, normalized size = 0.65 \[ \frac {7559 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{11907}-\frac {100 \sqrt {-2 x +1}}{81}-\frac {4 \left (-\frac {2801 \left (-2 x +1\right )^{\frac {5}{2}}}{84}+\frac {4093 \left (-2 x +1\right )^{\frac {3}{2}}}{27}-\frac {18613 \sqrt {-2 x +1}}{108}\right )}{3 \left (-6 x -4\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-100/81*(-2*x+1)^(1/2)-4/3*(-2801/84*(-2*x+1)^(5/2)+4093/27*(-2*x+1)^(3/2)-18613/108*(-2*x+1)^(1/2))/(-6*x-4)^
3+7559/11907*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)

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maxima [A]  time = 1.16, size = 101, normalized size = 1.00 \[ -\frac {7559}{23814} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {100}{81} \, \sqrt {-2 \, x + 1} - \frac {25209 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 114604 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 130291 \, \sqrt {-2 \, x + 1}}{567 \, {\left (27 \, {\left (2 \, x - 1\right )}^{3} + 189 \, {\left (2 \, x - 1\right )}^{2} + 882 \, x - 98\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7559/23814*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 100/81*sqrt(-2*x + 1)
- 1/567*(25209*(-2*x + 1)^(5/2) - 114604*(-2*x + 1)^(3/2) + 130291*sqrt(-2*x + 1))/(27*(2*x - 1)^3 + 189*(2*x
- 1)^2 + 882*x - 98)

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mupad [B]  time = 1.16, size = 81, normalized size = 0.80 \[ \frac {7559\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{11907}-\frac {100\,\sqrt {1-2\,x}}{81}-\frac {\frac {18613\,\sqrt {1-2\,x}}{2187}-\frac {16372\,{\left (1-2\,x\right )}^{3/2}}{2187}+\frac {2801\,{\left (1-2\,x\right )}^{5/2}}{1701}}{\frac {98\,x}{3}+7\,{\left (2\,x-1\right )}^2+{\left (2\,x-1\right )}^3-\frac {98}{27}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(7559*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/11907 - (100*(1 - 2*x)^(1/2))/81 - ((18613*(1 - 2*x)^(1/2)
)/2187 - (16372*(1 - 2*x)^(3/2))/2187 + (2801*(1 - 2*x)^(5/2))/1701)/((98*x)/3 + 7*(2*x - 1)^2 + (2*x - 1)^3 -
98/27)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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