3.1887 \(\int \frac {(1-2 x)^{3/2} (3+5 x)^3}{2+3 x} \, dx\)

Optimal. Leaf size=95 \[ -\frac {125}{108} (1-2 x)^{9/2}+\frac {400}{63} (1-2 x)^{7/2}-\frac {1027}{108} (1-2 x)^{5/2}-\frac {2}{243} (1-2 x)^{3/2}-\frac {14}{243} \sqrt {1-2 x}+\frac {14}{243} \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ) \]

[Out]

-2/243*(1-2*x)^(3/2)-1027/108*(1-2*x)^(5/2)+400/63*(1-2*x)^(7/2)-125/108*(1-2*x)^(9/2)+14/729*arctanh(1/7*21^(
1/2)*(1-2*x)^(1/2))*21^(1/2)-14/243*(1-2*x)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {88, 50, 63, 206} \[ -\frac {125}{108} (1-2 x)^{9/2}+\frac {400}{63} (1-2 x)^{7/2}-\frac {1027}{108} (1-2 x)^{5/2}-\frac {2}{243} (1-2 x)^{3/2}-\frac {14}{243} \sqrt {1-2 x}+\frac {14}{243} \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-14*Sqrt[1 - 2*x])/243 - (2*(1 - 2*x)^(3/2))/243 - (1027*(1 - 2*x)^(5/2))/108 + (400*(1 - 2*x)^(7/2))/63 - (1
25*(1 - 2*x)^(9/2))/108 + (14*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/243

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 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 {(1-2 x)^{3/2} (3+5 x)^3}{2+3 x} \, dx &=\int \left (\frac {5135}{108} (1-2 x)^{3/2}-\frac {400}{9} (1-2 x)^{5/2}+\frac {125}{12} (1-2 x)^{7/2}-\frac {(1-2 x)^{3/2}}{27 (2+3 x)}\right ) \, dx\\ &=-\frac {1027}{108} (1-2 x)^{5/2}+\frac {400}{63} (1-2 x)^{7/2}-\frac {125}{108} (1-2 x)^{9/2}-\frac {1}{27} \int \frac {(1-2 x)^{3/2}}{2+3 x} \, dx\\ &=-\frac {2}{243} (1-2 x)^{3/2}-\frac {1027}{108} (1-2 x)^{5/2}+\frac {400}{63} (1-2 x)^{7/2}-\frac {125}{108} (1-2 x)^{9/2}-\frac {7}{81} \int \frac {\sqrt {1-2 x}}{2+3 x} \, dx\\ &=-\frac {14}{243} \sqrt {1-2 x}-\frac {2}{243} (1-2 x)^{3/2}-\frac {1027}{108} (1-2 x)^{5/2}+\frac {400}{63} (1-2 x)^{7/2}-\frac {125}{108} (1-2 x)^{9/2}-\frac {49}{243} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=-\frac {14}{243} \sqrt {1-2 x}-\frac {2}{243} (1-2 x)^{3/2}-\frac {1027}{108} (1-2 x)^{5/2}+\frac {400}{63} (1-2 x)^{7/2}-\frac {125}{108} (1-2 x)^{9/2}+\frac {49}{243} \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {14}{243} \sqrt {1-2 x}-\frac {2}{243} (1-2 x)^{3/2}-\frac {1027}{108} (1-2 x)^{5/2}+\frac {400}{63} (1-2 x)^{7/2}-\frac {125}{108} (1-2 x)^{9/2}+\frac {14}{243} \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 61, normalized size = 0.64 \[ \frac {98 \sqrt {21} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-3 \sqrt {1-2 x} \left (31500 x^4+23400 x^3-17649 x^2-15679 x+7456\right )}{5103} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[1 - 2*x]*(7456 - 15679*x - 17649*x^2 + 23400*x^3 + 31500*x^4) + 98*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1
- 2*x]])/5103

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fricas [A]  time = 0.78, size = 67, normalized size = 0.71 \[ \frac {7}{729} \, \sqrt {7} \sqrt {3} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) - \frac {1}{1701} \, {\left (31500 \, x^{4} + 23400 \, x^{3} - 17649 \, x^{2} - 15679 \, x + 7456\right )} \sqrt {-2 \, x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/729*sqrt(7)*sqrt(3)*log(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x + 5)/(3*x + 2)) - 1/1701*(31500*x^4 + 23400*x
^3 - 17649*x^2 - 15679*x + 7456)*sqrt(-2*x + 1)

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giac [A]  time = 1.08, size = 106, normalized size = 1.12 \[ -\frac {125}{108} \, {\left (2 \, x - 1\right )}^{4} \sqrt {-2 \, x + 1} - \frac {400}{63} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} - \frac {1027}{108} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - \frac {2}{243} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {7}{729} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {14}{243} \, \sqrt {-2 \, x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125/108*(2*x - 1)^4*sqrt(-2*x + 1) - 400/63*(2*x - 1)^3*sqrt(-2*x + 1) - 1027/108*(2*x - 1)^2*sqrt(-2*x + 1)
- 2/243*(-2*x + 1)^(3/2) - 7/729*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x
+ 1))) - 14/243*sqrt(-2*x + 1)

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maple [A]  time = 0.01, size = 65, normalized size = 0.68 \[ \frac {14 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{729}-\frac {2 \left (-2 x +1\right )^{\frac {3}{2}}}{243}-\frac {1027 \left (-2 x +1\right )^{\frac {5}{2}}}{108}+\frac {400 \left (-2 x +1\right )^{\frac {7}{2}}}{63}-\frac {125 \left (-2 x +1\right )^{\frac {9}{2}}}{108}-\frac {14 \sqrt {-2 x +1}}{243} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/243*(-2*x+1)^(3/2)-1027/108*(-2*x+1)^(5/2)+400/63*(-2*x+1)^(7/2)-125/108*(-2*x+1)^(9/2)+14/729*arctanh(1/7*
21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)-14/243*(-2*x+1)^(1/2)

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maxima [A]  time = 1.21, size = 82, normalized size = 0.86 \[ -\frac {125}{108} \, {\left (-2 \, x + 1\right )}^{\frac {9}{2}} + \frac {400}{63} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - \frac {1027}{108} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - \frac {2}{243} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {7}{729} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {14}{243} \, \sqrt {-2 \, x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125/108*(-2*x + 1)^(9/2) + 400/63*(-2*x + 1)^(7/2) - 1027/108*(-2*x + 1)^(5/2) - 2/243*(-2*x + 1)^(3/2) - 7/7
29*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 14/243*sqrt(-2*x + 1)

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mupad [B]  time = 0.06, size = 66, normalized size = 0.69 \[ \frac {400\,{\left (1-2\,x\right )}^{7/2}}{63}-\frac {2\,{\left (1-2\,x\right )}^{3/2}}{243}-\frac {1027\,{\left (1-2\,x\right )}^{5/2}}{108}-\frac {14\,\sqrt {1-2\,x}}{243}-\frac {125\,{\left (1-2\,x\right )}^{9/2}}{108}-\frac {\sqrt {21}\,\mathrm {atan}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{7}\right )\,14{}\mathrm {i}}{729} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(400*(1 - 2*x)^(7/2))/63 - (14*(1 - 2*x)^(1/2))/243 - (2*(1 - 2*x)^(3/2))/243 - (1027*(1 - 2*x)^(5/2))/108 - (
21^(1/2)*atan((21^(1/2)*(1 - 2*x)^(1/2)*1i)/7)*14i)/729 - (125*(1 - 2*x)^(9/2))/108

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sympy [A]  time = 55.28, size = 126, normalized size = 1.33 \[ - \frac {125 \left (1 - 2 x\right )^{\frac {9}{2}}}{108} + \frac {400 \left (1 - 2 x\right )^{\frac {7}{2}}}{63} - \frac {1027 \left (1 - 2 x\right )^{\frac {5}{2}}}{108} - \frac {2 \left (1 - 2 x\right )^{\frac {3}{2}}}{243} - \frac {14 \sqrt {1 - 2 x}}{243} - \frac {98 \left (\begin {cases} - \frac {\sqrt {21} \operatorname {acoth}{\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} \right )}}{21} & \text {for}\: 2 x - 1 < - \frac {7}{3} \\- \frac {\sqrt {21} \operatorname {atanh}{\left (\frac {\sqrt {21} \sqrt {1 - 2 x}}{7} \right )}}{21} & \text {for}\: 2 x - 1 > - \frac {7}{3} \end {cases}\right )}{243} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125*(1 - 2*x)**(9/2)/108 + 400*(1 - 2*x)**(7/2)/63 - 1027*(1 - 2*x)**(5/2)/108 - 2*(1 - 2*x)**(3/2)/243 - 14*
sqrt(1 - 2*x)/243 - 98*Piecewise((-sqrt(21)*acoth(sqrt(21)*sqrt(1 - 2*x)/7)/21, 2*x - 1 < -7/3), (-sqrt(21)*at
anh(sqrt(21)*sqrt(1 - 2*x)/7)/21, 2*x - 1 > -7/3))/243

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