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

Optimal. Leaf size=94 \[ -\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{5/2}+\frac {3}{40} \sqrt {5 x+3} (1-2 x)^{3/2}+\frac {99}{400} \sqrt {5 x+3} \sqrt {1-2 x}+\frac {1089 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{400 \sqrt {10}} \]

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Rubi [A]  time = 0.02, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {80, 50, 54, 216} \begin {gather*} -\frac {1}{10} \sqrt {5 x+3} (1-2 x)^{5/2}+\frac {3}{40} \sqrt {5 x+3} (1-2 x)^{3/2}+\frac {99}{400} \sqrt {5 x+3} \sqrt {1-2 x}+\frac {1089 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{400 \sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(99*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/400 + (3*(1 - 2*x)^(3/2)*Sqrt[3 + 5*x])/40 - ((1 - 2*x)^(5/2)*Sqrt[3 + 5*x])/
10 + (1089*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(400*Sqrt[10])

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 54

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 80

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

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin {align*} \int \frac {(1-2 x)^{3/2} (2+3 x)}{\sqrt {3+5 x}} \, dx &=-\frac {1}{10} (1-2 x)^{5/2} \sqrt {3+5 x}+\frac {3}{4} \int \frac {(1-2 x)^{3/2}}{\sqrt {3+5 x}} \, dx\\ &=\frac {3}{40} (1-2 x)^{3/2} \sqrt {3+5 x}-\frac {1}{10} (1-2 x)^{5/2} \sqrt {3+5 x}+\frac {99}{80} \int \frac {\sqrt {1-2 x}}{\sqrt {3+5 x}} \, dx\\ &=\frac {99}{400} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {3}{40} (1-2 x)^{3/2} \sqrt {3+5 x}-\frac {1}{10} (1-2 x)^{5/2} \sqrt {3+5 x}+\frac {1089}{800} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=\frac {99}{400} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {3}{40} (1-2 x)^{3/2} \sqrt {3+5 x}-\frac {1}{10} (1-2 x)^{5/2} \sqrt {3+5 x}+\frac {1089 \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{400 \sqrt {5}}\\ &=\frac {99}{400} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {3}{40} (1-2 x)^{3/2} \sqrt {3+5 x}-\frac {1}{10} (1-2 x)^{5/2} \sqrt {3+5 x}+\frac {1089 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{400 \sqrt {10}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 69, normalized size = 0.73 \begin {gather*} \frac {10 \sqrt {5 x+3} \left (320 x^3-360 x^2-78 x+89\right )+1089 \sqrt {20 x-10} \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )}{4000 \sqrt {1-2 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(10*Sqrt[3 + 5*x]*(89 - 78*x - 360*x^2 + 320*x^3) + 1089*Sqrt[-10 + 20*x]*ArcSinh[Sqrt[5/11]*Sqrt[-1 + 2*x]])/
(4000*Sqrt[1 - 2*x])

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IntegrateAlgebraic [A]  time = 0.18, size = 109, normalized size = 1.16 \begin {gather*} -\frac {121 \sqrt {1-2 x} \left (\frac {65 (1-2 x)^2}{(5 x+3)^2}-\frac {240 (1-2 x)}{5 x+3}-36\right )}{400 \sqrt {5 x+3} \left (\frac {5 (1-2 x)}{5 x+3}+2\right )^3}-\frac {1089 \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}} \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )}{400 \sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-121*Sqrt[1 - 2*x]*(-36 + (65*(1 - 2*x)^2)/(3 + 5*x)^2 - (240*(1 - 2*x))/(3 + 5*x)))/(400*Sqrt[3 + 5*x]*(2 +
(5*(1 - 2*x))/(3 + 5*x))^3) - (1089*ArcTan[(Sqrt[5/2]*Sqrt[1 - 2*x])/Sqrt[3 + 5*x]])/(400*Sqrt[10])

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fricas [A]  time = 1.66, size = 67, normalized size = 0.71 \begin {gather*} -\frac {1}{400} \, {\left (160 \, x^{2} - 100 \, x - 89\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} - \frac {1089}{8000} \, \sqrt {10} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/400*(160*x^2 - 100*x - 89)*sqrt(5*x + 3)*sqrt(-2*x + 1) - 1089/8000*sqrt(10)*arctan(1/20*sqrt(10)*(20*x + 1
)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3))

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giac [B]  time = 1.08, size = 140, normalized size = 1.49 \begin {gather*} -\frac {1}{20000} \, \sqrt {5} {\left (2 \, {\left (4 \, {\left (40 \, x - 59\right )} {\left (5 \, x + 3\right )} + 1293\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + 4785 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} - \frac {1}{2000} \, \sqrt {5} {\left (2 \, {\left (20 \, x - 23\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - 143 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} + \frac {1}{25} \, \sqrt {5} {\left (11 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + 2 \, \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20000*sqrt(5)*(2*(4*(40*x - 59)*(5*x + 3) + 1293)*sqrt(5*x + 3)*sqrt(-10*x + 5) + 4785*sqrt(2)*arcsin(1/11*
sqrt(22)*sqrt(5*x + 3))) - 1/2000*sqrt(5)*(2*(20*x - 23)*sqrt(5*x + 3)*sqrt(-10*x + 5) - 143*sqrt(2)*arcsin(1/
11*sqrt(22)*sqrt(5*x + 3))) + 1/25*sqrt(5)*(11*sqrt(2)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 2*sqrt(5*x + 3)*s
qrt(-10*x + 5))

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maple [A]  time = 0.01, size = 87, normalized size = 0.93 \begin {gather*} \frac {\sqrt {-2 x +1}\, \sqrt {5 x +3}\, \left (-3200 \sqrt {-10 x^{2}-x +3}\, x^{2}+2000 \sqrt {-10 x^{2}-x +3}\, x +1089 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+1780 \sqrt {-10 x^{2}-x +3}\right )}{8000 \sqrt {-10 x^{2}-x +3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8000*(-2*x+1)^(1/2)*(5*x+3)^(1/2)*(-3200*(-10*x^2-x+3)^(1/2)*x^2+1089*10^(1/2)*arcsin(20/11*x+1/11)+2000*(-1
0*x^2-x+3)^(1/2)*x+1780*(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)

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maxima [A]  time = 1.17, size = 58, normalized size = 0.62 \begin {gather*} -\frac {2}{5} \, \sqrt {-10 \, x^{2} - x + 3} x^{2} + \frac {1}{4} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {1089}{8000} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) + \frac {89}{400} \, \sqrt {-10 \, x^{2} - x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*sqrt(-10*x^2 - x + 3)*x^2 + 1/4*sqrt(-10*x^2 - x + 3)*x - 1089/8000*sqrt(10)*arcsin(-20/11*x - 1/11) + 89
/400*sqrt(-10*x^2 - x + 3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (1-2\,x\right )}^{3/2}\,\left (3\,x+2\right )}{\sqrt {5\,x+3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 81.45, size = 223, normalized size = 2.37 \begin {gather*} - \frac {7 \sqrt {2} \left (\begin {cases} \frac {121 \sqrt {5} \left (\frac {\sqrt {5} \sqrt {1 - 2 x} \sqrt {10 x + 6} \left (20 x + 1\right )}{968} - \frac {\sqrt {5} \sqrt {1 - 2 x} \sqrt {10 x + 6}}{22} + \frac {3 \operatorname {asin}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{8}\right )}{125} & \text {for}\: x \leq \frac {1}{2} \wedge x > - \frac {3}{5} \end {cases}\right )}{2} + \frac {3 \sqrt {2} \left (\begin {cases} \frac {1331 \sqrt {5} \left (\frac {5 \sqrt {5} \left (1 - 2 x\right )^{\frac {3}{2}} \left (10 x + 6\right )^{\frac {3}{2}}}{7986} + \frac {3 \sqrt {5} \sqrt {1 - 2 x} \sqrt {10 x + 6} \left (20 x + 1\right )}{1936} - \frac {\sqrt {5} \sqrt {1 - 2 x} \sqrt {10 x + 6}}{22} + \frac {5 \operatorname {asin}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{16}\right )}{625} & \text {for}\: x \leq \frac {1}{2} \wedge x > - \frac {3}{5} \end {cases}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7*sqrt(2)*Piecewise((121*sqrt(5)*(sqrt(5)*sqrt(1 - 2*x)*sqrt(10*x + 6)*(20*x + 1)/968 - sqrt(5)*sqrt(1 - 2*x)
*sqrt(10*x + 6)/22 + 3*asin(sqrt(55)*sqrt(1 - 2*x)/11)/8)/125, (x <= 1/2) & (x > -3/5)))/2 + 3*sqrt(2)*Piecewi
se((1331*sqrt(5)*(5*sqrt(5)*(1 - 2*x)**(3/2)*(10*x + 6)**(3/2)/7986 + 3*sqrt(5)*sqrt(1 - 2*x)*sqrt(10*x + 6)*(
20*x + 1)/1936 - sqrt(5)*sqrt(1 - 2*x)*sqrt(10*x + 6)/22 + 5*asin(sqrt(55)*sqrt(1 - 2*x)/11)/16)/625, (x <= 1/
2) & (x > -3/5)))/2

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