3.2554 \(\int \frac {2+3 x}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx\)

Optimal. Leaf size=48 \[ \frac {7 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {3 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{\sqrt {10}} \]

[Out]

-3/10*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+7/11*(3+5*x)^(1/2)/(1-2*x)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {78, 54, 216} \[ \frac {7 \sqrt {5 x+3}}{11 \sqrt {1-2 x}}-\frac {3 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{\sqrt {10}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*Sqrt[3 + 5*x])/(11*Sqrt[1 - 2*x]) - (3*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/Sqrt[10]

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 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 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 {2+3 x}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx &=\frac {7 \sqrt {3+5 x}}{11 \sqrt {1-2 x}}-\frac {3}{2} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=\frac {7 \sqrt {3+5 x}}{11 \sqrt {1-2 x}}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{\sqrt {5}}\\ &=\frac {7 \sqrt {3+5 x}}{11 \sqrt {1-2 x}}-\frac {3 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{\sqrt {10}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 54, normalized size = 1.12 \[ \frac {70 \sqrt {5 x+3}-33 \sqrt {20 x-10} \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )}{110 \sqrt {1-2 x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(70*Sqrt[3 + 5*x] - 33*Sqrt[-10 + 20*x]*ArcSinh[Sqrt[5/11]*Sqrt[-1 + 2*x]])/(110*Sqrt[1 - 2*x])

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fricas [B]  time = 1.18, size = 71, normalized size = 1.48 \[ \frac {33 \, \sqrt {10} {\left (2 \, x - 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 140 \, \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{220 \, {\left (2 \, x - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/220*(33*sqrt(10)*(2*x - 1)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) -
140*sqrt(5*x + 3)*sqrt(-2*x + 1))/(2*x - 1)

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giac [A]  time = 0.90, size = 45, normalized size = 0.94 \[ -\frac {3}{10} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {7 \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{55 \, {\left (2 \, x - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/10*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 7/55*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)

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maple [B]  time = 0.01, size = 74, normalized size = 1.54 \[ -\frac {\left (66 \sqrt {10}\, x \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-33 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+140 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {5 x +3}\, \sqrt {-2 x +1}}{220 \left (2 x -1\right ) \sqrt {-10 x^{2}-x +3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/220*(66*10^(1/2)*x*arcsin(20/11*x+1/11)-33*10^(1/2)*arcsin(20/11*x+1/11)+140*(-10*x^2-x+3)^(1/2))*(5*x+3)^(
1/2)*(-2*x+1)^(1/2)/(2*x-1)/(-10*x^2-x+3)^(1/2)

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maxima [A]  time = 1.31, size = 36, normalized size = 0.75 \[ -\frac {3}{20} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {7 \, \sqrt {-10 \, x^{2} - x + 3}}{11 \, {\left (2 \, x - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/20*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 7/11*sqrt(-10*x^2 - x + 3)/(2*x - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {3\,x+2}{{\left (1-2\,x\right )}^{3/2}\,\sqrt {5\,x+3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {3 x + 2}{\left (1 - 2 x\right )^{\frac {3}{2}} \sqrt {5 x + 3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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