3.25.61 \(\int \frac {(2+3 x) \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx\) [2461]

Optimal. Leaf size=72 \[ -\frac {107}{80} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {3}{20} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {1177 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{80 \sqrt {10}} \]

[Out]

1177/800*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-3/20*(3+5*x)^(3/2)*(1-2*x)^(1/2)-107/80*(1-2*x)^(1/2)*(3
+5*x)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {81, 52, 56, 222} \begin {gather*} \frac {1177 \text {ArcSin}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{80 \sqrt {10}}-\frac {3}{20} \sqrt {1-2 x} (5 x+3)^{3/2}-\frac {107}{80} \sqrt {1-2 x} \sqrt {5 x+3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-107*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/80 - (3*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/20 + (1177*ArcSin[Sqrt[2/11]*Sqrt[3
+ 5*x]])/(80*Sqrt[10])

Rule 52

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 56

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 81

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 222

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) \sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx &=-\frac {3}{20} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {107}{40} \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx\\ &=-\frac {107}{80} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {3}{20} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {1177}{160} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=-\frac {107}{80} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {3}{20} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {1177 \text {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{80 \sqrt {5}}\\ &=-\frac {107}{80} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {3}{20} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {1177 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{80 \sqrt {10}}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 72, normalized size = 1.00 \begin {gather*} \frac {-\sqrt {5-10 x} \sqrt {3+5 x} (143+60 x)-1177 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {6+10 x}}{\sqrt {11}-\sqrt {5-10 x}}\right )}{80 \sqrt {5}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[5 - 10*x]*Sqrt[3 + 5*x]*(143 + 60*x)) - 1177*Sqrt[2]*ArcTan[Sqrt[6 + 10*x]/(Sqrt[11] - Sqrt[5 - 10*x])
])/(80*Sqrt[5])

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Maple [A]
time = 0.08, size = 70, normalized size = 0.97

method	result	size
default	\(\frac {\sqrt {3+5 x}\, \sqrt {1-2 x}\, \left (1177 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-1200 x \sqrt {-10 x^{2}-x +3}-2860 \sqrt {-10 x^{2}-x +3}\right )}{1600 \sqrt {-10 x^{2}-x +3}}\)	\(70\)
risch	\(\frac {\left (143+60 x \right ) \sqrt {3+5 x}\, \left (-1+2 x \right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{80 \sqrt {-\left (3+5 x \right ) \left (-1+2 x \right )}\, \sqrt {1-2 x}}+\frac {1177 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{1600 \sqrt {1-2 x}\, \sqrt {3+5 x}}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1600*(3+5*x)^(1/2)*(1-2*x)^(1/2)*(1177*10^(1/2)*arcsin(20/11*x+1/11)-1200*x*(-10*x^2-x+3)^(1/2)-2860*(-10*x^
2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)

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Maxima [A]
time = 0.50, size = 44, normalized size = 0.61 \begin {gather*} \frac {1177}{1600} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {3}{4} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {143}{80} \, \sqrt {-10 \, x^{2} - x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1177/1600*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 3/4*sqrt(-10*x^2 - x + 3)*x - 143/80*sqrt(-10*x^2 - x + 3)

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Fricas [A]
time = 0.40, size = 62, normalized size = 0.86 \begin {gather*} -\frac {1}{80} \, {\left (60 \, x + 143\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} - \frac {1177}{1600} \, \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((2+3*x)*(3+5*x)^(1/2)/(1-2*x)^(1/2),x, algorithm="fricas")

[Out]

-1/80*(60*x + 143)*sqrt(5*x + 3)*sqrt(-2*x + 1) - 1177/1600*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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Sympy [A]
time = 4.27, size = 207, normalized size = 2.88 \begin {gather*} \frac {2 \sqrt {5} \left (\begin {cases} \frac {11 \sqrt {2} \left (- \frac {\sqrt {2} \sqrt {5 - 10 x} \sqrt {5 x + 3}}{22} + \frac {\operatorname {asin}{\left (\frac {\sqrt {22} \sqrt {5 x + 3}}{11} \right )}}{2}\right )}{4} & \text {for}\: \sqrt {5 x + 3} > - \frac {\sqrt {22}}{2} \wedge \sqrt {5 x + 3} < \frac {\sqrt {22}}{2} \end {cases}\right )}{25} + \frac {6 \sqrt {5} \left (\begin {cases} \frac {121 \sqrt {2} \left (\frac {\sqrt {2} \sqrt {5 - 10 x} \left (- 20 x - 1\right ) \sqrt {5 x + 3}}{968} - \frac {\sqrt {2} \sqrt {5 - 10 x} \sqrt {5 x + 3}}{22} + \frac {3 \operatorname {asin}{\left (\frac {\sqrt {22} \sqrt {5 x + 3}}{11} \right )}}{8}\right )}{8} & \text {for}\: \sqrt {5 x + 3} > - \frac {\sqrt {22}}{2} \wedge \sqrt {5 x + 3} < \frac {\sqrt {22}}{2} \end {cases}\right )}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(5)*Piecewise((11*sqrt(2)*(-sqrt(2)*sqrt(5 - 10*x)*sqrt(5*x + 3)/22 + asin(sqrt(22)*sqrt(5*x + 3)/11)/2)
/4, (sqrt(5*x + 3) > -sqrt(22)/2) & (sqrt(5*x + 3) < sqrt(22)/2)))/25 + 6*sqrt(5)*Piecewise((121*sqrt(2)*(sqrt
(2)*sqrt(5 - 10*x)*(-20*x - 1)*sqrt(5*x + 3)/968 - sqrt(2)*sqrt(5 - 10*x)*sqrt(5*x + 3)/22 + 3*asin(sqrt(22)*s
qrt(5*x + 3)/11)/8)/8, (sqrt(5*x + 3) > -sqrt(22)/2) & (sqrt(5*x + 3) < sqrt(22)/2)))/25

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Giac [A]
time = 1.17, size = 45, normalized size = 0.62 \begin {gather*} -\frac {1}{800} \, \sqrt {5} {\left (2 \, {\left (60 \, x + 143\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - 1177 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/800*sqrt(5)*(2*(60*x + 143)*sqrt(5*x + 3)*sqrt(-10*x + 5) - 1177*sqrt(2)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)
))

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Mupad [B]
time = 12.10, size = 509, normalized size = 7.07 \begin {gather*} \frac {1177\,\sqrt {10}\,\mathrm {atan}\left (\frac {\sqrt {10}\,\left (\sqrt {1-2\,x}-1\right )}{2\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}\right )}{400}-\frac {\frac {297\,\left (\sqrt {1-2\,x}-1\right )}{3125\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}-\frac {2559\,{\left (\sqrt {1-2\,x}-1\right )}^3}{1250\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3}+\frac {2559\,{\left (\sqrt {1-2\,x}-1\right )}^5}{500\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^5}-\frac {297\,{\left (\sqrt {1-2\,x}-1\right )}^7}{200\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^7}+\frac {288\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^2}{625\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {192\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^4}{125\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {72\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^6}{25\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^6}}{\frac {32\,{\left (\sqrt {1-2\,x}-1\right )}^2}{125\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {24\,{\left (\sqrt {1-2\,x}-1\right )}^4}{25\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {8\,{\left (\sqrt {1-2\,x}-1\right )}^6}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^6}+\frac {{\left (\sqrt {1-2\,x}-1\right )}^8}{{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^8}+\frac {16}{625}}-\frac {\frac {2\,{\left (\sqrt {1-2\,x}-1\right )}^3}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^3}-\frac {4\,\left (\sqrt {1-2\,x}-1\right )}{25\,\left (\sqrt {3}-\sqrt {5\,x+3}\right )}+\frac {16\,\sqrt {3}\,{\left (\sqrt {1-2\,x}-1\right )}^2}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}}{\frac {4\,{\left (\sqrt {1-2\,x}-1\right )}^2}{5\,{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^2}+\frac {{\left (\sqrt {1-2\,x}-1\right )}^4}{{\left (\sqrt {3}-\sqrt {5\,x+3}\right )}^4}+\frac {4}{25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1177*10^(1/2)*atan((10^(1/2)*((1 - 2*x)^(1/2) - 1))/(2*(3^(1/2) - (5*x + 3)^(1/2)))))/400 - ((297*((1 - 2*x)^
(1/2) - 1))/(3125*(3^(1/2) - (5*x + 3)^(1/2))) - (2559*((1 - 2*x)^(1/2) - 1)^3)/(1250*(3^(1/2) - (5*x + 3)^(1/
2))^3) + (2559*((1 - 2*x)^(1/2) - 1)^5)/(500*(3^(1/2) - (5*x + 3)^(1/2))^5) - (297*((1 - 2*x)^(1/2) - 1)^7)/(2
00*(3^(1/2) - (5*x + 3)^(1/2))^7) + (288*3^(1/2)*((1 - 2*x)^(1/2) - 1)^2)/(625*(3^(1/2) - (5*x + 3)^(1/2))^2)
+ (192*3^(1/2)*((1 - 2*x)^(1/2) - 1)^4)/(125*(3^(1/2) - (5*x + 3)^(1/2))^4) + (72*3^(1/2)*((1 - 2*x)^(1/2) - 1
)^6)/(25*(3^(1/2) - (5*x + 3)^(1/2))^6))/((32*((1 - 2*x)^(1/2) - 1)^2)/(125*(3^(1/2) - (5*x + 3)^(1/2))^2) + (
24*((1 - 2*x)^(1/2) - 1)^4)/(25*(3^(1/2) - (5*x + 3)^(1/2))^4) + (8*((1 - 2*x)^(1/2) - 1)^6)/(5*(3^(1/2) - (5*
x + 3)^(1/2))^6) + ((1 - 2*x)^(1/2) - 1)^8/(3^(1/2) - (5*x + 3)^(1/2))^8 + 16/625) - ((2*((1 - 2*x)^(1/2) - 1)
^3)/(5*(3^(1/2) - (5*x + 3)^(1/2))^3) - (4*((1 - 2*x)^(1/2) - 1))/(25*(3^(1/2) - (5*x + 3)^(1/2))) + (16*3^(1/
2)*((1 - 2*x)^(1/2) - 1)^2)/(5*(3^(1/2) - (5*x + 3)^(1/2))^2))/((4*((1 - 2*x)^(1/2) - 1)^2)/(5*(3^(1/2) - (5*x
+ 3)^(1/2))^2) + ((1 - 2*x)^(1/2) - 1)^4/(3^(1/2) - (5*x + 3)^(1/2))^4 + 4/25)

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