3.26.83 \(\int \frac {(2+3 x) \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx\) [2583]

Optimal. Leaf size=74 \[ -\frac {3 \sqrt {3+5 x}}{2 \sqrt {1-2 x}}+\frac {7 (3+5 x)^{3/2}}{33 (1-2 x)^{3/2}}+\frac {3}{2} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right ) \]

[Out]

7/33*(3+5*x)^(3/2)/(1-2*x)^(3/2)+3/4*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-3/2*(3+5*x)^(1/2)/(1-2*x)^(1
/2)

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Rubi [A]
time = 0.01, antiderivative size = 74, 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 = {79, 49, 56, 222} \begin {gather*} \frac {3}{2} \sqrt {\frac {5}{2}} \text {ArcSin}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )+\frac {7 (5 x+3)^{3/2}}{33 (1-2 x)^{3/2}}-\frac {3 \sqrt {5 x+3}}{2 \sqrt {1-2 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[3 + 5*x])/(2*Sqrt[1 - 2*x]) + (7*(3 + 5*x)^(3/2))/(33*(1 - 2*x)^(3/2)) + (3*Sqrt[5/2]*ArcSin[Sqrt[2/1
1]*Sqrt[3 + 5*x]])/2

Rule 49

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 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 79

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] || L
tQ[p, n]))))

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}}{(1-2 x)^{5/2}} \, dx &=\frac {7 (3+5 x)^{3/2}}{33 (1-2 x)^{3/2}}-\frac {3}{2} \int \frac {\sqrt {3+5 x}}{(1-2 x)^{3/2}} \, dx\\ &=-\frac {3 \sqrt {3+5 x}}{2 \sqrt {1-2 x}}+\frac {7 (3+5 x)^{3/2}}{33 (1-2 x)^{3/2}}+\frac {15}{4} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=-\frac {3 \sqrt {3+5 x}}{2 \sqrt {1-2 x}}+\frac {7 (3+5 x)^{3/2}}{33 (1-2 x)^{3/2}}+\frac {1}{2} \left (3 \sqrt {5}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )\\ &=-\frac {3 \sqrt {3+5 x}}{2 \sqrt {1-2 x}}+\frac {7 (3+5 x)^{3/2}}{33 (1-2 x)^{3/2}}+\frac {3}{2} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 68, normalized size = 0.92 \begin {gather*} \frac {2 \sqrt {3+5 x} (-57+268 x)+99 \sqrt {10-20 x} (-1+2 x) \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{132 (1-2 x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[3 + 5*x]*(-57 + 268*x) + 99*Sqrt[10 - 20*x]*(-1 + 2*x)*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(132*(1
- 2*x)^(3/2))

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Maple [A]
time = 0.08, size = 103, normalized size = 1.39

method	result	size
default	\(\frac {\left (396 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}-396 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x +99 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+1072 x \sqrt {-10 x^{2}-x +3}-228 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}\, \sqrt {3+5 x}}{264 \left (-1+2 x \right )^{2} \sqrt {-10 x^{2}-x +3}}\)	\(103\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/264*(396*10^(1/2)*arcsin(20/11*x+1/11)*x^2-396*10^(1/2)*arcsin(20/11*x+1/11)*x+99*10^(1/2)*arcsin(20/11*x+1/
11)+1072*x*(-10*x^2-x+3)^(1/2)-228*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(-1+2*x)^2/(-10*x^2-x+3)^(
1/2)

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Maxima [A]
time = 0.52, size = 48, normalized size = 0.65 \begin {gather*} \frac {2 \, \sqrt {-10 \, x^{2} - x + 3}}{3 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac {10 \, \sqrt {-10 \, x^{2} - x + 3}}{33 \, {\left (2 \, x - 1\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)^(5/2),x, algorithm="maxima")

[Out]

2/3*sqrt(-10*x^2 - x + 3)/(4*x^2 - 4*x + 1) + 10/33*sqrt(-10*x^2 - x + 3)/(2*x - 1)

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Fricas [A]
time = 0.51, size = 92, normalized size = 1.24 \begin {gather*} -\frac {99 \, \sqrt {5} \sqrt {2} {\left (4 \, x^{2} - 4 \, x + 1\right )} \arctan \left (\frac {\sqrt {5} \sqrt {2} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 4 \, {\left (268 \, x - 57\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{264 \, {\left (4 \, x^{2} - 4 \, x + 1\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)^(5/2),x, algorithm="fricas")

[Out]

-1/264*(99*sqrt(5)*sqrt(2)*(4*x^2 - 4*x + 1)*arctan(1/20*sqrt(5)*sqrt(2)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x +
1)/(10*x^2 + x - 3)) - 4*(268*x - 57)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1)

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

[Out]

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

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Giac [A]
time = 0.67, size = 58, normalized size = 0.78 \begin {gather*} \frac {3}{4} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (268 \, \sqrt {5} {\left (5 \, x + 3\right )} - 1089 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{1650 \, {\left (2 \, x - 1\right )}^{2}} \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)^(5/2),x, algorithm="giac")

[Out]

3/4*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/1650*(268*sqrt(5)*(5*x + 3) - 1089*sqrt(5))*sqrt(5*x + 3)
*sqrt(-10*x + 5)/(2*x - 1)^2

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

[Out]

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

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