3.26.53 \(\int \frac {(2+3 x)^2}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx\) [2553]

Optimal. Leaf size=72 \[ \frac {49 \sqrt {3+5 x}}{22 \sqrt {1-2 x}}+\frac {9}{20} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {321 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{20 \sqrt {10}} \]

[Out]

-321/200*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+49/22*(3+5*x)^(1/2)/(1-2*x)^(1/2)+9/20*(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 = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {91, 81, 56, 222} \begin {gather*} -\frac {321 \text {ArcSin}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{20 \sqrt {10}}+\frac {9}{20} \sqrt {1-2 x} \sqrt {5 x+3}+\frac {49 \sqrt {5 x+3}}{22 \sqrt {1-2 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(49*Sqrt[3 + 5*x])/(22*Sqrt[1 - 2*x]) + (9*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/20 - (321*ArcSin[Sqrt[2/11]*Sqrt[3 + 5
*x]])/(20*Sqrt[10])

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 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c - a*d
)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d*e - c*f)*(n + 1))), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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)^2}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx &=\frac {49 \sqrt {3+5 x}}{22 \sqrt {1-2 x}}-\frac {1}{22} \int \frac {\frac {363}{2}+99 x}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=\frac {49 \sqrt {3+5 x}}{22 \sqrt {1-2 x}}+\frac {9}{20} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {321}{40} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=\frac {49 \sqrt {3+5 x}}{22 \sqrt {1-2 x}}+\frac {9}{20} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {321 \text {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{20 \sqrt {5}}\\ &=\frac {49 \sqrt {3+5 x}}{22 \sqrt {1-2 x}}+\frac {9}{20} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {321 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{20 \sqrt {10}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 63, normalized size = 0.88 \begin {gather*} \frac {10 (589-198 x) \sqrt {3+5 x}+3531 \sqrt {10-20 x} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{2200 \sqrt {1-2 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(10*(589 - 198*x)*Sqrt[3 + 5*x] + 3531*Sqrt[10 - 20*x]*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(2200*Sqrt[1 - 2
*x])

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Maple [A]
time = 0.08, size = 89, normalized size = 1.24

method	result	size
default	\(-\frac {\left (7062 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -3531 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-3960 x \sqrt {-10 x^{2}-x +3}+11780 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {3+5 x}\, \sqrt {1-2 x}}{4400 \left (-1+2 x \right ) \sqrt {-10 x^{2}-x +3}}\)	\(89\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4400*(7062*10^(1/2)*arcsin(20/11*x+1/11)*x-3531*10^(1/2)*arcsin(20/11*x+1/11)-3960*x*(-10*x^2-x+3)^(1/2)+11
780*(-10*x^2-x+3)^(1/2))*(3+5*x)^(1/2)*(1-2*x)^(1/2)/(-1+2*x)/(-10*x^2-x+3)^(1/2)

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Maxima [A]
time = 0.63, size = 50, normalized size = 0.69 \begin {gather*} -\frac {321}{400} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {9}{20} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {49 \, \sqrt {-10 \, x^{2} - x + 3}}{22 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-321/400*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) + 9/20*sqrt(-10*x^2 - x + 3) - 49/22*sqrt(-10*x^2 - x + 3)/(2*
x - 1)

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Fricas [A]
time = 0.81, size = 76, normalized size = 1.06 \begin {gather*} \frac {3531 \, \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 ) + 20 \, {\left (198 \, x - 589\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{4400 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4400*(3531*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))
+ 20*(198*x - 589)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(2*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 )^{2}}{\left (1 - 2 x\right )^{\frac {3}{2}} \sqrt {5 x + 3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.91, size = 58, normalized size = 0.81 \begin {gather*} -\frac {321}{200} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (198 \, \sqrt {5} {\left (5 \, x + 3\right )} - 3539 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{5500 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-321/200*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/5500*(198*sqrt(5)*(5*x + 3) - 3539*sqrt(5))*sqrt(5*x
+ 3)*sqrt(-10*x + 5)/(2*x - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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