3.2582 \(\int \frac {(2+3 x)^2 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx\)

Optimal. Leaf size=94 \[ -\frac {21 (5 x+3)^{3/2}}{11 \sqrt {1-2 x}}+\frac {49 (5 x+3)^{3/2}}{66 (1-2 x)^{3/2}}-\frac {519}{88} \sqrt {1-2 x} \sqrt {5 x+3}+\frac {519 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{8 \sqrt {10}} \]

[Out]

49/66*(3+5*x)^(3/2)/(1-2*x)^(3/2)+519/80*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-21/11*(3+5*x)^(3/2)/(1-2
*x)^(1/2)-519/88*(1-2*x)^(1/2)*(3+5*x)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {89, 78, 50, 54, 216} \[ -\frac {21 (5 x+3)^{3/2}}{11 \sqrt {1-2 x}}+\frac {49 (5 x+3)^{3/2}}{66 (1-2 x)^{3/2}}-\frac {519}{88} \sqrt {1-2 x} \sqrt {5 x+3}+\frac {519 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{8 \sqrt {10}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-519*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/88 + (49*(3 + 5*x)^(3/2))/(66*(1 - 2*x)^(3/2)) - (21*(3 + 5*x)^(3/2))/(11*S
qrt[1 - 2*x]) + (519*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(8*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 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 89

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 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)^2 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx &=\frac {49 (3+5 x)^{3/2}}{66 (1-2 x)^{3/2}}-\frac {1}{66} \int \frac {\sqrt {3+5 x} \left (\frac {1089}{2}+297 x\right )}{(1-2 x)^{3/2}} \, dx\\ &=\frac {49 (3+5 x)^{3/2}}{66 (1-2 x)^{3/2}}-\frac {21 (3+5 x)^{3/2}}{11 \sqrt {1-2 x}}+\frac {519}{44} \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx\\ &=-\frac {519}{88} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {49 (3+5 x)^{3/2}}{66 (1-2 x)^{3/2}}-\frac {21 (3+5 x)^{3/2}}{11 \sqrt {1-2 x}}+\frac {519}{16} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=-\frac {519}{88} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {49 (3+5 x)^{3/2}}{66 (1-2 x)^{3/2}}-\frac {21 (3+5 x)^{3/2}}{11 \sqrt {1-2 x}}+\frac {519 \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{8 \sqrt {5}}\\ &=-\frac {519}{88} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {49 (3+5 x)^{3/2}}{66 (1-2 x)^{3/2}}-\frac {21 (3+5 x)^{3/2}}{11 \sqrt {1-2 x}}+\frac {519 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{8 \sqrt {10}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 73, normalized size = 0.78 \[ -\frac {\sqrt {5 x+3} \left (1188 x^2-7712 x+2481\right )}{264 (1-2 x)^{3/2}}-\frac {519 \sqrt {1-2 x} \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )}{8 \sqrt {20 x-10}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/264*(Sqrt[3 + 5*x]*(2481 - 7712*x + 1188*x^2))/(1 - 2*x)^(3/2) - (519*Sqrt[1 - 2*x]*ArcSinh[Sqrt[5/11]*Sqrt
[-1 + 2*x]])/(8*Sqrt[-10 + 20*x])

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fricas [A]  time = 1.13, size = 91, normalized size = 0.97 \[ -\frac {17127 \, \sqrt {10} {\left (4 \, x^{2} - 4 \, x + 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (1188 \, x^{2} - 7712 \, x + 2481\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{5280 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5280*(17127*sqrt(10)*(4*x^2 - 4*x + 1)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2
+ x - 3)) + 20*(1188*x^2 - 7712*x + 2481)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1)

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giac [A]  time = 1.20, size = 71, normalized size = 0.76 \[ \frac {519}{80} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {{\left (4 \, {\left (297 \, \sqrt {5} {\left (5 \, x + 3\right )} - 11422 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 188397 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{33000 \, {\left (2 \, x - 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

519/80*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 1/33000*(4*(297*sqrt(5)*(5*x + 3) - 11422*sqrt(5))*(5*x
+ 3) + 188397*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2

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maple [A]  time = 0.01, size = 120, normalized size = 1.28 \[ \frac {\left (68508 \sqrt {10}\, x^{2} \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-23760 \sqrt {-10 x^{2}-x +3}\, x^{2}-68508 \sqrt {10}\, x \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+154240 \sqrt {-10 x^{2}-x +3}\, x +17127 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-49620 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {-2 x +1}\, \sqrt {5 x +3}}{5280 \left (2 x -1\right )^{2} \sqrt {-10 x^{2}-x +3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5280*(68508*10^(1/2)*x^2*arcsin(20/11*x+1/11)-68508*10^(1/2)*x*arcsin(20/11*x+1/11)-23760*(-10*x^2-x+3)^(1/2
)*x^2+17127*10^(1/2)*arcsin(20/11*x+1/11)+154240*(-10*x^2-x+3)^(1/2)*x-49620*(-10*x^2-x+3)^(1/2))*(-2*x+1)^(1/
2)*(5*x+3)^(1/2)/(2*x-1)^2/(-10*x^2-x+3)^(1/2)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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