3.2592 \(\int \frac{(2+3 x) (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx\)

Optimal. Leaf size=96 \[ \frac{7 (5 x+3)^{5/2}}{33 (1-2 x)^{3/2}}-\frac{169 (5 x+3)^{3/2}}{66 \sqrt{1-2 x}}-\frac{845}{88} \sqrt{1-2 x} \sqrt{5 x+3}+\frac{169}{8} \sqrt{\frac{5}{2}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right ) \]

[Out]

(-845*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/88 - (169*(3 + 5*x)^(3/2))/(66*Sqrt[1 - 2*x]) + (7*(3 + 5*x)^(5/2))/(33*(1
- 2*x)^(3/2)) + (169*Sqrt[5/2]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/8

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Rubi [A]  time = 0.0214043, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {78, 47, 50, 54, 216} \[ \frac{7 (5 x+3)^{5/2}}{33 (1-2 x)^{3/2}}-\frac{169 (5 x+3)^{3/2}}{66 \sqrt{1-2 x}}-\frac{845}{88} \sqrt{1-2 x} \sqrt{5 x+3}+\frac{169}{8} \sqrt{\frac{5}{2}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-845*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/88 - (169*(3 + 5*x)^(3/2))/(66*Sqrt[1 - 2*x]) + (7*(3 + 5*x)^(5/2))/(33*(1
- 2*x)^(3/2)) + (169*Sqrt[5/2]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/8

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 47

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 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 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) (3+5 x)^{3/2}}{(1-2 x)^{5/2}} \, dx &=\frac{7 (3+5 x)^{5/2}}{33 (1-2 x)^{3/2}}-\frac{169}{66} \int \frac{(3+5 x)^{3/2}}{(1-2 x)^{3/2}} \, dx\\ &=-\frac{169 (3+5 x)^{3/2}}{66 \sqrt{1-2 x}}+\frac{7 (3+5 x)^{5/2}}{33 (1-2 x)^{3/2}}+\frac{845}{44} \int \frac{\sqrt{3+5 x}}{\sqrt{1-2 x}} \, dx\\ &=-\frac{845}{88} \sqrt{1-2 x} \sqrt{3+5 x}-\frac{169 (3+5 x)^{3/2}}{66 \sqrt{1-2 x}}+\frac{7 (3+5 x)^{5/2}}{33 (1-2 x)^{3/2}}+\frac{845}{16} \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx\\ &=-\frac{845}{88} \sqrt{1-2 x} \sqrt{3+5 x}-\frac{169 (3+5 x)^{3/2}}{66 \sqrt{1-2 x}}+\frac{7 (3+5 x)^{5/2}}{33 (1-2 x)^{3/2}}+\frac{1}{8} \left (169 \sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{11-2 x^2}} \, dx,x,\sqrt{3+5 x}\right )\\ &=-\frac{845}{88} \sqrt{1-2 x} \sqrt{3+5 x}-\frac{169 (3+5 x)^{3/2}}{66 \sqrt{1-2 x}}+\frac{7 (3+5 x)^{5/2}}{33 (1-2 x)^{3/2}}+\frac{169}{8} \sqrt{\frac{5}{2}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )\\ \end{align*}

Mathematica [C]  time = 0.0316773, size = 56, normalized size = 0.58 \[ \frac{1859 \sqrt{22} (2 x-1) \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};\frac{5}{11} (1-2 x)\right )+56 (5 x+3)^{5/2}}{264 (1-2 x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(56*(3 + 5*x)^(5/2) + 1859*Sqrt[22]*(-1 + 2*x)*Hypergeometric2F1[-3/2, -1/2, 1/2, (5*(1 - 2*x))/11])/(264*(1 -
2*x)^(3/2))

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Maple [A]  time = 0.01, size = 120, normalized size = 1.3 \begin{align*}{\frac{1}{96\, \left ( 2\,x-1 \right ) ^{2}} \left ( 2028\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ){x}^{2}-2028\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) x-720\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}+507\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) +4544\,x\sqrt{-10\,{x}^{2}-x+3}-1476\,\sqrt{-10\,{x}^{2}-x+3} \right ) \sqrt{1-2\,x}\sqrt{3+5\,x}{\frac{1}{\sqrt{-10\,{x}^{2}-x+3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(2028*10^(1/2)*arcsin(20/11*x+1/11)*x^2-2028*10^(1/2)*arcsin(20/11*x+1/11)*x-720*x^2*(-10*x^2-x+3)^(1/2)+
507*10^(1/2)*arcsin(20/11*x+1/11)+4544*x*(-10*x^2-x+3)^(1/2)-1476*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)*(3+5*x)^(
1/2)/(2*x-1)^2/(-10*x^2-x+3)^(1/2)

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Maxima [A]  time = 4.53903, size = 161, normalized size = 1.68 \begin{align*} \frac{169}{32} \, \sqrt{5} \sqrt{2} \arcsin \left (\frac{20}{11} \, x + \frac{1}{11}\right ) - \frac{7 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}}{12 \,{\left (8 \, x^{3} - 12 \, x^{2} + 6 \, x - 1\right )}} + \frac{3 \,{\left (-10 \, x^{2} - x + 3\right )}^{\frac{3}{2}}}{4 \,{\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac{77 \, \sqrt{-10 \, x^{2} - x + 3}}{24 \,{\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac{271 \, \sqrt{-10 \, x^{2} - x + 3}}{12 \,{\left (2 \, x - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

169/32*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 7/12*(-10*x^2 - x + 3)^(3/2)/(8*x^3 - 12*x^2 + 6*x - 1) + 3/4*
(-10*x^2 - x + 3)^(3/2)/(4*x^2 - 4*x + 1) + 77/24*sqrt(-10*x^2 - x + 3)/(4*x^2 - 4*x + 1) + 271/12*sqrt(-10*x^
2 - x + 3)/(2*x - 1)

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Fricas [A]  time = 1.51612, size = 282, normalized size = 2.94 \begin{align*} -\frac{507 \, \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 (180 \, x^{2} - 1136 \, x + 369\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{96 \,{\left (4 \, x^{2} - 4 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(507*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*(180*x^2 - 1136*x + 369)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 2.11393, size = 96, normalized size = 1. \begin{align*} \frac{169}{16} \, \sqrt{10} \arcsin \left (\frac{1}{11} \, \sqrt{22} \sqrt{5 \, x + 3}\right ) - \frac{{\left (4 \,{\left (9 \, \sqrt{5}{\left (5 \, x + 3\right )} - 338 \, \sqrt{5}\right )}{\left (5 \, x + 3\right )} + 5577 \, \sqrt{5}\right )} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5}}{600 \,{\left (2 \, x - 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

169/16*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 1/600*(4*(9*sqrt(5)*(5*x + 3) - 338*sqrt(5))*(5*x + 3) +
5577*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2