3.2542 \(\int \frac{(2+3 x) (3+5 x)^{5/2}}{(1-2 x)^{3/2}} \, dx\)

Optimal. Leaf size=118 \[ \frac{7 (5 x+3)^{7/2}}{11 \sqrt{1-2 x}}+\frac{81}{44} \sqrt{1-2 x} (5 x+3)^{5/2}+\frac{405}{32} \sqrt{1-2 x} (5 x+3)^{3/2}+\frac{13365}{128} \sqrt{1-2 x} \sqrt{5 x+3}-\frac{29403}{128} \sqrt{\frac{5}{2}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right ) \]

[Out]

(13365*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/128 + (405*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/32 + (81*Sqrt[1 - 2*x]*(3 + 5*x)
^(5/2))/44 + (7*(3 + 5*x)^(7/2))/(11*Sqrt[1 - 2*x]) - (29403*Sqrt[5/2]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/128

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Rubi [A]  time = 0.0303091, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {78, 50, 54, 216} \[ \frac{7 (5 x+3)^{7/2}}{11 \sqrt{1-2 x}}+\frac{81}{44} \sqrt{1-2 x} (5 x+3)^{5/2}+\frac{405}{32} \sqrt{1-2 x} (5 x+3)^{3/2}+\frac{13365}{128} \sqrt{1-2 x} \sqrt{5 x+3}-\frac{29403}{128} \sqrt{\frac{5}{2}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(13365*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/128 + (405*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/32 + (81*Sqrt[1 - 2*x]*(3 + 5*x)
^(5/2))/44 + (7*(3 + 5*x)^(7/2))/(11*Sqrt[1 - 2*x]) - (29403*Sqrt[5/2]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/128

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 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)^{5/2}}{(1-2 x)^{3/2}} \, dx &=\frac{7 (3+5 x)^{7/2}}{11 \sqrt{1-2 x}}-\frac{243}{22} \int \frac{(3+5 x)^{5/2}}{\sqrt{1-2 x}} \, dx\\ &=\frac{81}{44} \sqrt{1-2 x} (3+5 x)^{5/2}+\frac{7 (3+5 x)^{7/2}}{11 \sqrt{1-2 x}}-\frac{405}{8} \int \frac{(3+5 x)^{3/2}}{\sqrt{1-2 x}} \, dx\\ &=\frac{405}{32} \sqrt{1-2 x} (3+5 x)^{3/2}+\frac{81}{44} \sqrt{1-2 x} (3+5 x)^{5/2}+\frac{7 (3+5 x)^{7/2}}{11 \sqrt{1-2 x}}-\frac{13365}{64} \int \frac{\sqrt{3+5 x}}{\sqrt{1-2 x}} \, dx\\ &=\frac{13365}{128} \sqrt{1-2 x} \sqrt{3+5 x}+\frac{405}{32} \sqrt{1-2 x} (3+5 x)^{3/2}+\frac{81}{44} \sqrt{1-2 x} (3+5 x)^{5/2}+\frac{7 (3+5 x)^{7/2}}{11 \sqrt{1-2 x}}-\frac{147015}{256} \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx\\ &=\frac{13365}{128} \sqrt{1-2 x} \sqrt{3+5 x}+\frac{405}{32} \sqrt{1-2 x} (3+5 x)^{3/2}+\frac{81}{44} \sqrt{1-2 x} (3+5 x)^{5/2}+\frac{7 (3+5 x)^{7/2}}{11 \sqrt{1-2 x}}-\frac{1}{128} \left (29403 \sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{11-2 x^2}} \, dx,x,\sqrt{3+5 x}\right )\\ &=\frac{13365}{128} \sqrt{1-2 x} \sqrt{3+5 x}+\frac{405}{32} \sqrt{1-2 x} (3+5 x)^{3/2}+\frac{81}{44} \sqrt{1-2 x} (3+5 x)^{5/2}+\frac{7 (3+5 x)^{7/2}}{11 \sqrt{1-2 x}}-\frac{29403}{128} \sqrt{\frac{5}{2}} \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0329608, size = 69, normalized size = 0.58 \[ \frac{29403 \sqrt{10-20 x} \sin ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )-2 \sqrt{5 x+3} \left (1600 x^3+6120 x^2+14526 x-22545\right )}{256 \sqrt{1-2 x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[3 + 5*x]*(-22545 + 14526*x + 6120*x^2 + 1600*x^3) + 29403*Sqrt[10 - 20*x]*ArcSin[Sqrt[5/11]*Sqrt[1 -
2*x]])/(256*Sqrt[1 - 2*x])

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Maple [A]  time = 0.011, size = 123, normalized size = 1. \begin{align*} -{\frac{1}{1024\,x-512} \left ( -6400\,{x}^{3}\sqrt{-10\,{x}^{2}-x+3}+58806\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) x-24480\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}-29403\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) -58104\,x\sqrt{-10\,{x}^{2}-x+3}+90180\,\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)^(5/2)/(1-2*x)^(3/2),x)

[Out]

-1/512*(-6400*x^3*(-10*x^2-x+3)^(1/2)+58806*10^(1/2)*arcsin(20/11*x+1/11)*x-24480*x^2*(-10*x^2-x+3)^(1/2)-2940
3*10^(1/2)*arcsin(20/11*x+1/11)-58104*x*(-10*x^2-x+3)^(1/2)+90180*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)*(3+5*x)^(
1/2)/(2*x-1)/(-10*x^2-x+3)^(1/2)

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Maxima [A]  time = 3.0932, size = 124, normalized size = 1.05 \begin{align*} -\frac{125 \, x^{4}}{2 \, \sqrt{-10 \, x^{2} - x + 3}} - \frac{4425 \, x^{3}}{16 \, \sqrt{-10 \, x^{2} - x + 3}} - \frac{45495 \, x^{2}}{64 \, \sqrt{-10 \, x^{2} - x + 3}} + \frac{29403}{512} \, \sqrt{10} \arcsin \left (-\frac{20}{11} \, x - \frac{1}{11}\right ) + \frac{69147 \, x}{128 \, \sqrt{-10 \, x^{2} - x + 3}} + \frac{67635}{128 \, \sqrt{-10 \, x^{2} - x + 3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125/2*x^4/sqrt(-10*x^2 - x + 3) - 4425/16*x^3/sqrt(-10*x^2 - x + 3) - 45495/64*x^2/sqrt(-10*x^2 - x + 3) + 29
403/512*sqrt(10)*arcsin(-20/11*x - 1/11) + 69147/128*x/sqrt(-10*x^2 - x + 3) + 67635/128/sqrt(-10*x^2 - x + 3)

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Fricas [A]  time = 1.76259, size = 284, normalized size = 2.41 \begin{align*} \frac{29403 \, \sqrt{5} \sqrt{2}{\left (2 \, 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 (1600 \, x^{3} + 6120 \, x^{2} + 14526 \, x - 22545\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{512 \,{\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)^(5/2)/(1-2*x)^(3/2),x, algorithm="fricas")

[Out]

1/512*(29403*sqrt(5)*sqrt(2)*(2*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*(1600*x^3 + 6120*x^2 + 14526*x - 22545)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(2*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)**(5/2)/(1-2*x)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.53529, size = 113, normalized size = 0.96 \begin{align*} -\frac{29403}{256} \, \sqrt{10} \arcsin \left (\frac{1}{11} \, \sqrt{22} \sqrt{5 \, x + 3}\right ) + \frac{{\left (2 \,{\left (4 \,{\left (8 \, \sqrt{5}{\left (5 \, x + 3\right )} + 81 \, \sqrt{5}\right )}{\left (5 \, x + 3\right )} + 4455 \, \sqrt{5}\right )}{\left (5 \, x + 3\right )} - 147015 \, \sqrt{5}\right )} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5}}{3200 \,{\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)^(5/2)/(1-2*x)^(3/2),x, algorithm="giac")

[Out]

-29403/256*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/3200*(2*(4*(8*sqrt(5)*(5*x + 3) + 81*sqrt(5))*(5*x
+ 3) + 4455*sqrt(5))*(5*x + 3) - 147015*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)