3.2572 \(\int \frac{(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx\)

Optimal. Leaf size=113 \[ \frac{7 (3 x+2)^3}{11 \sqrt{1-2 x} (5 x+3)^{3/2}}-\frac{107 \sqrt{1-2 x} (3 x+2)^2}{1815 (5 x+3)^{3/2}}+\frac{\sqrt{1-2 x} (1051875 x+627641)}{399300 \sqrt{5 x+3}}-\frac{621 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{100 \sqrt{10}} \]

[Out]

(-107*Sqrt[1 - 2*x]*(2 + 3*x)^2)/(1815*(3 + 5*x)^(3/2)) + (7*(2 + 3*x)^3)/(11*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) +
(Sqrt[1 - 2*x]*(627641 + 1051875*x))/(399300*Sqrt[3 + 5*x]) - (621*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(100*Sqr
t[10])

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Rubi [A]  time = 0.032144, antiderivative size = 113, normalized size of antiderivative = 1., 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 = {98, 150, 143, 54, 216} \[ \frac{7 (3 x+2)^3}{11 \sqrt{1-2 x} (5 x+3)^{3/2}}-\frac{107 \sqrt{1-2 x} (3 x+2)^2}{1815 (5 x+3)^{3/2}}+\frac{\sqrt{1-2 x} (1051875 x+627641)}{399300 \sqrt{5 x+3}}-\frac{621 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{5 x+3}\right )}{100 \sqrt{10}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x)^4/((1 - 2*x)^(3/2)*(3 + 5*x)^(5/2)),x]

[Out]

(-107*Sqrt[1 - 2*x]*(2 + 3*x)^2)/(1815*(3 + 5*x)^(3/2)) + (7*(2 + 3*x)^3)/(11*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) +
(Sqrt[1 - 2*x]*(627641 + 1051875*x))/(399300*Sqrt[3 + 5*x]) - (621*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(100*Sqr
t[10])

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 150

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]

Rule 143

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^2*d*e*g - a^2*d*f*h*m - a*b*(d*(f*g + e*h) - c*f*h*(m + 1)) + b*f*h*(b*c - a*d)*(m + 1)*x)*(a + b*x
)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)), x] + Dist[(a*d*f*h*m + b*(d*(f*g + e*h) - c*f*h*(m +
2)))/(b^2*d), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && EqQ[m
+ n + 2, 0] && NeQ[m, -1] &&  !(SumSimplerQ[n, 1] &&  !SumSimplerQ[m, 1])

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)^4}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx &=\frac{7 (2+3 x)^3}{11 \sqrt{1-2 x} (3+5 x)^{3/2}}-\frac{1}{11} \int \frac{(2+3 x)^2 \left (82+\frac{309 x}{2}\right )}{\sqrt{1-2 x} (3+5 x)^{5/2}} \, dx\\ &=-\frac{107 \sqrt{1-2 x} (2+3 x)^2}{1815 (3+5 x)^{3/2}}+\frac{7 (2+3 x)^3}{11 \sqrt{1-2 x} (3+5 x)^{3/2}}-\frac{2 \int \frac{(2+3 x) \left (\frac{9127}{2}+\frac{31875 x}{4}\right )}{\sqrt{1-2 x} (3+5 x)^{3/2}} \, dx}{1815}\\ &=-\frac{107 \sqrt{1-2 x} (2+3 x)^2}{1815 (3+5 x)^{3/2}}+\frac{7 (2+3 x)^3}{11 \sqrt{1-2 x} (3+5 x)^{3/2}}+\frac{\sqrt{1-2 x} (627641+1051875 x)}{399300 \sqrt{3+5 x}}-\frac{621}{200} \int \frac{1}{\sqrt{1-2 x} \sqrt{3+5 x}} \, dx\\ &=-\frac{107 \sqrt{1-2 x} (2+3 x)^2}{1815 (3+5 x)^{3/2}}+\frac{7 (2+3 x)^3}{11 \sqrt{1-2 x} (3+5 x)^{3/2}}+\frac{\sqrt{1-2 x} (627641+1051875 x)}{399300 \sqrt{3+5 x}}-\frac{621 \operatorname{Subst}\left (\int \frac{1}{\sqrt{11-2 x^2}} \, dx,x,\sqrt{3+5 x}\right )}{100 \sqrt{5}}\\ &=-\frac{107 \sqrt{1-2 x} (2+3 x)^2}{1815 (3+5 x)^{3/2}}+\frac{7 (2+3 x)^3}{11 \sqrt{1-2 x} (3+5 x)^{3/2}}+\frac{\sqrt{1-2 x} (627641+1051875 x)}{399300 \sqrt{3+5 x}}-\frac{621 \sin ^{-1}\left (\sqrt{\frac{2}{11}} \sqrt{3+5 x}\right )}{100 \sqrt{10}}\\ \end{align*}

Mathematica [A]  time = 0.0989986, size = 65, normalized size = 0.58 \[ \frac{-3234330 x^3+6746215 x^2+11581424 x+3821563}{399300 \sqrt{1-2 x} (5 x+3)^{3/2}}+\frac{621 \sin ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{100 \sqrt{10}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + 3*x)^4/((1 - 2*x)^(3/2)*(3 + 5*x)^(5/2)),x]

[Out]

(3821563 + 11581424*x + 6746215*x^2 - 3234330*x^3)/(399300*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) + (621*ArcSin[Sqrt[5
/11]*Sqrt[1 - 2*x]])/(100*Sqrt[10])

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Maple [A]  time = 0.013, size = 151, normalized size = 1.3 \begin{align*} -{\frac{1}{15972000\,x-7986000}\sqrt{1-2\,x} \left ( 123982650\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ){x}^{3}[phone]\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ){x}^{2}-64686600\,{x}^{3}\sqrt{-10\,{x}^{2}-x+3}-29755836\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) x+134924300\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}-22316877\,\sqrt{10}\arcsin \left ({\frac{20\,x}{11}}+1/11 \right ) [phone]\,x\sqrt{-10\,{x}^{2}-x+3}[phone]\,\sqrt{-10\,{x}^{2}-x+3} \right ){\frac{1}{\sqrt{-10\,{x}^{2}-x+3}}} \left ( 3+5\,x \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2+3*x)^4/(1-2*x)^(3/2)/(3+5*x)^(5/2),x)

[Out]

-1/7986000*(1-2*x)^(1/2)*(123982650*10^(1/2)*arcsin(20/11*x+1/11)*x^3+86787855*10^(1/2)*arcsin(20/11*x+1/11)*x
^2-64686600*x^3*(-10*x^2-x+3)^(1/2)-29755836*10^(1/2)*arcsin(20/11*x+1/11)*x+134924300*x^2*(-10*x^2-x+3)^(1/2)
-22316877*10^(1/2)*arcsin(20/11*x+1/11)[phone]*x*(-10*x^2-x+3)^(1/2)[phone]*(-10*x^2-x+3)^(1/2))/(2*x-1)/
(-10*x^2-x+3)^(1/2)/(3+5*x)^(3/2)

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Maxima [A]  time = 3.85106, size = 128, normalized size = 1.13 \begin{align*} -\frac{621}{2000} \, \sqrt{5} \sqrt{2} \arcsin \left (\frac{20}{11} \, x + \frac{1}{11}\right ) - \frac{81 \, x^{2}}{50 \, \sqrt{-10 \, x^{2} - x + 3}} + \frac{8686813 \, x}{1996500 \, \sqrt{-10 \, x^{2} - x + 3}} + \frac{31846681}{9982500 \, \sqrt{-10 \, x^{2} - x + 3}} - \frac{2}{20625 \,{\left (5 \, \sqrt{-10 \, x^{2} - x + 3} x + 3 \, \sqrt{-10 \, x^{2} - x + 3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4/(1-2*x)^(3/2)/(3+5*x)^(5/2),x, algorithm="maxima")

[Out]

-621/2000*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 81/50*x^2/sqrt(-10*x^2 - x + 3) + 8686813/1996500*x/sqrt(-1
0*x^2 - x + 3) + 31846681/9982500/sqrt(-10*x^2 - x + 3) - 2/20625/(5*sqrt(-10*x^2 - x + 3)*x + 3*sqrt(-10*x^2
- x + 3))

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Fricas [A]  time = 1.56681, size = 340, normalized size = 3.01 \begin{align*} \frac{2479653 \, \sqrt{10}{\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\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 (3234330 \, x^{3} - 6746215 \, x^{2} - 11581424 \, x - 3821563\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{7986000 \,{\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4/(1-2*x)^(3/2)/(3+5*x)^(5/2),x, algorithm="fricas")

[Out]

1/7986000*(2479653*sqrt(10)*(50*x^3 + 35*x^2 - 12*x - 9)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2
*x + 1)/(10*x^2 + x - 3)) + 20*(3234330*x^3 - 6746215*x^2 - 11581424*x - 3821563)*sqrt(5*x + 3)*sqrt(-2*x + 1)
)/(50*x^3 + 35*x^2 - 12*x - 9)

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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)**4/(1-2*x)**(3/2)/(3+5*x)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 2.77315, size = 247, normalized size = 2.19 \begin{align*} -\frac{\sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{3}}{39930000 \,{\left (5 \, x + 3\right )}^{\frac{3}{2}}} - \frac{621}{1000} \, \sqrt{10} \arcsin \left (\frac{1}{11} \, \sqrt{22} \sqrt{5 \, x + 3}\right ) + \frac{{\left (215622 \, \sqrt{5}{\left (5 \, x + 3\right )} - 4187171 \, \sqrt{5}\right )} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5}}{16637500 \,{\left (2 \, x - 1\right )}} - \frac{271 \, \sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}}{3327500 \, \sqrt{5 \, x + 3}} + \frac{{\left (\frac{813 \, \sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{2}}{5 \, x + 3} + 4 \, \sqrt{10}\right )}{\left (5 \, x + 3\right )}^{\frac{3}{2}}}{2495625 \,{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4/(1-2*x)^(3/2)/(3+5*x)^(5/2),x, algorithm="giac")

[Out]

-1/39930000*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) - 621/1000*sqrt(10)*arcsin(1/11*sq
rt(22)*sqrt(5*x + 3)) + 1/16637500*(215622*sqrt(5)*(5*x + 3) - 4187171*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/
(2*x - 1) - 271/3327500*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 1/2495625*(813*sqrt(10)*
(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) + 4*sqrt(10))*(5*x + 3)^(3/2)/(sqrt(2)*sqrt(-10*x + 5) - sqrt
(22))^3