3.2545 \(\int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2} (2+3 x)^2} \, dx\)

Optimal. Leaf size=122 \[ \frac {11 (5 x+3)^{3/2}}{7 \sqrt {1-2 x} (3 x+2)}+\frac {32 \sqrt {1-2 x} \sqrt {5 x+3}}{147 (3 x+2)}-\frac {25}{9} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {169 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{441 \sqrt {7}} \]

[Out]

-25/18*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-169/3087*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7
^(1/2)+11/7*(3+5*x)^(3/2)/(2+3*x)/(1-2*x)^(1/2)+32/147*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x)

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Rubi [A]  time = 0.04, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {98, 149, 157, 54, 216, 93, 204} \[ \frac {11 (5 x+3)^{3/2}}{7 \sqrt {1-2 x} (3 x+2)}+\frac {32 \sqrt {1-2 x} \sqrt {5 x+3}}{147 (3 x+2)}-\frac {25}{9} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {169 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{441 \sqrt {7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(147*(2 + 3*x)) + (11*(3 + 5*x)^(3/2))/(7*Sqrt[1 - 2*x]*(2 + 3*x)) - (25*Sqrt
[5/2]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/9 - (169*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(441*Sqrt[7])

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 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

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 149

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] && IntegerQ[m]

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[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 {(3+5 x)^{5/2}}{(1-2 x)^{3/2} (2+3 x)^2} \, dx &=\frac {11 (3+5 x)^{3/2}}{7 \sqrt {1-2 x} (2+3 x)}-\frac {1}{7} \int \frac {\sqrt {3+5 x} \left (69+\frac {175 x}{2}\right )}{\sqrt {1-2 x} (2+3 x)^2} \, dx\\ &=\frac {32 \sqrt {1-2 x} \sqrt {3+5 x}}{147 (2+3 x)}+\frac {11 (3+5 x)^{3/2}}{7 \sqrt {1-2 x} (2+3 x)}-\frac {1}{147} \int \frac {\frac {4027}{2}+\frac {6125 x}{2}}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=\frac {32 \sqrt {1-2 x} \sqrt {3+5 x}}{147 (2+3 x)}+\frac {11 (3+5 x)^{3/2}}{7 \sqrt {1-2 x} (2+3 x)}+\frac {169}{882} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx-\frac {125}{18} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=\frac {32 \sqrt {1-2 x} \sqrt {3+5 x}}{147 (2+3 x)}+\frac {11 (3+5 x)^{3/2}}{7 \sqrt {1-2 x} (2+3 x)}+\frac {169}{441} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )-\frac {1}{9} \left (25 \sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )\\ &=\frac {32 \sqrt {1-2 x} \sqrt {3+5 x}}{147 (2+3 x)}+\frac {11 (3+5 x)^{3/2}}{7 \sqrt {1-2 x} (2+3 x)}-\frac {25}{9} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )-\frac {169 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{441 \sqrt {7}}\\ \end {align*}

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Mathematica [C]  time = 0.24, size = 173, normalized size = 1.42 \[ \frac {-468930 (3 x+2) \sqrt {44 x-22} \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};-\frac {5}{11} (2 x-1)\right )-42 \sqrt {2 x-1} \sqrt {5 x+3} \left (30450 x^2-70643 x-60610\right )-771995 \sqrt {10} \left (6 x^2+x-2\right ) \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )-2704 \sqrt {7} \sqrt {-(1-2 x)^2} (3 x+2) \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{49392 \sqrt {-(1-2 x)^2} (3 x+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-42*Sqrt[-1 + 2*x]*Sqrt[3 + 5*x]*(-60610 - 70643*x + 30450*x^2) - 771995*Sqrt[10]*(-2 + x + 6*x^2)*ArcSinh[Sq
rt[5/11]*Sqrt[-1 + 2*x]] - 2704*Sqrt[7]*Sqrt[-(1 - 2*x)^2]*(2 + 3*x)*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*
x])] - 468930*(2 + 3*x)*Sqrt[-22 + 44*x]*Hypergeometric2F1[-3/2, -1/2, 1/2, (-5*(-1 + 2*x))/11])/(49392*Sqrt[-
(1 - 2*x)^2]*(2 + 3*x))

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fricas [A]  time = 0.78, size = 136, normalized size = 1.11 \[ \frac {8575 \, \sqrt {5} \sqrt {2} {\left (6 \, x^{2} + x - 2\right )} \arctan \left (\frac {\sqrt {5} \sqrt {2} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 338 \, \sqrt {7} {\left (6 \, x^{2} + x - 2\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 84 \, {\left (1091 \, x + 725\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{12348 \, {\left (6 \, x^{2} + x - 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12348*(8575*sqrt(5)*sqrt(2)*(6*x^2 + x - 2)*arctan(1/20*sqrt(5)*sqrt(2)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x +
1)/(10*x^2 + x - 3)) - 338*sqrt(7)*(6*x^2 + x - 2)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x +
1)/(10*x^2 + x - 3)) - 84*(1091*x + 725)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(6*x^2 + x - 2)

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giac [B]  time = 2.00, size = 286, normalized size = 2.34 \[ \frac {169}{61740} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {25}{36} \, \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {121 \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{245 \, {\left (2 \, x - 1\right )}} - \frac {22 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{147 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

169/61740*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))
^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) - 25/36*sqrt(10)*(pi + 2*arctan(-1/4*sqrt(5*x + 3)*((
sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) - 121/245*sqrt(5)*
sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1) - 22/147*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3)
- 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))/(((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) -
4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^2 + 280)

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maple [B]  time = 0.02, size = 198, normalized size = 1.62 \[ -\frac {\left (51450 \sqrt {10}\, x^{2} \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-2028 \sqrt {7}\, x^{2} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+8575 \sqrt {10}\, x \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-338 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+91644 \sqrt {-10 x^{2}-x +3}\, x -17150 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+676 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+60900 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {-2 x +1}\, \sqrt {5 x +3}}{12348 \left (3 x +2\right ) \left (2 x -1\right ) \sqrt {-10 x^{2}-x +3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12348*(51450*10^(1/2)*x^2*arcsin(20/11*x+1/11)-2028*7^(1/2)*x^2*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)
^(1/2))+8575*10^(1/2)*x*arcsin(20/11*x+1/11)-338*7^(1/2)*x*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-
17150*10^(1/2)*arcsin(20/11*x+1/11)+676*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+91644*(-10*
x^2-x+3)^(1/2)*x+60900*(-10*x^2-x+3)^(1/2))*(-2*x+1)^(1/2)*(5*x+3)^(1/2)/(3*x+2)/(2*x-1)/(-10*x^2-x+3)^(1/2)

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maxima [A]  time = 1.15, size = 103, normalized size = 0.84 \[ -\frac {25}{36} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {169}{6174} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {5455 \, x}{441 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {9784}{1323 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {1}{189 \, {\left (3 \, \sqrt {-10 \, x^{2} - x + 3} x + 2 \, \sqrt {-10 \, x^{2} - x + 3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/36*sqrt(10)*arcsin(20/11*x + 1/11) + 169/6174*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) +
5455/441*x/sqrt(-10*x^2 - x + 3) + 9784/1323/sqrt(-10*x^2 - x + 3) + 1/189/(3*sqrt(-10*x^2 - x + 3)*x + 2*sqrt
(-10*x^2 - x + 3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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