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

Optimal. Leaf size=108 \[ \frac {11 (5 x+3)^{3/2}}{7 \sqrt {1-2 x}}+\frac {505}{84} \sqrt {1-2 x} \sqrt {5 x+3}-\frac {475}{36} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )+\frac {2 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{63 \sqrt {7}} \]

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Rubi [A]  time = 0.04, antiderivative size = 108, 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, 154, 157, 54, 216, 93, 204} \begin {gather*} \frac {11 (5 x+3)^{3/2}}{7 \sqrt {1-2 x}}+\frac {505}{84} \sqrt {1-2 x} \sqrt {5 x+3}-\frac {475}{36} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )+\frac {2 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{63 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(505*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/84 + (11*(3 + 5*x)^(3/2))/(7*Sqrt[1 - 2*x]) - (475*Sqrt[5/2]*ArcSin[Sqrt[2/1
1]*Sqrt[3 + 5*x]])/36 + (2*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(63*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 154

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

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)} \, dx &=\frac {11 (3+5 x)^{3/2}}{7 \sqrt {1-2 x}}-\frac {1}{7} \int \frac {\sqrt {3+5 x} \left (168+\frac {505 x}{2}\right )}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=\frac {505}{84} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {11 (3+5 x)^{3/2}}{7 \sqrt {1-2 x}}+\frac {1}{42} \int \frac {-\frac {5543}{2}-\frac {16625 x}{4}}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=\frac {505}{84} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {11 (3+5 x)^{3/2}}{7 \sqrt {1-2 x}}-\frac {1}{63} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx-\frac {2375}{72} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=\frac {505}{84} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {11 (3+5 x)^{3/2}}{7 \sqrt {1-2 x}}-\frac {2}{63} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )-\frac {1}{36} \left (475 \sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )\\ &=\frac {505}{84} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {11 (3+5 x)^{3/2}}{7 \sqrt {1-2 x}}-\frac {475}{36} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )+\frac {2 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{63 \sqrt {7}}\\ \end {align*}

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Mathematica [C]  time = 0.07, size = 116, normalized size = 1.07 \begin {gather*} \frac {8085 \sqrt {22} \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};-\frac {5}{11} (2 x-1)\right )-924 \sqrt {5 x+3}+8 \sqrt {7-14 x} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )+490 \sqrt {20 x-10} \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )}{1764 \sqrt {1-2 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-924*Sqrt[3 + 5*x] + 490*Sqrt[-10 + 20*x]*ArcSinh[Sqrt[5/11]*Sqrt[-1 + 2*x]] + 8*Sqrt[7 - 14*x]*ArcTan[Sqrt[1
- 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])] + 8085*Sqrt[22]*Hypergeometric2F1[-3/2, -1/2, 1/2, (-5*(-1 + 2*x))/11])/(1764
*Sqrt[1 - 2*x])

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IntegrateAlgebraic [A]  time = 0.18, size = 129, normalized size = 1.19 \begin {gather*} \frac {11 \sqrt {5 x+3} \left (\frac {505 (1-2 x)}{5 x+3}+132\right )}{84 \sqrt {1-2 x} \left (\frac {5 (1-2 x)}{5 x+3}+2\right )}+\frac {475}{36} \sqrt {\frac {5}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}} \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )+\frac {2 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{63 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(11*Sqrt[3 + 5*x]*(132 + (505*(1 - 2*x))/(3 + 5*x)))/(84*Sqrt[1 - 2*x]*(2 + (5*(1 - 2*x))/(3 + 5*x))) + (475*S
qrt[5/2]*ArcTan[(Sqrt[5/2]*Sqrt[1 - 2*x])/Sqrt[3 + 5*x]])/36 + (2*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])
])/(63*Sqrt[7])

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fricas [A]  time = 1.30, size = 127, normalized size = 1.18 \begin {gather*} \frac {23275 \, \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 ) + 16 \, \sqrt {7} {\left (2 \, x - 1\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 (350 \, x - 901\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{7056 \, {\left (2 \, x - 1\right )}} \end {gather*}

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),x, algorithm="fricas")

[Out]

1/7056*(23275*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)/(1
0*x^2 + x - 3)) + 16*sqrt(7)*(2*x - 1)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 +
x - 3)) + 84*(350*x - 901)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(2*x - 1)

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giac [B]  time = 1.33, size = 180, normalized size = 1.67 \begin {gather*} -\frac {1}{4410} \, \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 {475}{144} \, \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 {{\left (70 \, \sqrt {5} {\left (5 \, x + 3\right )} - 1111 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{420 \, {\left (2 \, x - 1\right )}} \end {gather*}

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),x, algorithm="giac")

[Out]

-1/4410*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)))) - 475/144*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)))) + 1/420*(70*sqrt(5
)*(5*x + 3) - 1111*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)

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maple [A]  time = 0.02, size = 146, normalized size = 1.35 \begin {gather*} -\frac {\left (46550 \sqrt {10}\, x \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+32 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-29400 \sqrt {-10 x^{2}-x +3}\, x -23275 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-16 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+75684 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {-2 x +1}\, \sqrt {5 x +3}}{7056 \left (2 x -1\right ) \sqrt {-10 x^{2}-x +3}} \end {gather*}

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),x)

[Out]

-1/7056*(46550*10^(1/2)*x*arcsin(20/11*x+1/11)+32*7^(1/2)*x*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))
-23275*10^(1/2)*arcsin(20/11*x+1/11)-16*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-29400*(-10*
x^2-x+3)^(1/2)*x+75684*(-10*x^2-x+3)^(1/2))*(-2*x+1)^(1/2)*(5*x+3)^(1/2)/(2*x-1)/(-10*x^2-x+3)^(1/2)

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maxima [A]  time = 1.54, size = 86, normalized size = 0.80 \begin {gather*} -\frac {125 \, x^{2}}{6 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {475}{144} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {1}{441} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {3455 \, x}{84 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {901}{28 \, \sqrt {-10 \, x^{2} - x + 3}} \end {gather*}

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),x, algorithm="maxima")

[Out]

-125/6*x^2/sqrt(-10*x^2 - x + 3) - 475/144*sqrt(10)*arcsin(20/11*x + 1/11) - 1/441*sqrt(7)*arcsin(37/11*x/abs(
3*x + 2) + 20/11/abs(3*x + 2)) + 3455/84*x/sqrt(-10*x^2 - x + 3) + 901/28/sqrt(-10*x^2 - x + 3)

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

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)),x)

[Out]

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

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

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),x)

[Out]

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

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