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

Optimal. Leaf size=115 \[ \frac {\sqrt {1-2 x} (5 x+3)^{3/2}}{21 (3 x+2)}-\frac {185}{126} \sqrt {1-2 x} \sqrt {5 x+3}+\frac {125}{54} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {173 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{189 \sqrt {7}} \]

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Rubi [A]  time = 0.04, antiderivative size = 115, 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 {\sqrt {1-2 x} (5 x+3)^{3/2}}{21 (3 x+2)}-\frac {185}{126} \sqrt {1-2 x} \sqrt {5 x+3}+\frac {125}{54} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {173 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{189 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-185*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/126 + (Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(21*(2 + 3*x)) + (125*Sqrt[5/2]*ArcSi
n[Sqrt[2/11]*Sqrt[3 + 5*x]])/54 - (173*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(189*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}}{\sqrt {1-2 x} (2+3 x)^2} \, dx &=\frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{21 (2+3 x)}-\frac {1}{21} \int \frac {\left (-\frac {189}{2}-185 x\right ) \sqrt {3+5 x}}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=-\frac {185}{126} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{21 (2+3 x)}+\frac {1}{126} \int \frac {1516+\frac {4375 x}{2}}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=-\frac {185}{126} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{21 (2+3 x)}+\frac {173}{378} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx+\frac {625}{108} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=-\frac {185}{126} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{21 (2+3 x)}+\frac {173}{189} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )+\frac {1}{54} \left (125 \sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )\\ &=-\frac {185}{126} \sqrt {1-2 x} \sqrt {3+5 x}+\frac {\sqrt {1-2 x} (3+5 x)^{3/2}}{21 (2+3 x)}+\frac {125}{54} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )-\frac {173 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{189 \sqrt {7}}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 125, normalized size = 1.09 \begin {gather*} \frac {-42 \sqrt {-(1-2 x)^2} \sqrt {5 x+3} (525 x+352)-692 (3 x+2) \sqrt {14 x-7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )-6125 \sqrt {10-20 x} (3 x+2) \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )}{5292 \sqrt {2 x-1} (3 x+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-42*Sqrt[-(1 - 2*x)^2]*Sqrt[3 + 5*x]*(352 + 525*x) - 6125*Sqrt[10 - 20*x]*(2 + 3*x)*ArcSinh[Sqrt[5/11]*Sqrt[-
1 + 2*x]] - 692*(2 + 3*x)*Sqrt[-7 + 14*x]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(5292*Sqrt[-1 + 2*x]*
(2 + 3*x))

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IntegrateAlgebraic [A]  time = 0.20, size = 146, normalized size = 1.27 \begin {gather*} -\frac {11 \sqrt {1-2 x} \left (\frac {185 (1-2 x)}{5 x+3}+1229\right )}{126 \sqrt {5 x+3} \left (\frac {1-2 x}{5 x+3}+7\right ) \left (\frac {5 (1-2 x)}{5 x+3}+2\right )}-\frac {125}{54} \sqrt {\frac {5}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}} \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )-\frac {173 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{189 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*Sqrt[1 - 2*x]*(1229 + (185*(1 - 2*x))/(3 + 5*x)))/(126*Sqrt[3 + 5*x]*(7 + (1 - 2*x)/(3 + 5*x))*(2 + (5*(1
- 2*x))/(3 + 5*x))) - (125*Sqrt[5/2]*ArcTan[(Sqrt[5/2]*Sqrt[1 - 2*x])/Sqrt[3 + 5*x]])/54 - (173*ArcTan[Sqrt[1
- 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(189*Sqrt[7])

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fricas [A]  time = 1.16, size = 127, normalized size = 1.10 \begin {gather*} -\frac {6125 \, \sqrt {5} \sqrt {2} {\left (3 \, 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 ) + 692 \, \sqrt {7} {\left (3 \, 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 (525 \, x + 352\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{10584 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10584*(6125*sqrt(5)*sqrt(2)*(3*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)) + 692*sqrt(7)*(3*x + 2)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2
+ x - 3)) + 84*(525*x + 352)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(3*x + 2)

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giac [B]  time = 1.60, size = 279, normalized size = 2.43 \begin {gather*} \frac {173}{26460} \, \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 {125}{216} \, \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 {5}{18} \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - \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 )}}{63 \, {\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

173/26460*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)))) + 125/216*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)))) - 5/18*sqrt(5)*s
qrt(5*x + 3)*sqrt(-10*x + 5) - 22/63*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 [A]  time = 0.02, size = 146, normalized size = 1.27 \begin {gather*} \frac {\sqrt {5 x +3}\, \sqrt {-2 x +1}\, \left (18375 \sqrt {10}\, x \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+2076 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-44100 \sqrt {-10 x^{2}-x +3}\, x +12250 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+1384 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-29568 \sqrt {-10 x^{2}-x +3}\right )}{10584 \sqrt {-10 x^{2}-x +3}\, \left (3 x +2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10584*(5*x+3)^(1/2)*(-2*x+1)^(1/2)*(18375*10^(1/2)*x*arcsin(20/11*x+1/11)+2076*7^(1/2)*x*arctan(1/14*(37*x+2
0)*7^(1/2)/(-10*x^2-x+3)^(1/2))+12250*10^(1/2)*arcsin(20/11*x+1/11)+1384*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)
/(-10*x^2-x+3)^(1/2))-44100*(-10*x^2-x+3)^(1/2)*x-29568*(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)/(3*x+2)

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maxima [A]  time = 1.28, size = 75, normalized size = 0.65 \begin {gather*} \frac {125}{216} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {173}{2646} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) - \frac {25}{18} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {\sqrt {-10 \, x^{2} - x + 3}}{63 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125/216*sqrt(10)*arcsin(20/11*x + 1/11) + 173/2646*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) -
25/18*sqrt(-10*x^2 - x + 3) - 1/63*sqrt(-10*x^2 - x + 3)/(3*x + 2)

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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}}{\sqrt {1-2\,x}\,{\left (3\,x+2\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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