3.2482 \(\int \frac {(3+5 x)^{5/2}}{\sqrt {1-2 x} (2+3 x)} \, dx\)

Optimal. Leaf size=108 \[ -\frac {5}{12} \sqrt {1-2 x} (5 x+3)^{3/2}-\frac {455}{144} \sqrt {1-2 x} \sqrt {5 x+3}+\frac {3035}{432} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )+\frac {2 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{27 \sqrt {7}} \]

[Out]

3035/864*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+2/189*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^
(1/2)-5/12*(3+5*x)^(3/2)*(1-2*x)^(1/2)-455/144*(1-2*x)^(1/2)*(3+5*x)^(1/2)

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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 = {102, 154, 157, 54, 216, 93, 204} \[ -\frac {5}{12} \sqrt {1-2 x} (5 x+3)^{3/2}-\frac {455}{144} \sqrt {1-2 x} \sqrt {5 x+3}+\frac {3035}{432} \sqrt {\frac {5}{2}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )+\frac {2 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{27 \sqrt {7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-455*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/144 - (5*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/12 + (3035*Sqrt[5/2]*ArcSin[Sqrt[2/
11]*Sqrt[3 + 5*x]])/432 + (2*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(27*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 102

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

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)} \, dx &=-\frac {5}{12} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {1}{12} \int \frac {\left (-153-\frac {455 x}{2}\right ) \sqrt {3+5 x}}{\sqrt {1-2 x} (2+3 x)} \, dx\\ &=-\frac {455}{144} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {5}{12} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {1}{72} \int \frac {\frac {5053}{2}+\frac {15175 x}{4}}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=-\frac {455}{144} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {5}{12} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {1}{27} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx+\frac {15175}{864} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=-\frac {455}{144} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {5}{12} \sqrt {1-2 x} (3+5 x)^{3/2}-\frac {2}{27} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )+\frac {1}{432} \left (3035 \sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )\\ &=-\frac {455}{144} \sqrt {1-2 x} \sqrt {3+5 x}-\frac {5}{12} \sqrt {1-2 x} (3+5 x)^{3/2}+\frac {3035}{432} \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 )}{27 \sqrt {7}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 108, normalized size = 1.00 \[ \frac {-210 \sqrt {-(1-2 x)^2} \sqrt {5 x+3} (60 x+127)+64 \sqrt {14 x-7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )-21245 \sqrt {10-20 x} \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )}{6048 \sqrt {2 x-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-210*Sqrt[-(1 - 2*x)^2]*Sqrt[3 + 5*x]*(127 + 60*x) - 21245*Sqrt[10 - 20*x]*ArcSinh[Sqrt[5/11]*Sqrt[-1 + 2*x]]
+ 64*Sqrt[-7 + 14*x]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(6048*Sqrt[-1 + 2*x])

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fricas [A]  time = 0.87, size = 108, normalized size = 1.00 \[ -\frac {5}{144} \, {\left (60 \, x + 127\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} - \frac {3035}{1728} \, \sqrt {5} \sqrt {2} \arctan \left (\frac {\sqrt {5} \sqrt {2} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + \frac {1}{189} \, \sqrt {7} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/144*(60*x + 127)*sqrt(5*x + 3)*sqrt(-2*x + 1) - 3035/1728*sqrt(5)*sqrt(2)*arctan(1/20*sqrt(5)*sqrt(2)*(20*x
+ 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) + 1/189*sqrt(7)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x
+ 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3))

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giac [B]  time = 1.57, size = 173, normalized size = 1.60 \[ -\frac {1}{1890} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {1}{144} \, {\left (12 \, \sqrt {5} {\left (5 \, x + 3\right )} + 91 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + \frac {3035}{1728} \, \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1890*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)))) - 1/144*(12*sqrt(5)*(5*x + 3) + 91*sqrt(5))*sqrt(5*x +
3)*sqrt(-10*x + 5) + 3035/1728*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))))

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maple [A]  time = 0.01, size = 98, normalized size = 0.91 \[ \frac {\sqrt {5 x +3}\, \sqrt {-2 x +1}\, \left (-25200 \sqrt {-10 x^{2}-x +3}\, x +21245 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-64 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-53340 \sqrt {-10 x^{2}-x +3}\right )}{12096 \sqrt {-10 x^{2}-x +3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12096*(5*x+3)^(1/2)*(-2*x+1)^(1/2)*(21245*10^(1/2)*arcsin(20/11*x+1/11)-64*7^(1/2)*arctan(1/14*(37*x+20)*7^(
1/2)/(-10*x^2-x+3)^(1/2))-25200*(-10*x^2-x+3)^(1/2)*x-53340*(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)

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maxima [A]  time = 1.12, size = 69, normalized size = 0.64 \[ -\frac {25}{12} \, \sqrt {-10 \, x^{2} - x + 3} x + \frac {3035}{1728} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {1}{189} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) - \frac {635}{144} \, \sqrt {-10 \, x^{2} - x + 3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/12*sqrt(-10*x^2 - x + 3)*x + 3035/1728*sqrt(10)*arcsin(20/11*x + 1/11) - 1/189*sqrt(7)*arcsin(37/11*x/abs(
3*x + 2) + 20/11/abs(3*x + 2)) - 635/144*sqrt(-10*x^2 - x + 3)

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

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)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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