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

Optimal. Leaf size=137 \[ -\frac {57595 \sqrt {5 x+3}}{249018 \sqrt {1-2 x}}+\frac {51 \sqrt {5 x+3}}{28 (1-2 x)^{3/2} (3 x+2)}-\frac {1735 \sqrt {5 x+3}}{3234 (1-2 x)^{3/2}}+\frac {3 \sqrt {5 x+3}}{14 (1-2 x)^{3/2} (3 x+2)^2}-\frac {5805 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{1372 \sqrt {7}} \]

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Rubi [A]  time = 0.04, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {103, 151, 152, 12, 93, 204} \begin {gather*} -\frac {57595 \sqrt {5 x+3}}{249018 \sqrt {1-2 x}}+\frac {51 \sqrt {5 x+3}}{28 (1-2 x)^{3/2} (3 x+2)}-\frac {1735 \sqrt {5 x+3}}{3234 (1-2 x)^{3/2}}+\frac {3 \sqrt {5 x+3}}{14 (1-2 x)^{3/2} (3 x+2)^2}-\frac {5805 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{1372 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-1735*Sqrt[3 + 5*x])/(3234*(1 - 2*x)^(3/2)) - (57595*Sqrt[3 + 5*x])/(249018*Sqrt[1 - 2*x]) + (3*Sqrt[3 + 5*x]
)/(14*(1 - 2*x)^(3/2)*(2 + 3*x)^2) + (51*Sqrt[3 + 5*x])/(28*(1 - 2*x)^(3/2)*(2 + 3*x)) - (5805*ArcTan[Sqrt[1 -
2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(1372*Sqrt[7])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 103

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))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 151

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 + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 152

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 + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

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])

Rubi steps

\begin {align*} \int \frac {1}{(1-2 x)^{5/2} (2+3 x)^3 \sqrt {3+5 x}} \, dx &=\frac {3 \sqrt {3+5 x}}{14 (1-2 x)^{3/2} (2+3 x)^2}+\frac {1}{14} \int \frac {-\frac {1}{2}-90 x}{(1-2 x)^{5/2} (2+3 x)^2 \sqrt {3+5 x}} \, dx\\ &=\frac {3 \sqrt {3+5 x}}{14 (1-2 x)^{3/2} (2+3 x)^2}+\frac {51 \sqrt {3+5 x}}{28 (1-2 x)^{3/2} (2+3 x)}+\frac {1}{98} \int \frac {-\frac {5005}{4}-3570 x}{(1-2 x)^{5/2} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=-\frac {1735 \sqrt {3+5 x}}{3234 (1-2 x)^{3/2}}+\frac {3 \sqrt {3+5 x}}{14 (1-2 x)^{3/2} (2+3 x)^2}+\frac {51 \sqrt {3+5 x}}{28 (1-2 x)^{3/2} (2+3 x)}-\frac {\int \frac {\frac {38815}{8}+\frac {182175 x}{2}}{(1-2 x)^{3/2} (2+3 x) \sqrt {3+5 x}} \, dx}{11319}\\ &=-\frac {1735 \sqrt {3+5 x}}{3234 (1-2 x)^{3/2}}-\frac {57595 \sqrt {3+5 x}}{249018 \sqrt {1-2 x}}+\frac {3 \sqrt {3+5 x}}{14 (1-2 x)^{3/2} (2+3 x)^2}+\frac {51 \sqrt {3+5 x}}{28 (1-2 x)^{3/2} (2+3 x)}+\frac {2 \int \frac {14750505}{16 \sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx}{871563}\\ &=-\frac {1735 \sqrt {3+5 x}}{3234 (1-2 x)^{3/2}}-\frac {57595 \sqrt {3+5 x}}{249018 \sqrt {1-2 x}}+\frac {3 \sqrt {3+5 x}}{14 (1-2 x)^{3/2} (2+3 x)^2}+\frac {51 \sqrt {3+5 x}}{28 (1-2 x)^{3/2} (2+3 x)}+\frac {5805 \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx}{2744}\\ &=-\frac {1735 \sqrt {3+5 x}}{3234 (1-2 x)^{3/2}}-\frac {57595 \sqrt {3+5 x}}{249018 \sqrt {1-2 x}}+\frac {3 \sqrt {3+5 x}}{14 (1-2 x)^{3/2} (2+3 x)^2}+\frac {51 \sqrt {3+5 x}}{28 (1-2 x)^{3/2} (2+3 x)}+\frac {5805 \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )}{1372}\\ &=-\frac {1735 \sqrt {3+5 x}}{3234 (1-2 x)^{3/2}}-\frac {57595 \sqrt {3+5 x}}{249018 \sqrt {1-2 x}}+\frac {3 \sqrt {3+5 x}}{14 (1-2 x)^{3/2} (2+3 x)^2}+\frac {51 \sqrt {3+5 x}}{28 (1-2 x)^{3/2} (2+3 x)}-\frac {5805 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{1372 \sqrt {7}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 95, normalized size = 0.69 \begin {gather*} -\frac {-7 \sqrt {5 x+3} \left (2073420 x^3-676860 x^2-945629 x+391476\right )-2107215 \sqrt {7-14 x} (2 x-1) (3 x+2)^2 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{3486252 (1-2 x)^{3/2} (3 x+2)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3486252*(-7*Sqrt[3 + 5*x]*(391476 - 945629*x - 676860*x^2 + 2073420*x^3) - 2107215*Sqrt[7 - 14*x]*(-1 + 2*x
)*(2 + 3*x)^2*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/((1 - 2*x)^(3/2)*(2 + 3*x)^2)

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IntegrateAlgebraic [A]  time = 0.22, size = 122, normalized size = 0.89 \begin {gather*} \frac {(5 x+3)^{3/2} \left (\frac {3037785 (1-2 x)^3}{(5 x+3)^3}+\frac {14174825 (1-2 x)^2}{(5 x+3)^2}+\frac {181888 (1-2 x)}{5 x+3}+6272\right )}{498036 (1-2 x)^{3/2} \left (\frac {1-2 x}{5 x+3}+7\right )^2}-\frac {5805 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{1372 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3 + 5*x)^(3/2)*(6272 + (3037785*(1 - 2*x)^3)/(3 + 5*x)^3 + (14174825*(1 - 2*x)^2)/(3 + 5*x)^2 + (181888*(1 -
2*x))/(3 + 5*x)))/(498036*(1 - 2*x)^(3/2)*(7 + (1 - 2*x)/(3 + 5*x))^2) - (5805*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*
Sqrt[3 + 5*x])])/(1372*Sqrt[7])

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fricas [A]  time = 1.18, size = 116, normalized size = 0.85 \begin {gather*} -\frac {2107215 \, \sqrt {7} {\left (36 \, x^{4} + 12 \, x^{3} - 23 \, x^{2} - 4 \, x + 4\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 ) - 14 \, {\left (2073420 \, x^{3} - 676860 \, x^{2} - 945629 \, x + 391476\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{6972504 \, {\left (36 \, x^{4} + 12 \, x^{3} - 23 \, x^{2} - 4 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6972504*(2107215*sqrt(7)*(36*x^4 + 12*x^3 - 23*x^2 - 4*x + 4)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)
*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 14*(2073420*x^3 - 676860*x^2 - 945629*x + 391476)*sqrt(5*x + 3)*sqrt(-2*x
+ 1))/(36*x^4 + 12*x^3 - 23*x^2 - 4*x + 4)

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giac [B]  time = 2.58, size = 291, normalized size = 2.12 \begin {gather*} \frac {1161}{38416} \, \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 {32 \, {\left (367 \, \sqrt {5} {\left (5 \, x + 3\right )} - 2211 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{21789075 \, {\left (2 \, x - 1\right )}^{2}} + \frac {297 \, \sqrt {10} {\left (197 \, {\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 )}^{3} + \frac {36680 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} - \frac {146720 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{4802 \, {\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 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1161/38416*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)))) - 32/21789075*(367*sqrt(5)*(5*x + 3) - 2211*sqrt(5))
*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2 + 297/4802*sqrt(10)*(197*((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)))^3 + 36680*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22
))/sqrt(5*x + 3) - 146720*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))/(((sqrt(2)*sqrt(-10*x + 5) - sqr
t(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^2 + 280)^2

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maple [B]  time = 0.02, size = 257, normalized size = 1.88 \begin {gather*} \frac {\left (75859740 \sqrt {7}\, x^{4} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )[phone] \sqrt {7}\, x^{3} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )[phone] \sqrt {-10 x^{2}-x +3}\, x^{3}-48465945 \sqrt {7}\, x^{2} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-9476040 \sqrt {-10 x^{2}-x +3}\, x^{2}-8428860 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-13238806 \sqrt {-10 x^{2}-x +3}\, x +8428860 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+5480664 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {5 x +3}\, \sqrt {-2 x +1}}{6972504 \left (3 x +2\right )^{2} \left (2 x -1\right )^{2} \sqrt {-10 x^{2}-x +3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6972504*(75859740*7^(1/2)*x^4*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))[phone]*7^(1/2)*x^3*arctan
(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-48465945*7^(1/2)*x^2*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^
(1/2))[phone]*(-10*x^2-x+3)^(1/2)*x^3-8428860*7^(1/2)*x*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-9
476040*(-10*x^2-x+3)^(1/2)*x^2+8428860*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-13238806*(-1
0*x^2-x+3)^(1/2)*x+5480664*(-10*x^2-x+3)^(1/2))*(5*x+3)^(1/2)*(-2*x+1)^(1/2)/(3*x+2)^2/(2*x-1)^2/(-10*x^2-x+3)
^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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