3.2615 \(\int \frac {1}{(1-2 x)^{5/2} (2+3 x)^2 \sqrt {3+5 x}} \, dx\)

Optimal. Leaf size=108 \[ -\frac {4390 \sqrt {5 x+3}}{124509 \sqrt {1-2 x}}+\frac {3 \sqrt {5 x+3}}{7 (1-2 x)^{3/2} (3 x+2)}-\frac {190 \sqrt {5 x+3}}{1617 (1-2 x)^{3/2}}-\frac {405 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{343 \sqrt {7}} \]

[Out]

-405/2401*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)-190/1617*(3+5*x)^(1/2)/(1-2*x)^(3/2)+3/7*(3+
5*x)^(1/2)/(1-2*x)^(3/2)/(2+3*x)-4390/124509*(3+5*x)^(1/2)/(1-2*x)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {103, 152, 12, 93, 204} \[ -\frac {4390 \sqrt {5 x+3}}{124509 \sqrt {1-2 x}}+\frac {3 \sqrt {5 x+3}}{7 (1-2 x)^{3/2} (3 x+2)}-\frac {190 \sqrt {5 x+3}}{1617 (1-2 x)^{3/2}}-\frac {405 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{343 \sqrt {7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-190*Sqrt[3 + 5*x])/(1617*(1 - 2*x)^(3/2)) - (4390*Sqrt[3 + 5*x])/(124509*Sqrt[1 - 2*x]) + (3*Sqrt[3 + 5*x])/
(7*(1 - 2*x)^(3/2)*(2 + 3*x)) - (405*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(343*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 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)^2 \sqrt {3+5 x}} \, dx &=\frac {3 \sqrt {3+5 x}}{7 (1-2 x)^{3/2} (2+3 x)}+\frac {1}{7} \int \frac {-\frac {35}{2}-60 x}{(1-2 x)^{5/2} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=-\frac {190 \sqrt {3+5 x}}{1617 (1-2 x)^{3/2}}+\frac {3 \sqrt {3+5 x}}{7 (1-2 x)^{3/2} (2+3 x)}-\frac {2 \int \frac {-\frac {655}{4}+1425 x}{(1-2 x)^{3/2} (2+3 x) \sqrt {3+5 x}} \, dx}{1617}\\ &=-\frac {190 \sqrt {3+5 x}}{1617 (1-2 x)^{3/2}}-\frac {4390 \sqrt {3+5 x}}{124509 \sqrt {1-2 x}}+\frac {3 \sqrt {3+5 x}}{7 (1-2 x)^{3/2} (2+3 x)}+\frac {4 \int \frac {147015}{8 \sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx}{124509}\\ &=-\frac {190 \sqrt {3+5 x}}{1617 (1-2 x)^{3/2}}-\frac {4390 \sqrt {3+5 x}}{124509 \sqrt {1-2 x}}+\frac {3 \sqrt {3+5 x}}{7 (1-2 x)^{3/2} (2+3 x)}+\frac {405}{686} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=-\frac {190 \sqrt {3+5 x}}{1617 (1-2 x)^{3/2}}-\frac {4390 \sqrt {3+5 x}}{124509 \sqrt {1-2 x}}+\frac {3 \sqrt {3+5 x}}{7 (1-2 x)^{3/2} (2+3 x)}+\frac {405}{343} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )\\ &=-\frac {190 \sqrt {3+5 x}}{1617 (1-2 x)^{3/2}}-\frac {4390 \sqrt {3+5 x}}{124509 \sqrt {1-2 x}}+\frac {3 \sqrt {3+5 x}}{7 (1-2 x)^{3/2} (2+3 x)}-\frac {405 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{343 \sqrt {7}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 86, normalized size = 0.80 \[ -\frac {-7 \sqrt {5 x+3} \left (26340 x^2-39500 x+15321\right )-147015 \sqrt {7-14 x} \left (6 x^2+x-2\right ) \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{871563 (1-2 x)^{3/2} (3 x+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/871563*(-7*Sqrt[3 + 5*x]*(15321 - 39500*x + 26340*x^2) - 147015*Sqrt[7 - 14*x]*(-2 + x + 6*x^2)*ArcTan[Sqrt
[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/((1 - 2*x)^(3/2)*(2 + 3*x))

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fricas [A]  time = 0.84, size = 101, normalized size = 0.94 \[ -\frac {147015 \, \sqrt {7} {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (26340 \, x^{2} - 39500 \, x + 15321\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{1743126 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1743126*(147015*sqrt(7)*(12*x^3 - 4*x^2 - 5*x + 2)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x
+ 1)/(10*x^2 + x - 3)) - 14*(26340*x^2 - 39500*x + 15321)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(12*x^3 - 4*x^2 - 5*x
+ 2)

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giac [B]  time = 1.47, size = 232, normalized size = 2.15 \[ \frac {81}{9604} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {594 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{343 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}} - \frac {8 \, {\left (536 \, \sqrt {5} {\left (5 \, x + 3\right )} - 3333 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{3112725 \, {\left (2 \, x - 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

81/9604*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)))) + 594/343*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) - 8/3112725*(536*sqrt(5)*(5*x
+ 3) - 3333*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2

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maple [B]  time = 0.02, size = 209, normalized size = 1.94 \[ \frac {\left (1764180 \sqrt {7}\, x^{3} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-588060 \sqrt {7}\, x^{2} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+368760 \sqrt {-10 x^{2}-x +3}\, x^{2}-735075 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-553000 \sqrt {-10 x^{2}-x +3}\, x +294030 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+214494 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {5 x +3}\, \sqrt {-2 x +1}}{1743126 \left (3 x +2\right ) \left (2 x -1\right )^{2} \sqrt {-10 x^{2}-x +3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1743126*(1764180*7^(1/2)*x^3*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-588060*7^(1/2)*x^2*arctan(1/
14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-735075*7^(1/2)*x*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+
368760*(-10*x^2-x+3)^(1/2)*x^2+294030*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-553000*(-10*x
^2-x+3)^(1/2)*x+214494*(-10*x^2-x+3)^(1/2))*(5*x+3)^(1/2)*(-2*x+1)^(1/2)/(3*x+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 \[ \int \frac {1}{\sqrt {5 \, x + 3} {\left (3 \, x + 2\right )}^{2} {\left (-2 \, x + 1\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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