3.24.5 \(\int \frac {(1-2 x)^{5/2}}{(2+3 x)^4 (3+5 x)^{3/2}} \, dx\)

Optimal. Leaf size=144 \[ \frac {(1-2 x)^{7/2}}{7 (3 x+2)^3 \sqrt {5 x+3}}+\frac {81 (1-2 x)^{5/2}}{28 (3 x+2)^2 \sqrt {5 x+3}}+\frac {4455 (1-2 x)^{3/2}}{56 (3 x+2) \sqrt {5 x+3}}-\frac {147015 \sqrt {1-2 x}}{56 \sqrt {5 x+3}}+\frac {147015 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{8 \sqrt {7}} \]

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Rubi [A]  time = 0.04, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {96, 94, 93, 204} \begin {gather*} \frac {(1-2 x)^{7/2}}{7 (3 x+2)^3 \sqrt {5 x+3}}+\frac {81 (1-2 x)^{5/2}}{28 (3 x+2)^2 \sqrt {5 x+3}}+\frac {4455 (1-2 x)^{3/2}}{56 (3 x+2) \sqrt {5 x+3}}-\frac {147015 \sqrt {1-2 x}}{56 \sqrt {5 x+3}}+\frac {147015 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{8 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - 2*x)^(5/2)/((2 + 3*x)^4*(3 + 5*x)^(3/2)),x]

[Out]

(-147015*Sqrt[1 - 2*x])/(56*Sqrt[3 + 5*x]) + (1 - 2*x)^(7/2)/(7*(2 + 3*x)^3*Sqrt[3 + 5*x]) + (81*(1 - 2*x)^(5/
2))/(28*(2 + 3*x)^2*Sqrt[3 + 5*x]) + (4455*(1 - 2*x)^(3/2))/(56*(2 + 3*x)*Sqrt[3 + 5*x]) + (147015*ArcTan[Sqrt
[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(8*Sqrt[7])

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 94

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

Rule 96

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[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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 {(1-2 x)^{5/2}}{(2+3 x)^4 (3+5 x)^{3/2}} \, dx &=\frac {(1-2 x)^{7/2}}{7 (2+3 x)^3 \sqrt {3+5 x}}+\frac {81}{14} \int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 (3+5 x)^{3/2}} \, dx\\ &=\frac {(1-2 x)^{7/2}}{7 (2+3 x)^3 \sqrt {3+5 x}}+\frac {81 (1-2 x)^{5/2}}{28 (2+3 x)^2 \sqrt {3+5 x}}+\frac {4455}{56} \int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)^{3/2}} \, dx\\ &=\frac {(1-2 x)^{7/2}}{7 (2+3 x)^3 \sqrt {3+5 x}}+\frac {81 (1-2 x)^{5/2}}{28 (2+3 x)^2 \sqrt {3+5 x}}+\frac {4455 (1-2 x)^{3/2}}{56 (2+3 x) \sqrt {3+5 x}}+\frac {147015}{112} \int \frac {\sqrt {1-2 x}}{(2+3 x) (3+5 x)^{3/2}} \, dx\\ &=-\frac {147015 \sqrt {1-2 x}}{56 \sqrt {3+5 x}}+\frac {(1-2 x)^{7/2}}{7 (2+3 x)^3 \sqrt {3+5 x}}+\frac {81 (1-2 x)^{5/2}}{28 (2+3 x)^2 \sqrt {3+5 x}}+\frac {4455 (1-2 x)^{3/2}}{56 (2+3 x) \sqrt {3+5 x}}-\frac {147015}{16} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=-\frac {147015 \sqrt {1-2 x}}{56 \sqrt {3+5 x}}+\frac {(1-2 x)^{7/2}}{7 (2+3 x)^3 \sqrt {3+5 x}}+\frac {81 (1-2 x)^{5/2}}{28 (2+3 x)^2 \sqrt {3+5 x}}+\frac {4455 (1-2 x)^{3/2}}{56 (2+3 x) \sqrt {3+5 x}}-\frac {147015}{8} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )\\ &=-\frac {147015 \sqrt {1-2 x}}{56 \sqrt {3+5 x}}+\frac {(1-2 x)^{7/2}}{7 (2+3 x)^3 \sqrt {3+5 x}}+\frac {81 (1-2 x)^{5/2}}{28 (2+3 x)^2 \sqrt {3+5 x}}+\frac {4455 (1-2 x)^{3/2}}{56 (2+3 x) \sqrt {3+5 x}}+\frac {147015 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{8 \sqrt {7}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 79, normalized size = 0.55 \begin {gather*} \frac {147015 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{8 \sqrt {7}}-\frac {\sqrt {1-2 x} \left (578245 x^3+1143741 x^2+753654 x+165424\right )}{8 (3 x+2)^3 \sqrt {5 x+3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - 2*x)^(5/2)/((2 + 3*x)^4*(3 + 5*x)^(3/2)),x]

[Out]

-1/8*(Sqrt[1 - 2*x]*(165424 + 753654*x + 1143741*x^2 + 578245*x^3))/((2 + 3*x)^3*Sqrt[3 + 5*x]) + (147015*ArcT
an[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(8*Sqrt[7])

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IntegrateAlgebraic [A]  time = 0.27, size = 122, normalized size = 0.85 \begin {gather*} \frac {147015 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{8 \sqrt {7}}-\frac {121 \sqrt {1-2 x} \left (\frac {80 (1-2 x)^3}{(5 x+3)^3}+\frac {2673 (1-2 x)^2}{(5 x+3)^2}+\frac {22680 (1-2 x)}{5 x+3}+59535\right )}{8 \sqrt {5 x+3} \left (\frac {1-2 x}{5 x+3}+7\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 - 2*x)^(5/2)/((2 + 3*x)^4*(3 + 5*x)^(3/2)),x]

[Out]

(-121*Sqrt[1 - 2*x]*(59535 + (80*(1 - 2*x)^3)/(3 + 5*x)^3 + (2673*(1 - 2*x)^2)/(3 + 5*x)^2 + (22680*(1 - 2*x))
/(3 + 5*x)))/(8*Sqrt[3 + 5*x]*(7 + (1 - 2*x)/(3 + 5*x))^3) + (147015*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*
x])])/(8*Sqrt[7])

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fricas [A]  time = 1.29, size = 116, normalized size = 0.81 \begin {gather*} \frac {147015 \, \sqrt {7} {\left (135 \, x^{4} + 351 \, x^{3} + 342 \, x^{2} + 148 \, x + 24\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 (578245 \, x^{3} + 1143741 \, x^{2} + 753654 \, x + 165424\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{112 \, {\left (135 \, x^{4} + 351 \, x^{3} + 342 \, x^{2} + 148 \, x + 24\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(5/2)/(2+3*x)^4/(3+5*x)^(3/2),x, algorithm="fricas")

[Out]

1/112*(147015*sqrt(7)*(135*x^4 + 351*x^3 + 342*x^2 + 148*x + 24)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)
*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 14*(578245*x^3 + 1143741*x^2 + 753654*x + 165424)*sqrt(5*x + 3)*sqrt(-2*x
+ 1))/(135*x^4 + 351*x^3 + 342*x^2 + 148*x + 24)

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giac [B]  time = 3.27, size = 377, normalized size = 2.62 \begin {gather*} -\frac {29403}{224} \, \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 {121}{2} \, \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 )} - \frac {121 \, {\left (993 \, \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 )}^{5} + 436800 \, \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 )}^{3} + 51352000 \, \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 )}\right )}}{4 \, {\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 )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(5/2)/(2+3*x)^4/(3+5*x)^(3/2),x, algorithm="giac")

[Out]

-29403/224*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)))) - 121/2*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))) - 121/4*(993*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)))^5 + 436800*sqrt(1
0)*((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 + 51352000*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)^3

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maple [B]  time = 0.02, size = 250, normalized size = 1.74 \begin {gather*} -\frac {\left (19847025 \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 )+8095430 \sqrt {-10 x^{2}-x +3}\, x^{3}[phone] \sqrt {7}\, x^{2} \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^{2}[phone] \sqrt {7}\, x \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 +3528360 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+2315936 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {-2 x +1}}{112 \left (3 x +2\right )^{3} \sqrt {-10 x^{2}-x +3}\, \sqrt {5 x +3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x+1)^(5/2)/(3*x+2)^4/(5*x+3)^(3/2),x)

[Out]

-1/112*(19847025*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))[phone]*7^(1/2)*x^2*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/
2))+8095430*(-10*x^2-x+3)^(1/2)*x^3+21758220*7^(1/2)*x*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+1601
2374*(-10*x^2-x+3)^(1/2)*x^2+3528360*7^(1/2)*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+2315936*(-10*x^2-x+3)^(1/2))*(-2*x+1)^(1/2)/(3*x+2)^3/(-10*x^2-x+3)^(1/2)/(5*x+3)^(1/2)

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maxima [A]  time = 1.25, size = 211, normalized size = 1.47 \begin {gather*} -\frac {147015}{112} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {578245 \, x}{108 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {603743}{216 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {343}{81 \, {\left (27 \, \sqrt {-10 \, x^{2} - x + 3} x^{3} + 54 \, \sqrt {-10 \, x^{2} - x + 3} x^{2} + 36 \, \sqrt {-10 \, x^{2} - x + 3} x + 8 \, \sqrt {-10 \, x^{2} - x + 3}\right )}} + \frac {10339}{324 \, {\left (9 \, \sqrt {-10 \, x^{2} - x + 3} x^{2} + 12 \, \sqrt {-10 \, x^{2} - x + 3} x + 4 \, \sqrt {-10 \, x^{2} - x + 3}\right )}} + \frac {87199}{216 \, {\left (3 \, \sqrt {-10 \, x^{2} - x + 3} x + 2 \, \sqrt {-10 \, x^{2} - x + 3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(5/2)/(2+3*x)^4/(3+5*x)^(3/2),x, algorithm="maxima")

[Out]

-147015/112*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 578245/108*x/sqrt(-10*x^2 - x + 3) - 6
03743/216/sqrt(-10*x^2 - x + 3) + 343/81/(27*sqrt(-10*x^2 - x + 3)*x^3 + 54*sqrt(-10*x^2 - x + 3)*x^2 + 36*sqr
t(-10*x^2 - x + 3)*x + 8*sqrt(-10*x^2 - x + 3)) + 10339/324/(9*sqrt(-10*x^2 - x + 3)*x^2 + 12*sqrt(-10*x^2 - x
+ 3)*x + 4*sqrt(-10*x^2 - x + 3)) + 87199/216/(3*sqrt(-10*x^2 - x + 3)*x + 2*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 (1-2\,x\right )}^{5/2}}{{\left (3\,x+2\right )}^4\,{\left (5\,x+3\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - 2*x)^(5/2)/((3*x + 2)^4*(5*x + 3)^(3/2)),x)

[Out]

int((1 - 2*x)^(5/2)/((3*x + 2)^4*(5*x + 3)^(3/2)), 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((1-2*x)**(5/2)/(2+3*x)**4/(3+5*x)**(3/2),x)

[Out]

Timed out

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