3.2475 \(\int \frac {(3+5 x)^{3/2}}{\sqrt {1-2 x} (2+3 x)^4} \, dx\)

Optimal. Leaf size=122 \[ \frac {\sqrt {1-2 x} (5 x+3)^{5/2}}{7 (3 x+2)^3}-\frac {15 \sqrt {1-2 x} (5 x+3)^{3/2}}{196 (3 x+2)^2}-\frac {495 \sqrt {1-2 x} \sqrt {5 x+3}}{2744 (3 x+2)}-\frac {5445 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{2744 \sqrt {7}} \]

[Out]

-5445/19208*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)-15/196*(3+5*x)^(3/2)*(1-2*x)^(1/2)/(2+3*x)
^2+1/7*(3+5*x)^(5/2)*(1-2*x)^(1/2)/(2+3*x)^3-495/2744*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x)

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Rubi [A]  time = 0.03, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {96, 94, 93, 204} \[ \frac {\sqrt {1-2 x} (5 x+3)^{5/2}}{7 (3 x+2)^3}-\frac {15 \sqrt {1-2 x} (5 x+3)^{3/2}}{196 (3 x+2)^2}-\frac {495 \sqrt {1-2 x} \sqrt {5 x+3}}{2744 (3 x+2)}-\frac {5445 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{2744 \sqrt {7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-495*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(2744*(2 + 3*x)) - (15*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(196*(2 + 3*x)^2) + (
Sqrt[1 - 2*x]*(3 + 5*x)^(5/2))/(7*(2 + 3*x)^3) - (5445*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(2744*Sq
rt[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 {(3+5 x)^{3/2}}{\sqrt {1-2 x} (2+3 x)^4} \, dx &=\frac {\sqrt {1-2 x} (3+5 x)^{5/2}}{7 (2+3 x)^3}+\frac {15}{14} \int \frac {(3+5 x)^{3/2}}{\sqrt {1-2 x} (2+3 x)^3} \, dx\\ &=-\frac {15 \sqrt {1-2 x} (3+5 x)^{3/2}}{196 (2+3 x)^2}+\frac {\sqrt {1-2 x} (3+5 x)^{5/2}}{7 (2+3 x)^3}+\frac {495}{392} \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x} (2+3 x)^2} \, dx\\ &=-\frac {495 \sqrt {1-2 x} \sqrt {3+5 x}}{2744 (2+3 x)}-\frac {15 \sqrt {1-2 x} (3+5 x)^{3/2}}{196 (2+3 x)^2}+\frac {\sqrt {1-2 x} (3+5 x)^{5/2}}{7 (2+3 x)^3}+\frac {5445 \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx}{5488}\\ &=-\frac {495 \sqrt {1-2 x} \sqrt {3+5 x}}{2744 (2+3 x)}-\frac {15 \sqrt {1-2 x} (3+5 x)^{3/2}}{196 (2+3 x)^2}+\frac {\sqrt {1-2 x} (3+5 x)^{5/2}}{7 (2+3 x)^3}+\frac {5445 \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )}{2744}\\ &=-\frac {495 \sqrt {1-2 x} \sqrt {3+5 x}}{2744 (2+3 x)}-\frac {15 \sqrt {1-2 x} (3+5 x)^{3/2}}{196 (2+3 x)^2}+\frac {\sqrt {1-2 x} (3+5 x)^{5/2}}{7 (2+3 x)^3}-\frac {5445 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{2744 \sqrt {7}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 74, normalized size = 0.61 \[ \frac {\frac {7 \sqrt {1-2 x} \sqrt {5 x+3} \left (2195 x^2+1830 x+288\right )}{(3 x+2)^3}-5445 \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{19208} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((7*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(288 + 1830*x + 2195*x^2))/(2 + 3*x)^3 - 5445*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sq
rt[7]*Sqrt[3 + 5*x])])/19208

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fricas [A]  time = 0.89, size = 101, normalized size = 0.83 \[ -\frac {5445 \, \sqrt {7} {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (2195 \, x^{2} + 1830 \, x + 288\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{38416 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/38416*(5445*sqrt(7)*(27*x^3 + 54*x^2 + 36*x + 8)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x +
1)/(10*x^2 + x - 3)) - 14*(2195*x^2 + 1830*x + 288)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(27*x^3 + 54*x^2 + 36*x + 8)

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giac [B]  time = 2.07, size = 310, normalized size = 2.54 \[ \frac {1089}{76832} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {605 \, \sqrt {10} {\left (9 \, {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{5} + 6720 \, {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} - \frac {203840 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} + \frac {815360 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{1372 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1089/76832*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)))) - 605/1372*sqrt(10)*(9*((sqrt(2)*sqrt(-10*x + 5) - s
qrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^5 + 6720*((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 - 203840*(sqrt(2)*sqrt
(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 815360*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 = 202, normalized size = 1.66 \[ \frac {\sqrt {5 x +3}\, \sqrt {-2 x +1}\, \left (147015 \sqrt {7}\, x^{3} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+294030 \sqrt {7}\, x^{2} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+30730 \sqrt {-10 x^{2}-x +3}\, x^{2}+196020 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+25620 \sqrt {-10 x^{2}-x +3}\, x +43560 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+4032 \sqrt {-10 x^{2}-x +3}\right )}{38416 \sqrt {-10 x^{2}-x +3}\, \left (3 x +2\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/38416*(5*x+3)^(1/2)*(-2*x+1)^(1/2)*(147015*7^(1/2)*x^3*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+29
4030*7^(1/2)*x^2*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+196020*7^(1/2)*x*arctan(1/14*(37*x+20)*7^(
1/2)/(-10*x^2-x+3)^(1/2))+30730*(-10*x^2-x+3)^(1/2)*x^2+43560*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x
+3)^(1/2))+25620*(-10*x^2-x+3)^(1/2)*x+4032*(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)/(3*x+2)^3

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maxima [A]  time = 1.20, size = 107, normalized size = 0.88 \[ \frac {5445}{38416} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {\sqrt {-10 \, x^{2} - x + 3}}{63 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} - \frac {235 \, \sqrt {-10 \, x^{2} - x + 3}}{1764 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} + \frac {2195 \, \sqrt {-10 \, x^{2} - x + 3}}{24696 \, {\left (3 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5445/38416*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 1/63*sqrt(-10*x^2 - x + 3)/(27*x^3 + 54
*x^2 + 36*x + 8) - 235/1764*sqrt(-10*x^2 - x + 3)/(9*x^2 + 12*x + 4) + 2195/24696*sqrt(-10*x^2 - x + 3)/(3*x +
2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (5\,x+3\right )}^{3/2}}{\sqrt {1-2\,x}\,{\left (3\,x+2\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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