3.2486 \(\int \frac {(3+5 x)^{5/2}}{\sqrt {1-2 x} (2+3 x)^5} \, dx\)

Optimal. Leaf size=151 \[ \frac {3 \sqrt {1-2 x} (5 x+3)^{7/2}}{28 (3 x+2)^4}-\frac {\sqrt {1-2 x} (5 x+3)^{5/2}}{24 (3 x+2)^3}-\frac {55 \sqrt {1-2 x} (5 x+3)^{3/2}}{672 (3 x+2)^2}-\frac {605 \sqrt {1-2 x} \sqrt {5 x+3}}{3136 (3 x+2)}-\frac {6655 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{3136 \sqrt {7}} \]

[Out]

-6655/21952*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)-55/672*(3+5*x)^(3/2)*(1-2*x)^(1/2)/(2+3*x)
^2-1/24*(3+5*x)^(5/2)*(1-2*x)^(1/2)/(2+3*x)^3+3/28*(3+5*x)^(7/2)*(1-2*x)^(1/2)/(2+3*x)^4-605/3136*(1-2*x)^(1/2
)*(3+5*x)^(1/2)/(2+3*x)

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Rubi [A]  time = 0.04, antiderivative size = 151, 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} \[ \frac {3 \sqrt {1-2 x} (5 x+3)^{7/2}}{28 (3 x+2)^4}-\frac {\sqrt {1-2 x} (5 x+3)^{5/2}}{24 (3 x+2)^3}-\frac {55 \sqrt {1-2 x} (5 x+3)^{3/2}}{672 (3 x+2)^2}-\frac {605 \sqrt {1-2 x} \sqrt {5 x+3}}{3136 (3 x+2)}-\frac {6655 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{3136 \sqrt {7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-605*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(3136*(2 + 3*x)) - (55*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(672*(2 + 3*x)^2) - (
Sqrt[1 - 2*x]*(3 + 5*x)^(5/2))/(24*(2 + 3*x)^3) + (3*Sqrt[1 - 2*x]*(3 + 5*x)^(7/2))/(28*(2 + 3*x)^4) - (6655*A
rcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(3136*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 {(3+5 x)^{5/2}}{\sqrt {1-2 x} (2+3 x)^5} \, dx &=\frac {3 \sqrt {1-2 x} (3+5 x)^{7/2}}{28 (2+3 x)^4}+\frac {7}{8} \int \frac {(3+5 x)^{5/2}}{\sqrt {1-2 x} (2+3 x)^4} \, dx\\ &=-\frac {\sqrt {1-2 x} (3+5 x)^{5/2}}{24 (2+3 x)^3}+\frac {3 \sqrt {1-2 x} (3+5 x)^{7/2}}{28 (2+3 x)^4}+\frac {55}{48} \int \frac {(3+5 x)^{3/2}}{\sqrt {1-2 x} (2+3 x)^3} \, dx\\ &=-\frac {55 \sqrt {1-2 x} (3+5 x)^{3/2}}{672 (2+3 x)^2}-\frac {\sqrt {1-2 x} (3+5 x)^{5/2}}{24 (2+3 x)^3}+\frac {3 \sqrt {1-2 x} (3+5 x)^{7/2}}{28 (2+3 x)^4}+\frac {605}{448} \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x} (2+3 x)^2} \, dx\\ &=-\frac {605 \sqrt {1-2 x} \sqrt {3+5 x}}{3136 (2+3 x)}-\frac {55 \sqrt {1-2 x} (3+5 x)^{3/2}}{672 (2+3 x)^2}-\frac {\sqrt {1-2 x} (3+5 x)^{5/2}}{24 (2+3 x)^3}+\frac {3 \sqrt {1-2 x} (3+5 x)^{7/2}}{28 (2+3 x)^4}+\frac {6655 \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx}{6272}\\ &=-\frac {605 \sqrt {1-2 x} \sqrt {3+5 x}}{3136 (2+3 x)}-\frac {55 \sqrt {1-2 x} (3+5 x)^{3/2}}{672 (2+3 x)^2}-\frac {\sqrt {1-2 x} (3+5 x)^{5/2}}{24 (2+3 x)^3}+\frac {3 \sqrt {1-2 x} (3+5 x)^{7/2}}{28 (2+3 x)^4}+\frac {6655 \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )}{3136}\\ &=-\frac {605 \sqrt {1-2 x} \sqrt {3+5 x}}{3136 (2+3 x)}-\frac {55 \sqrt {1-2 x} (3+5 x)^{3/2}}{672 (2+3 x)^2}-\frac {\sqrt {1-2 x} (3+5 x)^{5/2}}{24 (2+3 x)^3}+\frac {3 \sqrt {1-2 x} (3+5 x)^{7/2}}{28 (2+3 x)^4}-\frac {6655 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{3136 \sqrt {7}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 79, normalized size = 0.52 \[ \frac {\frac {7 \sqrt {1-2 x} \sqrt {5 x+3} \left (12945 x^3+6920 x^2-6484 x-3600\right )}{(3 x+2)^4}-19965 \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{65856} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((7*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(-3600 - 6484*x + 6920*x^2 + 12945*x^3))/(2 + 3*x)^4 - 19965*Sqrt[7]*ArcTan[Sq
rt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/65856

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fricas [A]  time = 1.13, size = 116, normalized size = 0.77 \[ -\frac {19965 \, \sqrt {7} {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (12945 \, x^{3} + 6920 \, x^{2} - 6484 \, x - 3600\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{131712 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/131712*(19965*sqrt(7)*(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3
)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 14*(12945*x^3 + 6920*x^2 - 6484*x - 3600)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(
81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)

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giac [B]  time = 2.75, size = 368, normalized size = 2.44 \[ \frac {1331}{87808} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {6655 \, \sqrt {10} {\left (3 \, {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{7} + 3080 \, {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{5} + 1144640 \, {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} + \frac {2956800 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} - \frac {11827200 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{4704 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1331/87808*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)))) - 6655/4704*sqrt(10)*(3*((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)))^7 + 3080*((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 + 1144640*((sqrt(2)*s
qrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^3 + 2956800*(
sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 11827200*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)^4

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maple [B]  time = 0.02, size = 250, normalized size = 1.66 \[ \frac {\sqrt {5 x +3}\, \sqrt {-2 x +1}\, \left (1617165 \sqrt {7}\, x^{4} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+4312440 \sqrt {7}\, x^{3} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+181230 \sqrt {-10 x^{2}-x +3}\, x^{3}+4312440 \sqrt {7}\, x^{2} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+96880 \sqrt {-10 x^{2}-x +3}\, x^{2}+1916640 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-90776 \sqrt {-10 x^{2}-x +3}\, x +319440 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-50400 \sqrt {-10 x^{2}-x +3}\right )}{131712 \sqrt {-10 x^{2}-x +3}\, \left (3 x +2\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/131712*(5*x+3)^(1/2)*(-2*x+1)^(1/2)*(1617165*7^(1/2)*x^4*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+
4312440*7^(1/2)*x^3*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+4312440*7^(1/2)*x^2*arctan(1/14*(37*x+2
0)*7^(1/2)/(-10*x^2-x+3)^(1/2))+181230*(-10*x^2-x+3)^(1/2)*x^3+1916640*7^(1/2)*x*arctan(1/14*(37*x+20)*7^(1/2)
/(-10*x^2-x+3)^(1/2))+96880*(-10*x^2-x+3)^(1/2)*x^2+319440*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)
^(1/2))-90776*(-10*x^2-x+3)^(1/2)*x-50400*(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)/(3*x+2)^4

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maxima [A]  time = 1.27, size = 143, normalized size = 0.95 \[ \frac {6655}{43904} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) - \frac {\sqrt {-10 \, x^{2} - x + 3}}{252 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} + \frac {83 \, \sqrt {-10 \, x^{2} - x + 3}}{1512 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} - \frac {1355 \, \sqrt {-10 \, x^{2} - x + 3}}{6048 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} + \frac {4315 \, \sqrt {-10 \, x^{2} - x + 3}}{84672 \, {\left (3 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6655/43904*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) - 1/252*sqrt(-10*x^2 - x + 3)/(81*x^4 + 2
16*x^3 + 216*x^2 + 96*x + 16) + 83/1512*sqrt(-10*x^2 - x + 3)/(27*x^3 + 54*x^2 + 36*x + 8) - 1355/6048*sqrt(-1
0*x^2 - x + 3)/(9*x^2 + 12*x + 4) + 4315/84672*sqrt(-10*x^2 - x + 3)/(3*x + 2)

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

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)^5),x)

[Out]

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

[Out]

Timed out

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