3.2605 \(\int \frac {(3+5 x)^{5/2}}{(1-2 x)^{5/2} (2+3 x)^3} \, dx\)

Optimal. Leaf size=151 \[ \frac {4 (5 x+3)^{7/2}}{231 (1-2 x)^{3/2} (3 x+2)^2}+\frac {26 (5 x+3)^{5/2}}{231 \sqrt {1-2 x} (3 x+2)^2}+\frac {65 \sqrt {1-2 x} (5 x+3)^{3/2}}{3234 (3 x+2)^2}+\frac {65 \sqrt {1-2 x} \sqrt {5 x+3}}{1372 (3 x+2)}+\frac {715 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{1372 \sqrt {7}} \]

[Out]

4/231*(3+5*x)^(7/2)/(1-2*x)^(3/2)/(2+3*x)^2+715/9604*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)+2
6/231*(3+5*x)^(5/2)/(2+3*x)^2/(1-2*x)^(1/2)+65/3234*(3+5*x)^(3/2)*(1-2*x)^(1/2)/(2+3*x)^2+65/1372*(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} \[ \frac {4 (5 x+3)^{7/2}}{231 (1-2 x)^{3/2} (3 x+2)^2}+\frac {26 (5 x+3)^{5/2}}{231 \sqrt {1-2 x} (3 x+2)^2}+\frac {65 \sqrt {1-2 x} (5 x+3)^{3/2}}{3234 (3 x+2)^2}+\frac {65 \sqrt {1-2 x} \sqrt {5 x+3}}{1372 (3 x+2)}+\frac {715 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{1372 \sqrt {7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(65*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(1372*(2 + 3*x)) + (65*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/(3234*(2 + 3*x)^2) + (2
6*(3 + 5*x)^(5/2))/(231*Sqrt[1 - 2*x]*(2 + 3*x)^2) + (4*(3 + 5*x)^(7/2))/(231*(1 - 2*x)^(3/2)*(2 + 3*x)^2) + (
715*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(1372*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}}{(1-2 x)^{5/2} (2+3 x)^3} \, dx &=\frac {4 (3+5 x)^{7/2}}{231 (1-2 x)^{3/2} (2+3 x)^2}+\frac {13}{33} \int \frac {(3+5 x)^{5/2}}{(1-2 x)^{3/2} (2+3 x)^3} \, dx\\ &=\frac {26 (3+5 x)^{5/2}}{231 \sqrt {1-2 x} (2+3 x)^2}+\frac {4 (3+5 x)^{7/2}}{231 (1-2 x)^{3/2} (2+3 x)^2}-\frac {65}{231} \int \frac {(3+5 x)^{3/2}}{\sqrt {1-2 x} (2+3 x)^3} \, dx\\ &=\frac {65 \sqrt {1-2 x} (3+5 x)^{3/2}}{3234 (2+3 x)^2}+\frac {26 (3+5 x)^{5/2}}{231 \sqrt {1-2 x} (2+3 x)^2}+\frac {4 (3+5 x)^{7/2}}{231 (1-2 x)^{3/2} (2+3 x)^2}-\frac {65}{196} \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x} (2+3 x)^2} \, dx\\ &=\frac {65 \sqrt {1-2 x} \sqrt {3+5 x}}{1372 (2+3 x)}+\frac {65 \sqrt {1-2 x} (3+5 x)^{3/2}}{3234 (2+3 x)^2}+\frac {26 (3+5 x)^{5/2}}{231 \sqrt {1-2 x} (2+3 x)^2}+\frac {4 (3+5 x)^{7/2}}{231 (1-2 x)^{3/2} (2+3 x)^2}-\frac {715 \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx}{2744}\\ &=\frac {65 \sqrt {1-2 x} \sqrt {3+5 x}}{1372 (2+3 x)}+\frac {65 \sqrt {1-2 x} (3+5 x)^{3/2}}{3234 (2+3 x)^2}+\frac {26 (3+5 x)^{5/2}}{231 \sqrt {1-2 x} (2+3 x)^2}+\frac {4 (3+5 x)^{7/2}}{231 (1-2 x)^{3/2} (2+3 x)^2}-\frac {715 \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )}{1372}\\ &=\frac {65 \sqrt {1-2 x} \sqrt {3+5 x}}{1372 (2+3 x)}+\frac {65 \sqrt {1-2 x} (3+5 x)^{3/2}}{3234 (2+3 x)^2}+\frac {26 (3+5 x)^{5/2}}{231 \sqrt {1-2 x} (2+3 x)^2}+\frac {4 (3+5 x)^{7/2}}{231 (1-2 x)^{3/2} (2+3 x)^2}+\frac {715 \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 = 0.63 \[ -\frac {7 \sqrt {5 x+3} \left (10260 x^3+1620 x^2-13627 x-6732\right )+2145 \sqrt {7-14 x} (2 x-1) (3 x+2)^2 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{28812 (1-2 x)^{3/2} (3 x+2)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/28812*(7*Sqrt[3 + 5*x]*(-6732 - 13627*x + 1620*x^2 + 10260*x^3) + 2145*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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fricas [A]  time = 1.42, size = 116, normalized size = 0.77 \[ \frac {2145 \, \sqrt {7} {\left (36 \, x^{4} + 12 \, x^{3} - 23 \, x^{2} - 4 \, x + 4\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (10260 \, x^{3} + 1620 \, x^{2} - 13627 \, x - 6732\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{57624 \, {\left (36 \, x^{4} + 12 \, x^{3} - 23 \, x^{2} - 4 \, x + 4\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/57624*(2145*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*(10260*x^3 + 1620*x^2 - 13627*x - 6732)*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 = 2.68, size = 291, normalized size = 1.93 \[ -\frac {143}{38416} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {22 \, {\left (104 \, \sqrt {5} {\left (5 \, x + 3\right )} - 957 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{180075 \, {\left (2 \, x - 1\right )}^{2}} + \frac {11 \, \sqrt {10} {\left (223 \, {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} + \frac {80920 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} - \frac {323680 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{4802 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-143/38416*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)))) - 22/180075*(104*sqrt(5)*(5*x + 3) - 957*sqrt(5))*sq
rt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2 + 11/4802*sqrt(10)*(223*((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 + 80920*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/s
qrt(5*x + 3) - 323680*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)^2

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maple [B]  time = 0.02, size = 257, normalized size = 1.70 \[ -\frac {\left (77220 \sqrt {7}\, x^{4} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+25740 \sqrt {7}\, x^{3} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+143640 \sqrt {-10 x^{2}-x +3}\, x^{3}-49335 \sqrt {7}\, x^{2} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+22680 \sqrt {-10 x^{2}-x +3}\, x^{2}-8580 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-190778 \sqrt {-10 x^{2}-x +3}\, x +8580 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-94248 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {-2 x +1}\, \sqrt {5 x +3}}{57624 \left (3 x +2\right )^{2} \left (2 x -1\right )^{2} \sqrt {-10 x^{2}-x +3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/57624*(77220*7^(1/2)*x^4*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+25740*7^(1/2)*x^3*arctan(1/14*(
37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-49335*7^(1/2)*x^2*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+143
640*(-10*x^2-x+3)^(1/2)*x^3-8580*7^(1/2)*x*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+22680*(-10*x^2-x
+3)^(1/2)*x^2+8580*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-190778*(-10*x^2-x+3)^(1/2)*x-942
48*(-10*x^2-x+3)^(1/2))*(-2*x+1)^(1/2)*(5*x+3)^(1/2)/(3*x+2)^2/(2*x-1)^2/(-10*x^2-x+3)^(1/2)

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maxima [A]  time = 1.28, size = 172, normalized size = 1.14 \[ -\frac {715}{19208} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {475 \, x}{686 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {215}{4116 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {17375 \, x}{2646 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {1}{1134 \, {\left (9 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x^{2} + 12 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x + 4 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}\right )}} + \frac {1}{36 \, {\left (3 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x + 2 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}\right )}} + \frac {60695}{15876 \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-715/19208*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 475/686*x/sqrt(-10*x^2 - x + 3) - 215/4
116/sqrt(-10*x^2 - x + 3) + 17375/2646*x/(-10*x^2 - x + 3)^(3/2) - 1/1134/(9*(-10*x^2 - x + 3)^(3/2)*x^2 + 12*
(-10*x^2 - x + 3)^(3/2)*x + 4*(-10*x^2 - x + 3)^(3/2)) + 1/36/(3*(-10*x^2 - x + 3)^(3/2)*x + 2*(-10*x^2 - x +
3)^(3/2)) + 60695/15876/(-10*x^2 - x + 3)^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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