3.26.4 \(\int \frac {1}{\sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{3/2}} \, dx\) [2504]

Optimal. Leaf size=86 \[ -\frac {515 \sqrt {1-2 x}}{77 \sqrt {3+5 x}}+\frac {3 \sqrt {1-2 x}}{7 (2+3 x) \sqrt {3+5 x}}+\frac {321 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{7 \sqrt {7}} \]

[Out]

321/49*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)-515/77*(1-2*x)^(1/2)/(3+5*x)^(1/2)+3/7*(1-2*x)^
(1/2)/(2+3*x)/(3+5*x)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {105, 157, 12, 95, 210} \begin {gather*} \frac {321 \text {ArcTan}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{7 \sqrt {7}}-\frac {515 \sqrt {1-2 x}}{77 \sqrt {5 x+3}}+\frac {3 \sqrt {1-2 x}}{7 (3 x+2) \sqrt {5 x+3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-515*Sqrt[1 - 2*x])/(77*Sqrt[3 + 5*x]) + (3*Sqrt[1 - 2*x])/(7*(2 + 3*x)*Sqrt[3 + 5*x]) + (321*ArcTan[Sqrt[1 -
2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(7*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 95

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 105

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] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

Rule 157

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 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{3/2}} \, dx &=\frac {3 \sqrt {1-2 x}}{7 (2+3 x) \sqrt {3+5 x}}+\frac {1}{7} \int \frac {\frac {67}{2}-30 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)^{3/2}} \, dx\\ &=-\frac {515 \sqrt {1-2 x}}{77 \sqrt {3+5 x}}+\frac {3 \sqrt {1-2 x}}{7 (2+3 x) \sqrt {3+5 x}}-\frac {2}{77} \int \frac {3531}{4 \sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=-\frac {515 \sqrt {1-2 x}}{77 \sqrt {3+5 x}}+\frac {3 \sqrt {1-2 x}}{7 (2+3 x) \sqrt {3+5 x}}-\frac {321}{14} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=-\frac {515 \sqrt {1-2 x}}{77 \sqrt {3+5 x}}+\frac {3 \sqrt {1-2 x}}{7 (2+3 x) \sqrt {3+5 x}}-\frac {321}{7} \text {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )\\ &=-\frac {515 \sqrt {1-2 x}}{77 \sqrt {3+5 x}}+\frac {3 \sqrt {1-2 x}}{7 (2+3 x) \sqrt {3+5 x}}+\frac {321 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{7 \sqrt {7}}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 69, normalized size = 0.80 \begin {gather*} -\frac {\sqrt {1-2 x} (997+1545 x)}{77 (2+3 x) \sqrt {3+5 x}}+\frac {321 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{7 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/77*(Sqrt[1 - 2*x]*(997 + 1545*x))/((2 + 3*x)*Sqrt[3 + 5*x]) + (321*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5
*x])])/(7*Sqrt[7])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(153\) vs. \(2(65)=130\).
time = 0.08, size = 154, normalized size = 1.79

method	result	size
default	\(-\frac {\left (52965 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x^{2}+67089 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right ) x +21186 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+21630 x \sqrt {-10 x^{2}-x +3}+13958 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{1078 \left (2+3 x \right ) \sqrt {-10 x^{2}-x +3}\, \sqrt {3+5 x}}\)	\(154\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1078*(52965*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2+67089*7^(1/2)*arctan(1/14*(37*x+
20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x+21186*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+21630*x*(-
10*x^2-x+3)^(1/2)+13958*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)/(2+3*x)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.07, size = 86, normalized size = 1.00 \begin {gather*} \frac {3531 \, \sqrt {7} {\left (15 \, x^{2} + 19 \, x + 6\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 (1545 \, x + 997\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{1078 \, {\left (15 \, x^{2} + 19 \, x + 6\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1078*(3531*sqrt(7)*(15*x^2 + 19*x + 6)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2
+ x - 3)) - 14*(1545*x + 997)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(15*x^2 + 19*x + 6)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 252 vs. \(2 (65) = 130\).
time = 0.60, size = 252, normalized size = 2.93 \begin {gather*} -\frac {321}{980} \, \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 {5}{22} \, \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 {198 \, \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 )}}{7 \, {\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-321/980*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)))) - 5/22*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))) - 198/7*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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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