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

Optimal. Leaf size=55 \[ \frac {6 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{\sqrt {7}}-\frac {10 \sqrt {1-2 x}}{11 \sqrt {5 x+3}} \]

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Rubi [A]  time = 0.01, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {96, 93, 204} \begin {gather*} \frac {6 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{\sqrt {7}}-\frac {10 \sqrt {1-2 x}}{11 \sqrt {5 x+3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*Sqrt[1 - 2*x])/(11*Sqrt[3 + 5*x]) + (6*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/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 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}{\sqrt {1-2 x} (2+3 x) (3+5 x)^{3/2}} \, dx &=-\frac {10 \sqrt {1-2 x}}{11 \sqrt {3+5 x}}-3 \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=-\frac {10 \sqrt {1-2 x}}{11 \sqrt {3+5 x}}-6 \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )\\ &=-\frac {10 \sqrt {1-2 x}}{11 \sqrt {3+5 x}}+\frac {6 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{\sqrt {7}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 55, normalized size = 1.00 \begin {gather*} \frac {6 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{\sqrt {7}}-\frac {10 \sqrt {1-2 x}}{11 \sqrt {5 x+3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*Sqrt[1 - 2*x])/(11*Sqrt[3 + 5*x]) + (6*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/Sqrt[7]

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IntegrateAlgebraic [A]  time = 0.08, size = 55, normalized size = 1.00 \begin {gather*} \frac {6 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{\sqrt {7}}-\frac {10 \sqrt {1-2 x}}{11 \sqrt {5 x+3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*Sqrt[1 - 2*x])/(11*Sqrt[3 + 5*x]) + (6*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/Sqrt[7]

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fricas [A]  time = 1.22, size = 71, normalized size = 1.29 \begin {gather*} \frac {33 \, \sqrt {7} {\left (5 \, x + 3\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 ) - 70 \, \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{77 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/77*(33*sqrt(7)*(5*x + 3)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 70
*sqrt(5*x + 3)*sqrt(-2*x + 1))/(5*x + 3)

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giac [B]  time = 1.11, size = 133, normalized size = 2.42 \begin {gather*} -\frac {3}{70} \, \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 {1}{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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/70*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)))) - 1/22*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqr
t(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))

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maple [B]  time = 0.02, size = 101, normalized size = 1.84 \begin {gather*} -\frac {\left (165 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+99 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+70 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {-2 x +1}}{77 \sqrt {-10 x^{2}-x +3}\, \sqrt {5 x +3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/77*(165*7^(1/2)*x*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+99*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/
2)/(-10*x^2-x+3)^(1/2))+70*(-10*x^2-x+3)^(1/2))*(-2*x+1)^(1/2)/(-10*x^2-x+3)^(1/2)/(5*x+3)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{\sqrt {1-2\,x}\,\left (3\,x+2\right )\,{\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)*(5*x + 3)^(3/2)),x)

[Out]

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

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

[Out]

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

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