3.2575 \(\int \frac{2+3 x}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx\)

Optimal. Leaf size=67 \[ -\frac{428 \sqrt{1-2 x}}{3993 \sqrt{5 x+3}}-\frac{107 \sqrt{1-2 x}}{363 (5 x+3)^{3/2}}+\frac{7}{11 (5 x+3)^{3/2} \sqrt{1-2 x}} \]

[Out]

7/(11*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) - (107*Sqrt[1 - 2*x])/(363*(3 + 5*x)^(3/2)) - (428*Sqrt[1 - 2*x])/(3993*S
qrt[3 + 5*x])

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Rubi [A]  time = 0.0098582, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {78, 45, 37} \[ -\frac{428 \sqrt{1-2 x}}{3993 \sqrt{5 x+3}}-\frac{107 \sqrt{1-2 x}}{363 (5 x+3)^{3/2}}+\frac{7}{11 (5 x+3)^{3/2} \sqrt{1-2 x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

7/(11*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) - (107*Sqrt[1 - 2*x])/(363*(3 + 5*x)^(3/2)) - (428*Sqrt[1 - 2*x])/(3993*S
qrt[3 + 5*x])

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{2+3 x}{(1-2 x)^{3/2} (3+5 x)^{5/2}} \, dx &=\frac{7}{11 \sqrt{1-2 x} (3+5 x)^{3/2}}+\frac{107}{22} \int \frac{1}{\sqrt{1-2 x} (3+5 x)^{5/2}} \, dx\\ &=\frac{7}{11 \sqrt{1-2 x} (3+5 x)^{3/2}}-\frac{107 \sqrt{1-2 x}}{363 (3+5 x)^{3/2}}+\frac{214}{363} \int \frac{1}{\sqrt{1-2 x} (3+5 x)^{3/2}} \, dx\\ &=\frac{7}{11 \sqrt{1-2 x} (3+5 x)^{3/2}}-\frac{107 \sqrt{1-2 x}}{363 (3+5 x)^{3/2}}-\frac{428 \sqrt{1-2 x}}{3993 \sqrt{3+5 x}}\\ \end{align*}

Mathematica [A]  time = 0.0087768, size = 32, normalized size = 0.48 \[ \frac{2 \left (2140 x^2+1391 x+40\right )}{3993 \sqrt{1-2 x} (5 x+3)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(40 + 1391*x + 2140*x^2))/(3993*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))

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Maple [A]  time = 0.003, size = 27, normalized size = 0.4 \begin{align*}{\frac{4280\,{x}^{2}+2782\,x+80}{3993} \left ( 3+5\,x \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{1-2\,x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3993*(2140*x^2+1391*x+40)/(3+5*x)^(3/2)/(1-2*x)^(1/2)

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Maxima [A]  time = 2.24404, size = 86, normalized size = 1.28 \begin{align*} \frac{856 \, x}{3993 \, \sqrt{-10 \, x^{2} - x + 3}} + \frac{214}{19965 \, \sqrt{-10 \, x^{2} - x + 3}} - \frac{2}{165 \,{\left (5 \, \sqrt{-10 \, x^{2} - x + 3} x + 3 \, \sqrt{-10 \, x^{2} - x + 3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

856/3993*x/sqrt(-10*x^2 - x + 3) + 214/19965/sqrt(-10*x^2 - x + 3) - 2/165/(5*sqrt(-10*x^2 - x + 3)*x + 3*sqrt
(-10*x^2 - x + 3))

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Fricas [A]  time = 1.45509, size = 124, normalized size = 1.85 \begin{align*} -\frac{2 \,{\left (2140 \, x^{2} + 1391 \, x + 40\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{3993 \,{\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3993*(2140*x^2 + 1391*x + 40)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(50*x^3 + 35*x^2 - 12*x - 9)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 2.62905, size = 205, normalized size = 3.06 \begin{align*} -\frac{\sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{3}}{319440 \,{\left (5 \, x + 3\right )}^{\frac{3}{2}}} - \frac{73 \, \sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}}{26620 \, \sqrt{5 \, x + 3}} - \frac{28 \, \sqrt{5} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5}}{6655 \,{\left (2 \, x - 1\right )}} + \frac{{\left (\frac{219 \, \sqrt{10}{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{2}}{5 \, x + 3} + 4 \, \sqrt{10}\right )}{\left (5 \, x + 3\right )}^{\frac{3}{2}}}{19965 \,{\left (\sqrt{2} \sqrt{-10 \, x + 5} - \sqrt{22}\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/319440*sqrt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) - 73/26620*sqrt(10)*(sqrt(2)*sqrt(-1
0*x + 5) - sqrt(22))/sqrt(5*x + 3) - 28/6655*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1) + 1/19965*(219*sq
rt(10)*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) + 4*sqrt(10))*(5*x + 3)^(3/2)/(sqrt(2)*sqrt(-10*x + 5)
- sqrt(22))^3