3.583 \(\int \frac{1}{x^{7/2} (a+b x)^{3/2}} \, dx\)

Optimal. Leaf size=87 \[ -\frac{32 b^2 \sqrt{a+b x}}{5 a^4 \sqrt{x}}+\frac{16 b \sqrt{a+b x}}{5 a^3 x^{3/2}}-\frac{12 \sqrt{a+b x}}{5 a^2 x^{5/2}}+\frac{2}{a x^{5/2} \sqrt{a+b x}} \]

[Out]

2/(a*x^(5/2)*Sqrt[a + b*x]) - (12*Sqrt[a + b*x])/(5*a^2*x^(5/2)) + (16*b*Sqrt[a + b*x])/(5*a^3*x^(3/2)) - (32*
b^2*Sqrt[a + b*x])/(5*a^4*Sqrt[x])

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Rubi [A]  time = 0.016875, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {45, 37} \[ -\frac{32 b^2 \sqrt{a+b x}}{5 a^4 \sqrt{x}}+\frac{16 b \sqrt{a+b x}}{5 a^3 x^{3/2}}-\frac{12 \sqrt{a+b x}}{5 a^2 x^{5/2}}+\frac{2}{a x^{5/2} \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(7/2)*(a + b*x)^(3/2)),x]

[Out]

2/(a*x^(5/2)*Sqrt[a + b*x]) - (12*Sqrt[a + b*x])/(5*a^2*x^(5/2)) + (16*b*Sqrt[a + b*x])/(5*a^3*x^(3/2)) - (32*
b^2*Sqrt[a + b*x])/(5*a^4*Sqrt[x])

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{1}{x^{7/2} (a+b x)^{3/2}} \, dx &=\frac{2}{a x^{5/2} \sqrt{a+b x}}+\frac{6 \int \frac{1}{x^{7/2} \sqrt{a+b x}} \, dx}{a}\\ &=\frac{2}{a x^{5/2} \sqrt{a+b x}}-\frac{12 \sqrt{a+b x}}{5 a^2 x^{5/2}}-\frac{(24 b) \int \frac{1}{x^{5/2} \sqrt{a+b x}} \, dx}{5 a^2}\\ &=\frac{2}{a x^{5/2} \sqrt{a+b x}}-\frac{12 \sqrt{a+b x}}{5 a^2 x^{5/2}}+\frac{16 b \sqrt{a+b x}}{5 a^3 x^{3/2}}+\frac{\left (16 b^2\right ) \int \frac{1}{x^{3/2} \sqrt{a+b x}} \, dx}{5 a^3}\\ &=\frac{2}{a x^{5/2} \sqrt{a+b x}}-\frac{12 \sqrt{a+b x}}{5 a^2 x^{5/2}}+\frac{16 b \sqrt{a+b x}}{5 a^3 x^{3/2}}-\frac{32 b^2 \sqrt{a+b x}}{5 a^4 \sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.010603, size = 49, normalized size = 0.56 \[ -\frac{2 \left (-2 a^2 b x+a^3+8 a b^2 x^2+16 b^3 x^3\right )}{5 a^4 x^{5/2} \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(7/2)*(a + b*x)^(3/2)),x]

[Out]

(-2*(a^3 - 2*a^2*b*x + 8*a*b^2*x^2 + 16*b^3*x^3))/(5*a^4*x^(5/2)*Sqrt[a + b*x])

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Maple [A]  time = 0.006, size = 44, normalized size = 0.5 \begin{align*} -{\frac{32\,{b}^{3}{x}^{3}+16\,a{b}^{2}{x}^{2}-4\,{a}^{2}bx+2\,{a}^{3}}{5\,{a}^{4}}{x}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{bx+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(7/2)/(b*x+a)^(3/2),x)

[Out]

-2/5*(16*b^3*x^3+8*a*b^2*x^2-2*a^2*b*x+a^3)/x^(5/2)/(b*x+a)^(1/2)/a^4

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Maxima [A]  time = 1.03385, size = 86, normalized size = 0.99 \begin{align*} -\frac{2 \, b^{3} \sqrt{x}}{\sqrt{b x + a} a^{4}} - \frac{2 \,{\left (\frac{15 \, \sqrt{b x + a} b^{2}}{\sqrt{x}} - \frac{5 \,{\left (b x + a\right )}^{\frac{3}{2}} b}{x^{\frac{3}{2}}} + \frac{{\left (b x + a\right )}^{\frac{5}{2}}}{x^{\frac{5}{2}}}\right )}}{5 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

-2*b^3*sqrt(x)/(sqrt(b*x + a)*a^4) - 2/5*(15*sqrt(b*x + a)*b^2/sqrt(x) - 5*(b*x + a)^(3/2)*b/x^(3/2) + (b*x +
a)^(5/2)/x^(5/2))/a^4

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Fricas [A]  time = 2.04874, size = 128, normalized size = 1.47 \begin{align*} -\frac{2 \,{\left (16 \, b^{3} x^{3} + 8 \, a b^{2} x^{2} - 2 \, a^{2} b x + a^{3}\right )} \sqrt{b x + a} \sqrt{x}}{5 \,{\left (a^{4} b x^{4} + a^{5} x^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

-2/5*(16*b^3*x^3 + 8*a*b^2*x^2 - 2*a^2*b*x + a^3)*sqrt(b*x + a)*sqrt(x)/(a^4*b*x^4 + a^5*x^3)

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Sympy [B]  time = 106.382, size = 348, normalized size = 4. \begin{align*} - \frac{2 a^{5} b^{\frac{19}{2}} \sqrt{\frac{a}{b x} + 1}}{5 a^{7} b^{9} x^{2} + 15 a^{6} b^{10} x^{3} + 15 a^{5} b^{11} x^{4} + 5 a^{4} b^{12} x^{5}} - \frac{10 a^{3} b^{\frac{23}{2}} x^{2} \sqrt{\frac{a}{b x} + 1}}{5 a^{7} b^{9} x^{2} + 15 a^{6} b^{10} x^{3} + 15 a^{5} b^{11} x^{4} + 5 a^{4} b^{12} x^{5}} - \frac{60 a^{2} b^{\frac{25}{2}} x^{3} \sqrt{\frac{a}{b x} + 1}}{5 a^{7} b^{9} x^{2} + 15 a^{6} b^{10} x^{3} + 15 a^{5} b^{11} x^{4} + 5 a^{4} b^{12} x^{5}} - \frac{80 a b^{\frac{27}{2}} x^{4} \sqrt{\frac{a}{b x} + 1}}{5 a^{7} b^{9} x^{2} + 15 a^{6} b^{10} x^{3} + 15 a^{5} b^{11} x^{4} + 5 a^{4} b^{12} x^{5}} - \frac{32 b^{\frac{29}{2}} x^{5} \sqrt{\frac{a}{b x} + 1}}{5 a^{7} b^{9} x^{2} + 15 a^{6} b^{10} x^{3} + 15 a^{5} b^{11} x^{4} + 5 a^{4} b^{12} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(7/2)/(b*x+a)**(3/2),x)

[Out]

-2*a**5*b**(19/2)*sqrt(a/(b*x) + 1)/(5*a**7*b**9*x**2 + 15*a**6*b**10*x**3 + 15*a**5*b**11*x**4 + 5*a**4*b**12
*x**5) - 10*a**3*b**(23/2)*x**2*sqrt(a/(b*x) + 1)/(5*a**7*b**9*x**2 + 15*a**6*b**10*x**3 + 15*a**5*b**11*x**4
+ 5*a**4*b**12*x**5) - 60*a**2*b**(25/2)*x**3*sqrt(a/(b*x) + 1)/(5*a**7*b**9*x**2 + 15*a**6*b**10*x**3 + 15*a*
*5*b**11*x**4 + 5*a**4*b**12*x**5) - 80*a*b**(27/2)*x**4*sqrt(a/(b*x) + 1)/(5*a**7*b**9*x**2 + 15*a**6*b**10*x
**3 + 15*a**5*b**11*x**4 + 5*a**4*b**12*x**5) - 32*b**(29/2)*x**5*sqrt(a/(b*x) + 1)/(5*a**7*b**9*x**2 + 15*a**
6*b**10*x**3 + 15*a**5*b**11*x**4 + 5*a**4*b**12*x**5)

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Giac [A]  time = 1.10179, size = 147, normalized size = 1.69 \begin{align*} -\frac{4 \, b^{\frac{9}{2}}}{{\left ({\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )} a^{3}{\left | b \right |}} + \frac{{\left (\frac{15 \, a^{4}}{b} +{\left (\frac{11 \,{\left (b x + a\right )} a^{2}}{b} - \frac{25 \, a^{3}}{b}\right )}{\left (b x + a\right )}\right )} \sqrt{b x + a}}{40 \,{\left ({\left (b x + a\right )} b - a b\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/(b*x+a)^(3/2),x, algorithm="giac")

[Out]

-4*b^(9/2)/(((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b)*a^3*abs(b)) + 1/40*(15*a^4/b + (11*(b*
x + a)*a^2/b - 25*a^3/b)*(b*x + a))*sqrt(b*x + a)/((b*x + a)*b - a*b)^(5/2)