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

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

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Rubi [A]  time = 0.02, antiderivative size = 87, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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}} \end {gather*}

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 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]

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])

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*}

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Mathematica [A]  time = 0.01, size = 49, normalized size = 0.56 \begin {gather*} -\frac {2 \left (a^3-2 a^2 b x+8 a b^2 x^2+16 b^3 x^3\right )}{5 a^4 x^{5/2} \sqrt {a+b x}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.12, size = 49, normalized size = 0.56 \begin {gather*} -\frac {2 \left (a^3-2 a^2 b x+8 a b^2 x^2+16 b^3 x^3\right )}{5 a^4 x^{5/2} \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[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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fricas [A]  time = 1.18, size = 58, normalized size = 0.67 \begin {gather*} -\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 {gather*}

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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giac [A]  time = 1.21, size = 121, normalized size = 1.39 \begin {gather*} -\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 {2 \, {\left (\frac {15 \, b^{6}}{a^{2} {\left | b \right |}} + {\left (b x + a\right )} {\left (\frac {11 \, {\left (b x + a\right )} b^{6}}{a^{4} {\left | b \right |}} - \frac {25 \, b^{6}}{a^{3} {\left | b \right |}}\right )}\right )} \sqrt {b x + a}}{5 \, {\left ({\left (b x + a\right )} b - a b\right )}^{\frac {5}{2}}} \end {gather*}

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)) - 2/5*(15*b^6/(a^2*abs(b))
+ (b*x + a)*(11*(b*x + a)*b^6/(a^4*abs(b)) - 25*b^6/(a^3*abs(b))))*sqrt(b*x + a)/((b*x + a)*b - a*b)^(5/2)

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maple [A]  time = 0.00, size = 44, normalized size = 0.51 \begin {gather*} -\frac {2 \left (16 b^{3} x^{3}+8 a \,b^{2} x^{2}-2 a^{2} b x +a^{3}\right )}{5 \sqrt {b x +a}\, a^{4} x^{\frac {5}{2}}} \end {gather*}

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)/(b*x+a)^(1/2)/x^(5/2)/a^4

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maxima [A]  time = 1.31, size = 64, normalized size = 0.74 \begin {gather*} -\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 {gather*}

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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mupad [B]  time = 0.43, size = 58, normalized size = 0.67 \begin {gather*} -\frac {\sqrt {a+b\,x}\,\left (\frac {2}{5\,a\,b}-\frac {4\,x}{5\,a^2}+\frac {16\,b\,x^2}{5\,a^3}+\frac {32\,b^2\,x^3}{5\,a^4}\right )}{x^{7/2}+\frac {a\,x^{5/2}}{b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(7/2)*(a + b*x)^(3/2)),x)

[Out]

-((a + b*x)^(1/2)*(2/(5*a*b) - (4*x)/(5*a^2) + (16*b*x^2)/(5*a^3) + (32*b^2*x^3)/(5*a^4)))/(x^(7/2) + (a*x^(5/
2))/b)

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sympy [B]  time = 11.15, size = 348, normalized size = 4.00 \begin {gather*} - \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 {gather*}

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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