3.468 \(\int \frac {1}{x^{5/2} (a+b x)^3} \, dx\)

Optimal. Leaf size=95 \[ \frac {35 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{9/2}}+\frac {35 b}{4 a^4 \sqrt {x}}-\frac {35}{12 a^3 x^{3/2}}+\frac {7}{4 a^2 x^{3/2} (a+b x)}+\frac {1}{2 a x^{3/2} (a+b x)^2} \]

[Out]

-35/12/a^3/x^(3/2)+1/2/a/x^(3/2)/(b*x+a)^2+7/4/a^2/x^(3/2)/(b*x+a)+35/4*b^(3/2)*arctan(b^(1/2)*x^(1/2)/a^(1/2)
)/a^(9/2)+35/4*b/a^4/x^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {51, 63, 205} \[ \frac {35 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{9/2}}+\frac {7}{4 a^2 x^{3/2} (a+b x)}+\frac {35 b}{4 a^4 \sqrt {x}}-\frac {35}{12 a^3 x^{3/2}}+\frac {1}{2 a x^{3/2} (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*(a + b*x)^3),x]

[Out]

-35/(12*a^3*x^(3/2)) + (35*b)/(4*a^4*Sqrt[x]) + 1/(2*a*x^(3/2)*(a + b*x)^2) + 7/(4*a^2*x^(3/2)*(a + b*x)) + (3
5*b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*a^(9/2))

Rule 51

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*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 205

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

Rubi steps

\begin {align*} \int \frac {1}{x^{5/2} (a+b x)^3} \, dx &=\frac {1}{2 a x^{3/2} (a+b x)^2}+\frac {7 \int \frac {1}{x^{5/2} (a+b x)^2} \, dx}{4 a}\\ &=\frac {1}{2 a x^{3/2} (a+b x)^2}+\frac {7}{4 a^2 x^{3/2} (a+b x)}+\frac {35 \int \frac {1}{x^{5/2} (a+b x)} \, dx}{8 a^2}\\ &=-\frac {35}{12 a^3 x^{3/2}}+\frac {1}{2 a x^{3/2} (a+b x)^2}+\frac {7}{4 a^2 x^{3/2} (a+b x)}-\frac {(35 b) \int \frac {1}{x^{3/2} (a+b x)} \, dx}{8 a^3}\\ &=-\frac {35}{12 a^3 x^{3/2}}+\frac {35 b}{4 a^4 \sqrt {x}}+\frac {1}{2 a x^{3/2} (a+b x)^2}+\frac {7}{4 a^2 x^{3/2} (a+b x)}+\frac {\left (35 b^2\right ) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{8 a^4}\\ &=-\frac {35}{12 a^3 x^{3/2}}+\frac {35 b}{4 a^4 \sqrt {x}}+\frac {1}{2 a x^{3/2} (a+b x)^2}+\frac {7}{4 a^2 x^{3/2} (a+b x)}+\frac {\left (35 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{4 a^4}\\ &=-\frac {35}{12 a^3 x^{3/2}}+\frac {35 b}{4 a^4 \sqrt {x}}+\frac {1}{2 a x^{3/2} (a+b x)^2}+\frac {7}{4 a^2 x^{3/2} (a+b x)}+\frac {35 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 27, normalized size = 0.28 \[ -\frac {2 \, _2F_1\left (-\frac {3}{2},3;-\frac {1}{2};-\frac {b x}{a}\right )}{3 a^3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*(a + b*x)^3),x]

[Out]

(-2*Hypergeometric2F1[-3/2, 3, -1/2, -((b*x)/a)])/(3*a^3*x^(3/2))

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fricas [A]  time = 0.45, size = 250, normalized size = 2.63 \[ \left [\frac {105 \, {\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} + a^{2} b x^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 2 \, {\left (105 \, b^{3} x^{3} + 175 \, a b^{2} x^{2} + 56 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {x}}{24 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}, -\frac {105 \, {\left (b^{3} x^{4} + 2 \, a b^{2} x^{3} + a^{2} b x^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (105 \, b^{3} x^{3} + 175 \, a b^{2} x^{2} + 56 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {x}}{12 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(105*(b^3*x^4 + 2*a*b^2*x^3 + a^2*b*x^2)*sqrt(-b/a)*log((b*x + 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) +
2*(105*b^3*x^3 + 175*a*b^2*x^2 + 56*a^2*b*x - 8*a^3)*sqrt(x))/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2), -1/12*(10
5*(b^3*x^4 + 2*a*b^2*x^3 + a^2*b*x^2)*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (105*b^3*x^3 + 175*a*b^2*x^2
+ 56*a^2*b*x - 8*a^3)*sqrt(x))/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2)]

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giac [A]  time = 1.10, size = 71, normalized size = 0.75 \[ \frac {35 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{4}} + \frac {2 \, {\left (9 \, b x - a\right )}}{3 \, a^{4} x^{\frac {3}{2}}} + \frac {11 \, b^{3} x^{\frac {3}{2}} + 13 \, a b^{2} \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

35/4*b^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4) + 2/3*(9*b*x - a)/(a^4*x^(3/2)) + 1/4*(11*b^3*x^(3/2) + 1
3*a*b^2*sqrt(x))/((b*x + a)^2*a^4)

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maple [A]  time = 0.02, size = 79, normalized size = 0.83 \[ \frac {11 b^{3} x^{\frac {3}{2}}}{4 \left (b x +a \right )^{2} a^{4}}+\frac {13 b^{2} \sqrt {x}}{4 \left (b x +a \right )^{2} a^{3}}+\frac {35 b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, a^{4}}+\frac {6 b}{a^{4} \sqrt {x}}-\frac {2}{3 a^{3} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/a^3/x^(3/2)+6*b/a^4/x^(1/2)+11/4/a^4*b^3/(b*x+a)^2*x^(3/2)+13/4/a^3*b^2/(b*x+a)^2*x^(1/2)+35/4/a^4*b^2/(a
*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x^(1/2))

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maxima [A]  time = 2.98, size = 86, normalized size = 0.91 \[ \frac {105 \, b^{3} x^{3} + 175 \, a b^{2} x^{2} + 56 \, a^{2} b x - 8 \, a^{3}}{12 \, {\left (a^{4} b^{2} x^{\frac {7}{2}} + 2 \, a^{5} b x^{\frac {5}{2}} + a^{6} x^{\frac {3}{2}}\right )}} + \frac {35 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(105*b^3*x^3 + 175*a*b^2*x^2 + 56*a^2*b*x - 8*a^3)/(a^4*b^2*x^(7/2) + 2*a^5*b*x^(5/2) + a^6*x^(3/2)) + 35
/4*b^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4)

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mupad [B]  time = 0.16, size = 80, normalized size = 0.84 \[ \frac {\frac {175\,b^2\,x^2}{12\,a^3}-\frac {2}{3\,a}+\frac {35\,b^3\,x^3}{4\,a^4}+\frac {14\,b\,x}{3\,a^2}}{a^2\,x^{3/2}+b^2\,x^{7/2}+2\,a\,b\,x^{5/2}}+\frac {35\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{4\,a^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/2)*(a + b*x)^3),x)

[Out]

((175*b^2*x^2)/(12*a^3) - 2/(3*a) + (35*b^3*x^3)/(4*a^4) + (14*b*x)/(3*a^2))/(a^2*x^(3/2) + b^2*x^(7/2) + 2*a*
b*x^(5/2)) + (35*b^(3/2)*atan((b^(1/2)*x^(1/2))/a^(1/2)))/(4*a^(9/2))

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sympy [A]  time = 138.08, size = 962, normalized size = 10.13 \[ \begin {cases} \frac {\tilde {\infty }}{x^{\frac {9}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{3 a^{3} x^{\frac {3}{2}}} & \text {for}\: b = 0 \\- \frac {2}{9 b^{3} x^{\frac {9}{2}}} & \text {for}\: a = 0 \\- \frac {16 i a^{\frac {7}{2}} \sqrt {\frac {1}{b}}}{24 i a^{\frac {13}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 48 i a^{\frac {11}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}} + 24 i a^{\frac {9}{2}} b^{2} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}} + \frac {112 i a^{\frac {5}{2}} b x \sqrt {\frac {1}{b}}}{24 i a^{\frac {13}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 48 i a^{\frac {11}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}} + 24 i a^{\frac {9}{2}} b^{2} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}} + \frac {350 i a^{\frac {3}{2}} b^{2} x^{2} \sqrt {\frac {1}{b}}}{24 i a^{\frac {13}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 48 i a^{\frac {11}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}} + 24 i a^{\frac {9}{2}} b^{2} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}} + \frac {210 i \sqrt {a} b^{3} x^{3} \sqrt {\frac {1}{b}}}{24 i a^{\frac {13}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 48 i a^{\frac {11}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}} + 24 i a^{\frac {9}{2}} b^{2} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}} + \frac {105 a^{2} b x^{\frac {3}{2}} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{24 i a^{\frac {13}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 48 i a^{\frac {11}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}} + 24 i a^{\frac {9}{2}} b^{2} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}} - \frac {105 a^{2} b x^{\frac {3}{2}} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{24 i a^{\frac {13}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 48 i a^{\frac {11}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}} + 24 i a^{\frac {9}{2}} b^{2} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}} + \frac {210 a b^{2} x^{\frac {5}{2}} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{24 i a^{\frac {13}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 48 i a^{\frac {11}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}} + 24 i a^{\frac {9}{2}} b^{2} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}} - \frac {210 a b^{2} x^{\frac {5}{2}} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{24 i a^{\frac {13}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 48 i a^{\frac {11}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}} + 24 i a^{\frac {9}{2}} b^{2} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}} + \frac {105 b^{3} x^{\frac {7}{2}} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{24 i a^{\frac {13}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 48 i a^{\frac {11}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}} + 24 i a^{\frac {9}{2}} b^{2} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}} - \frac {105 b^{3} x^{\frac {7}{2}} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{24 i a^{\frac {13}{2}} x^{\frac {3}{2}} \sqrt {\frac {1}{b}} + 48 i a^{\frac {11}{2}} b x^{\frac {5}{2}} \sqrt {\frac {1}{b}} + 24 i a^{\frac {9}{2}} b^{2} x^{\frac {7}{2}} \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo/x**(9/2), Eq(a, 0) & Eq(b, 0)), (-2/(3*a**3*x**(3/2)), Eq(b, 0)), (-2/(9*b**3*x**(9/2)), Eq(a,
0)), (-16*I*a**(7/2)*sqrt(1/b)/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I
*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) + 112*I*a**(5/2)*b*x*sqrt(1/b)/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 48*I*a*
*(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) + 350*I*a**(3/2)*b**2*x**2*sqrt(1/b)/(24
*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(1/b))
+ 210*I*sqrt(a)*b**3*x**3*sqrt(1/b)/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/2)*sqrt(1/b)
+ 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) + 105*a**2*b*x**(3/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a**(1
3/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) - 105*a
**2*b*x**(3/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/
2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) + 210*a*b**2*x**(5/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x)
)/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(
1/b)) - 210*a*b**2*x**(5/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(1
1/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)) + 105*b**3*x**(7/2)*log(-I*sqrt(a)*sqrt(1/b
) + sqrt(x))/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 48*I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**
(7/2)*sqrt(1/b)) - 105*b**3*x**(7/2)*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(24*I*a**(13/2)*x**(3/2)*sqrt(1/b) + 4
8*I*a**(11/2)*b*x**(5/2)*sqrt(1/b) + 24*I*a**(9/2)*b**2*x**(7/2)*sqrt(1/b)), True))

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