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

Optimal. Leaf size=82 \[ -\frac {15 \sqrt {a} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{7/2}}-\frac {5 x^{3/2}}{4 b^2 (a+b x)}-\frac {x^{5/2}}{2 b (a+b x)^2}+\frac {15 \sqrt {x}}{4 b^3} \]

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Rubi [A]  time = 0.02, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {47, 50, 63, 205} \begin {gather*} -\frac {5 x^{3/2}}{4 b^2 (a+b x)}-\frac {15 \sqrt {a} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{7/2}}-\frac {x^{5/2}}{2 b (a+b x)^2}+\frac {15 \sqrt {x}}{4 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(a + b*x)^3,x]

[Out]

(15*Sqrt[x])/(4*b^3) - x^(5/2)/(2*b*(a + b*x)^2) - (5*x^(3/2))/(4*b^2*(a + b*x)) - (15*Sqrt[a]*ArcTan[(Sqrt[b]
*Sqrt[x])/Sqrt[a]])/(4*b^(7/2))

Rule 47

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 {x^{5/2}}{(a+b x)^3} \, dx &=-\frac {x^{5/2}}{2 b (a+b x)^2}+\frac {5 \int \frac {x^{3/2}}{(a+b x)^2} \, dx}{4 b}\\ &=-\frac {x^{5/2}}{2 b (a+b x)^2}-\frac {5 x^{3/2}}{4 b^2 (a+b x)}+\frac {15 \int \frac {\sqrt {x}}{a+b x} \, dx}{8 b^2}\\ &=\frac {15 \sqrt {x}}{4 b^3}-\frac {x^{5/2}}{2 b (a+b x)^2}-\frac {5 x^{3/2}}{4 b^2 (a+b x)}-\frac {(15 a) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{8 b^3}\\ &=\frac {15 \sqrt {x}}{4 b^3}-\frac {x^{5/2}}{2 b (a+b x)^2}-\frac {5 x^{3/2}}{4 b^2 (a+b x)}-\frac {(15 a) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{4 b^3}\\ &=\frac {15 \sqrt {x}}{4 b^3}-\frac {x^{5/2}}{2 b (a+b x)^2}-\frac {5 x^{3/2}}{4 b^2 (a+b x)}-\frac {15 \sqrt {a} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.00, size = 27, normalized size = 0.33 \begin {gather*} \frac {2 x^{7/2} \, _2F_1\left (3,\frac {7}{2};\frac {9}{2};-\frac {b x}{a}\right )}{7 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(a + b*x)^3,x]

[Out]

(2*x^(7/2)*Hypergeometric2F1[3, 7/2, 9/2, -((b*x)/a)])/(7*a^3)

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IntegrateAlgebraic [A]  time = 0.12, size = 76, normalized size = 0.93 \begin {gather*} \frac {15 a^2 \sqrt {x}+25 a b x^{3/2}+8 b^2 x^{5/2}}{4 b^3 (a+b x)^2}-\frac {15 \sqrt {a} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(5/2)/(a + b*x)^3,x]

[Out]

(15*a^2*Sqrt[x] + 25*a*b*x^(3/2) + 8*b^2*x^(5/2))/(4*b^3*(a + b*x)^2) - (15*Sqrt[a]*ArcTan[(Sqrt[b]*Sqrt[x])/S
qrt[a]])/(4*b^(7/2))

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fricas [A]  time = 0.59, size = 200, normalized size = 2.44 \begin {gather*} \left [\frac {15 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) + 2 \, {\left (8 \, b^{2} x^{2} + 25 \, a b x + 15 \, a^{2}\right )} \sqrt {x}}{8 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac {15 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (8 \, b^{2} x^{2} + 25 \, a b x + 15 \, a^{2}\right )} \sqrt {x}}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(15*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(-a/b)*log((b*x - 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) + 2*(8*b^2*x^2
+ 25*a*b*x + 15*a^2)*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3), -1/4*(15*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(a/b)*a
rctan(b*sqrt(x)*sqrt(a/b)/a) - (8*b^2*x^2 + 25*a*b*x + 15*a^2)*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)]

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giac [A]  time = 0.95, size = 59, normalized size = 0.72 \begin {gather*} -\frac {15 \, a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} + \frac {2 \, \sqrt {x}}{b^{3}} + \frac {9 \, a b x^{\frac {3}{2}} + 7 \, a^{2} \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4*a*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 2*sqrt(x)/b^3 + 1/4*(9*a*b*x^(3/2) + 7*a^2*sqrt(x))/((b*
x + a)^2*b^3)

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maple [A]  time = 0.02, size = 66, normalized size = 0.80 \begin {gather*} \frac {9 a \,x^{\frac {3}{2}}}{4 \left (b x +a \right )^{2} b^{2}}+\frac {7 a^{2} \sqrt {x}}{4 \left (b x +a \right )^{2} b^{3}}-\frac {15 a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, b^{3}}+\frac {2 \sqrt {x}}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x^(1/2)/b^3+9/4/b^2*a/(b*x+a)^2*x^(3/2)+7/4/b^3*a^2/(b*x+a)^2*x^(1/2)-15/4/b^3*a/(a*b)^(1/2)*arctan(1/(a*b)^
(1/2)*b*x^(1/2))

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maxima [A]  time = 2.94, size = 73, normalized size = 0.89 \begin {gather*} \frac {9 \, a b x^{\frac {3}{2}} + 7 \, a^{2} \sqrt {x}}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} - \frac {15 \, a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} + \frac {2 \, \sqrt {x}}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(9*a*b*x^(3/2) + 7*a^2*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3) - 15/4*a*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt
(a*b)*b^3) + 2*sqrt(x)/b^3

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mupad [B]  time = 0.14, size = 69, normalized size = 0.84 \begin {gather*} \frac {\frac {7\,a^2\,\sqrt {x}}{4}+\frac {9\,a\,b\,x^{3/2}}{4}}{a^2\,b^3+2\,a\,b^4\,x+b^5\,x^2}+\frac {2\,\sqrt {x}}{b^3}-\frac {15\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{4\,b^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(a + b*x)^3,x)

[Out]

((7*a^2*x^(1/2))/4 + (9*a*b*x^(3/2))/4)/(a^2*b^3 + b^5*x^2 + 2*a*b^4*x) + (2*x^(1/2))/b^3 - (15*a^(1/2)*atan((
b^(1/2)*x^(1/2))/a^(1/2)))/(4*b^(7/2))

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sympy [A]  time = 53.29, size = 816, normalized size = 9.95 \begin {gather*} \begin {cases} \tilde {\infty } \sqrt {x} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {7}{2}}}{7 a^{3}} & \text {for}\: b = 0 \\\frac {2 \sqrt {x}}{b^{3}} & \text {for}\: a = 0 \\\frac {30 i a^{\frac {5}{2}} b \sqrt {x} \sqrt {\frac {1}{b}}}{8 i a^{\frac {5}{2}} b^{4} \sqrt {\frac {1}{b}} + 16 i a^{\frac {3}{2}} b^{5} x \sqrt {\frac {1}{b}} + 8 i \sqrt {a} b^{6} x^{2} \sqrt {\frac {1}{b}}} + \frac {50 i a^{\frac {3}{2}} b^{2} x^{\frac {3}{2}} \sqrt {\frac {1}{b}}}{8 i a^{\frac {5}{2}} b^{4} \sqrt {\frac {1}{b}} + 16 i a^{\frac {3}{2}} b^{5} x \sqrt {\frac {1}{b}} + 8 i \sqrt {a} b^{6} x^{2} \sqrt {\frac {1}{b}}} + \frac {16 i \sqrt {a} b^{3} x^{\frac {5}{2}} \sqrt {\frac {1}{b}}}{8 i a^{\frac {5}{2}} b^{4} \sqrt {\frac {1}{b}} + 16 i a^{\frac {3}{2}} b^{5} x \sqrt {\frac {1}{b}} + 8 i \sqrt {a} b^{6} x^{2} \sqrt {\frac {1}{b}}} - \frac {15 a^{3} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 i a^{\frac {5}{2}} b^{4} \sqrt {\frac {1}{b}} + 16 i a^{\frac {3}{2}} b^{5} x \sqrt {\frac {1}{b}} + 8 i \sqrt {a} b^{6} x^{2} \sqrt {\frac {1}{b}}} + \frac {15 a^{3} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 i a^{\frac {5}{2}} b^{4} \sqrt {\frac {1}{b}} + 16 i a^{\frac {3}{2}} b^{5} x \sqrt {\frac {1}{b}} + 8 i \sqrt {a} b^{6} x^{2} \sqrt {\frac {1}{b}}} - \frac {30 a^{2} b x \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 i a^{\frac {5}{2}} b^{4} \sqrt {\frac {1}{b}} + 16 i a^{\frac {3}{2}} b^{5} x \sqrt {\frac {1}{b}} + 8 i \sqrt {a} b^{6} x^{2} \sqrt {\frac {1}{b}}} + \frac {30 a^{2} b x \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 i a^{\frac {5}{2}} b^{4} \sqrt {\frac {1}{b}} + 16 i a^{\frac {3}{2}} b^{5} x \sqrt {\frac {1}{b}} + 8 i \sqrt {a} b^{6} x^{2} \sqrt {\frac {1}{b}}} - \frac {15 a b^{2} x^{2} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 i a^{\frac {5}{2}} b^{4} \sqrt {\frac {1}{b}} + 16 i a^{\frac {3}{2}} b^{5} x \sqrt {\frac {1}{b}} + 8 i \sqrt {a} b^{6} x^{2} \sqrt {\frac {1}{b}}} + \frac {15 a b^{2} x^{2} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 i a^{\frac {5}{2}} b^{4} \sqrt {\frac {1}{b}} + 16 i a^{\frac {3}{2}} b^{5} x \sqrt {\frac {1}{b}} + 8 i \sqrt {a} b^{6} x^{2} \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*sqrt(x), Eq(a, 0) & Eq(b, 0)), (2*x**(7/2)/(7*a**3), Eq(b, 0)), (2*sqrt(x)/b**3, Eq(a, 0)), (30
*I*a**(5/2)*b*sqrt(x)*sqrt(1/b)/(8*I*a**(5/2)*b**4*sqrt(1/b) + 16*I*a**(3/2)*b**5*x*sqrt(1/b) + 8*I*sqrt(a)*b*
*6*x**2*sqrt(1/b)) + 50*I*a**(3/2)*b**2*x**(3/2)*sqrt(1/b)/(8*I*a**(5/2)*b**4*sqrt(1/b) + 16*I*a**(3/2)*b**5*x
*sqrt(1/b) + 8*I*sqrt(a)*b**6*x**2*sqrt(1/b)) + 16*I*sqrt(a)*b**3*x**(5/2)*sqrt(1/b)/(8*I*a**(5/2)*b**4*sqrt(1
/b) + 16*I*a**(3/2)*b**5*x*sqrt(1/b) + 8*I*sqrt(a)*b**6*x**2*sqrt(1/b)) - 15*a**3*log(-I*sqrt(a)*sqrt(1/b) + s
qrt(x))/(8*I*a**(5/2)*b**4*sqrt(1/b) + 16*I*a**(3/2)*b**5*x*sqrt(1/b) + 8*I*sqrt(a)*b**6*x**2*sqrt(1/b)) + 15*
a**3*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(5/2)*b**4*sqrt(1/b) + 16*I*a**(3/2)*b**5*x*sqrt(1/b) + 8*I*sq
rt(a)*b**6*x**2*sqrt(1/b)) - 30*a**2*b*x*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(5/2)*b**4*sqrt(1/b) + 16
*I*a**(3/2)*b**5*x*sqrt(1/b) + 8*I*sqrt(a)*b**6*x**2*sqrt(1/b)) + 30*a**2*b*x*log(I*sqrt(a)*sqrt(1/b) + sqrt(x
))/(8*I*a**(5/2)*b**4*sqrt(1/b) + 16*I*a**(3/2)*b**5*x*sqrt(1/b) + 8*I*sqrt(a)*b**6*x**2*sqrt(1/b)) - 15*a*b**
2*x**2*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(5/2)*b**4*sqrt(1/b) + 16*I*a**(3/2)*b**5*x*sqrt(1/b) + 8*I
*sqrt(a)*b**6*x**2*sqrt(1/b)) + 15*a*b**2*x**2*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(8*I*a**(5/2)*b**4*sqrt(1/b)
+ 16*I*a**(3/2)*b**5*x*sqrt(1/b) + 8*I*sqrt(a)*b**6*x**2*sqrt(1/b)), True))

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