3.5.62 \(\int \frac {x^{7/2}}{(a+b x)^3} \, dx\) [462]

Optimal. Leaf size=95 \[ -\frac {35 a \sqrt {x}}{4 b^4}+\frac {35 x^{3/2}}{12 b^3}-\frac {x^{7/2}}{2 b (a+b x)^2}-\frac {7 x^{5/2}}{4 b^2 (a+b x)}+\frac {35 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{9/2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {43, 52, 65, 211} \begin {gather*} \frac {35 a^{3/2} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{9/2}}-\frac {35 a \sqrt {x}}{4 b^4}-\frac {7 x^{5/2}}{4 b^2 (a+b x)}-\frac {x^{7/2}}{2 b (a+b x)^2}+\frac {35 x^{3/2}}{12 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(7/2)/(a + b*x)^3,x]

[Out]

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

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

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 65

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 211

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

Rubi steps

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

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Mathematica [A]
time = 0.13, size = 81, normalized size = 0.85 \begin {gather*} \frac {\sqrt {x} \left (-105 a^3-175 a^2 b x-56 a b^2 x^2+8 b^3 x^3\right )}{12 b^4 (a+b x)^2}+\frac {35 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)/(a + b*x)^3,x]

[Out]

(Sqrt[x]*(-105*a^3 - 175*a^2*b*x - 56*a*b^2*x^2 + 8*b^3*x^3))/(12*b^4*(a + b*x)^2) + (35*a^(3/2)*ArcTan[(Sqrt[
b]*Sqrt[x])/Sqrt[a]])/(4*b^(9/2))

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Maple [A]
time = 0.13, size = 68, normalized size = 0.72

method	result	size
derivativedivides	\(-\frac {2 \left (-\frac {b \,x^{\frac {3}{2}}}{3}+3 a \sqrt {x}\right )}{b^{4}}+\frac {2 a^{2} \left (\frac {-\frac {13 b \,x^{\frac {3}{2}}}{8}-\frac {11 a \sqrt {x}}{8}}{\left (b x +a \right )^{2}}+\frac {35 \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{4}}\)	\(68\)
default	\(-\frac {2 \left (-\frac {b \,x^{\frac {3}{2}}}{3}+3 a \sqrt {x}\right )}{b^{4}}+\frac {2 a^{2} \left (\frac {-\frac {13 b \,x^{\frac {3}{2}}}{8}-\frac {11 a \sqrt {x}}{8}}{\left (b x +a \right )^{2}}+\frac {35 \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{4}}\)	\(68\)
risch	\(-\frac {2 \left (-b x +9 a \right ) \sqrt {x}}{3 b^{4}}-\frac {13 a^{2} x^{\frac {3}{2}}}{4 b^{3} \left (b x +a \right )^{2}}-\frac {11 a^{3} \sqrt {x}}{4 b^{4} \left (b x +a \right )^{2}}+\frac {35 a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 b^{4} \sqrt {a b}}\)	\(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

-2/b^4*(-1/3*b*x^(3/2)+3*a*x^(1/2))+2/b^4*a^2*((-13/8*b*x^(3/2)-11/8*a*x^(1/2))/(b*x+a)^2+35/8/(a*b)^(1/2)*arc
tan(b*x^(1/2)/(a*b)^(1/2)))

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Maxima [A]
time = 0.52, size = 86, normalized size = 0.91 \begin {gather*} -\frac {13 \, a^{2} b x^{\frac {3}{2}} + 11 \, a^{3} \sqrt {x}}{4 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} + \frac {35 \, a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{4}} + \frac {2 \, {\left (b x^{\frac {3}{2}} - 9 \, a \sqrt {x}\right )}}{3 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(13*a^2*b*x^(3/2) + 11*a^3*sqrt(x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4) + 35/4*a^2*arctan(b*sqrt(x)/sqrt(a*b)
)/(sqrt(a*b)*b^4) + 2/3*(b*x^(3/2) - 9*a*sqrt(x))/b^4

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Fricas [A]
time = 0.56, size = 227, normalized size = 2.39 \begin {gather*} \left [\frac {105 \, {\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\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^{3} x^{3} - 56 \, a b^{2} x^{2} - 175 \, a^{2} b x - 105 \, a^{3}\right )} \sqrt {x}}{24 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, \frac {105 \, {\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (8 \, b^{3} x^{3} - 56 \, a b^{2} x^{2} - 175 \, a^{2} b x - 105 \, a^{3}\right )} \sqrt {x}}{12 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 762 vs. \(2 (87) = 174\).
time = 96.96, size = 762, normalized size = 8.02 \begin {gather*} \begin {cases} \tilde {\infty } x^{\frac {3}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {9}{2}}}{9 a^{3}} & \text {for}\: b = 0 \\\frac {2 x^{\frac {3}{2}}}{3 b^{3}} & \text {for}\: a = 0 \\\frac {105 a^{4} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{24 a^{2} b^{5} \sqrt {- \frac {a}{b}} + 48 a b^{6} x \sqrt {- \frac {a}{b}} + 24 b^{7} x^{2} \sqrt {- \frac {a}{b}}} - \frac {105 a^{4} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{24 a^{2} b^{5} \sqrt {- \frac {a}{b}} + 48 a b^{6} x \sqrt {- \frac {a}{b}} + 24 b^{7} x^{2} \sqrt {- \frac {a}{b}}} - \frac {210 a^{3} b \sqrt {x} \sqrt {- \frac {a}{b}}}{24 a^{2} b^{5} \sqrt {- \frac {a}{b}} + 48 a b^{6} x \sqrt {- \frac {a}{b}} + 24 b^{7} x^{2} \sqrt {- \frac {a}{b}}} + \frac {210 a^{3} b x \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{24 a^{2} b^{5} \sqrt {- \frac {a}{b}} + 48 a b^{6} x \sqrt {- \frac {a}{b}} + 24 b^{7} x^{2} \sqrt {- \frac {a}{b}}} - \frac {210 a^{3} b x \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{24 a^{2} b^{5} \sqrt {- \frac {a}{b}} + 48 a b^{6} x \sqrt {- \frac {a}{b}} + 24 b^{7} x^{2} \sqrt {- \frac {a}{b}}} - \frac {350 a^{2} b^{2} x^{\frac {3}{2}} \sqrt {- \frac {a}{b}}}{24 a^{2} b^{5} \sqrt {- \frac {a}{b}} + 48 a b^{6} x \sqrt {- \frac {a}{b}} + 24 b^{7} x^{2} \sqrt {- \frac {a}{b}}} + \frac {105 a^{2} b^{2} x^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{24 a^{2} b^{5} \sqrt {- \frac {a}{b}} + 48 a b^{6} x \sqrt {- \frac {a}{b}} + 24 b^{7} x^{2} \sqrt {- \frac {a}{b}}} - \frac {105 a^{2} b^{2} x^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{24 a^{2} b^{5} \sqrt {- \frac {a}{b}} + 48 a b^{6} x \sqrt {- \frac {a}{b}} + 24 b^{7} x^{2} \sqrt {- \frac {a}{b}}} - \frac {112 a b^{3} x^{\frac {5}{2}} \sqrt {- \frac {a}{b}}}{24 a^{2} b^{5} \sqrt {- \frac {a}{b}} + 48 a b^{6} x \sqrt {- \frac {a}{b}} + 24 b^{7} x^{2} \sqrt {- \frac {a}{b}}} + \frac {16 b^{4} x^{\frac {7}{2}} \sqrt {- \frac {a}{b}}}{24 a^{2} b^{5} \sqrt {- \frac {a}{b}} + 48 a b^{6} x \sqrt {- \frac {a}{b}} + 24 b^{7} x^{2} \sqrt {- \frac {a}{b}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x**(3/2), Eq(a, 0) & Eq(b, 0)), (2*x**(9/2)/(9*a**3), Eq(b, 0)), (2*x**(3/2)/(3*b**3), Eq(a, 0)
), (105*a**4*log(sqrt(x) - sqrt(-a/b))/(24*a**2*b**5*sqrt(-a/b) + 48*a*b**6*x*sqrt(-a/b) + 24*b**7*x**2*sqrt(-
a/b)) - 105*a**4*log(sqrt(x) + sqrt(-a/b))/(24*a**2*b**5*sqrt(-a/b) + 48*a*b**6*x*sqrt(-a/b) + 24*b**7*x**2*sq
rt(-a/b)) - 210*a**3*b*sqrt(x)*sqrt(-a/b)/(24*a**2*b**5*sqrt(-a/b) + 48*a*b**6*x*sqrt(-a/b) + 24*b**7*x**2*sqr
t(-a/b)) + 210*a**3*b*x*log(sqrt(x) - sqrt(-a/b))/(24*a**2*b**5*sqrt(-a/b) + 48*a*b**6*x*sqrt(-a/b) + 24*b**7*
x**2*sqrt(-a/b)) - 210*a**3*b*x*log(sqrt(x) + sqrt(-a/b))/(24*a**2*b**5*sqrt(-a/b) + 48*a*b**6*x*sqrt(-a/b) +
24*b**7*x**2*sqrt(-a/b)) - 350*a**2*b**2*x**(3/2)*sqrt(-a/b)/(24*a**2*b**5*sqrt(-a/b) + 48*a*b**6*x*sqrt(-a/b)
+ 24*b**7*x**2*sqrt(-a/b)) + 105*a**2*b**2*x**2*log(sqrt(x) - sqrt(-a/b))/(24*a**2*b**5*sqrt(-a/b) + 48*a*b**
6*x*sqrt(-a/b) + 24*b**7*x**2*sqrt(-a/b)) - 105*a**2*b**2*x**2*log(sqrt(x) + sqrt(-a/b))/(24*a**2*b**5*sqrt(-a
/b) + 48*a*b**6*x*sqrt(-a/b) + 24*b**7*x**2*sqrt(-a/b)) - 112*a*b**3*x**(5/2)*sqrt(-a/b)/(24*a**2*b**5*sqrt(-a
/b) + 48*a*b**6*x*sqrt(-a/b) + 24*b**7*x**2*sqrt(-a/b)) + 16*b**4*x**(7/2)*sqrt(-a/b)/(24*a**2*b**5*sqrt(-a/b)
+ 48*a*b**6*x*sqrt(-a/b) + 24*b**7*x**2*sqrt(-a/b)), True))

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Giac [A]
time = 1.38, size = 77, normalized size = 0.81 \begin {gather*} \frac {35 \, a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{4}} - \frac {13 \, a^{2} b x^{\frac {3}{2}} + 11 \, a^{3} \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} b^{4}} + \frac {2 \, {\left (b^{6} x^{\frac {3}{2}} - 9 \, a b^{5} \sqrt {x}\right )}}{3 \, b^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

35/4*a^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) - 1/4*(13*a^2*b*x^(3/2) + 11*a^3*sqrt(x))/((b*x + a)^2*b^
4) + 2/3*(b^6*x^(3/2) - 9*a*b^5*sqrt(x))/b^9

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Mupad [B]
time = 0.12, size = 81, normalized size = 0.85 \begin {gather*} \frac {2\,x^{3/2}}{3\,b^3}-\frac {\frac {11\,a^3\,\sqrt {x}}{4}+\frac {13\,a^2\,b\,x^{3/2}}{4}}{a^2\,b^4+2\,a\,b^5\,x+b^6\,x^2}-\frac {6\,a\,\sqrt {x}}{b^4}+\frac {35\,a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{4\,b^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)/(a + b*x)^3,x)

[Out]

(2*x^(3/2))/(3*b^3) - ((11*a^3*x^(1/2))/4 + (13*a^2*b*x^(3/2))/4)/(a^2*b^4 + b^6*x^2 + 2*a*b^5*x) - (6*a*x^(1/
2))/b^4 + (35*a^(3/2)*atan((b^(1/2)*x^(1/2))/a^(1/2)))/(4*b^(9/2))

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