3.5.55 \(\int \frac {1}{x^{7/2} (a+b x)} \, dx\) [455]

Optimal. Leaf size=68 \[ -\frac {2}{5 a x^{5/2}}+\frac {2 b}{3 a^2 x^{3/2}}-\frac {2 b^2}{a^3 \sqrt {x}}-\frac {2 b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {53, 65, 211} \begin {gather*} -\frac {2 b^{5/2} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}}-\frac {2 b^2}{a^3 \sqrt {x}}+\frac {2 b}{3 a^2 x^{3/2}}-\frac {2}{5 a x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(7/2)*(a + b*x)),x]

[Out]

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

Rule 53

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

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Mathematica [A]
time = 0.05, size = 61, normalized size = 0.90 \begin {gather*} -\frac {2 \left (3 a^2-5 a b x+15 b^2 x^2\right )}{15 a^3 x^{5/2}}-\frac {2 b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(7/2)*(a + b*x)),x]

[Out]

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

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Maple [A]
time = 0.11, size = 54, normalized size = 0.79

method	result	size
risch	\(-\frac {2 \left (15 x^{2} b^{2}-5 a b x +3 a^{2}\right )}{15 a^{3} x^{\frac {5}{2}}}-\frac {2 b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a^{3} \sqrt {a b}}\)	\(53\)
derivativedivides	\(-\frac {2 b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a^{3} \sqrt {a b}}-\frac {2}{5 a \,x^{\frac {5}{2}}}-\frac {2 b^{2}}{a^{3} \sqrt {x}}+\frac {2 b}{3 a^{2} x^{\frac {3}{2}}}\)	\(54\)
default	\(-\frac {2 b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{a^{3} \sqrt {a b}}-\frac {2}{5 a \,x^{\frac {5}{2}}}-\frac {2 b^{2}}{a^{3} \sqrt {x}}+\frac {2 b}{3 a^{2} x^{\frac {3}{2}}}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(7/2)/(b*x+a),x,method=_RETURNVERBOSE)

[Out]

-2*b^3/a^3/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2))-2/5/a/x^(5/2)-2*b^2/a^3/x^(1/2)+2/3*b/a^2/x^(3/2)

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Maxima [A]
time = 0.50, size = 52, normalized size = 0.76 \begin {gather*} -\frac {2 \, b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {2 \, {\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )}}{15 \, a^{3} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.61, size = 144, normalized size = 2.12 \begin {gather*} \left [\frac {15 \, b^{2} x^{3} \sqrt {-\frac {b}{a}} \log \left (\frac {b x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) - 2 \, {\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )} \sqrt {x}}{15 \, a^{3} x^{3}}, \frac {2 \, {\left (15 \, b^{2} x^{3} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - {\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )} \sqrt {x}\right )}}{15 \, a^{3} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 14.78, size = 126, normalized size = 1.85 \begin {gather*} \begin {cases} \frac {\tilde {\infty }}{x^{\frac {7}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{7 b x^{\frac {7}{2}}} & \text {for}\: a = 0 \\- \frac {2}{5 a x^{\frac {5}{2}}} & \text {for}\: b = 0 \\- \frac {2}{5 a x^{\frac {5}{2}}} + \frac {2 b}{3 a^{2} x^{\frac {3}{2}}} - \frac {b^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{a^{3} \sqrt {- \frac {a}{b}}} + \frac {b^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{a^{3} \sqrt {- \frac {a}{b}}} - \frac {2 b^{2}}{a^{3} \sqrt {x}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo/x**(7/2), Eq(a, 0) & Eq(b, 0)), (-2/(7*b*x**(7/2)), Eq(a, 0)), (-2/(5*a*x**(5/2)), Eq(b, 0)), (
-2/(5*a*x**(5/2)) + 2*b/(3*a**2*x**(3/2)) - b**2*log(sqrt(x) - sqrt(-a/b))/(a**3*sqrt(-a/b)) + b**2*log(sqrt(x
) + sqrt(-a/b))/(a**3*sqrt(-a/b)) - 2*b**2/(a**3*sqrt(x)), True))

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Giac [A]
time = 0.66, size = 52, normalized size = 0.76 \begin {gather*} -\frac {2 \, b^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {2 \, {\left (15 \, b^{2} x^{2} - 5 \, a b x + 3 \, a^{2}\right )}}{15 \, a^{3} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.11, size = 49, normalized size = 0.72 \begin {gather*} -\frac {\frac {2}{5\,a}+\frac {2\,b^2\,x^2}{a^3}-\frac {2\,b\,x}{3\,a^2}}{x^{5/2}}-\frac {2\,b^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{a^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(7/2)*(a + b*x)),x)

[Out]

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

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