3.379 \(\int \frac {A+B x}{x (a+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=76 \[ -\frac {A \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {3 A+2 B x}{3 a^2 \sqrt {a+c x^2}}+\frac {A+B x}{3 a \left (a+c x^2\right )^{3/2}} \]

[Out]

1/3*(B*x+A)/a/(c*x^2+a)^(3/2)-A*arctanh((c*x^2+a)^(1/2)/a^(1/2))/a^(5/2)+1/3*(2*B*x+3*A)/a^2/(c*x^2+a)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {823, 12, 266, 63, 208} \[ \frac {3 A+2 B x}{3 a^2 \sqrt {a+c x^2}}-\frac {A \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {A+B x}{3 a \left (a+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x*(a + c*x^2)^(5/2)),x]

[Out]

(A + B*x)/(3*a*(a + c*x^2)^(3/2)) + (3*A + 2*B*x)/(3*a^2*Sqrt[a + c*x^2]) - (A*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]
])/a^(5/2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 208

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rubi steps

\begin {align*} \int \frac {A+B x}{x \left (a+c x^2\right )^{5/2}} \, dx &=\frac {A+B x}{3 a \left (a+c x^2\right )^{3/2}}-\frac {\int \frac {-3 a A c-2 a B c x}{x \left (a+c x^2\right )^{3/2}} \, dx}{3 a^2 c}\\ &=\frac {A+B x}{3 a \left (a+c x^2\right )^{3/2}}+\frac {3 A+2 B x}{3 a^2 \sqrt {a+c x^2}}+\frac {\int \frac {3 a^2 A c^2}{x \sqrt {a+c x^2}} \, dx}{3 a^4 c^2}\\ &=\frac {A+B x}{3 a \left (a+c x^2\right )^{3/2}}+\frac {3 A+2 B x}{3 a^2 \sqrt {a+c x^2}}+\frac {A \int \frac {1}{x \sqrt {a+c x^2}} \, dx}{a^2}\\ &=\frac {A+B x}{3 a \left (a+c x^2\right )^{3/2}}+\frac {3 A+2 B x}{3 a^2 \sqrt {a+c x^2}}+\frac {A \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{2 a^2}\\ &=\frac {A+B x}{3 a \left (a+c x^2\right )^{3/2}}+\frac {3 A+2 B x}{3 a^2 \sqrt {a+c x^2}}+\frac {A \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )}{a^2 c}\\ &=\frac {A+B x}{3 a \left (a+c x^2\right )^{3/2}}+\frac {3 A+2 B x}{3 a^2 \sqrt {a+c x^2}}-\frac {A \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 69, normalized size = 0.91 \[ \frac {a (4 A+3 B x)+c x^2 (3 A+2 B x)}{3 a^2 \left (a+c x^2\right )^{3/2}}-\frac {A \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x*(a + c*x^2)^(5/2)),x]

[Out]

(c*x^2*(3*A + 2*B*x) + a*(4*A + 3*B*x))/(3*a^2*(a + c*x^2)^(3/2)) - (A*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/a^(5/
2)

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fricas [A]  time = 0.92, size = 239, normalized size = 3.14 \[ \left [\frac {3 \, {\left (A c^{2} x^{4} + 2 \, A a c x^{2} + A a^{2}\right )} \sqrt {a} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (2 \, B a c x^{3} + 3 \, A a c x^{2} + 3 \, B a^{2} x + 4 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{6 \, {\left (a^{3} c^{2} x^{4} + 2 \, a^{4} c x^{2} + a^{5}\right )}}, \frac {3 \, {\left (A c^{2} x^{4} + 2 \, A a c x^{2} + A a^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + {\left (2 \, B a c x^{3} + 3 \, A a c x^{2} + 3 \, B a^{2} x + 4 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{3 \, {\left (a^{3} c^{2} x^{4} + 2 \, a^{4} c x^{2} + a^{5}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x/(c*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(A*c^2*x^4 + 2*A*a*c*x^2 + A*a^2)*sqrt(a)*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(2*B
*a*c*x^3 + 3*A*a*c*x^2 + 3*B*a^2*x + 4*A*a^2)*sqrt(c*x^2 + a))/(a^3*c^2*x^4 + 2*a^4*c*x^2 + a^5), 1/3*(3*(A*c^
2*x^4 + 2*A*a*c*x^2 + A*a^2)*sqrt(-a)*arctan(sqrt(-a)/sqrt(c*x^2 + a)) + (2*B*a*c*x^3 + 3*A*a*c*x^2 + 3*B*a^2*
x + 4*A*a^2)*sqrt(c*x^2 + a))/(a^3*c^2*x^4 + 2*a^4*c*x^2 + a^5)]

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giac [A]  time = 0.20, size = 82, normalized size = 1.08 \[ \frac {{\left ({\left (\frac {2 \, B c x}{a^{2}} + \frac {3 \, A c}{a^{2}}\right )} x + \frac {3 \, B}{a}\right )} x + \frac {4 \, A}{a}}{3 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}}} + \frac {2 \, A \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x/(c*x^2+a)^(5/2),x, algorithm="giac")

[Out]

1/3*(((2*B*c*x/a^2 + 3*A*c/a^2)*x + 3*B/a)*x + 4*A/a)/(c*x^2 + a)^(3/2) + 2*A*arctan(-(sqrt(c)*x - sqrt(c*x^2
+ a))/sqrt(-a))/(sqrt(-a)*a^2)

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maple [A]  time = 0.05, size = 92, normalized size = 1.21 \[ \frac {B x}{3 \left (c \,x^{2}+a \right )^{\frac {3}{2}} a}+\frac {A}{3 \left (c \,x^{2}+a \right )^{\frac {3}{2}} a}+\frac {2 B x}{3 \sqrt {c \,x^{2}+a}\, a^{2}}-\frac {A \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{a^{\frac {5}{2}}}+\frac {A}{\sqrt {c \,x^{2}+a}\, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(c*x^2+a)^(5/2)/x,x)

[Out]

1/3*B*x/a/(c*x^2+a)^(3/2)+2/3*B/a^2*x/(c*x^2+a)^(1/2)+1/3*A/a/(c*x^2+a)^(3/2)+A/a^2/(c*x^2+a)^(1/2)-A/a^(5/2)*
ln((2*a+2*(c*x^2+a)^(1/2)*a^(1/2))/x)

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maxima [A]  time = 0.60, size = 80, normalized size = 1.05 \[ \frac {2 \, B x}{3 \, \sqrt {c x^{2} + a} a^{2}} + \frac {B x}{3 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} a} - \frac {A \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right )}{a^{\frac {5}{2}}} + \frac {A}{\sqrt {c x^{2} + a} a^{2}} + \frac {A}{3 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x/(c*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

2/3*B*x/(sqrt(c*x^2 + a)*a^2) + 1/3*B*x/((c*x^2 + a)^(3/2)*a) - A*arcsinh(a/(sqrt(a*c)*abs(x)))/a^(5/2) + A/(s
qrt(c*x^2 + a)*a^2) + 1/3*A/((c*x^2 + a)^(3/2)*a)

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mupad [B]  time = 1.51, size = 80, normalized size = 1.05 \[ \frac {\frac {A}{3\,a}+\frac {A\,\left (c\,x^2+a\right )}{a^2}}{{\left (c\,x^2+a\right )}^{3/2}}+\frac {2\,B\,x\,\left (c\,x^2+a\right )+B\,a\,x}{3\,a^2\,{\left (c\,x^2+a\right )}^{3/2}}-\frac {A\,\mathrm {atanh}\left (\frac {\sqrt {c\,x^2+a}}{\sqrt {a}}\right )}{a^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x*(a + c*x^2)^(5/2)),x)

[Out]

(A/(3*a) + (A*(a + c*x^2))/a^2)/(a + c*x^2)^(3/2) + (2*B*x*(a + c*x^2) + B*a*x)/(3*a^2*(a + c*x^2)^(3/2)) - (A
*atanh((a + c*x^2)^(1/2)/a^(1/2)))/a^(5/2)

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sympy [B]  time = 25.67, size = 840, normalized size = 11.05 \[ A \left (\frac {8 a^{7} \sqrt {1 + \frac {c x^{2}}{a}}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} c x^{2} + 18 a^{\frac {15}{2}} c^{2} x^{4} + 6 a^{\frac {13}{2}} c^{3} x^{6}} + \frac {3 a^{7} \log {\left (\frac {c x^{2}}{a} \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} c x^{2} + 18 a^{\frac {15}{2}} c^{2} x^{4} + 6 a^{\frac {13}{2}} c^{3} x^{6}} - \frac {6 a^{7} \log {\left (\sqrt {1 + \frac {c x^{2}}{a}} + 1 \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} c x^{2} + 18 a^{\frac {15}{2}} c^{2} x^{4} + 6 a^{\frac {13}{2}} c^{3} x^{6}} + \frac {14 a^{6} c x^{2} \sqrt {1 + \frac {c x^{2}}{a}}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} c x^{2} + 18 a^{\frac {15}{2}} c^{2} x^{4} + 6 a^{\frac {13}{2}} c^{3} x^{6}} + \frac {9 a^{6} c x^{2} \log {\left (\frac {c x^{2}}{a} \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} c x^{2} + 18 a^{\frac {15}{2}} c^{2} x^{4} + 6 a^{\frac {13}{2}} c^{3} x^{6}} - \frac {18 a^{6} c x^{2} \log {\left (\sqrt {1 + \frac {c x^{2}}{a}} + 1 \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} c x^{2} + 18 a^{\frac {15}{2}} c^{2} x^{4} + 6 a^{\frac {13}{2}} c^{3} x^{6}} + \frac {6 a^{5} c^{2} x^{4} \sqrt {1 + \frac {c x^{2}}{a}}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} c x^{2} + 18 a^{\frac {15}{2}} c^{2} x^{4} + 6 a^{\frac {13}{2}} c^{3} x^{6}} + \frac {9 a^{5} c^{2} x^{4} \log {\left (\frac {c x^{2}}{a} \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} c x^{2} + 18 a^{\frac {15}{2}} c^{2} x^{4} + 6 a^{\frac {13}{2}} c^{3} x^{6}} - \frac {18 a^{5} c^{2} x^{4} \log {\left (\sqrt {1 + \frac {c x^{2}}{a}} + 1 \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} c x^{2} + 18 a^{\frac {15}{2}} c^{2} x^{4} + 6 a^{\frac {13}{2}} c^{3} x^{6}} + \frac {3 a^{4} c^{3} x^{6} \log {\left (\frac {c x^{2}}{a} \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} c x^{2} + 18 a^{\frac {15}{2}} c^{2} x^{4} + 6 a^{\frac {13}{2}} c^{3} x^{6}} - \frac {6 a^{4} c^{3} x^{6} \log {\left (\sqrt {1 + \frac {c x^{2}}{a}} + 1 \right )}}{6 a^{\frac {19}{2}} + 18 a^{\frac {17}{2}} c x^{2} + 18 a^{\frac {15}{2}} c^{2} x^{4} + 6 a^{\frac {13}{2}} c^{3} x^{6}}\right ) + B \left (\frac {3 a x}{3 a^{\frac {7}{2}} \sqrt {1 + \frac {c x^{2}}{a}} + 3 a^{\frac {5}{2}} c x^{2} \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {2 c x^{3}}{3 a^{\frac {7}{2}} \sqrt {1 + \frac {c x^{2}}{a}} + 3 a^{\frac {5}{2}} c x^{2} \sqrt {1 + \frac {c x^{2}}{a}}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x/(c*x**2+a)**(5/2),x)

[Out]

A*(8*a**7*sqrt(1 + c*x**2/a)/(6*a**(19/2) + 18*a**(17/2)*c*x**2 + 18*a**(15/2)*c**2*x**4 + 6*a**(13/2)*c**3*x*
*6) + 3*a**7*log(c*x**2/a)/(6*a**(19/2) + 18*a**(17/2)*c*x**2 + 18*a**(15/2)*c**2*x**4 + 6*a**(13/2)*c**3*x**6
) - 6*a**7*log(sqrt(1 + c*x**2/a) + 1)/(6*a**(19/2) + 18*a**(17/2)*c*x**2 + 18*a**(15/2)*c**2*x**4 + 6*a**(13/
2)*c**3*x**6) + 14*a**6*c*x**2*sqrt(1 + c*x**2/a)/(6*a**(19/2) + 18*a**(17/2)*c*x**2 + 18*a**(15/2)*c**2*x**4
+ 6*a**(13/2)*c**3*x**6) + 9*a**6*c*x**2*log(c*x**2/a)/(6*a**(19/2) + 18*a**(17/2)*c*x**2 + 18*a**(15/2)*c**2*
x**4 + 6*a**(13/2)*c**3*x**6) - 18*a**6*c*x**2*log(sqrt(1 + c*x**2/a) + 1)/(6*a**(19/2) + 18*a**(17/2)*c*x**2
+ 18*a**(15/2)*c**2*x**4 + 6*a**(13/2)*c**3*x**6) + 6*a**5*c**2*x**4*sqrt(1 + c*x**2/a)/(6*a**(19/2) + 18*a**(
17/2)*c*x**2 + 18*a**(15/2)*c**2*x**4 + 6*a**(13/2)*c**3*x**6) + 9*a**5*c**2*x**4*log(c*x**2/a)/(6*a**(19/2) +
18*a**(17/2)*c*x**2 + 18*a**(15/2)*c**2*x**4 + 6*a**(13/2)*c**3*x**6) - 18*a**5*c**2*x**4*log(sqrt(1 + c*x**2
/a) + 1)/(6*a**(19/2) + 18*a**(17/2)*c*x**2 + 18*a**(15/2)*c**2*x**4 + 6*a**(13/2)*c**3*x**6) + 3*a**4*c**3*x*
*6*log(c*x**2/a)/(6*a**(19/2) + 18*a**(17/2)*c*x**2 + 18*a**(15/2)*c**2*x**4 + 6*a**(13/2)*c**3*x**6) - 6*a**4
*c**3*x**6*log(sqrt(1 + c*x**2/a) + 1)/(6*a**(19/2) + 18*a**(17/2)*c*x**2 + 18*a**(15/2)*c**2*x**4 + 6*a**(13/
2)*c**3*x**6)) + B*(3*a*x/(3*a**(7/2)*sqrt(1 + c*x**2/a) + 3*a**(5/2)*c*x**2*sqrt(1 + c*x**2/a)) + 2*c*x**3/(3
*a**(7/2)*sqrt(1 + c*x**2/a) + 3*a**(5/2)*c*x**2*sqrt(1 + c*x**2/a)))

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