3.4.73 \(\int \frac {A+B x}{x^4 (a+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=120 \[ \frac {3 B c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{5/2}}+\frac {8 A c \sqrt {a+c x^2}}{3 a^3 x}-\frac {4 A \sqrt {a+c x^2}}{3 a^2 x^3}-\frac {3 B \sqrt {a+c x^2}}{2 a^2 x^2}+\frac {A+B x}{a x^3 \sqrt {a+c x^2}} \]

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Rubi [A]  time = 0.11, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {823, 835, 807, 266, 63, 208} \begin {gather*} \frac {8 A c \sqrt {a+c x^2}}{3 a^3 x}-\frac {4 A \sqrt {a+c x^2}}{3 a^2 x^3}-\frac {3 B \sqrt {a+c x^2}}{2 a^2 x^2}+\frac {3 B c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{5/2}}+\frac {A+B x}{a x^3 \sqrt {a+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^4*(a + c*x^2)^(3/2)),x]

[Out]

(A + B*x)/(a*x^3*Sqrt[a + c*x^2]) - (4*A*Sqrt[a + c*x^2])/(3*a^2*x^3) - (3*B*Sqrt[a + c*x^2])/(2*a^2*x^2) + (8
*A*c*Sqrt[a + c*x^2])/(3*a^3*x) + (3*B*c*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/(2*a^(5/2))

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 807

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

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])

Rule 835

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

Rubi steps

\begin {align*} \int \frac {A+B x}{x^4 \left (a+c x^2\right )^{3/2}} \, dx &=\frac {A+B x}{a x^3 \sqrt {a+c x^2}}-\frac {\int \frac {-4 a A c-3 a B c x}{x^4 \sqrt {a+c x^2}} \, dx}{a^2 c}\\ &=\frac {A+B x}{a x^3 \sqrt {a+c x^2}}-\frac {4 A \sqrt {a+c x^2}}{3 a^2 x^3}+\frac {\int \frac {9 a^2 B c-8 a A c^2 x}{x^3 \sqrt {a+c x^2}} \, dx}{3 a^3 c}\\ &=\frac {A+B x}{a x^3 \sqrt {a+c x^2}}-\frac {4 A \sqrt {a+c x^2}}{3 a^2 x^3}-\frac {3 B \sqrt {a+c x^2}}{2 a^2 x^2}-\frac {\int \frac {16 a^2 A c^2+9 a^2 B c^2 x}{x^2 \sqrt {a+c x^2}} \, dx}{6 a^4 c}\\ &=\frac {A+B x}{a x^3 \sqrt {a+c x^2}}-\frac {4 A \sqrt {a+c x^2}}{3 a^2 x^3}-\frac {3 B \sqrt {a+c x^2}}{2 a^2 x^2}+\frac {8 A c \sqrt {a+c x^2}}{3 a^3 x}-\frac {(3 B c) \int \frac {1}{x \sqrt {a+c x^2}} \, dx}{2 a^2}\\ &=\frac {A+B x}{a x^3 \sqrt {a+c x^2}}-\frac {4 A \sqrt {a+c x^2}}{3 a^2 x^3}-\frac {3 B \sqrt {a+c x^2}}{2 a^2 x^2}+\frac {8 A c \sqrt {a+c x^2}}{3 a^3 x}-\frac {(3 B c) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{4 a^2}\\ &=\frac {A+B x}{a x^3 \sqrt {a+c x^2}}-\frac {4 A \sqrt {a+c x^2}}{3 a^2 x^3}-\frac {3 B \sqrt {a+c x^2}}{2 a^2 x^2}+\frac {8 A c \sqrt {a+c x^2}}{3 a^3 x}-\frac {(3 B) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )}{2 a^2}\\ &=\frac {A+B x}{a x^3 \sqrt {a+c x^2}}-\frac {4 A \sqrt {a+c x^2}}{3 a^2 x^3}-\frac {3 B \sqrt {a+c x^2}}{2 a^2 x^2}+\frac {8 A c \sqrt {a+c x^2}}{3 a^3 x}+\frac {3 B c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 90, normalized size = 0.75 \begin {gather*} \frac {-\frac {a^2 (2 A+3 B x)}{x^3}+a \left (\frac {8 A c}{x}-9 B c\right )+9 a B c \sqrt {\frac {c x^2}{a}+1} \tanh ^{-1}\left (\sqrt {\frac {c x^2}{a}+1}\right )+16 A c^2 x}{6 a^3 \sqrt {a+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^4*(a + c*x^2)^(3/2)),x]

[Out]

(a*(-9*B*c + (8*A*c)/x) + 16*A*c^2*x - (a^2*(2*A + 3*B*x))/x^3 + 9*a*B*c*Sqrt[1 + (c*x^2)/a]*ArcTanh[Sqrt[1 +
(c*x^2)/a]])/(6*a^3*Sqrt[a + c*x^2])

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IntegrateAlgebraic [A]  time = 0.55, size = 102, normalized size = 0.85 \begin {gather*} \frac {-2 a^2 A-3 a^2 B x+8 a A c x^2-9 a B c x^3+16 A c^2 x^4}{6 a^3 x^3 \sqrt {a+c x^2}}-\frac {3 B c \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x^4*(a + c*x^2)^(3/2)),x]

[Out]

(-2*a^2*A - 3*a^2*B*x + 8*a*A*c*x^2 - 9*a*B*c*x^3 + 16*A*c^2*x^4)/(6*a^3*x^3*Sqrt[a + c*x^2]) - (3*B*c*ArcTanh
[(Sqrt[c]*x)/Sqrt[a] - Sqrt[a + c*x^2]/Sqrt[a]])/a^(5/2)

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fricas [A]  time = 0.46, size = 232, normalized size = 1.93 \begin {gather*} \left [\frac {9 \, {\left (B c^{2} x^{5} + B a c x^{3}\right )} \sqrt {a} \log \left (-\frac {c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (16 \, A c^{2} x^{4} - 9 \, B a c x^{3} + 8 \, A a c x^{2} - 3 \, B a^{2} x - 2 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{12 \, {\left (a^{3} c x^{5} + a^{4} x^{3}\right )}}, -\frac {9 \, {\left (B c^{2} x^{5} + B a c x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (16 \, A c^{2} x^{4} - 9 \, B a c x^{3} + 8 \, A a c x^{2} - 3 \, B a^{2} x - 2 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{6 \, {\left (a^{3} c x^{5} + a^{4} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^4/(c*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

[1/12*(9*(B*c^2*x^5 + B*a*c*x^3)*sqrt(a)*log(-(c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(16*A*c^2*x^4
- 9*B*a*c*x^3 + 8*A*a*c*x^2 - 3*B*a^2*x - 2*A*a^2)*sqrt(c*x^2 + a))/(a^3*c*x^5 + a^4*x^3), -1/6*(9*(B*c^2*x^5
+ B*a*c*x^3)*sqrt(-a)*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - (16*A*c^2*x^4 - 9*B*a*c*x^3 + 8*A*a*c*x^2 - 3*B*a^2*
x - 2*A*a^2)*sqrt(c*x^2 + a))/(a^3*c*x^5 + a^4*x^3)]

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giac [B]  time = 0.21, size = 203, normalized size = 1.69 \begin {gather*} \frac {\frac {A c^{2} x}{a^{3}} - \frac {B c}{a^{2}}}{\sqrt {c x^{2} + a}} - \frac {3 \, B c \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{5} B c - 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} A c^{\frac {3}{2}} + 24 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} A a c^{\frac {3}{2}} - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} B a^{2} c - 10 \, A a^{2} c^{\frac {3}{2}}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{3} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^4/(c*x^2+a)^(3/2),x, algorithm="giac")

[Out]

(A*c^2*x/a^3 - B*c/a^2)/sqrt(c*x^2 + a) - 3*B*c*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^2)
+ 1/3*(3*(sqrt(c)*x - sqrt(c*x^2 + a))^5*B*c - 6*(sqrt(c)*x - sqrt(c*x^2 + a))^4*A*c^(3/2) + 24*(sqrt(c)*x -
sqrt(c*x^2 + a))^2*A*a*c^(3/2) - 3*(sqrt(c)*x - sqrt(c*x^2 + a))*B*a^2*c - 10*A*a^2*c^(3/2))/(((sqrt(c)*x - sq
rt(c*x^2 + a))^2 - a)^3*a^2)

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maple [A]  time = 0.05, size = 122, normalized size = 1.02 \begin {gather*} \frac {8 A \,c^{2} x}{3 \sqrt {c \,x^{2}+a}\, a^{3}}+\frac {3 B c \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2 a^{\frac {5}{2}}}-\frac {3 B c}{2 \sqrt {c \,x^{2}+a}\, a^{2}}+\frac {4 A c}{3 \sqrt {c \,x^{2}+a}\, a^{2} x}-\frac {B}{2 \sqrt {c \,x^{2}+a}\, a \,x^{2}}-\frac {A}{3 \sqrt {c \,x^{2}+a}\, a \,x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^4/(c*x^2+a)^(3/2),x)

[Out]

-1/3*A/a/x^3/(c*x^2+a)^(1/2)+4/3*A*c/a^2/x/(c*x^2+a)^(1/2)+8/3*A*c^2/a^3*x/(c*x^2+a)^(1/2)-1/2*B/a/x^2/(c*x^2+
a)^(1/2)-3/2*B*c/a^2/(c*x^2+a)^(1/2)+3/2*B*c/a^(5/2)*ln((2*a+2*(c*x^2+a)^(1/2)*a^(1/2))/x)

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maxima [A]  time = 0.50, size = 110, normalized size = 0.92 \begin {gather*} \frac {8 \, A c^{2} x}{3 \, \sqrt {c x^{2} + a} a^{3}} + \frac {3 \, B c \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right )}{2 \, a^{\frac {5}{2}}} - \frac {3 \, B c}{2 \, \sqrt {c x^{2} + a} a^{2}} + \frac {4 \, A c}{3 \, \sqrt {c x^{2} + a} a^{2} x} - \frac {B}{2 \, \sqrt {c x^{2} + a} a x^{2}} - \frac {A}{3 \, \sqrt {c x^{2} + a} a x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^4/(c*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

8/3*A*c^2*x/(sqrt(c*x^2 + a)*a^3) + 3/2*B*c*arcsinh(a/(sqrt(a*c)*abs(x)))/a^(5/2) - 3/2*B*c/(sqrt(c*x^2 + a)*a
^2) + 4/3*A*c/(sqrt(c*x^2 + a)*a^2*x) - 1/2*B/(sqrt(c*x^2 + a)*a*x^2) - 1/3*A/(sqrt(c*x^2 + a)*a*x^3)

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mupad [B]  time = 1.85, size = 95, normalized size = 0.79 \begin {gather*} \frac {3\,B\,c\,\mathrm {atanh}\left (\frac {\sqrt {c\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{5/2}}-\frac {B}{2\,a\,x^2\,\sqrt {c\,x^2+a}}-\frac {3\,B\,c}{2\,a^2\,\sqrt {c\,x^2+a}}+\frac {A\,\left (-a^2+4\,a\,c\,x^2+8\,c^2\,x^4\right )}{3\,a^3\,x^3\,\sqrt {c\,x^2+a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^4*(a + c*x^2)^(3/2)),x)

[Out]

(3*B*c*atanh((a + c*x^2)^(1/2)/a^(1/2)))/(2*a^(5/2)) - B/(2*a*x^2*(a + c*x^2)^(1/2)) - (3*B*c)/(2*a^2*(a + c*x
^2)^(1/2)) + (A*(8*c^2*x^4 - a^2 + 4*a*c*x^2))/(3*a^3*x^3*(a + c*x^2)^(1/2))

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sympy [B]  time = 28.08, size = 311, normalized size = 2.59 \begin {gather*} A \left (- \frac {a^{3} c^{\frac {9}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{3 a^{5} c^{4} x^{2} + 6 a^{4} c^{5} x^{4} + 3 a^{3} c^{6} x^{6}} + \frac {3 a^{2} c^{\frac {11}{2}} x^{2} \sqrt {\frac {a}{c x^{2}} + 1}}{3 a^{5} c^{4} x^{2} + 6 a^{4} c^{5} x^{4} + 3 a^{3} c^{6} x^{6}} + \frac {12 a c^{\frac {13}{2}} x^{4} \sqrt {\frac {a}{c x^{2}} + 1}}{3 a^{5} c^{4} x^{2} + 6 a^{4} c^{5} x^{4} + 3 a^{3} c^{6} x^{6}} + \frac {8 c^{\frac {15}{2}} x^{6} \sqrt {\frac {a}{c x^{2}} + 1}}{3 a^{5} c^{4} x^{2} + 6 a^{4} c^{5} x^{4} + 3 a^{3} c^{6} x^{6}}\right ) + B \left (- \frac {1}{2 a \sqrt {c} x^{3} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {3 \sqrt {c}}{2 a^{2} x \sqrt {\frac {a}{c x^{2}} + 1}} + \frac {3 c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{2 a^{\frac {5}{2}}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**4/(c*x**2+a)**(3/2),x)

[Out]

A*(-a**3*c**(9/2)*sqrt(a/(c*x**2) + 1)/(3*a**5*c**4*x**2 + 6*a**4*c**5*x**4 + 3*a**3*c**6*x**6) + 3*a**2*c**(1
1/2)*x**2*sqrt(a/(c*x**2) + 1)/(3*a**5*c**4*x**2 + 6*a**4*c**5*x**4 + 3*a**3*c**6*x**6) + 12*a*c**(13/2)*x**4*
sqrt(a/(c*x**2) + 1)/(3*a**5*c**4*x**2 + 6*a**4*c**5*x**4 + 3*a**3*c**6*x**6) + 8*c**(15/2)*x**6*sqrt(a/(c*x**
2) + 1)/(3*a**5*c**4*x**2 + 6*a**4*c**5*x**4 + 3*a**3*c**6*x**6)) + B*(-1/(2*a*sqrt(c)*x**3*sqrt(a/(c*x**2) +
1)) - 3*sqrt(c)/(2*a**2*x*sqrt(a/(c*x**2) + 1)) + 3*c*asinh(sqrt(a)/(sqrt(c)*x))/(2*a**(5/2)))

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