3.4.36 \(\int \frac {(A+B x) (a+c x^2)^{3/2}}{x^6} \, dx\)

Optimal. Leaf size=93 \[ -\frac {A \left (a+c x^2\right )^{5/2}}{5 a x^5}-\frac {3 B c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{8 \sqrt {a}}-\frac {3 B c \sqrt {a+c x^2}}{8 x^2}-\frac {B \left (a+c x^2\right )^{3/2}}{4 x^4} \]

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Rubi [A]  time = 0.06, antiderivative size = 93, 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 = {807, 266, 47, 63, 208} \begin {gather*} -\frac {A \left (a+c x^2\right )^{5/2}}{5 a x^5}-\frac {3 B c^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{8 \sqrt {a}}-\frac {B \left (a+c x^2\right )^{3/2}}{4 x^4}-\frac {3 B c \sqrt {a+c x^2}}{8 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^(3/2))/x^6,x]

[Out]

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 103, normalized size = 1.11 \begin {gather*} \frac {-\frac {\left (a+c x^2\right ) \left (2 a^2 (4 A+5 B x)+a c x^2 (16 A+25 B x)+8 A c^2 x^4\right )}{a x^5}-15 B c^2 \sqrt {\frac {c x^2}{a}+1} \tanh ^{-1}\left (\sqrt {\frac {c x^2}{a}+1}\right )}{40 \sqrt {a+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2)^(3/2))/x^6,x]

[Out]

(-(((a + c*x^2)*(8*A*c^2*x^4 + 2*a^2*(4*A + 5*B*x) + a*c*x^2*(16*A + 25*B*x)))/(a*x^5)) - 15*B*c^2*Sqrt[1 + (c
*x^2)/a]*ArcTanh[Sqrt[1 + (c*x^2)/a]])/(40*Sqrt[a + c*x^2])

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IntegrateAlgebraic [A]  time = 0.63, size = 106, normalized size = 1.14 \begin {gather*} \frac {\sqrt {a+c x^2} \left (-8 a^2 A-10 a^2 B x-16 a A c x^2-25 a B c x^3-8 A c^2 x^4\right )}{40 a x^5}+\frac {3 B c^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{4 \sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(a + c*x^2)^(3/2))/x^6,x]

[Out]

(Sqrt[a + c*x^2]*(-8*a^2*A - 10*a^2*B*x - 16*a*A*c*x^2 - 25*a*B*c*x^3 - 8*A*c^2*x^4))/(40*a*x^5) + (3*B*c^2*Ar
cTanh[(Sqrt[c]*x)/Sqrt[a] - Sqrt[a + c*x^2]/Sqrt[a]])/(4*Sqrt[a])

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fricas [A]  time = 0.45, size = 190, normalized size = 2.04 \begin {gather*} \left [\frac {15 \, B \sqrt {a} c^{2} x^{5} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (8 \, A c^{2} x^{4} + 25 \, B a c x^{3} + 16 \, A a c x^{2} + 10 \, B a^{2} x + 8 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{80 \, a x^{5}}, \frac {15 \, B \sqrt {-a} c^{2} x^{5} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - {\left (8 \, A c^{2} x^{4} + 25 \, B a c x^{3} + 16 \, A a c x^{2} + 10 \, B a^{2} x + 8 \, A a^{2}\right )} \sqrt {c x^{2} + a}}{40 \, a x^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^6,x, algorithm="fricas")

[Out]

[1/80*(15*B*sqrt(a)*c^2*x^5*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(8*A*c^2*x^4 + 25*B*a*c*x^
3 + 16*A*a*c*x^2 + 10*B*a^2*x + 8*A*a^2)*sqrt(c*x^2 + a))/(a*x^5), 1/40*(15*B*sqrt(-a)*c^2*x^5*arctan(sqrt(-a)
/sqrt(c*x^2 + a)) - (8*A*c^2*x^4 + 25*B*a*c*x^3 + 16*A*a*c*x^2 + 10*B*a^2*x + 8*A*a^2)*sqrt(c*x^2 + a))/(a*x^5
)]

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giac [B]  time = 0.22, size = 232, normalized size = 2.49 \begin {gather*} \frac {3 \, B c^{2} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a}} + \frac {25 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{9} B c^{2} + 40 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{8} A c^{\frac {5}{2}} - 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{7} B a c^{2} + 80 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{4} A a^{2} c^{\frac {5}{2}} + 10 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{3} B a^{3} c^{2} - 25 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} B a^{4} c^{2} + 8 \, A a^{4} c^{\frac {5}{2}}}{20 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )}^{2} - a\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^6,x, algorithm="giac")

[Out]

3/4*B*c^2*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/sqrt(-a) + 1/20*(25*(sqrt(c)*x - sqrt(c*x^2 + a))^9*
B*c^2 + 40*(sqrt(c)*x - sqrt(c*x^2 + a))^8*A*c^(5/2) - 10*(sqrt(c)*x - sqrt(c*x^2 + a))^7*B*a*c^2 + 80*(sqrt(c
)*x - sqrt(c*x^2 + a))^4*A*a^2*c^(5/2) + 10*(sqrt(c)*x - sqrt(c*x^2 + a))^3*B*a^3*c^2 - 25*(sqrt(c)*x - sqrt(c
*x^2 + a))*B*a^4*c^2 + 8*A*a^4*c^(5/2))/((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^5

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maple [A]  time = 0.06, size = 125, normalized size = 1.34 \begin {gather*} -\frac {3 B \,c^{2} \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{8 \sqrt {a}}+\frac {3 \sqrt {c \,x^{2}+a}\, B \,c^{2}}{8 a}+\frac {\left (c \,x^{2}+a \right )^{\frac {3}{2}} B \,c^{2}}{8 a^{2}}-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} B c}{8 a^{2} x^{2}}-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} B}{4 a \,x^{4}}-\frac {\left (c \,x^{2}+a \right )^{\frac {5}{2}} A}{5 a \,x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^(3/2)/x^6,x)

[Out]

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

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maxima [A]  time = 0.65, size = 113, normalized size = 1.22 \begin {gather*} -\frac {3 \, B c^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a c} {\left | x \right |}}\right )}{8 \, \sqrt {a}} + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} B c^{2}}{8 \, a^{2}} + \frac {3 \, \sqrt {c x^{2} + a} B c^{2}}{8 \, a} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} B c}{8 \, a^{2} x^{2}} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} B}{4 \, a x^{4}} - \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} A}{5 \, a x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^6,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.76, size = 73, normalized size = 0.78 \begin {gather*} \frac {3\,B\,a\,\sqrt {c\,x^2+a}}{8\,x^4}-\frac {3\,B\,c^2\,\mathrm {atanh}\left (\frac {\sqrt {c\,x^2+a}}{\sqrt {a}}\right )}{8\,\sqrt {a}}-\frac {5\,B\,{\left (c\,x^2+a\right )}^{3/2}}{8\,x^4}-\frac {A\,{\left (c\,x^2+a\right )}^{5/2}}{5\,a\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)^(3/2)*(A + B*x))/x^6,x)

[Out]

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

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sympy [B]  time = 8.95, size = 199, normalized size = 2.14 \begin {gather*} - \frac {A a \sqrt {c} \sqrt {\frac {a}{c x^{2}} + 1}}{5 x^{4}} - \frac {2 A c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{5 x^{2}} - \frac {A c^{\frac {5}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{5 a} - \frac {B a^{2}}{4 \sqrt {c} x^{5} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {3 B a \sqrt {c}}{8 x^{3} \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {B c^{\frac {3}{2}} \sqrt {\frac {a}{c x^{2}} + 1}}{2 x} - \frac {B c^{\frac {3}{2}}}{8 x \sqrt {\frac {a}{c x^{2}} + 1}} - \frac {3 B c^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x} \right )}}{8 \sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**(3/2)/x**6,x)

[Out]

-A*a*sqrt(c)*sqrt(a/(c*x**2) + 1)/(5*x**4) - 2*A*c**(3/2)*sqrt(a/(c*x**2) + 1)/(5*x**2) - A*c**(5/2)*sqrt(a/(c
*x**2) + 1)/(5*a) - B*a**2/(4*sqrt(c)*x**5*sqrt(a/(c*x**2) + 1)) - 3*B*a*sqrt(c)/(8*x**3*sqrt(a/(c*x**2) + 1))
- B*c**(3/2)*sqrt(a/(c*x**2) + 1)/(2*x) - B*c**(3/2)/(8*x*sqrt(a/(c*x**2) + 1)) - 3*B*c**2*asinh(sqrt(a)/(sqr
t(c)*x))/(8*sqrt(a))

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