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

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

[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])

________________________________________________________________________________________

Rubi [A]  time = 0.0568545, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {807, 266, 47, 63, 208} \[ -\frac{A \left (a+c x^2\right )^{5/2}}{5 a x^5}-\frac{3 B c^2 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{8 \sqrt{a}}-\frac{B \left (a+c x^2\right )^{3/2}}{4 x^4}-\frac{3 B c \sqrt{a+c x^2}}{8 x^2} \]

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 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 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 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]

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*}

Mathematica [A]  time = 0.0855196, size = 103, normalized size = 1.11 \[ \frac{-\frac{\left (a+c x^2\right ) \left (2 a^2 (4 A+5 B x)+a c x^2 (16 A+25 B x)+8 A c^2 x^4\right )}{a x^5}-15 B c^2 \sqrt{\frac{c x^2}{a}+1} \tanh ^{-1}\left (\sqrt{\frac{c x^2}{a}+1}\right )}{40 \sqrt{a+c x^2}} \]

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])

________________________________________________________________________________________

Maple [A]  time = 0.012, size = 125, normalized size = 1.3 \begin{align*} -{\frac{B}{4\,a{x}^{4}} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{Bc}{8\,{a}^{2}{x}^{2}} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{B{c}^{2}}{8\,{a}^{2}} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{3\,B{c}^{2}}{8}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{c{x}^{2}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}+{\frac{3\,B{c}^{2}}{8\,a}\sqrt{c{x}^{2}+a}}-{\frac{A}{5\,a{x}^{5}} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}} \end{align*}

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]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.64154, size = 454, normalized size = 4.88 \begin{align*} \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{align*}

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
)]

________________________________________________________________________________________

Sympy [B]  time = 11.1437, size = 199, normalized size = 2.14 \begin{align*} - \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{align*}

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))

________________________________________________________________________________________

Giac [B]  time = 1.13682, size = 313, normalized size = 3.37 \begin{align*} \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{align*}

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