3.335 \(\int \frac{(A+B x) (a+c x^2)^{3/2}}{x^5} \, dx\)

Optimal. Leaf size=111 \[ -\frac{c \sqrt{a+c x^2} (3 A+8 B x)}{8 x^2}-\frac{\left (a+c x^2\right )^{3/2} (3 A+4 B x)}{12 x^4}-\frac{3 A c^2 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{8 \sqrt{a}}+B c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right ) \]

[Out]

-(c*(3*A + 8*B*x)*Sqrt[a + c*x^2])/(8*x^2) - ((3*A + 4*B*x)*(a + c*x^2)^(3/2))/(12*x^4) + B*c^(3/2)*ArcTanh[(S
qrt[c]*x)/Sqrt[a + c*x^2]] - (3*A*c^2*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/(8*Sqrt[a])

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Rubi [A]  time = 0.087298, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {811, 844, 217, 206, 266, 63, 208} \[ -\frac{c \sqrt{a+c x^2} (3 A+8 B x)}{8 x^2}-\frac{\left (a+c x^2\right )^{3/2} (3 A+4 B x)}{12 x^4}-\frac{3 A c^2 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{8 \sqrt{a}}+B c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^(3/2))/x^5,x]

[Out]

-(c*(3*A + 8*B*x)*Sqrt[a + c*x^2])/(8*x^2) - ((3*A + 4*B*x)*(a + c*x^2)^(3/2))/(12*x^4) + B*c^(3/2)*ArcTanh[(S
qrt[c]*x)/Sqrt[a + c*x^2]] - (3*A*c^2*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/(8*Sqrt[a])

Rule 811

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

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 0]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 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^5} \, dx &=-\frac{(3 A+4 B x) \left (a+c x^2\right )^{3/2}}{12 x^4}-\frac{\int \frac{(-6 a A c-8 a B c x) \sqrt{a+c x^2}}{x^3} \, dx}{8 a}\\ &=-\frac{c (3 A+8 B x) \sqrt{a+c x^2}}{8 x^2}-\frac{(3 A+4 B x) \left (a+c x^2\right )^{3/2}}{12 x^4}+\frac{\int \frac{12 a^2 A c^2+32 a^2 B c^2 x}{x \sqrt{a+c x^2}} \, dx}{32 a^2}\\ &=-\frac{c (3 A+8 B x) \sqrt{a+c x^2}}{8 x^2}-\frac{(3 A+4 B x) \left (a+c x^2\right )^{3/2}}{12 x^4}+\frac{1}{8} \left (3 A c^2\right ) \int \frac{1}{x \sqrt{a+c x^2}} \, dx+\left (B c^2\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx\\ &=-\frac{c (3 A+8 B x) \sqrt{a+c x^2}}{8 x^2}-\frac{(3 A+4 B x) \left (a+c x^2\right )^{3/2}}{12 x^4}+\frac{1}{16} \left (3 A c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+c x}} \, dx,x,x^2\right )+\left (B c^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )\\ &=-\frac{c (3 A+8 B x) \sqrt{a+c x^2}}{8 x^2}-\frac{(3 A+4 B x) \left (a+c x^2\right )^{3/2}}{12 x^4}+B c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )+\frac{1}{8} (3 A c) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{c}+\frac{x^2}{c}} \, dx,x,\sqrt{a+c x^2}\right )\\ &=-\frac{c (3 A+8 B x) \sqrt{a+c x^2}}{8 x^2}-\frac{(3 A+4 B x) \left (a+c x^2\right )^{3/2}}{12 x^4}+B c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )-\frac{3 A c^2 \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{8 \sqrt{a}}\\ \end{align*}

Mathematica [C]  time = 0.0864605, size = 114, normalized size = 1.03 \[ -\frac{\sqrt{a+c x^2} \left (8 a^2 B x \, _2F_1\left (-\frac{3}{2},-\frac{3}{2};-\frac{1}{2};-\frac{c x^2}{a}\right )+9 A c^2 x^4 \tanh ^{-1}\left (\sqrt{\frac{c x^2}{a}+1}\right )+3 a A \left (2 a+5 c x^2\right ) \sqrt{\frac{c x^2}{a}+1}\right )}{24 a x^4 \sqrt{\frac{c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2)^(3/2))/x^5,x]

[Out]

-(Sqrt[a + c*x^2]*(3*a*A*(2*a + 5*c*x^2)*Sqrt[1 + (c*x^2)/a] + 9*A*c^2*x^4*ArcTanh[Sqrt[1 + (c*x^2)/a]] + 8*a^
2*B*x*Hypergeometric2F1[-3/2, -3/2, -1/2, -((c*x^2)/a)]))/(24*a*x^4*Sqrt[1 + (c*x^2)/a])

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Maple [B]  time = 0.01, size = 202, normalized size = 1.8 \begin{align*} -{\frac{B}{3\,a{x}^{3}} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{2\,Bc}{3\,{a}^{2}x} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{2\,B{c}^{2}x}{3\,{a}^{2}} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{B{c}^{2}x}{a}\sqrt{c{x}^{2}+a}}+B{c}^{{\frac{3}{2}}}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ) -{\frac{A}{4\,a{x}^{4}} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{Ac}{8\,{a}^{2}{x}^{2}} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{A{c}^{2}}{8\,{a}^{2}} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{3\,A{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\,A{c}^{2}}{8\,a}\sqrt{c{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^(3/2)/x^5,x)

[Out]

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

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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^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.80599, size = 1169, normalized size = 10.53 \begin{align*} \left [\frac{24 \, B a c^{\frac{3}{2}} x^{4} \log \left (-2 \, c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) + 9 \, A \sqrt{a} c^{2} x^{4} \log \left (-\frac{c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) - 2 \,{\left (32 \, B a c x^{3} + 15 \, A a c x^{2} + 8 \, B a^{2} x + 6 \, A a^{2}\right )} \sqrt{c x^{2} + a}}{48 \, a x^{4}}, -\frac{48 \, B a \sqrt{-c} c x^{4} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) - 9 \, A \sqrt{a} c^{2} x^{4} \log \left (-\frac{c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (32 \, B a c x^{3} + 15 \, A a c x^{2} + 8 \, B a^{2} x + 6 \, A a^{2}\right )} \sqrt{c x^{2} + a}}{48 \, a x^{4}}, \frac{9 \, A \sqrt{-a} c^{2} x^{4} \arctan \left (\frac{\sqrt{-a}}{\sqrt{c x^{2} + a}}\right ) + 12 \, B a c^{\frac{3}{2}} x^{4} \log \left (-2 \, c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) -{\left (32 \, B a c x^{3} + 15 \, A a c x^{2} + 8 \, B a^{2} x + 6 \, A a^{2}\right )} \sqrt{c x^{2} + a}}{24 \, a x^{4}}, -\frac{24 \, B a \sqrt{-c} c x^{4} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) - 9 \, A \sqrt{-a} c^{2} x^{4} \arctan \left (\frac{\sqrt{-a}}{\sqrt{c x^{2} + a}}\right ) +{\left (32 \, B a c x^{3} + 15 \, A a c x^{2} + 8 \, B a^{2} x + 6 \, A a^{2}\right )} \sqrt{c x^{2} + a}}{24 \, a x^{4}}\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^5,x, algorithm="fricas")

[Out]

[1/48*(24*B*a*c^(3/2)*x^4*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 9*A*sqrt(a)*c^2*x^4*log(-(c*x^2 -
2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(32*B*a*c*x^3 + 15*A*a*c*x^2 + 8*B*a^2*x + 6*A*a^2)*sqrt(c*x^2 + a))
/(a*x^4), -1/48*(48*B*a*sqrt(-c)*c*x^4*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - 9*A*sqrt(a)*c^2*x^4*log(-(c*x^2 -
2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(32*B*a*c*x^3 + 15*A*a*c*x^2 + 8*B*a^2*x + 6*A*a^2)*sqrt(c*x^2 + a))
/(a*x^4), 1/24*(9*A*sqrt(-a)*c^2*x^4*arctan(sqrt(-a)/sqrt(c*x^2 + a)) + 12*B*a*c^(3/2)*x^4*log(-2*c*x^2 - 2*sq
rt(c*x^2 + a)*sqrt(c)*x - a) - (32*B*a*c*x^3 + 15*A*a*c*x^2 + 8*B*a^2*x + 6*A*a^2)*sqrt(c*x^2 + a))/(a*x^4), -
1/24*(24*B*a*sqrt(-c)*c*x^4*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - 9*A*sqrt(-a)*c^2*x^4*arctan(sqrt(-a)/sqrt(c*x
^2 + a)) + (32*B*a*c*x^3 + 15*A*a*c*x^2 + 8*B*a^2*x + 6*A*a^2)*sqrt(c*x^2 + a))/(a*x^4)]

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Sympy [B]  time = 11.3187, size = 236, normalized size = 2.13 \begin{align*} - \frac{A a^{2}}{4 \sqrt{c} x^{5} \sqrt{\frac{a}{c x^{2}} + 1}} - \frac{3 A a \sqrt{c}}{8 x^{3} \sqrt{\frac{a}{c x^{2}} + 1}} - \frac{A c^{\frac{3}{2}} \sqrt{\frac{a}{c x^{2}} + 1}}{2 x} - \frac{A c^{\frac{3}{2}}}{8 x \sqrt{\frac{a}{c x^{2}} + 1}} - \frac{3 A c^{2} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{c} x} \right )}}{8 \sqrt{a}} - \frac{B \sqrt{a} c}{x \sqrt{1 + \frac{c x^{2}}{a}}} - \frac{B a \sqrt{c} \sqrt{\frac{a}{c x^{2}} + 1}}{3 x^{2}} - \frac{B c^{\frac{3}{2}} \sqrt{\frac{a}{c x^{2}} + 1}}{3} + B c^{\frac{3}{2}} \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{a}} \right )} - \frac{B c^{2} x}{\sqrt{a} \sqrt{1 + \frac{c x^{2}}{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**5,x)

[Out]

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

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Giac [B]  time = 1.17331, size = 385, normalized size = 3.47 \begin{align*} \frac{3 \, A c^{2} \arctan \left (-\frac{\sqrt{c} x - \sqrt{c x^{2} + a}}{\sqrt{-a}}\right )}{4 \, \sqrt{-a}} - B c^{\frac{3}{2}} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right ) + \frac{15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{7} A c^{2} + 48 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{6} B a c^{\frac{3}{2}} + 9 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{5} A a c^{2} - 96 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{4} B a^{2} c^{\frac{3}{2}} + 9 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{3} A a^{2} c^{2} + 80 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} B a^{3} c^{\frac{3}{2}} + 15 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} A a^{3} c^{2} - 32 \, B a^{4} c^{\frac{3}{2}}}{12 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} - a\right )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^(3/2)/x^5,x, algorithm="giac")

[Out]

3/4*A*c^2*arctan(-(sqrt(c)*x - sqrt(c*x^2 + a))/sqrt(-a))/sqrt(-a) - B*c^(3/2)*log(abs(-sqrt(c)*x + sqrt(c*x^2
+ a))) + 1/12*(15*(sqrt(c)*x - sqrt(c*x^2 + a))^7*A*c^2 + 48*(sqrt(c)*x - sqrt(c*x^2 + a))^6*B*a*c^(3/2) + 9*
(sqrt(c)*x - sqrt(c*x^2 + a))^5*A*a*c^2 - 96*(sqrt(c)*x - sqrt(c*x^2 + a))^4*B*a^2*c^(3/2) + 9*(sqrt(c)*x - sq
rt(c*x^2 + a))^3*A*a^2*c^2 + 80*(sqrt(c)*x - sqrt(c*x^2 + a))^2*B*a^3*c^(3/2) + 15*(sqrt(c)*x - sqrt(c*x^2 + a
))*A*a^3*c^2 - 32*B*a^4*c^(3/2))/((sqrt(c)*x - sqrt(c*x^2 + a))^2 - a)^4