3.13 \(\int \frac{(A+B x) (a+b x^2)^{3/2}}{x^2} \, dx\)

Optimal. Leaf size=108 \[ a^{3/2} (-B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )-\frac{\left (a+b x^2\right )^{3/2} (3 A-B x)}{3 x}+\frac{1}{2} \sqrt{a+b x^2} (2 a B+3 A b x)+\frac{3}{2} a A \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0890723, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {813, 815, 844, 217, 206, 266, 63, 208} \[ a^{3/2} (-B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )-\frac{\left (a+b x^2\right )^{3/2} (3 A-B x)}{3 x}+\frac{1}{2} \sqrt{a+b x^2} (2 a B+3 A b x)+\frac{3}{2} a A \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 813

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

Rule 815

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

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

Mathematica [C]  time = 0.166813, size = 105, normalized size = 0.97 \[ -\frac{a^2 A \sqrt{\frac{b x^2}{a}+1} \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};-\frac{b x^2}{a}\right )}{x \sqrt{a+b x^2}}-a^{3/2} B \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )+\frac{1}{3} B \sqrt{a+b x^2} \left (4 a+b x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*Sqrt[a + b*x^2]*(4*a + b*x^2))/3 - a^(3/2)*B*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]] - (a^2*A*Sqrt[1 + (b*x^2)/a]*
Hypergeometric2F1[-3/2, -1/2, 1/2, -((b*x^2)/a)])/(x*Sqrt[a + b*x^2])

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Maple [A]  time = 0.007, size = 126, normalized size = 1.2 \begin{align*}{\frac{B}{3} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-B{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ) +B\sqrt{b{x}^{2}+a}a-{\frac{A}{ax} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{Abx}{a} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,Abx}{2}\sqrt{b{x}^{2}+a}}+{\frac{3\,Aa}{2}\sqrt{b}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*B*(b*x^2+a)^(3/2)-B*a^(3/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)+B*(b*x^2+a)^(1/2)*a-A/a/x*(b*x^2+a)^(5/2
)+A*b/a*x*(b*x^2+a)^(3/2)+3/2*A*b*x*(b*x^2+a)^(1/2)+3/2*A*b^(1/2)*a*ln(x*b^(1/2)+(b*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)*(b*x^2+a)^(3/2)/x^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.67224, size = 1033, normalized size = 9.56 \begin{align*} \left [\frac{9 \, A a \sqrt{b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + 6 \, B a^{\frac{3}{2}} x \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (2 \, B b x^{3} + 3 \, A b x^{2} + 8 \, B a x - 6 \, A a\right )} \sqrt{b x^{2} + a}}{12 \, x}, -\frac{9 \, A a \sqrt{-b} x \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) - 3 \, B a^{\frac{3}{2}} x \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) -{\left (2 \, B b x^{3} + 3 \, A b x^{2} + 8 \, B a x - 6 \, A a\right )} \sqrt{b x^{2} + a}}{6 \, x}, \frac{12 \, B \sqrt{-a} a x \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) + 9 \, A a \sqrt{b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + 2 \,{\left (2 \, B b x^{3} + 3 \, A b x^{2} + 8 \, B a x - 6 \, A a\right )} \sqrt{b x^{2} + a}}{12 \, x}, -\frac{9 \, A a \sqrt{-b} x \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) - 6 \, B \sqrt{-a} a x \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) -{\left (2 \, B b x^{3} + 3 \, A b x^{2} + 8 \, B a x - 6 \, A a\right )} \sqrt{b x^{2} + a}}{6 \, x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(9*A*a*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 6*B*a^(3/2)*x*log(-(b*x^2 - 2*sqrt(b*
x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(2*B*b*x^3 + 3*A*b*x^2 + 8*B*a*x - 6*A*a)*sqrt(b*x^2 + a))/x, -1/6*(9*A*a*sqr
t(-b)*x*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 3*B*a^(3/2)*x*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2)
- (2*B*b*x^3 + 3*A*b*x^2 + 8*B*a*x - 6*A*a)*sqrt(b*x^2 + a))/x, 1/12*(12*B*sqrt(-a)*a*x*arctan(sqrt(-a)/sqrt(
b*x^2 + a)) + 9*A*a*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(2*B*b*x^3 + 3*A*b*x^2 + 8*B
*a*x - 6*A*a)*sqrt(b*x^2 + a))/x, -1/6*(9*A*a*sqrt(-b)*x*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - 6*B*sqrt(-a)*a*x
*arctan(sqrt(-a)/sqrt(b*x^2 + a)) - (2*B*b*x^3 + 3*A*b*x^2 + 8*B*a*x - 6*A*a)*sqrt(b*x^2 + a))/x]

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Sympy [A]  time = 5.77296, size = 184, normalized size = 1.7 \begin{align*} - \frac{A a^{\frac{3}{2}}}{x \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{A \sqrt{a} b x \sqrt{1 + \frac{b x^{2}}{a}}}{2} - \frac{A \sqrt{a} b x}{\sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 A a \sqrt{b} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{2} - B a^{\frac{3}{2}} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )} + \frac{B a^{2}}{\sqrt{b} x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{B a \sqrt{b} x}{\sqrt{\frac{a}{b x^{2}} + 1}} + B b \left (\begin{cases} \frac{\sqrt{a} x^{2}}{2} & \text{for}\: b = 0 \\\frac{\left (a + b x^{2}\right )^{\frac{3}{2}}}{3 b} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-A*a**(3/2)/(x*sqrt(1 + b*x**2/a)) + A*sqrt(a)*b*x*sqrt(1 + b*x**2/a)/2 - A*sqrt(a)*b*x/sqrt(1 + b*x**2/a) + 3
*A*a*sqrt(b)*asinh(sqrt(b)*x/sqrt(a))/2 - B*a**(3/2)*asinh(sqrt(a)/(sqrt(b)*x)) + B*a**2/(sqrt(b)*x*sqrt(a/(b*
x**2) + 1)) + B*a*sqrt(b)*x/sqrt(a/(b*x**2) + 1) + B*b*Piecewise((sqrt(a)*x**2/2, Eq(b, 0)), ((a + b*x**2)**(3
/2)/(3*b), True))

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Giac [A]  time = 1.21717, size = 167, normalized size = 1.55 \begin{align*} \frac{2 \, B a^{2} \arctan \left (-\frac{\sqrt{b} x - \sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{3}{2} \, A a \sqrt{b} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right ) + \frac{2 \, A a^{2} \sqrt{b}}{{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a} + \frac{1}{6} \, \sqrt{b x^{2} + a}{\left (8 \, B a +{\left (2 \, B b x + 3 \, A b\right )} x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*B*a^2*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/sqrt(-a) - 3/2*A*a*sqrt(b)*log(abs(-sqrt(b)*x + sqrt(b
*x^2 + a))) + 2*A*a^2*sqrt(b)/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a) + 1/6*sqrt(b*x^2 + a)*(8*B*a + (2*B*b*x +
3*A*b)*x)