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

Optimal. Leaf size=109 \[ \frac {x \left (a+b x^2\right )^{3/2} (a B+4 A b)}{4 a}+\frac {3}{8} x \sqrt {a+b x^2} (a B+4 A b)+\frac {3 a (a B+4 A b) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 \sqrt {b}}-\frac {A \left (a+b x^2\right )^{5/2}}{a x} \]

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Rubi [A]  time = 0.04, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {453, 195, 217, 206} \begin {gather*} \frac {x \left (a+b x^2\right )^{3/2} (a B+4 A b)}{4 a}+\frac {3}{8} x \sqrt {a+b x^2} (a B+4 A b)+\frac {3 a (a B+4 A b) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 \sqrt {b}}-\frac {A \left (a+b x^2\right )^{5/2}}{a x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 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 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 87, normalized size = 0.80 \begin {gather*} \frac {1}{8} \sqrt {a+b x^2} \left (\frac {3 \sqrt {a} (a B+4 A b) \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b} \sqrt {\frac {b x^2}{a}+1}}-\frac {8 a A}{x}+5 a B x+4 A b x+2 b B x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*((-8*a*A)/x + 4*A*b*x + 5*a*B*x + 2*b*B*x^3 + (3*Sqrt[a]*(4*A*b + a*B)*ArcSinh[(Sqrt[b]*x)/Sq
rt[a]])/(Sqrt[b]*Sqrt[1 + (b*x^2)/a])))/8

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IntegrateAlgebraic [A]  time = 0.21, size = 86, normalized size = 0.79 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-8 a A+5 a B x^2+4 A b x^2+2 b B x^4\right )}{8 x}-\frac {3 \left (a^2 B+4 a A b\right ) \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{8 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.71, size = 182, normalized size = 1.67 \begin {gather*} \left [\frac {3 \, {\left (B a^{2} + 4 \, A a b\right )} \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^{2} x^{4} - 8 \, A a b + {\left (5 \, B a b + 4 \, A b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{16 \, b x}, -\frac {3 \, {\left (B a^{2} + 4 \, A a b\right )} \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, B b^{2} x^{4} - 8 \, A a b + {\left (5 \, B a b + 4 \, A b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{8 \, b x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*(B*a^2 + 4*A*a*b)*sqrt(b)*x*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(2*B*b^2*x^4 - 8*A*a*
b + (5*B*a*b + 4*A*b^2)*x^2)*sqrt(b*x^2 + a))/(b*x), -1/8*(3*(B*a^2 + 4*A*a*b)*sqrt(-b)*x*arctan(sqrt(-b)*x/sq
rt(b*x^2 + a)) - (2*B*b^2*x^4 - 8*A*a*b + (5*B*a*b + 4*A*b^2)*x^2)*sqrt(b*x^2 + a))/(b*x)]

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giac [A]  time = 0.44, size = 114, normalized size = 1.05 \begin {gather*} \frac {2 \, A a^{2} \sqrt {b}}{{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a} + \frac {1}{8} \, {\left (2 \, B b x^{2} + \frac {5 \, B a b^{2} + 4 \, A b^{3}}{b^{2}}\right )} \sqrt {b x^{2} + a} x - \frac {3 \, {\left (B a^{2} \sqrt {b} + 4 \, A a b^{\frac {3}{2}}\right )} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right )}{16 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a^2*sqrt(b)/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a) + 1/8*(2*B*b*x^2 + (5*B*a*b^2 + 4*A*b^3)/b^2)*sqrt(b*x^2
+ a)*x - 3/16*(B*a^2*sqrt(b) + 4*A*a*b^(3/2))*log((sqrt(b)*x - sqrt(b*x^2 + a))^2)/b

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maple [A]  time = 0.01, size = 125, normalized size = 1.15 \begin {gather*} \frac {3 A a \sqrt {b}\, \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2}+\frac {3 B \,a^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 \sqrt {b}}+\frac {3 \sqrt {b \,x^{2}+a}\, A b x}{2}+\frac {3 \sqrt {b \,x^{2}+a}\, B a x}{8}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A b x}{a}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B x}{4}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A}{a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.14, size = 91, normalized size = 0.83 \begin {gather*} \frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B x + \frac {3}{8} \, \sqrt {b x^{2} + a} B a x + \frac {3}{2} \, \sqrt {b x^{2} + a} A b x + \frac {3 \, B a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} + \frac {3}{2} \, A a \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.50, size = 80, normalized size = 0.73 \begin {gather*} \frac {B\,x\,{\left (b\,x^2+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{3/2}}-\frac {A\,{\left (b\,x^2+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {b\,x^2}{a}\right )}{x\,{\left (\frac {b\,x^2}{a}+1\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*x*(a + b*x^2)^(3/2)*hypergeom([-3/2, 1/2], 3/2, -(b*x^2)/a))/((b*x^2)/a + 1)^(3/2) - (A*(a + b*x^2)^(3/2)*h
ypergeom([-3/2, -1/2], 1/2, -(b*x^2)/a))/(x*((b*x^2)/a + 1)^(3/2))

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sympy [B]  time = 12.20, size = 216, normalized size = 1.98 \begin {gather*} - \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} + \frac {B a^{\frac {3}{2}} x \sqrt {1 + \frac {b x^{2}}{a}}}{2} + \frac {B a^{\frac {3}{2}} x}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 B \sqrt {a} b x^{3}}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 B a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8 \sqrt {b}} + \frac {B b^{2} x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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