\(\int \frac {(a+b x^2)^{3/2} (A+B x^2)}{x^2} \, dx\) [528]

Optimal result

Integrand size = 22, antiderivative size = 109 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^2} \, dx=\frac {3}{8} (4 A b+a B) x \sqrt {a+b x^2}+\frac {(4 A b+a B) x \left (a+b x^2\right )^{3/2}}{4 a}-\frac {A \left (a+b x^2\right )^{5/2}}{a x}+\frac {3 a (4 A b+a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 \sqrt {b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {464, 201, 223, 212} \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^2} \, dx=\frac {3 a (a B+4 A b) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 \sqrt {b}}+\frac {x \left (a+b x^2\right )^{3/2} (a B+4 A b)}{4 a}+\frac {3}{8} x \sqrt {a+b x^2} (a B+4 A b)-\frac {A \left (a+b x^2\right )^{5/2}}{a x} \]

[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 201

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 464

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*} \text {integral}& = -\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)) \text {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*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^2} \, dx=\frac {\sqrt {a+b x^2} \left (-8 a A+4 A b x^2+5 a B x^2+2 b B x^4\right )}{8 x}+\frac {3 a (4 A b+a B) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{4 \sqrt {b}} \]

[In]

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

Maple [A] (verified)

Time = 2.83 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.66

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-2 b B \,x^{4}-4 A b \,x^{2}-5 B a \,x^{2}+8 A a \right )}{8 x}+\frac {3 a \left (4 A b +B a \right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{8 \sqrt {b}}\)	\(72\)
pseudoelliptic	\(\frac {\frac {3 a x \left (A b +\frac {B a}{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )}{2}-\left (-\frac {x^{2} \left (\frac {x^{2} B}{2}+A \right ) b^{\frac {3}{2}}}{2}+a \sqrt {b}\, \left (-\frac {5 x^{2} B}{8}+A \right )\right ) \sqrt {b \,x^{2}+a}}{x \sqrt {b}}\)	\(79\)
default	\(B \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )+A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{a x}+\frac {4 b \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{a}\right )\)	\(132\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.67 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^2} \, dx=\left [\frac {3 \, {\left (B a^{2} + 4 \, A a b\right )} \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (2 \, B b^{2} x^{4} - 8 \, A a b + {\left (5 \, B a b + 4 \, A b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{16 \, b x}, -\frac {3 \, {\left (B a^{2} + 4 \, A a b\right )} \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, B b^{2} x^{4} - 8 \, A a b + {\left (5 \, B a b + 4 \, A b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{8 \, b x}\right ] \]

[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)]

Sympy [A] (verification not implemented)

Time = 1.66 (sec) , antiderivative size = 298, normalized size of antiderivative = 2.73 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^2} \, dx=- \frac {A a^{\frac {3}{2}}}{x \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {A \sqrt {a} b x}{\sqrt {1 + \frac {b x^{2}}{a}}} + A a \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} + A b \left (\begin {cases} \frac {a \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {a + b x^{2}}}{2} & \text {for}\: b \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) + B a \left (\begin {cases} \frac {a \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {a + b x^{2}}}{2} & \text {for}\: b \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) + B b \left (\begin {cases} - \frac {a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{8 b} + \frac {a x \sqrt {a + b x^{2}}}{8 b} + \frac {x^{3} \sqrt {a + b x^{2}}}{4} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{3}}{3} & \text {otherwise} \end {cases}\right ) \]

[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) + A*a*sqrt(b)*asinh(sqrt(b)*x/sqrt(a)) +
A*b*Piecewise((a*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2
), True))/2 + x*sqrt(a + b*x**2)/2, Ne(b, 0)), (sqrt(a)*x, True)) + B*a*Piecewise((a*Piecewise((log(2*sqrt(b)*
sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), True))/2 + x*sqrt(a + b*x**2)/2, Ne(b, 0
)), (sqrt(a)*x, True)) + B*b*Piecewise((-a**2*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a
, 0)), (x*log(x)/sqrt(b*x**2), True))/(8*b) + a*x*sqrt(a + b*x**2)/(8*b) + x**3*sqrt(a + b*x**2)/4, Ne(b, 0)),
(sqrt(a)*x**3/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^2} \, dx=\frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B x + \frac {3}{8} \, \sqrt {b x^{2} + a} B a x + \frac {3}{2} \, \sqrt {b x^{2} + a} A b x + \frac {3 \, B a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} + \frac {3}{2} \, A a \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{x} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^2} \, dx=\frac {2 \, A a^{2} \sqrt {b}}{{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a} + \frac {1}{8} \, {\left (2 \, B b x^{2} + \frac {5 \, B a b^{2} + 4 \, A b^{3}}{b^{2}}\right )} \sqrt {b x^{2} + a} x - \frac {3 \, {\left (B a^{2} + 4 \, A a b\right )} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right )}{16 \, \sqrt {b}} \]

[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 + 4*A*a*b)*log((sqrt(b)*x - sqrt(b*x^2 + a))^2)/sqrt(b)

Mupad [B] (verification not implemented)

Time = 6.06 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.73 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^2} \, dx=\frac {B\,x\,{\left (b\,x^2+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{3/2}}-\frac {A\,{\left (b\,x^2+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {b\,x^2}{a}\right )}{x\,{\left (\frac {b\,x^2}{a}+1\right )}^{3/2}} \]

[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))