\(\int \frac {(a+b x^2)^{3/2} (A+B x^2)}{x^4} \, dx\) [530]

Optimal result

Integrand size = 22, antiderivative size = 119 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^4} \, dx=\frac {b (2 A b+3 a B) x \sqrt {a+b x^2}}{2 a}-\frac {(2 A b+3 a B) \left (a+b x^2\right )^{3/2}}{3 a x}-\frac {A \left (a+b x^2\right )^{5/2}}{3 a x^3}+\frac {1}{2} \sqrt {b} (2 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {464, 283, 201, 223, 212} \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^4} \, dx=\frac {1}{2} \sqrt {b} (3 a B+2 A b) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {\left (a+b x^2\right )^{3/2} (3 a B+2 A b)}{3 a x}+\frac {b x \sqrt {a+b x^2} (3 a B+2 A b)}{2 a}-\frac {A \left (a+b x^2\right )^{5/2}}{3 a x^3} \]

[In]

Int[((a + b*x^2)^(3/2)*(A + B*x^2))/x^4,x]

[Out]

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

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 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.75 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^4} \, dx=\frac {\sqrt {a+b x^2} \left (-2 a A-8 A b x^2-6 a B x^2+3 b B x^4\right )}{6 x^3}+\sqrt {b} (2 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right ) \]

[In]

Integrate[((a + b*x^2)^(3/2)*(A + B*x^2))/x^4,x]

[Out]

(Sqrt[a + b*x^2]*(-2*a*A - 8*A*b*x^2 - 6*a*B*x^2 + 3*b*B*x^4))/(6*x^3) + Sqrt[b]*(2*A*b + 3*a*B)*ArcTanh[(Sqrt
[b]*x)/(-Sqrt[a] + Sqrt[a + b*x^2])]

Maple [A] (verified)

Time = 2.84 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.61

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-3 b B \,x^{4}+8 A b \,x^{2}+6 B a \,x^{2}+2 A a \right )}{6 x^{3}}+\frac {\left (2 A b +3 B a \right ) \sqrt {b}\, \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2}\)	\(72\)
pseudoelliptic	\(\frac {b \,x^{3} \left (A b +\frac {3 B a}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )-\frac {\left (\left (-\frac {3}{2} x^{4} B +4 A \,x^{2}\right ) b^{\frac {3}{2}}+a \sqrt {b}\, \left (3 x^{2} B +A \right )\right ) \sqrt {b \,x^{2}+a}}{3}}{\sqrt {b}\, x^{3}}\)	\(81\)
default	\(A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{3 a \,x^{3}}+\frac {2 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{a x}+\frac {4 b \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{a}\right )}{3 a}\right )+B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{a x}+\frac {4 b \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{a}\right )\)	\(180\)

[In]

int((b*x^2+a)^(3/2)*(B*x^2+A)/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/6*(b*x^2+a)^(1/2)*(-3*B*b*x^4+8*A*b*x^2+6*B*a*x^2+2*A*a)/x^3+1/2*(2*A*b+3*B*a)*b^(1/2)*ln(x*b^(1/2)+(b*x^2+
a)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.39 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^4} \, dx=\left [\frac {3 \, {\left (3 \, B a + 2 \, A b\right )} \sqrt {b} x^{3} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (3 \, B b x^{4} - 2 \, {\left (3 \, B a + 4 \, A b\right )} x^{2} - 2 \, A a\right )} \sqrt {b x^{2} + a}}{12 \, x^{3}}, -\frac {3 \, {\left (3 \, B a + 2 \, A b\right )} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (3 \, B b x^{4} - 2 \, {\left (3 \, B a + 4 \, A b\right )} x^{2} - 2 \, A a\right )} \sqrt {b x^{2} + a}}{6 \, x^{3}}\right ] \]

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^4,x, algorithm="fricas")

[Out]

[1/12*(3*(3*B*a + 2*A*b)*sqrt(b)*x^3*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(3*B*b*x^4 - 2*(3*B*a
+ 4*A*b)*x^2 - 2*A*a)*sqrt(b*x^2 + a))/x^3, -1/6*(3*(3*B*a + 2*A*b)*sqrt(-b)*x^3*arctan(sqrt(-b)*x/sqrt(b*x^2
+ a)) - (3*B*b*x^4 - 2*(3*B*a + 4*A*b)*x^2 - 2*A*a)*sqrt(b*x^2 + a))/x^3]

Sympy [A] (verification not implemented)

Time = 2.27 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.06 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^4} \, dx=- \frac {A \sqrt {a} b}{x \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {A a \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{3 x^{2}} - \frac {A b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{3} + A b^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} - \frac {A b^{2} x}{\sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {B a^{\frac {3}{2}}}{x \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {B \sqrt {a} b x}{\sqrt {1 + \frac {b x^{2}}{a}}} + B a \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} + B b \left (\begin {cases} \frac {a \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {a + b x^{2}}}{2} & \text {for}\: b \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((b*x**2+a)**(3/2)*(B*x**2+A)/x**4,x)

[Out]

-A*sqrt(a)*b/(x*sqrt(1 + b*x**2/a)) - A*a*sqrt(b)*sqrt(a/(b*x**2) + 1)/(3*x**2) - A*b**(3/2)*sqrt(a/(b*x**2) +
1)/3 + A*b**(3/2)*asinh(sqrt(b)*x/sqrt(a)) - A*b**2*x/(sqrt(a)*sqrt(1 + b*x**2/a)) - B*a**(3/2)/(x*sqrt(1 + b
*x**2/a)) - B*sqrt(a)*b*x/sqrt(1 + b*x**2/a) + B*a*sqrt(b)*asinh(sqrt(b)*x/sqrt(a)) + B*b*Piecewise((a*Piecewi
se((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))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^4} \, dx=\frac {3}{2} \, \sqrt {b x^{2} + a} B b x + \frac {\sqrt {b x^{2} + a} A b^{2} x}{a} + \frac {3}{2} \, B a \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) + A b^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B}{x} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b}{3 \, a x} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A}{3 \, a x^{3}} \]

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^4,x, algorithm="maxima")

[Out]

3/2*sqrt(b*x^2 + a)*B*b*x + sqrt(b*x^2 + a)*A*b^2*x/a + 3/2*B*a*sqrt(b)*arcsinh(b*x/sqrt(a*b)) + A*b^(3/2)*arc
sinh(b*x/sqrt(a*b)) - (b*x^2 + a)^(3/2)*B/x - 2/3*(b*x^2 + a)^(3/2)*A*b/(a*x) - 1/3*(b*x^2 + a)^(5/2)*A/(a*x^3
)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 207 vs. \(2 (99) = 198\).

Time = 0.33 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.74 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^4} \, dx=\frac {1}{2} \, \sqrt {b x^{2} + a} B b x - \frac {1}{4} \, {\left (3 \, B a \sqrt {b} + 2 \, A b^{\frac {3}{2}}\right )} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right ) + \frac {2 \, {\left (3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} B a^{2} \sqrt {b} + 6 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} A a b^{\frac {3}{2}} - 6 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{3} \sqrt {b} - 6 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} A a^{2} b^{\frac {3}{2}} + 3 \, B a^{4} \sqrt {b} + 4 \, A a^{3} b^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{3}} \]

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^4,x, algorithm="giac")

[Out]

1/2*sqrt(b*x^2 + a)*B*b*x - 1/4*(3*B*a*sqrt(b) + 2*A*b^(3/2))*log((sqrt(b)*x - sqrt(b*x^2 + a))^2) + 2/3*(3*(s
qrt(b)*x - sqrt(b*x^2 + a))^4*B*a^2*sqrt(b) + 6*(sqrt(b)*x - sqrt(b*x^2 + a))^4*A*a*b^(3/2) - 6*(sqrt(b)*x - s
qrt(b*x^2 + a))^2*B*a^3*sqrt(b) - 6*(sqrt(b)*x - sqrt(b*x^2 + a))^2*A*a^2*b^(3/2) + 3*B*a^4*sqrt(b) + 4*A*a^3*
b^(3/2))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^3

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^4} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (b\,x^2+a\right )}^{3/2}}{x^4} \,d x \]

[In]

int(((A + B*x^2)*(a + b*x^2)^(3/2))/x^4,x)

[Out]

int(((A + B*x^2)*(a + b*x^2)^(3/2))/x^4, x)