\(\int \frac {(a+b x^2)^{5/2} (A+B x^2)}{x} \, dx\) [544]

Optimal result

Integrand size = 22, antiderivative size = 95 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x} \, dx=a^2 A \sqrt {a+b x^2}+\frac {1}{3} a A \left (a+b x^2\right )^{3/2}+\frac {1}{5} A \left (a+b x^2\right )^{5/2}+\frac {B \left (a+b x^2\right )^{7/2}}{7 b}-a^{5/2} A \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]

[Out]

1/3*a*A*(b*x^2+a)^(3/2)+1/5*A*(b*x^2+a)^(5/2)+1/7*B*(b*x^2+a)^(7/2)/b-a^(5/2)*A*arctanh((b*x^2+a)^(1/2)/a^(1/2
))+a^2*A*(b*x^2+a)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {457, 81, 52, 65, 214} \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x} \, dx=-a^{5/2} A \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+a^2 A \sqrt {a+b x^2}+\frac {1}{5} A \left (a+b x^2\right )^{5/2}+\frac {1}{3} a A \left (a+b x^2\right )^{3/2}+\frac {B \left (a+b x^2\right )^{7/2}}{7 b} \]

[In]

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

[Out]

a^2*A*Sqrt[a + b*x^2] + (a*A*(a + b*x^2)^(3/2))/3 + (A*(a + b*x^2)^(5/2))/5 + (B*(a + b*x^2)^(7/2))/(7*b) - a^
(5/2)*A*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

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 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 214

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

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(a+b x)^{5/2} (A+B x)}{x} \, dx,x,x^2\right ) \\ & = \frac {B \left (a+b x^2\right )^{7/2}}{7 b}+\frac {1}{2} A \text {Subst}\left (\int \frac {(a+b x)^{5/2}}{x} \, dx,x,x^2\right ) \\ & = \frac {1}{5} A \left (a+b x^2\right )^{5/2}+\frac {B \left (a+b x^2\right )^{7/2}}{7 b}+\frac {1}{2} (a A) \text {Subst}\left (\int \frac {(a+b x)^{3/2}}{x} \, dx,x,x^2\right ) \\ & = \frac {1}{3} a A \left (a+b x^2\right )^{3/2}+\frac {1}{5} A \left (a+b x^2\right )^{5/2}+\frac {B \left (a+b x^2\right )^{7/2}}{7 b}+\frac {1}{2} \left (a^2 A\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,x^2\right ) \\ & = a^2 A \sqrt {a+b x^2}+\frac {1}{3} a A \left (a+b x^2\right )^{3/2}+\frac {1}{5} A \left (a+b x^2\right )^{5/2}+\frac {B \left (a+b x^2\right )^{7/2}}{7 b}+\frac {1}{2} \left (a^3 A\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right ) \\ & = a^2 A \sqrt {a+b x^2}+\frac {1}{3} a A \left (a+b x^2\right )^{3/2}+\frac {1}{5} A \left (a+b x^2\right )^{5/2}+\frac {B \left (a+b x^2\right )^{7/2}}{7 b}+\frac {\left (a^3 A\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{b} \\ & = a^2 A \sqrt {a+b x^2}+\frac {1}{3} a A \left (a+b x^2\right )^{3/2}+\frac {1}{5} A \left (a+b x^2\right )^{5/2}+\frac {B \left (a+b x^2\right )^{7/2}}{7 b}-a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x} \, dx=\frac {\sqrt {a+b x^2} \left (15 a^3 B+3 b^3 x^4 \left (7 A+5 B x^2\right )+a b^2 x^2 \left (77 A+45 B x^2\right )+a^2 b \left (161 A+45 B x^2\right )\right )}{105 b}-a^{5/2} A \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]

[In]

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

[Out]

(Sqrt[a + b*x^2]*(15*a^3*B + 3*b^3*x^4*(7*A + 5*B*x^2) + a*b^2*x^2*(77*A + 45*B*x^2) + a^2*b*(161*A + 45*B*x^2
)))/(105*b) - a^(5/2)*A*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]]

Maple [A] (verified)

Time = 2.81 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89

method	result	size
default	\(\frac {B \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{7 b}+A \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )\)	\(85\)
pseudoelliptic	\(\frac {-15 A \,a^{\frac {5}{2}} b \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )+23 \left (\frac {3 x^{4} \left (\frac {5 x^{2} B}{7}+A \right ) b^{3}}{23}+\frac {11 x^{2} \left (\frac {45 x^{2} B}{77}+A \right ) a \,b^{2}}{23}+a^{2} \left (\frac {45 x^{2} B}{161}+A \right ) b +\frac {15 a^{3} B}{161}\right ) \sqrt {b \,x^{2}+a}}{15 b}\)	\(92\)

[In]

int((b*x^2+a)^(5/2)*(B*x^2+A)/x,x,method=_RETURNVERBOSE)

[Out]

1/7*B*(b*x^2+a)^(7/2)/b+A*(1/5*(b*x^2+a)^(5/2)+a*(1/3*(b*x^2+a)^(3/2)+a*((b*x^2+a)^(1/2)-a^(1/2)*ln((2*a+2*a^(
1/2)*(b*x^2+a)^(1/2))/x))))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.32 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x} \, dx=\left [\frac {105 \, A a^{\frac {5}{2}} b \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (15 \, B b^{3} x^{6} + 3 \, {\left (15 \, B a b^{2} + 7 \, A b^{3}\right )} x^{4} + 15 \, B a^{3} + 161 \, A a^{2} b + {\left (45 \, B a^{2} b + 77 \, A a b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{210 \, b}, \frac {105 \, A \sqrt {-a} a^{2} b \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (15 \, B b^{3} x^{6} + 3 \, {\left (15 \, B a b^{2} + 7 \, A b^{3}\right )} x^{4} + 15 \, B a^{3} + 161 \, A a^{2} b + {\left (45 \, B a^{2} b + 77 \, A a b^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{105 \, b}\right ] \]

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x,x, algorithm="fricas")

[Out]

[1/210*(105*A*a^(5/2)*b*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(15*B*b^3*x^6 + 3*(15*B*a*b^2
+ 7*A*b^3)*x^4 + 15*B*a^3 + 161*A*a^2*b + (45*B*a^2*b + 77*A*a*b^2)*x^2)*sqrt(b*x^2 + a))/b, 1/105*(105*A*sqrt
(-a)*a^2*b*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (15*B*b^3*x^6 + 3*(15*B*a*b^2 + 7*A*b^3)*x^4 + 15*B*a^3 + 161*A*
a^2*b + (45*B*a^2*b + 77*A*a*b^2)*x^2)*sqrt(b*x^2 + a))/b]

Sympy [A] (verification not implemented)

Time = 13.91 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.35 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x} \, dx=\frac {\begin {cases} \frac {2 A a^{3} \operatorname {atan}{\left (\frac {\sqrt {a + b x^{2}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 A a^{2} \sqrt {a + b x^{2}} + \frac {2 A a \left (a + b x^{2}\right )^{\frac {3}{2}}}{3} + \frac {2 A \left (a + b x^{2}\right )^{\frac {5}{2}}}{5} + \frac {2 B \left (a + b x^{2}\right )^{\frac {7}{2}}}{7 b} & \text {for}\: b \neq 0 \\A a^{\frac {5}{2}} \log {\left (B a^{\frac {5}{2}} x^{2} \right )} + B a^{\frac {5}{2}} x^{2} & \text {otherwise} \end {cases}}{2} \]

[In]

integrate((b*x**2+a)**(5/2)*(B*x**2+A)/x,x)

[Out]

Piecewise((2*A*a**3*atan(sqrt(a + b*x**2)/sqrt(-a))/sqrt(-a) + 2*A*a**2*sqrt(a + b*x**2) + 2*A*a*(a + b*x**2)*
*(3/2)/3 + 2*A*(a + b*x**2)**(5/2)/5 + 2*B*(a + b*x**2)**(7/2)/(7*b), Ne(b, 0)), (A*a**(5/2)*log(B*a**(5/2)*x*
*2) + B*a**(5/2)*x**2, True))/2

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.77 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x} \, dx=-A a^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {1}{5} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A + \frac {1}{3} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a + \sqrt {b x^{2} + a} A a^{2} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B}{7 \, b} \]

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x,x, algorithm="maxima")

[Out]

-A*a^(5/2)*arcsinh(a/(sqrt(a*b)*abs(x))) + 1/5*(b*x^2 + a)^(5/2)*A + 1/3*(b*x^2 + a)^(3/2)*A*a + sqrt(b*x^2 +
a)*A*a^2 + 1/7*(b*x^2 + a)^(7/2)*B/b

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x} \, dx=\frac {A a^{3} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {15 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B b^{6} + 21 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b^{7} + 35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a b^{7} + 105 \, \sqrt {b x^{2} + a} A a^{2} b^{7}}{105 \, b^{7}} \]

[In]

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

[Out]

A*a^3*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + 1/105*(15*(b*x^2 + a)^(7/2)*B*b^6 + 21*(b*x^2 + a)^(5/2)*A*b
^7 + 35*(b*x^2 + a)^(3/2)*A*a*b^7 + 105*sqrt(b*x^2 + a)*A*a^2*b^7)/b^7

Mupad [B] (verification not implemented)

Time = 5.40 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.82 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x} \, dx=\frac {A\,{\left (b\,x^2+a\right )}^{5/2}}{5}+A\,a^2\,\sqrt {b\,x^2+a}+\frac {B\,{\left (b\,x^2+a\right )}^{7/2}}{7\,b}+\frac {A\,a\,{\left (b\,x^2+a\right )}^{3/2}}{3}+A\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i} \]

[In]

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

[Out]

(A*(a + b*x^2)^(5/2))/5 + A*a^2*(a + b*x^2)^(1/2) + (B*(a + b*x^2)^(7/2))/(7*b) + A*a^(5/2)*atan(((a + b*x^2)^
(1/2)*1i)/a^(1/2))*1i + (A*a*(a + b*x^2)^(3/2))/3