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

Optimal. Leaf size=95 \[ -a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+a^2 A \sqrt {a+b x^2}+\frac {1}{5} A \left (a+b x^2\right )^{5/2}+\frac {1}{3} a A \left (a+b x^2\right )^{3/2}+\frac {B \left (a+b x^2\right )^{7/2}}{7 b} \]

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Rubi [A]  time = 0.06, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {446, 80, 50, 63, 208} \begin {gather*} a^2 A \sqrt {a+b x^2}-a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\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} \end {gather*}

Antiderivative was successfully verified.

[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 50

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 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 80

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 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]

Rule 446

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*} \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x} \, dx &=\frac {1}{2} \operatorname {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 \operatorname {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) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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*}

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Mathematica [A]  time = 0.10, size = 88, normalized size = 0.93 \begin {gather*} -a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )+\frac {1}{5} A \left (a+b x^2\right )^{5/2}+\frac {1}{3} a A \left (4 a+b x^2\right ) \sqrt {a+b x^2}+\frac {B \left (a+b x^2\right )^{7/2}}{7 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(A*(a + b*x^2)^(5/2))/5 + (B*(a + b*x^2)^(7/2))/(7*b) + (a*A*Sqrt[a + b*x^2]*(4*a + b*x^2))/3 - a^(5/2)*A*ArcT
anh[Sqrt[a + b*x^2]/Sqrt[a]]

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IntegrateAlgebraic [A]  time = 0.08, size = 107, normalized size = 1.13 \begin {gather*} \frac {\sqrt {a+b x^2} \left (15 a^3 B+161 a^2 A b+45 a^2 b B x^2+77 a A b^2 x^2+45 a b^2 B x^4+21 A b^3 x^4+15 b^3 B x^6\right )}{105 b}-a^{5/2} A \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(161*a^2*A*b + 15*a^3*B + 77*a*A*b^2*x^2 + 45*a^2*b*B*x^2 + 21*A*b^3*x^4 + 45*a*b^2*B*x^4 + 1
5*b^3*B*x^6))/(105*b) - a^(5/2)*A*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]]

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fricas [A]  time = 1.11, size = 220, normalized size = 2.32 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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giac [A]  time = 0.37, size = 97, normalized size = 1.02 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.01, size = 85, normalized size = 0.89 \begin {gather*} -A \,a^{\frac {5}{2}} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )+\sqrt {b \,x^{2}+a}\, A \,a^{2}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A a}{3}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A}{5}+\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} B}{7 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.06, size = 73, normalized size = 0.77 \begin {gather*} -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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 1.03, size = 78, normalized size = 0.82 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 82.47, size = 88, normalized size = 0.93 \begin {gather*} \frac {A a^{3} \operatorname {atan}{\left (\frac {\sqrt {a + b x^{2}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + A a^{2} \sqrt {a + b x^{2}} + \frac {A a \left (a + b x^{2}\right )^{\frac {3}{2}}}{3} + \frac {A \left (a + b x^{2}\right )^{\frac {5}{2}}}{5} + \frac {B \left (a + b x^{2}\right )^{\frac {7}{2}}}{7 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a**3*atan(sqrt(a + b*x**2)/sqrt(-a))/sqrt(-a) + A*a**2*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)

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