3.51 \(\int \frac {(a+b x^2) (A+B x^2) \sqrt {c+d x^2}}{x} \, dx\)

Optimal. Leaf size=84 \[ -\frac {\left (c+d x^2\right )^{3/2} \left (-5 d (a B+A b)+2 b B c-3 b B d x^2\right )}{15 d^2}+a A \sqrt {c+d x^2}-a A \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \]

[Out]

-1/15*(d*x^2+c)^(3/2)*(2*b*B*c-5*(A*b+B*a)*d-3*b*B*d*x^2)/d^2-a*A*arctanh((d*x^2+c)^(1/2)/c^(1/2))*c^(1/2)+a*A
*(d*x^2+c)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {573, 147, 50, 63, 208} \[ -\frac {\left (c+d x^2\right )^{3/2} \left (-5 d (a B+A b)+2 b B c-3 b B d x^2\right )}{15 d^2}+a A \sqrt {c+d x^2}-a A \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*A*Sqrt[c + d*x^2] - ((c + d*x^2)^(3/2)*(2*b*B*c - 5*(A*b + a*B)*d - 3*b*B*d*x^2))/(15*d^2) - a*A*Sqrt[c]*Arc
Tanh[Sqrt[c + d*x^2]/Sqrt[c]]

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 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 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 573

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

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right ) \left (A+B x^2\right ) \sqrt {c+d x^2}}{x} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x) (A+B x) \sqrt {c+d x}}{x} \, dx,x,x^2\right )\\ &=-\frac {\left (c+d x^2\right )^{3/2} \left (2 b B c-5 (A b+a B) d-3 b B d x^2\right )}{15 d^2}+\frac {1}{2} (a A) \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{x} \, dx,x,x^2\right )\\ &=a A \sqrt {c+d x^2}-\frac {\left (c+d x^2\right )^{3/2} \left (2 b B c-5 (A b+a B) d-3 b B d x^2\right )}{15 d^2}+\frac {1}{2} (a A c) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )\\ &=a A \sqrt {c+d x^2}-\frac {\left (c+d x^2\right )^{3/2} \left (2 b B c-5 (A b+a B) d-3 b B d x^2\right )}{15 d^2}+\frac {(a A c) \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{d}\\ &=a A \sqrt {c+d x^2}-\frac {\left (c+d x^2\right )^{3/2} \left (2 b B c-5 (A b+a B) d-3 b B d x^2\right )}{15 d^2}-a A \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 91, normalized size = 1.08 \[ \frac {\sqrt {c+d x^2} \left (5 a d \left (3 A d+B \left (c+d x^2\right )\right )-b \left (c+d x^2\right ) \left (-5 A d+2 B c-3 B d x^2\right )\right )}{15 d^2}-a A \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^2]*(-(b*(c + d*x^2)*(2*B*c - 5*A*d - 3*B*d*x^2)) + 5*a*d*(3*A*d + B*(c + d*x^2))))/(15*d^2) - a*
A*Sqrt[c]*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]]

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fricas [A]  time = 0.66, size = 217, normalized size = 2.58 \[ \left [\frac {15 \, A a \sqrt {c} d^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (3 \, B b d^{2} x^{4} - 2 \, B b c^{2} + 15 \, A a d^{2} + 5 \, {\left (B a + A b\right )} c d + {\left (B b c d + 5 \, {\left (B a + A b\right )} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{30 \, d^{2}}, \frac {15 \, A a \sqrt {-c} d^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (3 \, B b d^{2} x^{4} - 2 \, B b c^{2} + 15 \, A a d^{2} + 5 \, {\left (B a + A b\right )} c d + {\left (B b c d + 5 \, {\left (B a + A b\right )} d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{15 \, d^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(B*x^2+A)*(d*x^2+c)^(1/2)/x,x, algorithm="fricas")

[Out]

[1/30*(15*A*a*sqrt(c)*d^2*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 2*(3*B*b*d^2*x^4 - 2*B*b*c^2 +
15*A*a*d^2 + 5*(B*a + A*b)*c*d + (B*b*c*d + 5*(B*a + A*b)*d^2)*x^2)*sqrt(d*x^2 + c))/d^2, 1/15*(15*A*a*sqrt(-
c)*d^2*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + (3*B*b*d^2*x^4 - 2*B*b*c^2 + 15*A*a*d^2 + 5*(B*a + A*b)*c*d + (B*b*c
*d + 5*(B*a + A*b)*d^2)*x^2)*sqrt(d*x^2 + c))/d^2]

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giac [A]  time = 0.57, size = 113, normalized size = 1.35 \[ \frac {A a c \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} B b d^{8} - 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} B b c d^{8} + 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} B a d^{9} + 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} A b d^{9} + 15 \, \sqrt {d x^{2} + c} A a d^{10}}{15 \, d^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(B*x^2+A)*(d*x^2+c)^(1/2)/x,x, algorithm="giac")

[Out]

A*a*c*arctan(sqrt(d*x^2 + c)/sqrt(-c))/sqrt(-c) + 1/15*(3*(d*x^2 + c)^(5/2)*B*b*d^8 - 5*(d*x^2 + c)^(3/2)*B*b*
c*d^8 + 5*(d*x^2 + c)^(3/2)*B*a*d^9 + 5*(d*x^2 + c)^(3/2)*A*b*d^9 + 15*sqrt(d*x^2 + c)*A*a*d^10)/d^10

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maple [A]  time = 0.01, size = 112, normalized size = 1.33 \[ -A a \sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {d \,x^{2}+c}\, \sqrt {c}}{x}\right )+\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} B b \,x^{2}}{5 d}+\sqrt {d \,x^{2}+c}\, A a +\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} A b}{3 d}+\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} B a}{3 d}-\frac {2 \left (d \,x^{2}+c \right )^{\frac {3}{2}} B b c}{15 d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)*(B*x^2+A)*(d*x^2+c)^(1/2)/x,x)

[Out]

1/5*b*B*x^2*(d*x^2+c)^(3/2)/d-2/15*b*B*c/d^2*(d*x^2+c)^(3/2)+1/3*A*b*(d*x^2+c)^(3/2)/d+1/3*B*a*(d*x^2+c)^(3/2)
/d-A*c^(1/2)*ln((2*c+2*(d*x^2+c)^(1/2)*c^(1/2))/x)*a+a*A*(d*x^2+c)^(1/2)

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maxima [A]  time = 1.37, size = 100, normalized size = 1.19 \[ \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} B b x^{2}}{5 \, d} - A a \sqrt {c} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) + \sqrt {d x^{2} + c} A a - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} B b c}{15 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} B a}{3 \, d} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} A b}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(B*x^2+A)*(d*x^2+c)^(1/2)/x,x, algorithm="maxima")

[Out]

1/5*(d*x^2 + c)^(3/2)*B*b*x^2/d - A*a*sqrt(c)*arcsinh(c/(sqrt(c*d)*abs(x))) + sqrt(d*x^2 + c)*A*a - 2/15*(d*x^
2 + c)^(3/2)*B*b*c/d^2 + 1/3*(d*x^2 + c)^(3/2)*B*a/d + 1/3*(d*x^2 + c)^(3/2)*A*b/d

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mupad [B]  time = 1.48, size = 101, normalized size = 1.20 \[ \sqrt {d\,x^2+c}\,\left (\frac {B\,b\,x^4}{5}+\frac {B\,c\,\left (5\,a\,d-2\,b\,c\right )}{15\,d^2}+\frac {B\,x^2\,\left (5\,a\,d+b\,c\right )}{15\,d}\right )+A\,a\,\sqrt {d\,x^2+c}-A\,a\,\sqrt {c}\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )+\frac {A\,b\,{\left (d\,x^2+c\right )}^{3/2}}{3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(a + b*x^2)*(c + d*x^2)^(1/2))/x,x)

[Out]

(c + d*x^2)^(1/2)*((B*b*x^4)/5 + (B*c*(5*a*d - 2*b*c))/(15*d^2) + (B*x^2*(5*a*d + b*c))/(15*d)) + A*a*(c + d*x
^2)^(1/2) - A*a*c^(1/2)*atanh((c + d*x^2)^(1/2)/c^(1/2)) + (A*b*(c + d*x^2)^(3/2))/(3*d)

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sympy [A]  time = 54.08, size = 97, normalized size = 1.15 \[ \frac {A a c \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + A a \sqrt {c + d x^{2}} + \frac {B b \left (c + d x^{2}\right )^{\frac {5}{2}}}{5 d^{2}} + \frac {\left (c + d x^{2}\right )^{\frac {3}{2}} \left (2 A b d + 2 B a d - 2 B b c\right )}{6 d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)*(B*x**2+A)*(d*x**2+c)**(1/2)/x,x)

[Out]

A*a*c*atan(sqrt(c + d*x**2)/sqrt(-c))/sqrt(-c) + A*a*sqrt(c + d*x**2) + B*b*(c + d*x**2)**(5/2)/(5*d**2) + (c
+ d*x**2)**(3/2)*(2*A*b*d + 2*B*a*d - 2*B*b*c)/(6*d**2)

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