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

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

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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} \begin {gather*} -\frac {\left (a+b x^2\right )^{3/2} \left (2 a B d-5 b (A d+B c)-3 b B d x^2\right )}{15 b^2}+A c \sqrt {a+b x^2}-\sqrt {a} A c \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.14, size = 91, normalized size = 1.08 \begin {gather*} \frac {\sqrt {a+b x^2} \left (5 A b \left (a d+3 b c+b d x^2\right )-B \left (a+b x^2\right ) \left (2 a d-5 b c-3 b d x^2\right )\right )}{15 b^2}-\sqrt {a} A c \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.09, size = 111, normalized size = 1.32 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-2 a^2 B d+5 a A b d+5 a b B c+a b B d x^2+15 A b^2 c+5 A b^2 d x^2+5 b^2 B c x^2+3 b^2 B d x^4\right )}{15 b^2}-\sqrt {a} A c \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(15*A*b^2*c + 5*a*b*B*c + 5*a*A*b*d - 2*a^2*B*d + 5*b^2*B*c*x^2 + 5*A*b^2*d*x^2 + a*b*B*d*x^2
+ 3*b^2*B*d*x^4))/(15*b^2) - Sqrt[a]*A*c*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]]

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fricas [A]  time = 0.95, size = 231, normalized size = 2.75 \begin {gather*} \left [\frac {15 \, A \sqrt {a} b^{2} c \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (3 \, B b^{2} d x^{4} + {\left (5 \, B b^{2} c + {\left (B a b + 5 \, A b^{2}\right )} d\right )} x^{2} + 5 \, {\left (B a b + 3 \, A b^{2}\right )} c - {\left (2 \, B a^{2} - 5 \, A a b\right )} d\right )} \sqrt {b x^{2} + a}}{30 \, b^{2}}, \frac {15 \, A \sqrt {-a} b^{2} c \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, B b^{2} d x^{4} + {\left (5 \, B b^{2} c + {\left (B a b + 5 \, A b^{2}\right )} d\right )} x^{2} + 5 \, {\left (B a b + 3 \, A b^{2}\right )} c - {\left (2 \, B a^{2} - 5 \, A a b\right )} d\right )} \sqrt {b x^{2} + a}}{15 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.43, size = 113, normalized size = 1.35 \begin {gather*} \frac {A a c \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{9} c + 15 \, \sqrt {b x^{2} + a} A b^{10} c + 3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B b^{8} d - 5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a b^{8} d + 5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{9} d}{15 \, b^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.38, size = 100, normalized size = 1.19 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B d x^{2}}{5 \, b} - A \sqrt {a} c \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \sqrt {b x^{2} + a} A c + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B c}{3 \, b} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a d}{15 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A d}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.51, size = 101, normalized size = 1.20 \begin {gather*} \sqrt {b\,x^2+a}\,\left (\frac {B\,d\,x^4}{5}-\frac {B\,a\,\left (2\,a\,d-5\,b\,c\right )}{15\,b^2}+\frac {B\,x^2\,\left (a\,d+5\,b\,c\right )}{15\,b}\right )+A\,c\,\sqrt {b\,x^2+a}+\frac {A\,d\,{\left (b\,x^2+a\right )}^{3/2}}{3\,b}-A\,\sqrt {a}\,c\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 59.29, size = 97, normalized size = 1.15 \begin {gather*} \frac {A a c \operatorname {atan}{\left (\frac {\sqrt {a + b x^{2}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + A c \sqrt {a + b x^{2}} + \frac {B d \left (a + b x^{2}\right )^{\frac {5}{2}}}{5 b^{2}} + \frac {\left (a + b x^{2}\right )^{\frac {3}{2}} \left (2 A b d - 2 B a d + 2 B b c\right )}{6 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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