3.1.94 \(\int \frac {(A+B x^2) \sqrt {b x^2+c x^4}}{x^3} \, dx\) [94]

Optimal. Leaf size=97 \[ \frac {(b B+2 A c) \sqrt {b x^2+c x^4}}{2 b}-\frac {A \left (b x^2+c x^4\right )^{3/2}}{b x^4}+\frac {(b B+2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{2 \sqrt {c}} \]

[Out]

-A*(c*x^4+b*x^2)^(3/2)/b/x^4+1/2*(2*A*c+B*b)*arctanh(x^2*c^(1/2)/(c*x^4+b*x^2)^(1/2))/c^(1/2)+1/2*(2*A*c+B*b)*
(c*x^4+b*x^2)^(1/2)/b

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Rubi [A]
time = 0.14, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2059, 806, 678, 634, 212} \begin {gather*} \frac {\sqrt {b x^2+c x^4} (2 A c+b B)}{2 b}+\frac {(2 A c+b B) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{2 \sqrt {c}}-\frac {A \left (b x^2+c x^4\right )^{3/2}}{b x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 634

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 678

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x] - Dist[p*((2*c*d - b*e)/(e^2*(m + 2*p + 1))), Int[(d + e*x)^(m + 1)*
(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a
*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rule 2059

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n
, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x]
/; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && IntegerQ[Simplify[j/n]] && Integ
erQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

\begin {align*} \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^3} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {(A+B x) \sqrt {b x+c x^2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {A \left (b x^2+c x^4\right )^{3/2}}{b x^4}+\frac {\left (-2 (-b B+A c)+\frac {3}{2} (-b B+2 A c)\right ) \text {Subst}\left (\int \frac {\sqrt {b x+c x^2}}{x} \, dx,x,x^2\right )}{b}\\ &=\frac {(b B+2 A c) \sqrt {b x^2+c x^4}}{2 b}-\frac {A \left (b x^2+c x^4\right )^{3/2}}{b x^4}+\frac {1}{4} (b B+2 A c) \text {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )\\ &=\frac {(b B+2 A c) \sqrt {b x^2+c x^4}}{2 b}-\frac {A \left (b x^2+c x^4\right )^{3/2}}{b x^4}+\frac {1}{2} (b B+2 A c) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x^2}{\sqrt {b x^2+c x^4}}\right )\\ &=\frac {(b B+2 A c) \sqrt {b x^2+c x^4}}{2 b}-\frac {A \left (b x^2+c x^4\right )^{3/2}}{b x^4}+\frac {(b B+2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{2 \sqrt {c}}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 90, normalized size = 0.93 \begin {gather*} \frac {\sqrt {c} \left (-2 A+B x^2\right ) \left (b+c x^2\right )-(b B+2 A c) x \sqrt {b+c x^2} \log \left (-\sqrt {c} x+\sqrt {b+c x^2}\right )}{2 \sqrt {c} \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*(-2*A + B*x^2)*(b + c*x^2) - (b*B + 2*A*c)*x*Sqrt[b + c*x^2]*Log[-(Sqrt[c]*x) + Sqrt[b + c*x^2]])/(2*
Sqrt[c]*Sqrt[x^2*(b + c*x^2)])

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Maple [A]
time = 0.38, size = 132, normalized size = 1.36

method	result	size
risch	\(-\frac {\left (-B \,x^{2}+2 A \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{2 x^{2}}+\frac {\left (\ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) \sqrt {c}\, A +\frac {\ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) B b}{2 \sqrt {c}}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{x \sqrt {c \,x^{2}+b}}\)	\(101\)
default	\(-\frac {\sqrt {x^{4} c +b \,x^{2}}\, \left (-2 A \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}} x^{2}-B \sqrt {c \,x^{2}+b}\, \sqrt {c}\, b \,x^{2}+2 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {c}-2 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b c x -B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{2} x \right )}{2 x^{2} \sqrt {c \,x^{2}+b}\, b \sqrt {c}}\)	\(132\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 105, normalized size = 1.08 \begin {gather*} \frac {1}{2} \, {\left (\sqrt {c} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - \frac {2 \, \sqrt {c x^{4} + b x^{2}}}{x^{2}}\right )} A + \frac {1}{4} \, {\left (\frac {b \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{\sqrt {c}} + 2 \, \sqrt {c x^{4} + b x^{2}}\right )} B \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^3,x, algorithm="maxima")

[Out]

1/2*(sqrt(c)*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*sqrt(c*x^4 + b*x^2)/x^2)*A + 1/4*(b*log(2*c*
x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/sqrt(c) + 2*sqrt(c*x^4 + b*x^2))*B

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Fricas [A]
time = 1.80, size = 161, normalized size = 1.66 \begin {gather*} \left [\frac {{\left (B b + 2 \, A c\right )} \sqrt {c} x^{2} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} {\left (B c x^{2} - 2 \, A c\right )}}{4 \, c x^{2}}, -\frac {{\left (B b + 2 \, A c\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - \sqrt {c x^{4} + b x^{2}} {\left (B c x^{2} - 2 \, A c\right )}}{2 \, c x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^3,x, algorithm="fricas")

[Out]

[1/4*((B*b + 2*A*c)*sqrt(c)*x^2*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) + 2*sqrt(c*x^4 + b*x^2)*(B*c
*x^2 - 2*A*c))/(c*x^2), -1/2*((B*b + 2*A*c)*sqrt(-c)*x^2*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) - sq
rt(c*x^4 + b*x^2)*(B*c*x^2 - 2*A*c))/(c*x^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )}{x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**(1/2)/x**3,x)

[Out]

Integral(sqrt(x**2*(b + c*x**2))*(A + B*x**2)/x**3, x)

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Giac [A]
time = 1.69, size = 92, normalized size = 0.95 \begin {gather*} \frac {1}{2} \, \sqrt {c x^{2} + b} B x \mathrm {sgn}\left (x\right ) + \frac {2 \, A b \sqrt {c} \mathrm {sgn}\left (x\right )}{{\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b} - \frac {{\left (B b \sqrt {c} \mathrm {sgn}\left (x\right ) + 2 \, A c^{\frac {3}{2}} \mathrm {sgn}\left (x\right )\right )} \log \left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2}\right )}{4 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^3,x, algorithm="giac")

[Out]

1/2*sqrt(c*x^2 + b)*B*x*sgn(x) + 2*A*b*sqrt(c)*sgn(x)/((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b) - 1/4*(B*b*sqrt(c)
*sgn(x) + 2*A*c^(3/2)*sgn(x))*log((sqrt(c)*x - sqrt(c*x^2 + b))^2)/c

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (B\,x^2+A\right )\,\sqrt {c\,x^4+b\,x^2}}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^3,x)

[Out]

int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^3, x)

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