3.148 \(\int \frac {x^3 (A+B x^2)}{(b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=67 \[ \frac {B \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{c^{3/2}}-\frac {x^2 (b B-A c)}{b c \sqrt {b x^2+c x^4}} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2034, 777, 620, 206} \[ \frac {B \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{c^{3/2}}-\frac {x^2 (b B-A c)}{b c \sqrt {b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 620

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 777

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((2
*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (b^2*e*g - b*c*(e*f + d*g) + 2*c*(c*d*f - a*e*g))*x)*(a + b*x + c*x^2)^
(p + 1))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p
+ 3))/(c*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && N
eQ[b^2 - 4*a*c, 0] && LtQ[p, -1]

Rule 2034

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

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Mathematica [A]  time = 0.11, size = 75, normalized size = 1.12 \[ \frac {x \left (\sqrt {c} x (A c-b B)+b^{3/2} B \sqrt {\frac {c x^2}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )\right )}{b c^{3/2} \sqrt {x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(Sqrt[c]*(-(b*B) + A*c)*x + b^(3/2)*B*Sqrt[1 + (c*x^2)/b]*ArcSinh[(Sqrt[c]*x)/Sqrt[b]]))/(b*c^(3/2)*Sqrt[x^
2*(b + c*x^2)])

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fricas [A]  time = 0.92, size = 188, normalized size = 2.81 \[ \left [\frac {{\left (B b c x^{2} + B b^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, \sqrt {c x^{4} + b x^{2}} {\left (B b c - A c^{2}\right )}}{2 \, {\left (b c^{3} x^{2} + b^{2} c^{2}\right )}}, -\frac {{\left (B b c x^{2} + B b^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + \sqrt {c x^{4} + b x^{2}} {\left (B b c - A c^{2}\right )}}{b c^{3} x^{2} + b^{2} c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((B*b*c*x^2 + B*b^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)*(B
*b*c - A*c^2))/(b*c^3*x^2 + b^2*c^2), -((B*b*c*x^2 + B*b^2)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^
2 + b)) + sqrt(c*x^4 + b*x^2)*(B*b*c - A*c^2))/(b*c^3*x^2 + b^2*c^2)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to divide, perhaps due to rounding error%%%{%%%{-2,[1]%%%},[2,2]%%%}+%%%{%%{[4,0]:[1,0,%%%{-1,[1]%%%}]%%},[
1,3]%%%}+%%%{-2,[0,4]%%%} / %%%{%%%{1,[2]%%%},[2,0]%%%}+%%%{%%{[%%%{-2,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[1,1
]%%%}+%%%{%%%{1,[1]%%%},[0,2]%%%} Error: Bad Argument Value

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maple [A]  time = 0.05, size = 75, normalized size = 1.12 \[ \frac {\left (c \,x^{2}+b \right ) \left (A \,c^{\frac {5}{2}} x -B b \,c^{\frac {3}{2}} x +\sqrt {c \,x^{2}+b}\, B b c \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )\right ) x^{3}}{\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b \,c^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^3*(c*x^2+b)*(A*c^(5/2)*x-B*b*c^(3/2)*x+(c*x^2+b)^(1/2)*B*b*c*ln(c^(1/2)*x+(c*x^2+b)^(1/2)))/(c*x^4+b*x^2)^(3
/2)/c^(5/2)/b

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maxima [A]  time = 1.48, size = 79, normalized size = 1.18 \[ -\frac {1}{2} \, B {\left (\frac {2 \, x^{2}}{\sqrt {c x^{4} + b x^{2}} c} - \frac {\log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {3}{2}}}\right )} + \frac {A x^{2}}{\sqrt {c x^{4} + b x^{2}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.63, size = 78, normalized size = 1.16 \[ \frac {B\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{2\,c^{3/2}}+\frac {A\,x^2}{b\,\sqrt {c\,x^4+b\,x^2}}-\frac {B\,x^2}{c\,\sqrt {c\,x^4+b\,x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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