3.1.92 \(\int x (A+B x^2) \sqrt {b x^2+c x^4} \, dx\)

Optimal. Leaf size=107 \[ \frac {b^2 (b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{5/2}}-\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4} (b B-2 A c)}{16 c^2}+\frac {B \left (b x^2+c x^4\right )^{3/2}}{6 c} \]

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Rubi [A]  time = 0.15, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2034, 640, 612, 620, 206} \begin {gather*} \frac {b^2 (b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{5/2}}-\frac {\left (b+2 c x^2\right ) \sqrt {b x^2+c x^4} (b B-2 A c)}{16 c^2}+\frac {B \left (b x^2+c x^4\right )^{3/2}}{6 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*B - 2*A*c)*(b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(16*c^2) + (B*(b*x^2 + c*x^4)^(3/2))/(6*c) + (b^2*(b*B - 2*
A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(16*c^(5/2))

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 612

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

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 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[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 x \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int (A+B x) \sqrt {b x+c x^2} \, dx,x,x^2\right )\\ &=\frac {B \left (b x^2+c x^4\right )^{3/2}}{6 c}+\frac {(-b B+2 A c) \operatorname {Subst}\left (\int \sqrt {b x+c x^2} \, dx,x,x^2\right )}{4 c}\\ &=-\frac {(b B-2 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{16 c^2}+\frac {B \left (b x^2+c x^4\right )^{3/2}}{6 c}+\frac {\left (b^2 (b B-2 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{32 c^2}\\ &=-\frac {(b B-2 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{16 c^2}+\frac {B \left (b x^2+c x^4\right )^{3/2}}{6 c}+\frac {\left (b^2 (b B-2 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^2}\\ &=-\frac {(b B-2 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{16 c^2}+\frac {B \left (b x^2+c x^4\right )^{3/2}}{6 c}+\frac {b^2 (b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 129, normalized size = 1.21 \begin {gather*} \frac {\sqrt {x^2 \left (b+c x^2\right )} \left (3 b^{3/2} (b B-2 A c) \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )+\sqrt {c} x \sqrt {\frac {c x^2}{b}+1} \left (2 b c \left (3 A+B x^2\right )+4 c^2 x^2 \left (3 A+2 B x^2\right )-3 b^2 B\right )\right )}{48 c^{5/2} x \sqrt {\frac {c x^2}{b}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x^2*(b + c*x^2)]*(Sqrt[c]*x*Sqrt[1 + (c*x^2)/b]*(-3*b^2*B + 2*b*c*(3*A + B*x^2) + 4*c^2*x^2*(3*A + 2*B*x
^2)) + 3*b^(3/2)*(b*B - 2*A*c)*ArcSinh[(Sqrt[c]*x)/Sqrt[b]]))/(48*c^(5/2)*x*Sqrt[1 + (c*x^2)/b])

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IntegrateAlgebraic [A]  time = 0.59, size = 115, normalized size = 1.07 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (6 A b c+12 A c^2 x^2-3 b^2 B+2 b B c x^2+8 B c^2 x^4\right )}{48 c^2}+\frac {\left (2 A b^2 c-b^3 B\right ) \log \left (-2 \sqrt {c} \sqrt {b x^2+c x^4}+b+2 c x^2\right )}{32 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x^2 + c*x^4]*(-3*b^2*B + 6*A*b*c + 2*b*B*c*x^2 + 12*A*c^2*x^2 + 8*B*c^2*x^4))/(48*c^2) + ((-(b^3*B) +
2*A*b^2*c)*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[b*x^2 + c*x^4]])/(32*c^(5/2))

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fricas [A]  time = 0.43, size = 223, normalized size = 2.08 \begin {gather*} \left [-\frac {3 \, {\left (B b^{3} - 2 \, A b^{2} c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (8 \, B c^{3} x^{4} - 3 \, B b^{2} c + 6 \, A b c^{2} + 2 \, {\left (B b c^{2} + 6 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{96 \, c^{3}}, -\frac {3 \, {\left (B b^{3} - 2 \, A b^{2} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - {\left (8 \, B c^{3} x^{4} - 3 \, B b^{2} c + 6 \, A b c^{2} + 2 \, {\left (B b c^{2} + 6 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{48 \, c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(3*(B*b^3 - 2*A*b^2*c)*sqrt(c)*log(-2*c*x^2 - b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*(8*B*c^3*x^4 - 3*B
*b^2*c + 6*A*b*c^2 + 2*(B*b*c^2 + 6*A*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^3, -1/48*(3*(B*b^3 - 2*A*b^2*c)*sqrt(-c
)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) - (8*B*c^3*x^4 - 3*B*b^2*c + 6*A*b*c^2 + 2*(B*b*c^2 + 6*A*c
^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^3]

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giac [A]  time = 0.19, size = 140, normalized size = 1.31 \begin {gather*} \frac {1}{48} \, {\left (2 \, {\left (4 \, B x^{2} \mathrm {sgn}\relax (x) + \frac {B b c^{3} \mathrm {sgn}\relax (x) + 6 \, A c^{4} \mathrm {sgn}\relax (x)}{c^{4}}\right )} x^{2} - \frac {3 \, {\left (B b^{2} c^{2} \mathrm {sgn}\relax (x) - 2 \, A b c^{3} \mathrm {sgn}\relax (x)\right )}}{c^{4}}\right )} \sqrt {c x^{2} + b} x - \frac {{\left (B b^{3} \mathrm {sgn}\relax (x) - 2 \, A b^{2} c \mathrm {sgn}\relax (x)\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{16 \, c^{\frac {5}{2}}} + \frac {{\left (B b^{3} \log \left ({\left | b \right |}\right ) - 2 \, A b^{2} c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\relax (x)}{32 \, c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(2*(4*B*x^2*sgn(x) + (B*b*c^3*sgn(x) + 6*A*c^4*sgn(x))/c^4)*x^2 - 3*(B*b^2*c^2*sgn(x) - 2*A*b*c^3*sgn(x))
/c^4)*sqrt(c*x^2 + b)*x - 1/16*(B*b^3*sgn(x) - 2*A*b^2*c*sgn(x))*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/c^(5/2
) + 1/32*(B*b^3*log(abs(b)) - 2*A*b^2*c*log(abs(b)))*sgn(x)/c^(5/2)

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maple [A]  time = 0.05, size = 164, normalized size = 1.53 \begin {gather*} \frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (8 \left (c \,x^{2}+b \right )^{\frac {3}{2}} B \,c^{\frac {3}{2}} x^{3}-6 A \,b^{2} c \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )+3 B \,b^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )-6 \sqrt {c \,x^{2}+b}\, A b \,c^{\frac {3}{2}} x +3 \sqrt {c \,x^{2}+b}\, B \,b^{2} \sqrt {c}\, x +12 \left (c \,x^{2}+b \right )^{\frac {3}{2}} A \,c^{\frac {3}{2}} x -6 \left (c \,x^{2}+b \right )^{\frac {3}{2}} B b \sqrt {c}\, x \right )}{48 \sqrt {c \,x^{2}+b}\, c^{\frac {5}{2}} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(c*x^4+b*x^2)^(1/2)*(8*B*c^(3/2)*(c*x^2+b)^(3/2)*x^3+12*A*c^(3/2)*(c*x^2+b)^(3/2)*x-6*B*c^(1/2)*(c*x^2+b)
^(3/2)*x*b-6*A*c^(3/2)*(c*x^2+b)^(1/2)*x*b+3*B*c^(1/2)*(c*x^2+b)^(1/2)*x*b^2-6*A*ln(c^(1/2)*x+(c*x^2+b)^(1/2))
*b^2*c+3*B*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*b^3)/x/(c*x^2+b)^(1/2)/c^(5/2)

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maxima [A]  time = 1.42, size = 177, normalized size = 1.65 \begin {gather*} \frac {1}{16} \, {\left (4 \, \sqrt {c x^{4} + b x^{2}} x^{2} - \frac {b^{2} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {3}{2}}} + \frac {2 \, \sqrt {c x^{4} + b x^{2}} b}{c}\right )} A - \frac {1}{96} \, {\left (\frac {12 \, \sqrt {c x^{4} + b x^{2}} b x^{2}}{c} - \frac {3 \, b^{3} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {5}{2}}} + \frac {6 \, \sqrt {c x^{4} + b x^{2}} b^{2}}{c^{2}} - \frac {16 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{c}\right )} B \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*sqrt(c*x^4 + b*x^2)*x^2 - b^2*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(3/2) + 2*sqrt(c*x^4
+ b*x^2)*b/c)*A - 1/96*(12*sqrt(c*x^4 + b*x^2)*b*x^2/c - 3*b^3*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)
)/c^(5/2) + 6*sqrt(c*x^4 + b*x^2)*b^2/c^2 - 16*(c*x^4 + b*x^2)^(3/2)/c)*B

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mupad [B]  time = 0.75, size = 140, normalized size = 1.31 \begin {gather*} \frac {A\,\left (\frac {b}{4\,c}+\frac {x^2}{2}\right )\,\sqrt {c\,x^4+b\,x^2}}{2}+\frac {B\,b^3\,\ln \left (b+2\,c\,x^2+2\,\sqrt {c}\,\relax |x|\,\sqrt {c\,x^2+b}\right )}{32\,c^{5/2}}-\frac {A\,b^2\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{16\,c^{3/2}}+\frac {B\,\sqrt {c\,x^4+b\,x^2}\,\left (-3\,b^2+2\,b\,c\,x^2+8\,c^2\,x^4\right )}{48\,c^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*(b/(4*c) + x^2/2)*(b*x^2 + c*x^4)^(1/2))/2 + (B*b^3*log(b + 2*c*x^2 + 2*c^(1/2)*abs(x)*(b + c*x^2)^(1/2)))/
(32*c^(5/2)) - (A*b^2*log((b/2 + c*x^2)/c^(1/2) + (b*x^2 + c*x^4)^(1/2)))/(16*c^(3/2)) + (B*(b*x^2 + c*x^4)^(1
/2)*(8*c^2*x^4 - 3*b^2 + 2*b*c*x^2))/(48*c^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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