3.91 \(\int x^3 (A+B x^2) \sqrt {b x^2+c x^4} \, dx\)

Optimal. Leaf size=125 \[ -\frac {b^3 (5 b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{128 c^{7/2}}+\frac {b \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4} (5 b B-8 A c)}{128 c^3}-\frac {\left (b x^2+c x^4\right )^{3/2} \left (-8 A c+5 b B-6 B c x^2\right )}{48 c^2} \]

[Out]

-1/48*(-6*B*c*x^2-8*A*c+5*B*b)*(c*x^4+b*x^2)^(3/2)/c^2-1/128*b^3*(-8*A*c+5*B*b)*arctanh(x^2*c^(1/2)/(c*x^4+b*x
^2)^(1/2))/c^(7/2)+1/128*b*(-8*A*c+5*B*b)*(2*c*x^2+b)*(c*x^4+b*x^2)^(1/2)/c^3

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Rubi [A]  time = 0.20, antiderivative size = 125, 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 = {2034, 779, 612, 620, 206} \[ -\frac {b^3 (5 b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{128 c^{7/2}}+\frac {b \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4} (5 b B-8 A c)}{128 c^3}-\frac {\left (b x^2+c x^4\right )^{3/2} \left (-8 A c+5 b B-6 B c x^2\right )}{48 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(5*b*B - 8*A*c)*(b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(128*c^3) - ((5*b*B - 8*A*c - 6*B*c*x^2)*(b*x^2 + c*x^4)
^(3/2))/(48*c^2) - (b^3*(5*b*B - 8*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(128*c^(7/2))

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 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), 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))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[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^3 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x (A+B x) \sqrt {b x+c x^2} \, dx,x,x^2\right )\\ &=-\frac {\left (5 b B-8 A c-6 B c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac {(b (5 b B-8 A c)) \operatorname {Subst}\left (\int \sqrt {b x+c x^2} \, dx,x,x^2\right )}{32 c^2}\\ &=\frac {b (5 b B-8 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{128 c^3}-\frac {\left (5 b B-8 A c-6 B c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{48 c^2}-\frac {\left (b^3 (5 b B-8 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{256 c^3}\\ &=\frac {b (5 b B-8 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{128 c^3}-\frac {\left (5 b B-8 A c-6 B c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{48 c^2}-\frac {\left (b^3 (5 b B-8 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 )}{128 c^3}\\ &=\frac {b (5 b B-8 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{128 c^3}-\frac {\left (5 b B-8 A c-6 B c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{48 c^2}-\frac {b^3 (5 b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{128 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 151, normalized size = 1.21 \[ \frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {c} x \sqrt {\frac {c x^2}{b}+1} \left (-2 b^2 c \left (12 A+5 B x^2\right )+8 b c^2 x^2 \left (2 A+B x^2\right )+16 c^3 x^4 \left (4 A+3 B x^2\right )+15 b^3 B\right )-3 b^{5/2} (5 b B-8 A c) \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )\right )}{384 c^{7/2} x \sqrt {\frac {c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(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]*(15*b^3*B + 8*b*c^2*x^2*(2*A + B*x^2) + 16*c^3*x^4*(4*A
+ 3*B*x^2) - 2*b^2*c*(12*A + 5*B*x^2)) - 3*b^(5/2)*(5*b*B - 8*A*c)*ArcSinh[(Sqrt[c]*x)/Sqrt[b]]))/(384*c^(7/2)
*x*Sqrt[1 + (c*x^2)/b])

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fricas [A]  time = 0.88, size = 272, normalized size = 2.18 \[ \left [-\frac {3 \, {\left (5 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (48 \, B c^{4} x^{6} + 15 \, B b^{3} c - 24 \, A b^{2} c^{2} + 8 \, {\left (B b c^{3} + 8 \, A c^{4}\right )} x^{4} - 2 \, {\left (5 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{768 \, c^{4}}, \frac {3 \, {\left (5 \, B b^{4} - 8 \, A b^{3} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (48 \, B c^{4} x^{6} + 15 \, B b^{3} c - 24 \, A b^{2} c^{2} + 8 \, {\left (B b c^{3} + 8 \, A c^{4}\right )} x^{4} - 2 \, {\left (5 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{384 \, c^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/768*(3*(5*B*b^4 - 8*A*b^3*c)*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*(48*B*c^4*x^6 +
15*B*b^3*c - 24*A*b^2*c^2 + 8*(B*b*c^3 + 8*A*c^4)*x^4 - 2*(5*B*b^2*c^2 - 8*A*b*c^3)*x^2)*sqrt(c*x^4 + b*x^2))
/c^4, 1/384*(3*(5*B*b^4 - 8*A*b^3*c)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) + (48*B*c^4*x^6
+ 15*B*b^3*c - 24*A*b^2*c^2 + 8*(B*b*c^3 + 8*A*c^4)*x^4 - 2*(5*B*b^2*c^2 - 8*A*b*c^3)*x^2)*sqrt(c*x^4 + b*x^2
))/c^4]

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giac [A]  time = 0.22, size = 177, normalized size = 1.42 \[ \frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, B x^{2} \mathrm {sgn}\relax (x) + \frac {B b c^{5} \mathrm {sgn}\relax (x) + 8 \, A c^{6} \mathrm {sgn}\relax (x)}{c^{6}}\right )} x^{2} - \frac {5 \, B b^{2} c^{4} \mathrm {sgn}\relax (x) - 8 \, A b c^{5} \mathrm {sgn}\relax (x)}{c^{6}}\right )} x^{2} + \frac {3 \, {\left (5 \, B b^{3} c^{3} \mathrm {sgn}\relax (x) - 8 \, A b^{2} c^{4} \mathrm {sgn}\relax (x)\right )}}{c^{6}}\right )} \sqrt {c x^{2} + b} x + \frac {{\left (5 \, B b^{4} \mathrm {sgn}\relax (x) - 8 \, A b^{3} c \mathrm {sgn}\relax (x)\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{128 \, c^{\frac {7}{2}}} - \frac {{\left (5 \, B b^{4} \log \left ({\left | b \right |}\right ) - 8 \, A b^{3} c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\relax (x)}{256 \, c^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(2*(4*(6*B*x^2*sgn(x) + (B*b*c^5*sgn(x) + 8*A*c^6*sgn(x))/c^6)*x^2 - (5*B*b^2*c^4*sgn(x) - 8*A*b*c^5*sgn
(x))/c^6)*x^2 + 3*(5*B*b^3*c^3*sgn(x) - 8*A*b^2*c^4*sgn(x))/c^6)*sqrt(c*x^2 + b)*x + 1/128*(5*B*b^4*sgn(x) - 8
*A*b^3*c*sgn(x))*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/c^(7/2) - 1/256*(5*B*b^4*log(abs(b)) - 8*A*b^3*c*log(a
bs(b)))*sgn(x)/c^(7/2)

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maple [A]  time = 0.05, size = 206, normalized size = 1.65 \[ \frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (48 \left (c \,x^{2}+b \right )^{\frac {3}{2}} B \,c^{\frac {5}{2}} x^{5}+64 \left (c \,x^{2}+b \right )^{\frac {3}{2}} A \,c^{\frac {5}{2}} x^{3}-40 \left (c \,x^{2}+b \right )^{\frac {3}{2}} B b \,c^{\frac {3}{2}} x^{3}+24 A \,b^{3} c \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )-15 B \,b^{4} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )+24 \sqrt {c \,x^{2}+b}\, A \,b^{2} c^{\frac {3}{2}} x -15 \sqrt {c \,x^{2}+b}\, B \,b^{3} \sqrt {c}\, x -48 \left (c \,x^{2}+b \right )^{\frac {3}{2}} A b \,c^{\frac {3}{2}} x +30 \left (c \,x^{2}+b \right )^{\frac {3}{2}} B \,b^{2} \sqrt {c}\, x \right )}{384 \sqrt {c \,x^{2}+b}\, c^{\frac {7}{2}} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(c*x^4+b*x^2)^(1/2)*(48*B*(c*x^2+b)^(3/2)*c^(5/2)*x^5+64*A*(c*x^2+b)^(3/2)*c^(5/2)*x^3-40*B*(c*x^2+b)^(3
/2)*c^(3/2)*x^3*b-48*A*(c*x^2+b)^(3/2)*c^(3/2)*x*b+30*B*(c*x^2+b)^(3/2)*c^(1/2)*x*b^2+24*A*(c*x^2+b)^(1/2)*c^(
3/2)*x*b^2-15*B*(c*x^2+b)^(1/2)*c^(1/2)*x*b^3+24*A*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*b^3*c-15*B*ln(c^(1/2)*x+(c*x^
2+b)^(1/2))*b^4)/x/(c*x^2+b)^(1/2)/c^(7/2)

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maxima [B]  time = 1.49, size = 225, normalized size = 1.80 \[ -\frac {1}{96} \, {\left (\frac {12 \, \sqrt {c x^{4} + b x^{2}} b x^{2}}{c} - \frac {3 \, b^{3} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {5}{2}}} + \frac {6 \, \sqrt {c x^{4} + b x^{2}} b^{2}}{c^{2}} - \frac {16 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{c}\right )} A + \frac {1}{768} \, {\left (\frac {60 \, \sqrt {c x^{4} + b x^{2}} b^{2} x^{2}}{c^{2}} + \frac {96 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{2}}{c} - \frac {15 \, b^{4} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {7}{2}}} + \frac {30 \, \sqrt {c x^{4} + b x^{2}} b^{3}}{c^{3}} - \frac {80 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b}{c^{2}}\right )} B \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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*sqr
t(c*x^4 + b*x^2)*b^2/c^2 - 16*(c*x^4 + b*x^2)^(3/2)/c)*A + 1/768*(60*sqrt(c*x^4 + b*x^2)*b^2*x^2/c^2 + 96*(c*x
^4 + b*x^2)^(3/2)*x^2/c - 15*b^4*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(7/2) + 30*sqrt(c*x^4 + b*
x^2)*b^3/c^3 - 80*(c*x^4 + b*x^2)^(3/2)*b/c^2)*B

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mupad [B]  time = 0.74, size = 177, normalized size = 1.42 \[ \frac {B\,x^2\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{8\,c}-\frac {5\,B\,b\,\left (\frac {b^3\,\ln \left (b+2\,c\,x^2+2\,\sqrt {c}\,\relax |x|\,\sqrt {c\,x^2+b}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^4+b\,x^2}\,\left (-3\,b^2+2\,b\,c\,x^2+8\,c^2\,x^4\right )}{24\,c^2}\right )}{16\,c}+\frac {A\,b^3\,\ln \left (b+2\,c\,x^2+2\,\sqrt {c}\,\relax |x|\,\sqrt {c\,x^2+b}\right )}{32\,c^{5/2}}+\frac {A\,\sqrt {c\,x^4+b\,x^2}\,\left (-3\,b^2+2\,b\,c\,x^2+8\,c^2\,x^4\right )}{48\,c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*x^2*(b*x^2 + c*x^4)^(3/2))/(8*c) - (5*B*b*((b^3*log(b + 2*c*x^2 + 2*c^(1/2)*abs(x)*(b + c*x^2)^(1/2)))/(16*
c^(5/2)) + ((b*x^2 + c*x^4)^(1/2)*(8*c^2*x^4 - 3*b^2 + 2*b*c*x^2))/(24*c^2)))/(16*c) + (A*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*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 \[ \int x^{3} \sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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