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

Optimal. Leaf size=148 \[ -\frac {3 b^4 (b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{256 c^{7/2}}+\frac {3 b^2 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4} (b B-2 A c)}{256 c^3}-\frac {\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2} (b B-2 A c)}{32 c^2}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{10 c} \]

[Out]

-1/32*(-2*A*c+B*b)*(2*c*x^2+b)*(c*x^4+b*x^2)^(3/2)/c^2+1/10*B*(c*x^4+b*x^2)^(5/2)/c-3/256*b^4*(-2*A*c+B*b)*arc
tanh(x^2*c^(1/2)/(c*x^4+b*x^2)^(1/2))/c^(7/2)+3/256*b^2*(-2*A*c+B*b)*(2*c*x^2+b)*(c*x^4+b*x^2)^(1/2)/c^3

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Rubi [A]  time = 0.19, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ \frac {3 b^2 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4} (b B-2 A c)}{256 c^3}-\frac {3 b^4 (b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{256 c^{7/2}}-\frac {\left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2} (b B-2 A c)}{32 c^2}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{10 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b^2*(b*B - 2*A*c)*(b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(256*c^3) - ((b*B - 2*A*c)*(b + 2*c*x^2)*(b*x^2 + c*x^
4)^(3/2))/(32*c^2) + (B*(b*x^2 + c*x^4)^(5/2))/(10*c) - (3*b^4*(b*B - 2*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2
+ c*x^4]])/(256*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 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 ) \left (b x^2+c x^4\right )^{3/2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int (A+B x) \left (b x+c x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac {B \left (b x^2+c x^4\right )^{5/2}}{10 c}+\frac {(-b B+2 A c) \operatorname {Subst}\left (\int \left (b x+c x^2\right )^{3/2} \, dx,x,x^2\right )}{4 c}\\ &=-\frac {(b B-2 A c) \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{10 c}+\frac {\left (3 b^2 (b B-2 A c)\right ) \operatorname {Subst}\left (\int \sqrt {b x+c x^2} \, dx,x,x^2\right )}{64 c^2}\\ &=\frac {3 b^2 (b B-2 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{256 c^3}-\frac {(b B-2 A c) \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{10 c}-\frac {\left (3 b^4 (b B-2 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{512 c^3}\\ &=\frac {3 b^2 (b B-2 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{256 c^3}-\frac {(b B-2 A c) \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{10 c}-\frac {\left (3 b^4 (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 )}{256 c^3}\\ &=\frac {3 b^2 (b B-2 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{256 c^3}-\frac {(b B-2 A c) \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{32 c^2}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{10 c}-\frac {3 b^4 (b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{256 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 171, normalized size = 1.16 \[ \frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {c} x \sqrt {\frac {c x^2}{b}+1} \left (-10 b^3 c \left (3 A+B x^2\right )+4 b^2 c^2 x^2 \left (5 A+2 B x^2\right )+16 b c^3 x^4 \left (15 A+11 B x^2\right )+32 c^4 x^6 \left (5 A+4 B x^2\right )+15 b^4 B\right )-15 b^{7/2} (b B-2 A c) \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )\right )}{1280 c^{7/2} x \sqrt {\frac {c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x^2*(b + c*x^2)]*(Sqrt[c]*x*Sqrt[1 + (c*x^2)/b]*(15*b^4*B - 10*b^3*c*(3*A + B*x^2) + 4*b^2*c^2*x^2*(5*A
+ 2*B*x^2) + 32*c^4*x^6*(5*A + 4*B*x^2) + 16*b*c^3*x^4*(15*A + 11*B*x^2)) - 15*b^(7/2)*(b*B - 2*A*c)*ArcSinh[(
Sqrt[c]*x)/Sqrt[b]]))/(1280*c^(7/2)*x*Sqrt[1 + (c*x^2)/b])

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fricas [A]  time = 1.12, size = 316, normalized size = 2.14 \[ \left [-\frac {15 \, {\left (B b^{5} - 2 \, A b^{4} c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (128 \, B c^{5} x^{8} + 16 \, {\left (11 \, B b c^{4} + 10 \, A c^{5}\right )} x^{6} + 15 \, B b^{4} c - 30 \, A b^{3} c^{2} + 8 \, {\left (B b^{2} c^{3} + 30 \, A b c^{4}\right )} x^{4} - 10 \, {\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{2560 \, c^{4}}, \frac {15 \, {\left (B b^{5} - 2 \, A b^{4} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (128 \, B c^{5} x^{8} + 16 \, {\left (11 \, B b c^{4} + 10 \, A c^{5}\right )} x^{6} + 15 \, B b^{4} c - 30 \, A b^{3} c^{2} + 8 \, {\left (B b^{2} c^{3} + 30 \, A b c^{4}\right )} x^{4} - 10 \, {\left (B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{1280 \, c^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2560*(15*(B*b^5 - 2*A*b^4*c)*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*(128*B*c^5*x^8
+ 16*(11*B*b*c^4 + 10*A*c^5)*x^6 + 15*B*b^4*c - 30*A*b^3*c^2 + 8*(B*b^2*c^3 + 30*A*b*c^4)*x^4 - 10*(B*b^3*c^2
- 2*A*b^2*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^4, 1/1280*(15*(B*b^5 - 2*A*b^4*c)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^
2)*sqrt(-c)/(c*x^2 + b)) + (128*B*c^5*x^8 + 16*(11*B*b*c^4 + 10*A*c^5)*x^6 + 15*B*b^4*c - 30*A*b^3*c^2 + 8*(B*
b^2*c^3 + 30*A*b*c^4)*x^4 - 10*(B*b^3*c^2 - 2*A*b^2*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^4]

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giac [A]  time = 0.22, size = 207, normalized size = 1.40 \[ \frac {1}{1280} \, {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, B c x^{2} \mathrm {sgn}\relax (x) + \frac {11 \, B b c^{8} \mathrm {sgn}\relax (x) + 10 \, A c^{9} \mathrm {sgn}\relax (x)}{c^{8}}\right )} x^{2} + \frac {B b^{2} c^{7} \mathrm {sgn}\relax (x) + 30 \, A b c^{8} \mathrm {sgn}\relax (x)}{c^{8}}\right )} x^{2} - \frac {5 \, {\left (B b^{3} c^{6} \mathrm {sgn}\relax (x) - 2 \, A b^{2} c^{7} \mathrm {sgn}\relax (x)\right )}}{c^{8}}\right )} x^{2} + \frac {15 \, {\left (B b^{4} c^{5} \mathrm {sgn}\relax (x) - 2 \, A b^{3} c^{6} \mathrm {sgn}\relax (x)\right )}}{c^{8}}\right )} \sqrt {c x^{2} + b} x + \frac {3 \, {\left (B b^{5} \mathrm {sgn}\relax (x) - 2 \, A b^{4} c \mathrm {sgn}\relax (x)\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{256 \, c^{\frac {7}{2}}} - \frac {3 \, {\left (B b^{5} \log \left ({\left | b \right |}\right ) - 2 \, A b^{4} c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\relax (x)}{512 \, c^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1280*(2*(4*(2*(8*B*c*x^2*sgn(x) + (11*B*b*c^8*sgn(x) + 10*A*c^9*sgn(x))/c^8)*x^2 + (B*b^2*c^7*sgn(x) + 30*A*
b*c^8*sgn(x))/c^8)*x^2 - 5*(B*b^3*c^6*sgn(x) - 2*A*b^2*c^7*sgn(x))/c^8)*x^2 + 15*(B*b^4*c^5*sgn(x) - 2*A*b^3*c
^6*sgn(x))/c^8)*sqrt(c*x^2 + b)*x + 3/256*(B*b^5*sgn(x) - 2*A*b^4*c*sgn(x))*log(abs(-sqrt(c)*x + sqrt(c*x^2 +
b)))/c^(7/2) - 3/512*(B*b^5*log(abs(b)) - 2*A*b^4*c*log(abs(b)))*sgn(x)/c^(7/2)

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maple [A]  time = 0.06, size = 244, normalized size = 1.65 \[ \frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (128 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B \,c^{\frac {5}{2}} x^{5}+30 A \,b^{4} c \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )-15 B \,b^{5} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )+30 \sqrt {c \,x^{2}+b}\, A \,b^{3} c^{\frac {3}{2}} x +160 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A \,c^{\frac {5}{2}} x^{3}-15 \sqrt {c \,x^{2}+b}\, B \,b^{4} \sqrt {c}\, x -80 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B b \,c^{\frac {3}{2}} x^{3}+20 \left (c \,x^{2}+b \right )^{\frac {3}{2}} A \,b^{2} c^{\frac {3}{2}} x -10 \left (c \,x^{2}+b \right )^{\frac {3}{2}} B \,b^{3} \sqrt {c}\, x -80 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A b \,c^{\frac {3}{2}} x +40 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B \,b^{2} \sqrt {c}\, x \right )}{1280 \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {7}{2}} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1280*(c*x^4+b*x^2)^(3/2)*(128*B*(c*x^2+b)^(5/2)*c^(5/2)*x^5+160*A*(c*x^2+b)^(5/2)*c^(5/2)*x^3-80*B*(c*x^2+b)
^(5/2)*c^(3/2)*x^3*b-80*A*(c*x^2+b)^(5/2)*c^(3/2)*x*b+40*B*(c*x^2+b)^(5/2)*c^(1/2)*x*b^2+20*(c*x^2+b)^(3/2)*A*
b^2*c^(3/2)*x-10*(c*x^2+b)^(3/2)*B*b^3*c^(1/2)*x+30*(c*x^2+b)^(1/2)*A*b^3*c^(3/2)*x-15*(c*x^2+b)^(1/2)*B*b^4*c
^(1/2)*x+30*A*b^4*c*ln(c^(1/2)*x+(c*x^2+b)^(1/2))-15*B*b^5*ln(c^(1/2)*x+(c*x^2+b)^(1/2)))/x^3/(c*x^2+b)^(3/2)/
c^(7/2)

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maxima [B]  time = 1.51, size = 267, normalized size = 1.80 \[ \frac {1}{256} \, {\left (32 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{2} - \frac {12 \, \sqrt {c x^{4} + b x^{2}} b^{2} x^{2}}{c} + \frac {3 \, b^{4} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {5}{2}}} - \frac {6 \, \sqrt {c x^{4} + b x^{2}} b^{3}}{c^{2}} + \frac {16 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b}{c}\right )} A + \frac {1}{2560} \, {\left (\frac {60 \, \sqrt {c x^{4} + b x^{2}} b^{3} x^{2}}{c^{2}} - \frac {160 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b x^{2}}{c} - \frac {15 \, b^{5} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {7}{2}}} + \frac {30 \, \sqrt {c x^{4} + b x^{2}} b^{4}}{c^{3}} - \frac {80 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b^{2}}{c^{2}} + \frac {256 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {5}{2}}}{c}\right )} B \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*(32*(c*x^4 + b*x^2)^(3/2)*x^2 - 12*sqrt(c*x^4 + b*x^2)*b^2*x^2/c + 3*b^4*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^3/c^2 + 16*(c*x^4 + b*x^2)^(3/2)*b/c)*A + 1/2560*(60*sqrt(
c*x^4 + b*x^2)*b^3*x^2/c^2 - 160*(c*x^4 + b*x^2)^(3/2)*b*x^2/c - 15*b^5*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^4/c^3 - 80*(c*x^4 + b*x^2)^(3/2)*b^2/c^2 + 256*(c*x^4 + b*x^2)^(
5/2)/c)*B

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mupad [B]  time = 1.01, size = 236, normalized size = 1.59 \[ \frac {B\,{\left (c\,x^4+b\,x^2\right )}^{5/2}}{10\,c}+\frac {A\,{\left (c\,x^4+b\,x^2\right )}^{3/2}\,\left (c\,x^2+\frac {b}{2}\right )}{8\,c}-\frac {3\,A\,b^2\,\left (\left (\frac {b}{4\,c}+\frac {x^2}{2}\right )\,\sqrt {c\,x^4+b\,x^2}-\frac {b^2\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{8\,c^{3/2}}\right )}{32\,c}-\frac {B\,b\,\left (\frac {x^2\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{4}-\frac {3\,b^2\,\left (\frac {\left (2\,c\,x^2+b\right )\,\sqrt {c\,x^4+b\,x^2}}{4\,c}-\frac {b^2\,\ln \left (\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}+\sqrt {c\,x^4+b\,x^2}\right )}{8\,c^{3/2}}\right )}{16\,c}+\frac {b\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{8\,c}\right )}{4\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*(b*x^2 + c*x^4)^(5/2))/(10*c) + (A*(b*x^2 + c*x^4)^(3/2)*(b/2 + c*x^2))/(8*c) - (3*A*b^2*((b/(4*c) + x^2/2)
*(b*x^2 + c*x^4)^(1/2) - (b^2*log((b/2 + c*x^2)/c^(1/2) + (b*x^2 + c*x^4)^(1/2)))/(8*c^(3/2))))/(32*c) - (B*b*
((x^2*(b*x^2 + c*x^4)^(3/2))/4 - (3*b^2*(((b + 2*c*x^2)*(b*x^2 + c*x^4)^(1/2))/(4*c) - (b^2*log((b/2 + c*x^2)/
c^(1/2) + (b*x^2 + c*x^4)^(1/2)))/(8*c^(3/2))))/(16*c) + (b*(b*x^2 + c*x^4)^(3/2))/(8*c)))/(4*c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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