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

Optimal. Leaf size=144 \[ \frac {b^3 (3 b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{128 c^{5/2}}-\frac {b \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4} (3 b B-8 A c)}{128 c^2}-\frac {\left (b x^2+c x^4\right )^{3/2} (3 b B-8 A c)}{48 c}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{8 c x^2} \]

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Rubi [A]  time = 0.27, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2034, 794, 664, 612, 620, 206} \begin {gather*} \frac {b^3 (3 b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{128 c^{5/2}}-\frac {b \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4} (3 b B-8 A c)}{128 c^2}-\frac {\left (b x^2+c x^4\right )^{3/2} (3 b B-8 A c)}{48 c}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{8 c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(3*b*B - 8*A*c)*(b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(128*c^2) - ((3*b*B - 8*A*c)*(b*x^2 + c*x^4)^(3/2))/(48
*c) + (B*(b*x^2 + c*x^4)^(5/2))/(8*c*x^2) + (b^3*(3*b*B - 8*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(
128*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 664

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 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(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] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

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 {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac {B \left (b x^2+c x^4\right )^{5/2}}{8 c x^2}+\frac {\left (b B-A c+\frac {5}{2} (-b B+2 A c)\right ) \operatorname {Subst}\left (\int \frac {\left (b x+c x^2\right )^{3/2}}{x} \, dx,x,x^2\right )}{8 c}\\ &=-\frac {(3 b B-8 A c) \left (b x^2+c x^4\right )^{3/2}}{48 c}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{8 c x^2}-\frac {(b (3 b B-8 A c)) \operatorname {Subst}\left (\int \sqrt {b x+c x^2} \, dx,x,x^2\right )}{32 c}\\ &=-\frac {b (3 b B-8 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{128 c^2}-\frac {(3 b B-8 A c) \left (b x^2+c x^4\right )^{3/2}}{48 c}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{8 c x^2}+\frac {\left (b^3 (3 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^2}\\ &=-\frac {b (3 b B-8 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{128 c^2}-\frac {(3 b B-8 A c) \left (b x^2+c x^4\right )^{3/2}}{48 c}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{8 c x^2}+\frac {\left (b^3 (3 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^2}\\ &=-\frac {b (3 b B-8 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{128 c^2}-\frac {(3 b B-8 A c) \left (b x^2+c x^4\right )^{3/2}}{48 c}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{8 c x^2}+\frac {b^3 (3 b B-8 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{128 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 151, normalized size = 1.05 \begin {gather*} \frac {\sqrt {x^2 \left (b+c x^2\right )} \left (3 b^{5/2} (3 b B-8 A c) \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )+\sqrt {c} x \sqrt {\frac {c x^2}{b}+1} \left (6 b^2 c \left (4 A+B x^2\right )+8 b c^2 x^2 \left (14 A+9 B x^2\right )+16 c^3 x^4 \left (4 A+3 B x^2\right )-9 b^3 B\right )\right )}{384 c^{5/2} x \sqrt {\frac {c x^2}{b}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x^2*(b + c*x^2)]*(Sqrt[c]*x*Sqrt[1 + (c*x^2)/b]*(-9*b^3*B + 6*b^2*c*(4*A + B*x^2) + 16*c^3*x^4*(4*A + 3*
B*x^2) + 8*b*c^2*x^2*(14*A + 9*B*x^2)) + 3*b^(5/2)*(3*b*B - 8*A*c)*ArcSinh[(Sqrt[c]*x)/Sqrt[b]]))/(384*c^(5/2)
*x*Sqrt[1 + (c*x^2)/b])

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IntegrateAlgebraic [A]  time = 0.71, size = 139, normalized size = 0.97 \begin {gather*} \frac {\left (8 A b^3 c-3 b^4 B\right ) \log \left (-2 \sqrt {c} \sqrt {b x^2+c x^4}+b+2 c x^2\right )}{256 c^{5/2}}+\frac {\sqrt {b x^2+c x^4} \left (24 A b^2 c+112 A b c^2 x^2+64 A c^3 x^4-9 b^3 B+6 b^2 B c x^2+72 b B c^2 x^4+48 B c^3 x^6\right )}{384 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x^2 + c*x^4]*(-9*b^3*B + 24*A*b^2*c + 6*b^2*B*c*x^2 + 112*A*b*c^2*x^2 + 72*b*B*c^2*x^4 + 64*A*c^3*x^4
+ 48*B*c^3*x^6))/(384*c^2) + ((-3*b^4*B + 8*A*b^3*c)*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[b*x^2 + c*x^4]])/(256*c^
(5/2))

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fricas [A]  time = 0.47, size = 275, normalized size = 1.91 \begin {gather*} \left [-\frac {3 \, {\left (3 \, 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} - 9 \, B b^{3} c + 24 \, A b^{2} c^{2} + 8 \, {\left (9 \, B b c^{3} + 8 \, A c^{4}\right )} x^{4} + 2 \, {\left (3 \, B b^{2} c^{2} + 56 \, A b c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{768 \, c^{3}}, -\frac {3 \, {\left (3 \, 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} - 9 \, B b^{3} c + 24 \, A b^{2} c^{2} + 8 \, {\left (9 \, B b c^{3} + 8 \, A c^{4}\right )} x^{4} + 2 \, {\left (3 \, B b^{2} c^{2} + 56 \, A b c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{384 \, c^{3}}\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)^(3/2)/x,x, algorithm="fricas")

[Out]

[-1/768*(3*(3*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 -
9*B*b^3*c + 24*A*b^2*c^2 + 8*(9*B*b*c^3 + 8*A*c^4)*x^4 + 2*(3*B*b^2*c^2 + 56*A*b*c^3)*x^2)*sqrt(c*x^4 + b*x^2
))/c^3, -1/384*(3*(3*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 - 9*B*b^3*c + 24*A*b^2*c^2 + 8*(9*B*b*c^3 + 8*A*c^4)*x^4 + 2*(3*B*b^2*c^2 + 56*A*b*c^3)*x^2)*sqrt(c*x^4 +
b*x^2))/c^3]

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giac [A]  time = 0.20, size = 178, normalized size = 1.24 \begin {gather*} \frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, B c x^{2} \mathrm {sgn}\relax (x) + \frac {9 \, B b c^{6} \mathrm {sgn}\relax (x) + 8 \, A c^{7} \mathrm {sgn}\relax (x)}{c^{6}}\right )} x^{2} + \frac {3 \, B b^{2} c^{5} \mathrm {sgn}\relax (x) + 56 \, A b c^{6} \mathrm {sgn}\relax (x)}{c^{6}}\right )} x^{2} - \frac {3 \, {\left (3 \, B b^{3} c^{4} \mathrm {sgn}\relax (x) - 8 \, A b^{2} c^{5} \mathrm {sgn}\relax (x)\right )}}{c^{6}}\right )} \sqrt {c x^{2} + b} x - \frac {{\left (3 \, 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 {5}{2}}} + \frac {{\left (3 \, 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 {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(2*(4*(6*B*c*x^2*sgn(x) + (9*B*b*c^6*sgn(x) + 8*A*c^7*sgn(x))/c^6)*x^2 + (3*B*b^2*c^5*sgn(x) + 56*A*b*c^
6*sgn(x))/c^6)*x^2 - 3*(3*B*b^3*c^4*sgn(x) - 8*A*b^2*c^5*sgn(x))/c^6)*sqrt(c*x^2 + b)*x - 1/128*(3*B*b^4*sgn(x
) - 8*A*b^3*c*sgn(x))*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/c^(5/2) + 1/256*(3*B*b^4*log(abs(b)) - 8*A*b^3*c*
log(abs(b)))*sgn(x)/c^(5/2)

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maple [A]  time = 0.06, size = 202, normalized size = 1.40 \begin {gather*} \frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (-24 A \,b^{3} c \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )+9 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 +9 \sqrt {c \,x^{2}+b}\, B \,b^{3} \sqrt {c}\, x +48 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B \,c^{\frac {3}{2}} x^{3}-16 \left (c \,x^{2}+b \right )^{\frac {3}{2}} A b \,c^{\frac {3}{2}} x +6 \left (c \,x^{2}+b \right )^{\frac {3}{2}} B \,b^{2} \sqrt {c}\, x +64 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A \,c^{\frac {3}{2}} x -24 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B b \sqrt {c}\, x \right )}{384 \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {5}{2}} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(c*x^4+b*x^2)^(3/2)*(48*B*c^(3/2)*(c*x^2+b)^(5/2)*x^3+64*A*c^(3/2)*(c*x^2+b)^(5/2)*x-24*B*c^(1/2)*(c*x^2
+b)^(5/2)*x*b-16*(c*x^2+b)^(3/2)*A*b*c^(3/2)*x+6*(c*x^2+b)^(3/2)*B*b^2*c^(1/2)*x-24*(c*x^2+b)^(1/2)*A*b^2*c^(3
/2)*x+9*(c*x^2+b)^(1/2)*B*b^3*c^(1/2)*x-24*A*b^3*c*ln(c^(1/2)*x+(c*x^2+b)^(1/2))+9*B*b^4*ln(c^(1/2)*x+(c*x^2+b
)^(1/2)))/x^3/(c*x^2+b)^(3/2)/c^(5/2)

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maxima [A]  time = 1.49, size = 216, normalized size = 1.50 \begin {gather*} \frac {1}{96} \, {\left (12 \, \sqrt {c x^{4} + b x^{2}} b x^{2} - \frac {3 \, b^{3} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {3}{2}}} + 16 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} + \frac {6 \, \sqrt {c x^{4} + b x^{2}} b^{2}}{c}\right )} A + \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 )} 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)^(3/2)/x,x, algorithm="maxima")

[Out]

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

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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 )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x} \,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)^(3/2))/x,x)

[Out]

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

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

[Out]

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

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