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

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

[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))

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Rubi [A]  time = 0.26578, antiderivative size = 144, normalized size of antiderivative = 1., 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} \[ \frac{b^3 (3 b B-8 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{128 c^{5/2}}-\frac{b \left (b+2 c x^2\right ) \sqrt{b x^2+c x^4} (3 b B-8 A c)}{128 c^2}-\frac{\left (b x^2+c x^4\right )^{3/2} (3 b B-8 A c)}{48 c}+\frac{B \left (b x^2+c x^4\right )^{5/2}}{8 c x^2} \]

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 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]

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 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 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 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])

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*}

Mathematica [A]  time = 0.218564, size = 151, normalized size = 1.05 \[ \frac{\sqrt{x^2 \left (b+c x^2\right )} \left (\sqrt{c} x \sqrt{\frac{c x^2}{b}+1} \left (6 b^2 c \left (4 A+B x^2\right )+8 b c^2 x^2 \left (14 A+9 B x^2\right )+16 c^3 x^4 \left (4 A+3 B x^2\right )-9 b^3 B\right )+3 b^{5/2} (3 b B-8 A c) \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )\right )}{384 c^{5/2} x \sqrt{\frac{c x^2}{b}+1}} \]

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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Maple [A]  time = 0.009, size = 202, normalized size = 1.4 \begin{align*}{\frac{1}{384\,{x}^{3}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( 48\,B{c}^{3/2} \left ( c{x}^{2}+b \right ) ^{5/2}{x}^{3}+64\,A{c}^{3/2} \left ( c{x}^{2}+b \right ) ^{5/2}x-24\,B\sqrt{c} \left ( c{x}^{2}+b \right ) ^{5/2}xb-16\,A{c}^{3/2} \left ( c{x}^{2}+b \right ) ^{3/2}xb+6\,B\sqrt{c} \left ( c{x}^{2}+b \right ) ^{3/2}x{b}^{2}-24\,A{c}^{3/2}\sqrt{c{x}^{2}+b}x{b}^{2}+9\,B\sqrt{c}\sqrt{c{x}^{2}+b}x{b}^{3}-24\,A\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ){b}^{3}c+9\,B\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ){b}^{4} \right ) \left ( c{x}^{2}+b \right ) ^{-{\frac{3}{2}}}{c}^{-{\frac{5}{2}}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.2829, size = 616, normalized size = 4.28 \begin{align*} \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{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac{3}{2}} \left (A + B x^{2}\right )}{x}\, dx \end{align*}

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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Giac [A]  time = 1.19221, size = 240, normalized size = 1.67 \begin{align*} \frac{1}{384} \,{\left (2 \,{\left (4 \,{\left (6 \, B c x^{2} \mathrm{sgn}\left (x\right ) + \frac{9 \, B b c^{6} \mathrm{sgn}\left (x\right ) + 8 \, A c^{7} \mathrm{sgn}\left (x\right )}{c^{6}}\right )} x^{2} + \frac{3 \, B b^{2} c^{5} \mathrm{sgn}\left (x\right ) + 56 \, A b c^{6} \mathrm{sgn}\left (x\right )}{c^{6}}\right )} x^{2} - \frac{3 \,{\left (3 \, B b^{3} c^{4} \mathrm{sgn}\left (x\right ) - 8 \, A b^{2} c^{5} \mathrm{sgn}\left (x\right )\right )}}{c^{6}}\right )} \sqrt{c x^{2} + b} x - \frac{{\left (3 \, B b^{4} \mathrm{sgn}\left (x\right ) - 8 \, A b^{3} c \mathrm{sgn}\left (x\right )\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}\left (x\right )}{256 \, c^{\frac{5}{2}}} \end{align*}

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)