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

Optimal. Leaf size=61 \[ \frac{B \left (b x^2+c x^4\right )^{5/2}}{7 c x^3}-\frac{\left (b x^2+c x^4\right )^{5/2} (2 b B-7 A c)}{35 c^2 x^5} \]

[Out]

-((2*b*B - 7*A*c)*(b*x^2 + c*x^4)^(5/2))/(35*c^2*x^5) + (B*(b*x^2 + c*x^4)^(5/2))/(7*c*x^3)

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Rubi [A]  time = 0.159236, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {2039, 2014} \[ \frac{B \left (b x^2+c x^4\right )^{5/2}}{7 c x^3}-\frac{\left (b x^2+c x^4\right )^{5/2} (2 b B-7 A c)}{35 c^2 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*b*B - 7*A*c)*(b*x^2 + c*x^4)^(5/2))/(35*c^2*x^5) + (B*(b*x^2 + c*x^4)^(5/2))/(7*c*x^3)

Rule 2039

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[(d*e^(j - 1)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(b*(m + n + p*(j + n) + 1)), x] - Dist[(a*d*(m
+ j*p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1)), Int[(e*x)^m*(a*x^j + b*x^(j + n))^p, x
], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[
m + n + p*(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \frac{\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^2} \, dx &=\frac{B \left (b x^2+c x^4\right )^{5/2}}{7 c x^3}-\frac{(2 b B-7 A c) \int \frac{\left (b x^2+c x^4\right )^{3/2}}{x^2} \, dx}{7 c}\\ &=-\frac{(2 b B-7 A c) \left (b x^2+c x^4\right )^{5/2}}{35 c^2 x^5}+\frac{B \left (b x^2+c x^4\right )^{5/2}}{7 c x^3}\\ \end{align*}

Mathematica [A]  time = 0.0312384, size = 48, normalized size = 0.79 \[ \frac{x \left (b+c x^2\right )^3 \left (7 A c-2 b B+5 B c x^2\right )}{35 c^2 \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(b + c*x^2)^3*(-2*b*B + 7*A*c + 5*B*c*x^2))/(35*c^2*Sqrt[x^2*(b + c*x^2)])

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Maple [A]  time = 0.003, size = 45, normalized size = 0.7 \begin{align*}{\frac{ \left ( c{x}^{2}+b \right ) \left ( 5\,Bc{x}^{2}+7\,Ac-2\,Bb \right ) }{35\,{c}^{2}{x}^{3}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{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^2,x)

[Out]

1/35*(c*x^2+b)*(5*B*c*x^2+7*A*c-2*B*b)*(c*x^4+b*x^2)^(3/2)/c^2/x^3

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Maxima [A]  time = 1.24526, size = 108, normalized size = 1.77 \begin{align*} \frac{{\left (c^{2} x^{4} + 2 \, b c x^{2} + b^{2}\right )} \sqrt{c x^{2} + b} A}{5 \, c} + \frac{{\left (5 \, c^{3} x^{6} + 8 \, b c^{2} x^{4} + b^{2} c x^{2} - 2 \, b^{3}\right )} \sqrt{c x^{2} + b} B}{35 \, c^{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^2,x, algorithm="maxima")

[Out]

1/5*(c^2*x^4 + 2*b*c*x^2 + b^2)*sqrt(c*x^2 + b)*A/c + 1/35*(5*c^3*x^6 + 8*b*c^2*x^4 + b^2*c*x^2 - 2*b^3)*sqrt(
c*x^2 + b)*B/c^2

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Fricas [A]  time = 1.10781, size = 171, normalized size = 2.8 \begin{align*} \frac{{\left (5 \, B c^{3} x^{6} +{\left (8 \, B b c^{2} + 7 \, A c^{3}\right )} x^{4} - 2 \, B b^{3} + 7 \, A b^{2} c +{\left (B b^{2} c + 14 \, A b c^{2}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{35 \, c^{2} x} \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^2,x, algorithm="fricas")

[Out]

1/35*(5*B*c^3*x^6 + (8*B*b*c^2 + 7*A*c^3)*x^4 - 2*B*b^3 + 7*A*b^2*c + (B*b^2*c + 14*A*b*c^2)*x^2)*sqrt(c*x^4 +
b*x^2)/(c^2*x)

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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^{2}}\, 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**2,x)

[Out]

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

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Giac [B]  time = 1.12936, size = 203, normalized size = 3.33 \begin{align*} \frac{35 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} A b \mathrm{sgn}\left (x\right ) + 7 \,{\left (3 \,{\left (c x^{2} + b\right )}^{\frac{5}{2}} - 5 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} b\right )} A \mathrm{sgn}\left (x\right ) + \frac{7 \,{\left (3 \,{\left (c x^{2} + b\right )}^{\frac{5}{2}} - 5 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} b\right )} B b \mathrm{sgn}\left (x\right )}{c} + \frac{{\left (15 \,{\left (c x^{2} + b\right )}^{\frac{7}{2}} - 42 \,{\left (c x^{2} + b\right )}^{\frac{5}{2}} b + 35 \,{\left (c x^{2} + b\right )}^{\frac{3}{2}} b^{2}\right )} B \mathrm{sgn}\left (x\right )}{c}}{105 \, c} + \frac{{\left (2 \, B b^{\frac{7}{2}} - 7 \, A b^{\frac{5}{2}} c\right )} \mathrm{sgn}\left (x\right )}{35 \, c^{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^2,x, algorithm="giac")

[Out]

1/105*(35*(c*x^2 + b)^(3/2)*A*b*sgn(x) + 7*(3*(c*x^2 + b)^(5/2) - 5*(c*x^2 + b)^(3/2)*b)*A*sgn(x) + 7*(3*(c*x^
2 + b)^(5/2) - 5*(c*x^2 + b)^(3/2)*b)*B*b*sgn(x)/c + (15*(c*x^2 + b)^(7/2) - 42*(c*x^2 + b)^(5/2)*b + 35*(c*x^
2 + b)^(3/2)*b^2)*B*sgn(x)/c)/c + 1/35*(2*B*b^(7/2) - 7*A*b^(5/2)*c)*sgn(x)/c^2