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

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

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Rubi [A]  time = 0.16, antiderivative size = 61, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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])

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])

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*}

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Mathematica [A]  time = 0.04, size = 48, normalized size = 0.79 \begin {gather*} \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 )}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.56, size = 41, normalized size = 0.67 \begin {gather*} \frac {\left (b x^2+c x^4\right )^{5/2} \left (7 A c-2 b B+5 B c x^2\right )}{35 c^2 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 80, normalized size = 1.31 \begin {gather*} \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 {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^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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giac [A]  time = 0.21, size = 72, normalized size = 1.18 \begin {gather*} \frac {{\left (2 \, B b^{\frac {7}{2}} - 7 \, A b^{\frac {5}{2}} c\right )} \mathrm {sgn}\relax (x)}{35 \, c^{2}} + \frac {5 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} B \mathrm {sgn}\relax (x) - 7 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B b \mathrm {sgn}\relax (x) + 7 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} A c \mathrm {sgn}\relax (x)}{35 \, c^{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^2,x, algorithm="giac")

[Out]

1/35*(2*B*b^(7/2) - 7*A*b^(5/2)*c)*sgn(x)/c^2 + 1/35*(5*(c*x^2 + b)^(7/2)*B*sgn(x) - 7*(c*x^2 + b)^(5/2)*B*b*s
gn(x) + 7*(c*x^2 + b)^(5/2)*A*c*sgn(x))/c^2

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maple [A]  time = 0.05, size = 45, normalized size = 0.74 \begin {gather*} \frac {\left (c \,x^{2}+b \right ) \left (5 B c \,x^{2}+7 A c -2 b B \right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{35 c^{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^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 = 1.55, size = 80, normalized size = 1.31 \begin {gather*} \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 {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^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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mupad [B]  time = 0.23, size = 83, normalized size = 1.36 \begin {gather*} \frac {\sqrt {c\,x^4+b\,x^2}\,\left (\frac {x^4\,\left (7\,A\,c^3+8\,B\,b\,c^2\right )}{35\,c^2}-\frac {2\,B\,b^3-7\,A\,b^2\,c}{35\,c^2}+\frac {B\,c\,x^6}{7}+\frac {b\,x^2\,\left (14\,A\,c+B\,b\right )}{35\,c}\right )}{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^2,x)

[Out]

((b*x^2 + c*x^4)^(1/2)*((x^4*(7*A*c^3 + 8*B*b*c^2))/(35*c^2) - (2*B*b^3 - 7*A*b^2*c)/(35*c^2) + (B*c*x^6)/7 +
(b*x^2*(14*A*c + B*b))/(35*c)))/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^{2}}\, 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**2,x)

[Out]

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

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