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

Optimal. Leaf size=102 \[ -A b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )+\frac {A b \sqrt {b x^2+c x^4}}{x}+\frac {A \left (b x^2+c x^4\right )^{3/2}}{3 x^3}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{5 c x^5} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2039, 2021, 2008, 206} \[ -A b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )+\frac {A \left (b x^2+c x^4\right )^{3/2}}{3 x^3}+\frac {A b \sqrt {b x^2+c x^4}}{x}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{5 c x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2008

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[2/(2 - n), Subst[Int[1/(1 - a*x^2), x], x, x/Sq
rt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]

Rule 2021

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b
*x^n)^p)/(c*(m + n*p + 1)), x] + Dist[(a*(n - j)*p)/(c^j*(m + n*p + 1)), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 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^4} \, dx &=\frac {B \left (b x^2+c x^4\right )^{5/2}}{5 c x^5}+A \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^4} \, dx\\ &=\frac {A \left (b x^2+c x^4\right )^{3/2}}{3 x^3}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{5 c x^5}+(A b) \int \frac {\sqrt {b x^2+c x^4}}{x^2} \, dx\\ &=\frac {A b \sqrt {b x^2+c x^4}}{x}+\frac {A \left (b x^2+c x^4\right )^{3/2}}{3 x^3}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{5 c x^5}+\left (A b^2\right ) \int \frac {1}{\sqrt {b x^2+c x^4}} \, dx\\ &=\frac {A b \sqrt {b x^2+c x^4}}{x}+\frac {A \left (b x^2+c x^4\right )^{3/2}}{3 x^3}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{5 c x^5}-\left (A b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {b x^2+c x^4}}\right )\\ &=\frac {A b \sqrt {b x^2+c x^4}}{x}+\frac {A \left (b x^2+c x^4\right )^{3/2}}{3 x^3}+\frac {B \left (b x^2+c x^4\right )^{5/2}}{5 c x^5}-A b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 109, normalized size = 1.07 \[ \frac {\left (x^2 \left (b+c x^2\right )\right )^{3/2} \left (-15 A b^{3/2} c \tanh ^{-1}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )+5 A c \left (b+c x^2\right )^{3/2}+15 A b c \sqrt {b+c x^2}+3 B \left (b+c x^2\right )^{5/2}\right )}{15 c x^3 \left (b+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((x^2*(b + c*x^2))^(3/2)*(15*A*b*c*Sqrt[b + c*x^2] + 5*A*c*(b + c*x^2)^(3/2) + 3*B*(b + c*x^2)^(5/2) - 15*A*b^
(3/2)*c*ArcTanh[Sqrt[b + c*x^2]/Sqrt[b]]))/(15*c*x^3*(b + c*x^2)^(3/2))

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fricas [A]  time = 0.87, size = 206, normalized size = 2.02 \[ \left [\frac {15 \, A b^{\frac {3}{2}} c x \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, {\left (3 \, B c^{2} x^{4} + 3 \, B b^{2} + 20 \, A b c + {\left (6 \, B b c + 5 \, A c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{30 \, c x}, \frac {15 \, A \sqrt {-b} b c x \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + {\left (3 \, B c^{2} x^{4} + 3 \, B b^{2} + 20 \, A b c + {\left (6 \, B b c + 5 \, A c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15 \, c x}\right ] \]

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^4,x, algorithm="fricas")

[Out]

[1/30*(15*A*b^(3/2)*c*x*log(-(c*x^3 + 2*b*x - 2*sqrt(c*x^4 + b*x^2)*sqrt(b))/x^3) + 2*(3*B*c^2*x^4 + 3*B*b^2 +
20*A*b*c + (6*B*b*c + 5*A*c^2)*x^2)*sqrt(c*x^4 + b*x^2))/(c*x), 1/15*(15*A*sqrt(-b)*b*c*x*arctan(sqrt(c*x^4 +
b*x^2)*sqrt(-b)/(c*x^3 + b*x)) + (3*B*c^2*x^4 + 3*B*b^2 + 20*A*b*c + (6*B*b*c + 5*A*c^2)*x^2)*sqrt(c*x^4 + b*
x^2))/(c*x)]

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giac [A]  time = 0.18, size = 140, normalized size = 1.37 \[ \frac {A b^{2} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\relax (x)}{\sqrt {-b}} - \frac {{\left (15 \, A b^{2} c \arctan \left (\frac {\sqrt {b}}{\sqrt {-b}}\right ) + 3 \, B \sqrt {-b} b^{\frac {5}{2}} + 20 \, A \sqrt {-b} b^{\frac {3}{2}} c\right )} \mathrm {sgn}\relax (x)}{15 \, \sqrt {-b} c} + \frac {3 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B c^{4} \mathrm {sgn}\relax (x) + 5 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A c^{5} \mathrm {sgn}\relax (x) + 15 \, \sqrt {c x^{2} + b} A b c^{5} \mathrm {sgn}\relax (x)}{15 \, c^{5}} \]

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^4,x, algorithm="giac")

[Out]

A*b^2*arctan(sqrt(c*x^2 + b)/sqrt(-b))*sgn(x)/sqrt(-b) - 1/15*(15*A*b^2*c*arctan(sqrt(b)/sqrt(-b)) + 3*B*sqrt(
-b)*b^(5/2) + 20*A*sqrt(-b)*b^(3/2)*c)*sgn(x)/(sqrt(-b)*c) + 1/15*(3*(c*x^2 + b)^(5/2)*B*c^4*sgn(x) + 5*(c*x^2
+ b)^(3/2)*A*c^5*sgn(x) + 15*sqrt(c*x^2 + b)*A*b*c^5*sgn(x))/c^5

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maple [A]  time = 0.05, size = 99, normalized size = 0.97 \[ -\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (15 A \,b^{\frac {3}{2}} c \ln \left (\frac {2 b +2 \sqrt {c \,x^{2}+b}\, \sqrt {b}}{x}\right )-15 \sqrt {c \,x^{2}+b}\, A b c -5 \left (c \,x^{2}+b \right )^{\frac {3}{2}} A c -3 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B \right )}{15 \left (c \,x^{2}+b \right )^{\frac {3}{2}} c \,x^{3}} \]

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^4,x)

[Out]

-1/15*(c*x^4+b*x^2)^(3/2)*(-3*B*(c*x^2+b)^(5/2)+15*A*b^(3/2)*ln(2*(b+(c*x^2+b)^(1/2)*b^(1/2))/x)*c-5*A*(c*x^2+
b)^(3/2)*c-15*A*(c*x^2+b)^(1/2)*b*c)/x^3/(c*x^2+b)^(3/2)/c

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{4}}\,{d x} \]

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^4,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^4} \,d x \]

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^4,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{4}}\, dx \]

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**4,x)

[Out]

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

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