3.243 \(\int \frac {(b x^2+c x^4)^{3/2}}{x^7} \, dx\)

Optimal. Leaf size=75 \[ c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )-\frac {c \sqrt {b x^2+c x^4}}{x^2}-\frac {\left (b x^2+c x^4\right )^{3/2}}{3 x^6} \]

[Out]

-1/3*(c*x^4+b*x^2)^(3/2)/x^6+c^(3/2)*arctanh(x^2*c^(1/2)/(c*x^4+b*x^2)^(1/2))-c*(c*x^4+b*x^2)^(1/2)/x^2

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2018, 662, 620, 206} \[ c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )-\frac {c \sqrt {b x^2+c x^4}}{x^2}-\frac {\left (b x^2+c x^4\right )^{3/2}}{3 x^6} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^7,x]

[Out]

-((c*Sqrt[b*x^2 + c*x^4])/x^2) - (b*x^2 + c*x^4)^(3/2)/(3*x^6) + c^(3/2)*ArcTanh[(Sqrt[c]*x^2)/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 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 662

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 + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(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] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 2018

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

\begin {align*} \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^7} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (b x+c x^2\right )^{3/2}}{x^4} \, dx,x,x^2\right )\\ &=-\frac {\left (b x^2+c x^4\right )^{3/2}}{3 x^6}+\frac {1}{2} c \operatorname {Subst}\left (\int \frac {\sqrt {b x+c x^2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {c \sqrt {b x^2+c x^4}}{x^2}-\frac {\left (b x^2+c x^4\right )^{3/2}}{3 x^6}+\frac {1}{2} c^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )\\ &=-\frac {c \sqrt {b x^2+c x^4}}{x^2}-\frac {\left (b x^2+c x^4\right )^{3/2}}{3 x^6}+c^2 \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x^2}{\sqrt {b x^2+c x^4}}\right )\\ &=-\frac {c \sqrt {b x^2+c x^4}}{x^2}-\frac {\left (b x^2+c x^4\right )^{3/2}}{3 x^6}+c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.02, size = 56, normalized size = 0.75 \[ -\frac {b \sqrt {x^2 \left (b+c x^2\right )} \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};-\frac {c x^2}{b}\right )}{3 x^4 \sqrt {\frac {c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(3/2)/x^7,x]

[Out]

-1/3*(b*Sqrt[x^2*(b + c*x^2)]*Hypergeometric2F1[-3/2, -3/2, -1/2, -((c*x^2)/b)])/(x^4*Sqrt[1 + (c*x^2)/b])

________________________________________________________________________________________

fricas [A]  time = 0.58, size = 135, normalized size = 1.80 \[ \left [\frac {3 \, c^{\frac {3}{2}} x^{4} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, \sqrt {c x^{4} + b x^{2}} {\left (4 \, c x^{2} + b\right )}}{6 \, x^{4}}, -\frac {3 \, \sqrt {-c} c x^{4} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + \sqrt {c x^{4} + b x^{2}} {\left (4 \, c x^{2} + b\right )}}{3 \, x^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^7,x, algorithm="fricas")

[Out]

[1/6*(3*c^(3/2)*x^4*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*sqrt(c*x^4 + b*x^2)*(4*c*x^2 + b))/x
^4, -1/3*(3*sqrt(-c)*c*x^4*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) + sqrt(c*x^4 + b*x^2)*(4*c*x^2 + b
))/x^4]

________________________________________________________________________________________

giac [A]  time = 0.44, size = 122, normalized size = 1.63 \[ -\frac {1}{2} \, c^{\frac {3}{2}} \log \left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2}\right ) \mathrm {sgn}\relax (x) + \frac {4 \, {\left (3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} b c^{\frac {3}{2}} \mathrm {sgn}\relax (x) - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} b^{2} c^{\frac {3}{2}} \mathrm {sgn}\relax (x) + 2 \, b^{3} c^{\frac {3}{2}} \mathrm {sgn}\relax (x)\right )}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^7,x, algorithm="giac")

[Out]

-1/2*c^(3/2)*log((sqrt(c)*x - sqrt(c*x^2 + b))^2)*sgn(x) + 4/3*(3*(sqrt(c)*x - sqrt(c*x^2 + b))^4*b*c^(3/2)*sg
n(x) - 3*(sqrt(c)*x - sqrt(c*x^2 + b))^2*b^2*c^(3/2)*sgn(x) + 2*b^3*c^(3/2)*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 +
b))^2 - b)^3

________________________________________________________________________________________

maple [B]  time = 0.01, size = 129, normalized size = 1.72 \[ \frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (3 b^{2} c^{2} x^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )+3 \sqrt {c \,x^{2}+b}\, b \,c^{\frac {5}{2}} x^{4}+2 \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {5}{2}} x^{4}-2 \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {3}{2}} x^{2}-\left (c \,x^{2}+b \right )^{\frac {5}{2}} b \sqrt {c}\right )}{3 \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{2} \sqrt {c}\, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^(3/2)/x^7,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.51, size = 89, normalized size = 1.19 \[ \frac {1}{2} \, c^{\frac {3}{2}} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - \frac {7 \, \sqrt {c x^{4} + b x^{2}} c}{6 \, x^{2}} - \frac {\sqrt {c x^{4} + b x^{2}} b}{6 \, x^{4}} - \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{6 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2)^(3/2)/x^7,x, algorithm="maxima")

[Out]

1/2*c^(3/2)*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 7/6*sqrt(c*x^4 + b*x^2)*c/x^2 - 1/6*sqrt(c*x^4
+ b*x^2)*b/x^4 - 1/6*(c*x^4 + b*x^2)^(3/2)/x^6

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^7} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^(3/2)/x^7,x)

[Out]

int((b*x^2 + c*x^4)^(3/2)/x^7, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{7}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(3/2)/x**7,x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)/x**7, x)

________________________________________________________________________________________