3.260 \(\int \frac {x^7}{\sqrt {b x^2+c x^4}} \, dx\)

Optimal. Leaf size=114 \[ -\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{7/2}}+\frac {5 b^2 \sqrt {b x^2+c x^4}}{16 c^3}-\frac {5 b x^2 \sqrt {b x^2+c x^4}}{24 c^2}+\frac {x^4 \sqrt {b x^2+c x^4}}{6 c} \]

[Out]

-5/16*b^3*arctanh(x^2*c^(1/2)/(c*x^4+b*x^2)^(1/2))/c^(7/2)+5/16*b^2*(c*x^4+b*x^2)^(1/2)/c^3-5/24*b*x^2*(c*x^4+
b*x^2)^(1/2)/c^2+1/6*x^4*(c*x^4+b*x^2)^(1/2)/c

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Rubi [A]  time = 0.13, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2018, 670, 640, 620, 206} \[ \frac {5 b^2 \sqrt {b x^2+c x^4}}{16 c^3}-\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{7/2}}-\frac {5 b x^2 \sqrt {b x^2+c x^4}}{24 c^2}+\frac {x^4 \sqrt {b x^2+c x^4}}{6 c} \]

Antiderivative was successfully verified.

[In]

Int[x^7/Sqrt[b*x^2 + c*x^4],x]

[Out]

(5*b^2*Sqrt[b*x^2 + c*x^4])/(16*c^3) - (5*b*x^2*Sqrt[b*x^2 + c*x^4])/(24*c^2) + (x^4*Sqrt[b*x^2 + c*x^4])/(6*c
) - (5*b^3*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(16*c^(7/2))

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 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 670

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[((m + p)*(2*c*d - b*e))/(c*(m + 2*p + 1)), Int[(d + e
*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*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 {x^7}{\sqrt {b x^2+c x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )\\ &=\frac {x^4 \sqrt {b x^2+c x^4}}{6 c}-\frac {(5 b) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{12 c}\\ &=-\frac {5 b x^2 \sqrt {b x^2+c x^4}}{24 c^2}+\frac {x^4 \sqrt {b x^2+c x^4}}{6 c}+\frac {\left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{16 c^2}\\ &=\frac {5 b^2 \sqrt {b x^2+c x^4}}{16 c^3}-\frac {5 b x^2 \sqrt {b x^2+c x^4}}{24 c^2}+\frac {x^4 \sqrt {b x^2+c x^4}}{6 c}-\frac {\left (5 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{32 c^3}\\ &=\frac {5 b^2 \sqrt {b x^2+c x^4}}{16 c^3}-\frac {5 b x^2 \sqrt {b x^2+c x^4}}{24 c^2}+\frac {x^4 \sqrt {b x^2+c x^4}}{6 c}-\frac {\left (5 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^3}\\ &=\frac {5 b^2 \sqrt {b x^2+c x^4}}{16 c^3}-\frac {5 b x^2 \sqrt {b x^2+c x^4}}{24 c^2}+\frac {x^4 \sqrt {b x^2+c x^4}}{6 c}-\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 100, normalized size = 0.88 \[ \frac {x \left (\sqrt {c} x \left (15 b^3+5 b^2 c x^2-2 b c^2 x^4+8 c^3 x^6\right )-15 b^3 \sqrt {b+c x^2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b+c x^2}}\right )\right )}{48 c^{7/2} \sqrt {x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^7/Sqrt[b*x^2 + c*x^4],x]

[Out]

(x*(Sqrt[c]*x*(15*b^3 + 5*b^2*c*x^2 - 2*b*c^2*x^4 + 8*c^3*x^6) - 15*b^3*Sqrt[b + c*x^2]*ArcTanh[(Sqrt[c]*x)/Sq
rt[b + c*x^2]]))/(48*c^(7/2)*Sqrt[x^2*(b + c*x^2)])

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fricas [A]  time = 0.52, size = 166, normalized size = 1.46 \[ \left [\frac {15 \, b^{3} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, {\left (8 \, c^{3} x^{4} - 10 \, b c^{2} x^{2} + 15 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}}}{96 \, c^{4}}, \frac {15 \, b^{3} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (8 \, c^{3} x^{4} - 10 \, b c^{2} x^{2} + 15 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}}}{48 \, c^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/96*(15*b^3*sqrt(c)*log(-2*c*x^2 - b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) + 2*(8*c^3*x^4 - 10*b*c^2*x^2 + 15*b^2
*c)*sqrt(c*x^4 + b*x^2))/c^4, 1/48*(15*b^3*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) + (8*c^3*
x^4 - 10*b*c^2*x^2 + 15*b^2*c)*sqrt(c*x^4 + b*x^2))/c^4]

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giac [A]  time = 0.22, size = 87, normalized size = 0.76 \[ \frac {1}{48} \, \sqrt {c x^{4} + b x^{2}} {\left (2 \, x^{2} {\left (\frac {4 \, x^{2}}{c} - \frac {5 \, b}{c^{2}}\right )} + \frac {15 \, b^{2}}{c^{3}}\right )} + \frac {5 \, b^{3} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2}}\right )} \sqrt {c} - b \right |}\right )}{32 \, c^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

1/48*sqrt(c*x^4 + b*x^2)*(2*x^2*(4*x^2/c - 5*b/c^2) + 15*b^2/c^3) + 5/32*b^3*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*
x^4 + b*x^2))*sqrt(c) - b))/c^(7/2)

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maple [A]  time = 0.01, size = 105, normalized size = 0.92 \[ \frac {\sqrt {c \,x^{2}+b}\, \left (8 \sqrt {c \,x^{2}+b}\, c^{\frac {7}{2}} x^{5}-10 \sqrt {c \,x^{2}+b}\, b \,c^{\frac {5}{2}} x^{3}-15 b^{3} c \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )+15 \sqrt {c \,x^{2}+b}\, b^{2} c^{\frac {3}{2}} x \right ) x}{48 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(c*x^4+b*x^2)^(1/2),x)

[Out]

1/48*x*(c*x^2+b)^(1/2)*(8*x^5*(c*x^2+b)^(1/2)*c^(7/2)-10*c^(5/2)*(c*x^2+b)^(1/2)*x^3*b+15*c^(3/2)*(c*x^2+b)^(1
/2)*x*b^2-15*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*b^3*c)/(c*x^4+b*x^2)^(1/2)/c^(9/2)

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maxima [A]  time = 1.47, size = 100, normalized size = 0.88 \[ \frac {\sqrt {c x^{4} + b x^{2}} x^{4}}{6 \, c} - \frac {5 \, \sqrt {c x^{4} + b x^{2}} b x^{2}}{24 \, c^{2}} - \frac {5 \, b^{3} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{32 \, c^{\frac {7}{2}}} + \frac {5 \, \sqrt {c x^{4} + b x^{2}} b^{2}}{16 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

1/6*sqrt(c*x^4 + b*x^2)*x^4/c - 5/24*sqrt(c*x^4 + b*x^2)*b*x^2/c^2 - 5/32*b^3*log(2*c*x^2 + b + 2*sqrt(c*x^4 +
b*x^2)*sqrt(c))/c^(7/2) + 5/16*sqrt(c*x^4 + b*x^2)*b^2/c^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^7}{\sqrt {c\,x^4+b\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(b*x^2 + c*x^4)^(1/2),x)

[Out]

int(x^7/(b*x^2 + c*x^4)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7/(c*x**4+b*x**2)**(1/2),x)

[Out]

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

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