\(\int \frac {x^7}{\sqrt {b x^2+c x^4}} \, dx\) [260]

Optimal result

Integrand size = 19, antiderivative size = 114 \[ \int \frac {x^7}{\sqrt {b x^2+c x^4}} \, dx=\frac {5 b^2 \sqrt {b x^2+c x^4}}{16 c^3}-\frac {5 b x^2 \sqrt {b x^2+c x^4}}{24 c^2}+\frac {x^4 \sqrt {b x^2+c x^4}}{6 c}-\frac {5 b^3 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{7/2}} \]

[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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2043, 684, 654, 634, 212} \[ \int \frac {x^7}{\sqrt {b x^2+c x^4}} \, dx=-\frac {5 b^3 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{7/2}}+\frac {5 b^2 \sqrt {b x^2+c x^4}}{16 c^3}-\frac {5 b x^2 \sqrt {b x^2+c x^4}}{24 c^2}+\frac {x^4 \sqrt {b x^2+c x^4}}{6 c} \]

[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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 634

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 654

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 684

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 2043

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*} \text {integral}& = \frac {1}{2} \text {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) \text {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 ) \text {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 ) \text {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 ) \text {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*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.96 \[ \int \frac {x^7}{\sqrt {b x^2+c x^4}} \, dx=\frac {x \left (\sqrt {c} x \left (15 b^3+5 b^2 c x^2-2 b c^2 x^4+8 c^3 x^6\right )+30 b^3 \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b}-\sqrt {b+c x^2}}\right )\right )}{48 c^{7/2} \sqrt {x^2 \left (b+c x^2\right )}} \]

[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) + 30*b^3*Sqrt[b + c*x^2]*ArcTanh[(Sqrt[c]*x)/(S
qrt[b] - Sqrt[b + c*x^2])]))/(48*c^(7/2)*Sqrt[x^2*(b + c*x^2)])

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.78

method	result	size
pseudoelliptic	\(-\frac {5 \left (\ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b^{3}+\left (-\frac {16 c^{\frac {5}{2}} x^{4}}{15}+\frac {4 c^{\frac {3}{2}} b \,x^{2}}{3}-2 \sqrt {c}\, b^{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}-\ln \left (2\right ) b^{3}\right )}{32 c^{\frac {7}{2}}}\)	\(89\)
risch	\(\frac {x^{2} \left (8 c^{2} x^{4}-10 b c \,x^{2}+15 b^{2}\right ) \left (c \,x^{2}+b \right )}{48 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {5 b^{3} \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) x \sqrt {c \,x^{2}+b}}{16 c^{\frac {7}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(98\)
default	\(\frac {x \sqrt {c \,x^{2}+b}\, \left (8 x^{5} \sqrt {c \,x^{2}+b}\, c^{\frac {7}{2}}-10 \sqrt {c \,x^{2}+b}\, c^{\frac {5}{2}} b \,x^{3}+15 \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}} b^{2} x -15 \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) b^{3} c \right )}{48 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{\frac {9}{2}}}\)	\(105\)

[In]

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

[Out]

-5/32/c^(7/2)*(ln((2*c*x^2+2*(x^2*(c*x^2+b))^(1/2)*c^(1/2)+b)/c^(1/2))*b^3+(-16/15*c^(5/2)*x^4+4/3*c^(3/2)*b*x
^2-2*c^(1/2)*b^2)*(x^2*(c*x^2+b))^(1/2)-ln(2)*b^3)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.46 \[ \int \frac {x^7}{\sqrt {b x^2+c x^4}} \, dx=\left [\frac {15 \, b^{3} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, {\left (8 \, c^{3} x^{4} - 10 \, b c^{2} x^{2} + 15 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}}}{96 \, c^{4}}, \frac {15 \, b^{3} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (8 \, c^{3} x^{4} - 10 \, b c^{2} x^{2} + 15 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}}}{48 \, c^{4}}\right ] \]

[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]

Sympy [F]

\[ \int \frac {x^7}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {x^{7}}{\sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.88 \[ \int \frac {x^7}{\sqrt {b x^2+c x^4}} \, dx=\frac {\sqrt {c x^{4} + b x^{2}} x^{4}}{6 \, c} - \frac {5 \, \sqrt {c x^{4} + b x^{2}} b x^{2}}{24 \, c^{2}} - \frac {5 \, b^{3} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{32 \, c^{\frac {7}{2}}} + \frac {5 \, \sqrt {c x^{4} + b x^{2}} b^{2}}{16 \, c^{3}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.85 \[ \int \frac {x^7}{\sqrt {b x^2+c x^4}} \, dx=\frac {1}{48} \, \sqrt {c x^{2} + b} {\left (2 \, x^{2} {\left (\frac {4 \, x^{2}}{c \mathrm {sgn}\left (x\right )} - \frac {5 \, b}{c^{2} \mathrm {sgn}\left (x\right )}\right )} + \frac {15 \, b^{2}}{c^{3} \mathrm {sgn}\left (x\right )}\right )} x - \frac {5 \, b^{3} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{32 \, c^{\frac {7}{2}}} + \frac {5 \, b^{3} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{16 \, c^{\frac {7}{2}} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

1/48*sqrt(c*x^2 + b)*(2*x^2*(4*x^2/(c*sgn(x)) - 5*b/(c^2*sgn(x))) + 15*b^2/(c^3*sgn(x)))*x - 5/32*b^3*log(abs(
b))*sgn(x)/c^(7/2) + 5/16*b^3*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/(c^(7/2)*sgn(x))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^7}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {x^7}{\sqrt {c\,x^4+b\,x^2}} \,d x \]

[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)