Integrand size = 19, antiderivative size = 114 \[ \int \frac {x^7}{\sqrt {b x^2+c x^4}} \, dx=\frac {5 b^2 \sqrt {b x^2+c x^4}}{16 c^3}-\frac {5 b x^2 \sqrt {b x^2+c x^4}}{24 c^2}+\frac {x^4 \sqrt {b x^2+c x^4}}{6 c}-\frac {5 b^3 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{16 c^{7/2}} \] Output:
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-5/16*b^3*arctanh(c^(1/2)*x^2/(c*x^4+b*x^2)^(1/2))/ c^(7/2)
Time = 0.22 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.96 \[ \int \frac {x^7}{\sqrt {b x^2+c x^4}} \, dx=\frac {x \left (\sqrt {c} x \left (15 b^3+5 b^2 c x^2-2 b c^2 x^4+8 c^3 x^6\right )+30 b^3 \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b}-\sqrt {b+c x^2}}\right )\right )}{48 c^{7/2} \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[x^7/Sqrt[b*x^2 + c*x^4],x]
Output:
(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*Sq rt[b + c*x^2]*ArcTanh[(Sqrt[c]*x)/(Sqrt[b] - Sqrt[b + c*x^2])]))/(48*c^(7/ 2)*Sqrt[x^2*(b + c*x^2)])
Time = 0.45 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1424, 1134, 1134, 1160, 1091, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^7}{\sqrt {b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1424 |
\(\displaystyle \frac {1}{2} \int \frac {x^6}{\sqrt {c x^4+b x^2}}dx^2\) |
\(\Big \downarrow \) 1134 |
\(\displaystyle \frac {1}{2} \left (\frac {x^4 \sqrt {b x^2+c x^4}}{3 c}-\frac {5 b \int \frac {x^4}{\sqrt {c x^4+b x^2}}dx^2}{6 c}\right )\) |
\(\Big \downarrow \) 1134 |
\(\displaystyle \frac {1}{2} \left (\frac {x^4 \sqrt {b x^2+c x^4}}{3 c}-\frac {5 b \left (\frac {x^2 \sqrt {b x^2+c x^4}}{2 c}-\frac {3 b \int \frac {x^2}{\sqrt {c x^4+b x^2}}dx^2}{4 c}\right )}{6 c}\right )\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{2} \left (\frac {x^4 \sqrt {b x^2+c x^4}}{3 c}-\frac {5 b \left (\frac {x^2 \sqrt {b x^2+c x^4}}{2 c}-\frac {3 b \left (\frac {\sqrt {b x^2+c x^4}}{c}-\frac {b \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2}{2 c}\right )}{4 c}\right )}{6 c}\right )\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {1}{2} \left (\frac {x^4 \sqrt {b x^2+c x^4}}{3 c}-\frac {5 b \left (\frac {x^2 \sqrt {b x^2+c x^4}}{2 c}-\frac {3 b \left (\frac {\sqrt {b x^2+c x^4}}{c}-\frac {b \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}}{c}\right )}{4 c}\right )}{6 c}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {x^4 \sqrt {b x^2+c x^4}}{3 c}-\frac {5 b \left (\frac {x^2 \sqrt {b x^2+c x^4}}{2 c}-\frac {3 b \left (\frac {\sqrt {b x^2+c x^4}}{c}-\frac {b \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{c^{3/2}}\right )}{4 c}\right )}{6 c}\right )\) |
Input:
Int[x^7/Sqrt[b*x^2 + c*x^4],x]
Output:
((x^4*Sqrt[b*x^2 + c*x^4])/(3*c) - (5*b*((x^2*Sqrt[b*x^2 + c*x^4])/(2*c) - (3*b*(Sqrt[b*x^2 + c*x^4]/c - (b*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4 ]])/c^(3/2)))/(4*c)))/(6*c))/2
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[(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] && EqQ[ c*d^2 - b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2 *p]
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] + Simp[(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[p, -1]
Int[(x_)^(m_.)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{b, c, m, p}, x] && !IntegerQ[p] && IntegerQ[(m - 1)/2]
Time = 0.20 (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 | \(\frac {x^{2} \left (8 c^{2} x^{4}-10 b c \,x^{2}+15 b^{2}\right ) \left (c \,x^{2}+b \right )}{48 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {5 b^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) x \sqrt {c \,x^{2}+b}}{16 c^{\frac {7}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(98\) |
default | \(\frac {x \sqrt {c \,x^{2}+b}\, \left (8 x^{5} \sqrt {c \,x^{2}+b}\, c^{\frac {7}{2}}-10 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b}\, b \,x^{3}+15 c^{\frac {3}{2}} \sqrt {c \,x^{2}+b}\, b^{2} x -15 \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{3} c \right )}{48 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{\frac {9}{2}}}\) | \(105\) |
Input:
int(x^7/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-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)
Time = 0.09 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.46 \[ \int \frac {x^7}{\sqrt {b x^2+c x^4}} \, dx=\left [\frac {15 \, b^{3} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, {\left (8 \, c^{3} x^{4} - 10 \, b c^{2} x^{2} + 15 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}}}{96 \, c^{4}}, \frac {15 \, b^{3} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (8 \, c^{3} x^{4} - 10 \, b c^{2} x^{2} + 15 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}}}{48 \, c^{4}}\right ] \] Input:
integrate(x^7/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
Output:
[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]
\[ \int \frac {x^7}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {x^{7}}{\sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \] Input:
integrate(x**7/(c*x**4+b*x**2)**(1/2),x)
Output:
Integral(x**7/sqrt(x**2*(b + c*x**2)), x)
Time = 0.04 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.88 \[ \int \frac {x^7}{\sqrt {b x^2+c x^4}} \, dx=\frac {\sqrt {c x^{4} + b x^{2}} x^{4}}{6 \, c} - \frac {5 \, \sqrt {c x^{4} + b x^{2}} b x^{2}}{24 \, c^{2}} - \frac {5 \, b^{3} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{32 \, c^{\frac {7}{2}}} + \frac {5 \, \sqrt {c x^{4} + b x^{2}} b^{2}}{16 \, c^{3}} \] Input:
integrate(x^7/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
Output:
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
Time = 0.16 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.85 \[ \int \frac {x^7}{\sqrt {b x^2+c x^4}} \, dx=\frac {1}{48} \, \sqrt {c x^{2} + b} {\left (2 \, x^{2} {\left (\frac {4 \, x^{2}}{c \mathrm {sgn}\left (x\right )} - \frac {5 \, b}{c^{2} \mathrm {sgn}\left (x\right )}\right )} + \frac {15 \, b^{2}}{c^{3} \mathrm {sgn}\left (x\right )}\right )} x - \frac {5 \, b^{3} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{32 \, c^{\frac {7}{2}}} + \frac {5 \, b^{3} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{16 \, c^{\frac {7}{2}} \mathrm {sgn}\left (x\right )} \] Input:
integrate(x^7/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
Output:
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))
Timed out. \[ \int \frac {x^7}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {x^7}{\sqrt {c\,x^4+b\,x^2}} \,d x \] Input:
int(x^7/(b*x^2 + c*x^4)^(1/2),x)
Output:
int(x^7/(b*x^2 + c*x^4)^(1/2), x)
Time = 0.18 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.70 \[ \int \frac {x^7}{\sqrt {b x^2+c x^4}} \, dx=\frac {15 \sqrt {c \,x^{2}+b}\, b^{2} c x -10 \sqrt {c \,x^{2}+b}\, b \,c^{2} x^{3}+8 \sqrt {c \,x^{2}+b}\, c^{3} x^{5}-15 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{3}}{48 c^{4}} \] Input:
int(x^7/(c*x^4+b*x^2)^(1/2),x)
Output:
(15*sqrt(b + c*x**2)*b**2*c*x - 10*sqrt(b + c*x**2)*b*c**2*x**3 + 8*sqrt(b + c*x**2)*c**3*x**5 - 15*sqrt(c)*log((sqrt(b + c*x**2) + sqrt(c)*x)/sqrt( b))*b**3)/(48*c**4)