Integrand size = 15, antiderivative size = 115 \[ \int x^4 \left (a+b x^2\right )^{3/2} \, dx=-\frac {3 a^3 x \sqrt {a+b x^2}}{128 b^2}+\frac {a^2 x^3 \sqrt {a+b x^2}}{64 b}+\frac {1}{16} a x^5 \sqrt {a+b x^2}+\frac {1}{8} x^5 \left (a+b x^2\right )^{3/2}+\frac {3 a^4 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{5/2}} \] Output:
-3/128*a^3*x*(b*x^2+a)^(1/2)/b^2+1/64*a^2*x^3*(b*x^2+a)^(1/2)/b+1/16*a*x^5 *(b*x^2+a)^(1/2)+1/8*x^5*(b*x^2+a)^(3/2)+3/128*a^4*arctanh(b^(1/2)*x/(b*x^ 2+a)^(1/2))/b^(5/2)
Time = 0.27 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.81 \[ \int x^4 \left (a+b x^2\right )^{3/2} \, dx=\frac {\sqrt {a+b x^2} \left (-3 a^3 x+2 a^2 b x^3+24 a b^2 x^5+16 b^3 x^7\right )}{128 b^2}+\frac {3 a^4 \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{64 b^{5/2}} \] Input:
Integrate[x^4*(a + b*x^2)^(3/2),x]
Output:
(Sqrt[a + b*x^2]*(-3*a^3*x + 2*a^2*b*x^3 + 24*a*b^2*x^5 + 16*b^3*x^7))/(12 8*b^2) + (3*a^4*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + Sqrt[a + b*x^2])])/(64*b^( 5/2))
Time = 0.21 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {248, 248, 262, 262, 224, 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.
\(\displaystyle \int x^4 \left (a+b x^2\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {3}{8} a \int x^4 \sqrt {b x^2+a}dx+\frac {1}{8} x^5 \left (a+b x^2\right )^{3/2}\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {3}{8} a \left (\frac {1}{6} a \int \frac {x^4}{\sqrt {b x^2+a}}dx+\frac {1}{6} x^5 \sqrt {a+b x^2}\right )+\frac {1}{8} x^5 \left (a+b x^2\right )^{3/2}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {3}{8} a \left (\frac {1}{6} a \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \int \frac {x^2}{\sqrt {b x^2+a}}dx}{4 b}\right )+\frac {1}{6} x^5 \sqrt {a+b x^2}\right )+\frac {1}{8} x^5 \left (a+b x^2\right )^{3/2}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {3}{8} a \left (\frac {1}{6} a \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}\right )}{4 b}\right )+\frac {1}{6} x^5 \sqrt {a+b x^2}\right )+\frac {1}{8} x^5 \left (a+b x^2\right )^{3/2}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {3}{8} a \left (\frac {1}{6} a \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}\right )}{4 b}\right )+\frac {1}{6} x^5 \sqrt {a+b x^2}\right )+\frac {1}{8} x^5 \left (a+b x^2\right )^{3/2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {3}{8} a \left (\frac {1}{6} a \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}}\right )}{4 b}\right )+\frac {1}{6} x^5 \sqrt {a+b x^2}\right )+\frac {1}{8} x^5 \left (a+b x^2\right )^{3/2}\) |
Input:
Int[x^4*(a + b*x^2)^(3/2),x]
Output:
(x^5*(a + b*x^2)^(3/2))/8 + (3*a*((x^5*Sqrt[a + b*x^2])/6 + (a*((x^3*Sqrt[ a + b*x^2])/(4*b) - (3*a*((x*Sqrt[a + b*x^2])/(2*b) - (a*ArcTanh[(Sqrt[b]* x)/Sqrt[a + b*x^2]])/(2*b^(3/2))))/(4*b)))/6))/8
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[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ p, 0] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Time = 0.36 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.63
method | result | size |
risch | \(-\frac {x \left (-16 b^{3} x^{6}-24 a \,b^{2} x^{4}-2 a^{2} b \,x^{2}+3 a^{3}\right ) \sqrt {b \,x^{2}+a}}{128 b^{2}}+\frac {3 a^{4} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{128 b^{\frac {5}{2}}}\) | \(73\) |
pseudoelliptic | \(\frac {\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right ) a^{4}}{128}-\frac {3 x \left (-\frac {16 b^{\frac {7}{2}} x^{6}}{3}-8 a \,b^{\frac {5}{2}} x^{4}-\frac {2 a^{2} b^{\frac {3}{2}} x^{2}}{3}+a^{3} \sqrt {b}\right ) \sqrt {b \,x^{2}+a}}{128}}{b^{\frac {5}{2}}}\) | \(76\) |
default | \(\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 b}-\frac {3 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )}{8 b}\) | \(98\) |
Input:
int(x^4*(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/128*x*(-16*b^3*x^6-24*a*b^2*x^4-2*a^2*b*x^2+3*a^3)*(b*x^2+a)^(1/2)/b^2+ 3/128/b^(5/2)*a^4*ln(b^(1/2)*x+(b*x^2+a)^(1/2))
Time = 0.13 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.46 \[ \int x^4 \left (a+b x^2\right )^{3/2} \, dx=\left [\frac {3 \, a^{4} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (16 \, b^{4} x^{7} + 24 \, a b^{3} x^{5} + 2 \, a^{2} b^{2} x^{3} - 3 \, a^{3} b x\right )} \sqrt {b x^{2} + a}}{256 \, b^{3}}, -\frac {3 \, a^{4} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (16 \, b^{4} x^{7} + 24 \, a b^{3} x^{5} + 2 \, a^{2} b^{2} x^{3} - 3 \, a^{3} b x\right )} \sqrt {b x^{2} + a}}{128 \, b^{3}}\right ] \] Input:
integrate(x^4*(b*x^2+a)^(3/2),x, algorithm="fricas")
Output:
[1/256*(3*a^4*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2* (16*b^4*x^7 + 24*a*b^3*x^5 + 2*a^2*b^2*x^3 - 3*a^3*b*x)*sqrt(b*x^2 + a))/b ^3, -1/128*(3*a^4*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (16*b^4*x^ 7 + 24*a*b^3*x^5 + 2*a^2*b^2*x^3 - 3*a^3*b*x)*sqrt(b*x^2 + a))/b^3]
Time = 10.43 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.29 \[ \int x^4 \left (a+b x^2\right )^{3/2} \, dx=- \frac {3 a^{\frac {7}{2}} x}{128 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {a^{\frac {5}{2}} x^{3}}{128 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {13 a^{\frac {3}{2}} x^{5}}{64 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {5 \sqrt {a} b x^{7}}{16 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 a^{4} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{128 b^{\frac {5}{2}}} + \frac {b^{2} x^{9}}{8 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \] Input:
integrate(x**4*(b*x**2+a)**(3/2),x)
Output:
-3*a**(7/2)*x/(128*b**2*sqrt(1 + b*x**2/a)) - a**(5/2)*x**3/(128*b*sqrt(1 + b*x**2/a)) + 13*a**(3/2)*x**5/(64*sqrt(1 + b*x**2/a)) + 5*sqrt(a)*b*x**7 /(16*sqrt(1 + b*x**2/a)) + 3*a**4*asinh(sqrt(b)*x/sqrt(a))/(128*b**(5/2)) + b**2*x**9/(8*sqrt(a)*sqrt(1 + b*x**2/a))
Time = 0.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.76 \[ \int x^4 \left (a+b x^2\right )^{3/2} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} x^{3}}{8 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} a x}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} x}{64 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} a^{3} x}{128 \, b^{2}} + \frac {3 \, a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {5}{2}}} \] Input:
integrate(x^4*(b*x^2+a)^(3/2),x, algorithm="maxima")
Output:
1/8*(b*x^2 + a)^(5/2)*x^3/b - 1/16*(b*x^2 + a)^(5/2)*a*x/b^2 + 1/64*(b*x^2 + a)^(3/2)*a^2*x/b^2 + 3/128*sqrt(b*x^2 + a)*a^3*x/b^2 + 3/128*a^4*arcsin h(b*x/sqrt(a*b))/b^(5/2)
Time = 0.13 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.66 \[ \int x^4 \left (a+b x^2\right )^{3/2} \, dx=\frac {1}{128} \, {\left (2 \, {\left (4 \, {\left (2 \, b x^{2} + 3 \, a\right )} x^{2} + \frac {a^{2}}{b}\right )} x^{2} - \frac {3 \, a^{3}}{b^{2}}\right )} \sqrt {b x^{2} + a} x - \frac {3 \, a^{4} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{128 \, b^{\frac {5}{2}}} \] Input:
integrate(x^4*(b*x^2+a)^(3/2),x, algorithm="giac")
Output:
1/128*(2*(4*(2*b*x^2 + 3*a)*x^2 + a^2/b)*x^2 - 3*a^3/b^2)*sqrt(b*x^2 + a)* x - 3/128*a^4*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)
Timed out. \[ \int x^4 \left (a+b x^2\right )^{3/2} \, dx=\int x^4\,{\left (b\,x^2+a\right )}^{3/2} \,d x \] Input:
int(x^4*(a + b*x^2)^(3/2),x)
Output:
int(x^4*(a + b*x^2)^(3/2), x)
Time = 0.19 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.86 \[ \int x^4 \left (a+b x^2\right )^{3/2} \, dx=\frac {-3 \sqrt {b \,x^{2}+a}\, a^{3} b x +2 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} x^{3}+24 \sqrt {b \,x^{2}+a}\, a \,b^{3} x^{5}+16 \sqrt {b \,x^{2}+a}\, b^{4} x^{7}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{4}}{128 b^{3}} \] Input:
int(x^4*(b*x^2+a)^(3/2),x)
Output:
( - 3*sqrt(a + b*x**2)*a**3*b*x + 2*sqrt(a + b*x**2)*a**2*b**2*x**3 + 24*s qrt(a + b*x**2)*a*b**3*x**5 + 16*sqrt(a + b*x**2)*b**4*x**7 + 3*sqrt(b)*lo g((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**4)/(128*b**3)