\(\int x^4 (a+b x^2)^{5/2} \, dx\) [407]

Optimal result

Integrand size = 15, antiderivative size = 136 \[ \int x^4 \left (a+b x^2\right )^{5/2} \, dx=-\frac {3 a^4 x \sqrt {a+b x^2}}{256 b^2}+\frac {a^3 x^3 \sqrt {a+b x^2}}{128 b}+\frac {1}{32} a^2 x^5 \sqrt {a+b x^2}+\frac {1}{16} a x^5 \left (a+b x^2\right )^{3/2}+\frac {1}{10} x^5 \left (a+b x^2\right )^{5/2}+\frac {3 a^5 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{256 b^{5/2}} \] Output:

-3/256*a^4*x*(b*x^2+a)^(1/2)/b^2+1/128*a^3*x^3*(b*x^2+a)^(1/2)/b+1/32*a^2* 
x^5*(b*x^2+a)^(1/2)+1/16*a*x^5*(b*x^2+a)^(3/2)+1/10*x^5*(b*x^2+a)^(5/2)+3/ 
256*a^5*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.76 \[ \int x^4 \left (a+b x^2\right )^{5/2} \, dx=\frac {\sqrt {a+b x^2} \left (-15 a^4 x+10 a^3 b x^3+248 a^2 b^2 x^5+336 a b^3 x^7+128 b^4 x^9\right )}{1280 b^2}+\frac {3 a^5 \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{128 b^{5/2}} \] Input:

Integrate[x^4*(a + b*x^2)^(5/2),x]
 

Output:

(Sqrt[a + b*x^2]*(-15*a^4*x + 10*a^3*b*x^3 + 248*a^2*b^2*x^5 + 336*a*b^3*x 
^7 + 128*b^4*x^9))/(1280*b^2) + (3*a^5*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + Sqr 
t[a + b*x^2])])/(128*b^(5/2))
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {248, 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.

\(\displaystyle \int x^4 \left (a+b x^2\right )^{5/2} \, dx\)

\(\Big \downarrow \) 248

\(\displaystyle \frac {1}{2} a \int x^4 \left (b x^2+a\right )^{3/2}dx+\frac {1}{10} x^5 \left (a+b x^2\right )^{5/2}\)

\(\Big \downarrow \) 248

\(\displaystyle \frac {1}{2} a \left (\frac {3}{8} a \int x^4 \sqrt {b x^2+a}dx+\frac {1}{8} x^5 \left (a+b x^2\right )^{3/2}\right )+\frac {1}{10} x^5 \left (a+b x^2\right )^{5/2}\)

\(\Big \downarrow \) 248

\(\displaystyle \frac {1}{2} a \left (\frac {3}{8} a \left (\frac {1}{6} a \int \frac {x^4}{\sqrt {b x^2+a}}dx+\frac {1}{6} x^5 \sqrt {a+b x^2}\right )+\frac {1}{8} x^5 \left (a+b x^2\right )^{3/2}\right )+\frac {1}{10} x^5 \left (a+b x^2\right )^{5/2}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {1}{2} a \left (\frac {3}{8} a \left (\frac {1}{6} a \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \int \frac {x^2}{\sqrt {b x^2+a}}dx}{4 b}\right )+\frac {1}{6} x^5 \sqrt {a+b x^2}\right )+\frac {1}{8} x^5 \left (a+b x^2\right )^{3/2}\right )+\frac {1}{10} x^5 \left (a+b x^2\right )^{5/2}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {1}{2} a \left (\frac {3}{8} a \left (\frac {1}{6} a \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}\right )}{4 b}\right )+\frac {1}{6} x^5 \sqrt {a+b x^2}\right )+\frac {1}{8} x^5 \left (a+b x^2\right )^{3/2}\right )+\frac {1}{10} x^5 \left (a+b x^2\right )^{5/2}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {1}{2} a \left (\frac {3}{8} a \left (\frac {1}{6} a \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}\right )}{4 b}\right )+\frac {1}{6} x^5 \sqrt {a+b x^2}\right )+\frac {1}{8} x^5 \left (a+b x^2\right )^{3/2}\right )+\frac {1}{10} x^5 \left (a+b x^2\right )^{5/2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} a \left (\frac {3}{8} a \left (\frac {1}{6} a \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}}\right )}{4 b}\right )+\frac {1}{6} x^5 \sqrt {a+b x^2}\right )+\frac {1}{8} x^5 \left (a+b x^2\right )^{3/2}\right )+\frac {1}{10} x^5 \left (a+b x^2\right )^{5/2}\)

Input:

Int[x^4*(a + b*x^2)^(5/2),x]
 

Output:

(x^5*(a + b*x^2)^(5/2))/10 + (a*((x^5*(a + b*x^2)^(3/2))/8 + (3*a*((x^5*Sq 
rt[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))/2
 

Defintions of rubi rules used

rule 219

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])
 

rule 224

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]
 

rule 248

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]
 

rule 262

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]
 

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.62

method	result	size
risch	\(-\frac {x \left (-128 b^{4} x^{8}-336 a \,b^{3} x^{6}-248 a^{2} b^{2} x^{4}-10 a^{3} b \,x^{2}+15 a^{4}\right ) \sqrt {b \,x^{2}+a}}{1280 b^{2}}+\frac {3 a^{5} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{256 b^{\frac {5}{2}}}\)	\(84\)
pseudoelliptic	\(\frac {\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right ) a^{5}}{256}-\frac {3 x \left (-\frac {128 b^{\frac {9}{2}} x^{8}}{15}-\frac {112 a \,b^{\frac {7}{2}} x^{6}}{5}-\frac {248 a^{2} b^{\frac {5}{2}} x^{4}}{15}-\frac {2 a^{3} b^{\frac {3}{2}} x^{2}}{3}+a^{4} \sqrt {b}\right ) \sqrt {b \,x^{2}+a}}{256}}{b^{\frac {5}{2}}}\)	\(87\)
default	\(\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{10 b}-\frac {3 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{8 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{8 b}\right )}{10 b}\)	\(114\)

Input:

int(x^4*(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/1280*x*(-128*b^4*x^8-336*a*b^3*x^6-248*a^2*b^2*x^4-10*a^3*b*x^2+15*a^4) 
*(b*x^2+a)^(1/2)/b^2+3/256*a^5/b^(5/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.40 \[ \int x^4 \left (a+b x^2\right )^{5/2} \, dx=\left [\frac {15 \, a^{5} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (128 \, b^{5} x^{9} + 336 \, a b^{4} x^{7} + 248 \, a^{2} b^{3} x^{5} + 10 \, a^{3} b^{2} x^{3} - 15 \, a^{4} b x\right )} \sqrt {b x^{2} + a}}{2560 \, b^{3}}, -\frac {15 \, a^{5} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (128 \, b^{5} x^{9} + 336 \, a b^{4} x^{7} + 248 \, a^{2} b^{3} x^{5} + 10 \, a^{3} b^{2} x^{3} - 15 \, a^{4} b x\right )} \sqrt {b x^{2} + a}}{1280 \, b^{3}}\right ] \] Input:

integrate(x^4*(b*x^2+a)^(5/2),x, algorithm="fricas")
 

Output:

[1/2560*(15*a^5*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 
2*(128*b^5*x^9 + 336*a*b^4*x^7 + 248*a^2*b^3*x^5 + 10*a^3*b^2*x^3 - 15*a^4 
*b*x)*sqrt(b*x^2 + a))/b^3, -1/1280*(15*a^5*sqrt(-b)*arctan(sqrt(-b)*x/sqr 
t(b*x^2 + a)) - (128*b^5*x^9 + 336*a*b^4*x^7 + 248*a^2*b^3*x^5 + 10*a^3*b^ 
2*x^3 - 15*a^4*b*x)*sqrt(b*x^2 + a))/b^3]
 

Sympy [A] (verification not implemented)

Time = 29.56 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.29 \[ \int x^4 \left (a+b x^2\right )^{5/2} \, dx=- \frac {3 a^{\frac {9}{2}} x}{256 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {a^{\frac {7}{2}} x^{3}}{256 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {129 a^{\frac {5}{2}} x^{5}}{640 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {73 a^{\frac {3}{2}} b x^{7}}{160 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {29 \sqrt {a} b^{2} x^{9}}{80 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 a^{5} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{256 b^{\frac {5}{2}}} + \frac {b^{3} x^{11}}{10 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \] Input:

integrate(x**4*(b*x**2+a)**(5/2),x)
 

Output:

-3*a**(9/2)*x/(256*b**2*sqrt(1 + b*x**2/a)) - a**(7/2)*x**3/(256*b*sqrt(1 
+ b*x**2/a)) + 129*a**(5/2)*x**5/(640*sqrt(1 + b*x**2/a)) + 73*a**(3/2)*b* 
x**7/(160*sqrt(1 + b*x**2/a)) + 29*sqrt(a)*b**2*x**9/(80*sqrt(1 + b*x**2/a 
)) + 3*a**5*asinh(sqrt(b)*x/sqrt(a))/(256*b**(5/2)) + b**3*x**11/(10*sqrt( 
a)*sqrt(1 + b*x**2/a))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.77 \[ \int x^4 \left (a+b x^2\right )^{5/2} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} x^{3}}{10 \, b} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a x}{80 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{2} x}{160 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3} x}{128 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} a^{4} x}{256 \, b^{2}} + \frac {3 \, a^{5} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{256 \, b^{\frac {5}{2}}} \] Input:

integrate(x^4*(b*x^2+a)^(5/2),x, algorithm="maxima")
 

Output:

1/10*(b*x^2 + a)^(7/2)*x^3/b - 3/80*(b*x^2 + a)^(7/2)*a*x/b^2 + 1/160*(b*x 
^2 + a)^(5/2)*a^2*x/b^2 + 1/128*(b*x^2 + a)^(3/2)*a^3*x/b^2 + 3/256*sqrt(b 
*x^2 + a)*a^4*x/b^2 + 3/256*a^5*arcsinh(b*x/sqrt(a*b))/b^(5/2)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.67 \[ \int x^4 \left (a+b x^2\right )^{5/2} \, dx=-\frac {3 \, a^{5} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{256 \, b^{\frac {5}{2}}} + \frac {1}{1280} \, {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, b^{2} x^{2} + 21 \, a b\right )} x^{2} + 31 \, a^{2}\right )} x^{2} + \frac {5 \, a^{3}}{b}\right )} x^{2} - \frac {15 \, a^{4}}{b^{2}}\right )} \sqrt {b x^{2} + a} x \] Input:

integrate(x^4*(b*x^2+a)^(5/2),x, algorithm="giac")
 

Output:

-3/256*a^5*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2) + 1/1280*(2*(4*( 
2*(8*b^2*x^2 + 21*a*b)*x^2 + 31*a^2)*x^2 + 5*a^3/b)*x^2 - 15*a^4/b^2)*sqrt 
(b*x^2 + a)*x
 

Mupad [F(-1)]

Timed out. \[ \int x^4 \left (a+b x^2\right )^{5/2} \, dx=\int x^4\,{\left (b\,x^2+a\right )}^{5/2} \,d x \] Input:

int(x^4*(a + b*x^2)^(5/2),x)
 

Output:

int(x^4*(a + b*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.87 \[ \int x^4 \left (a+b x^2\right )^{5/2} \, dx=\frac {-15 \sqrt {b \,x^{2}+a}\, a^{4} b x +10 \sqrt {b \,x^{2}+a}\, a^{3} b^{2} x^{3}+248 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} x^{5}+336 \sqrt {b \,x^{2}+a}\, a \,b^{4} x^{7}+128 \sqrt {b \,x^{2}+a}\, b^{5} x^{9}+15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{5}}{1280 b^{3}} \] Input:

int(x^4*(b*x^2+a)^(5/2),x)
 

Output:

( - 15*sqrt(a + b*x**2)*a**4*b*x + 10*sqrt(a + b*x**2)*a**3*b**2*x**3 + 24 
8*sqrt(a + b*x**2)*a**2*b**3*x**5 + 336*sqrt(a + b*x**2)*a*b**4*x**7 + 128 
*sqrt(a + b*x**2)*b**5*x**9 + 15*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x 
)/sqrt(a))*a**5)/(1280*b**3)