3.3.5 \(\int \frac {(a-b x^2)^{5/2}}{\sqrt {a^2-b^2 x^4}} \, dx\) [205]

3.3.5.1 Optimal result

Integrand size = 29, antiderivative size = 152 \[ \int \frac {\left (a-b x^2\right )^{5/2}}{\sqrt {a^2-b^2 x^4}} \, dx=-\frac {9 a x \sqrt {a-b x^2} \left (a+b x^2\right )}{8 \sqrt {a^2-b^2 x^4}}-\frac {x \left (a-b x^2\right )^{3/2} \left (a+b x^2\right )}{4 \sqrt {a^2-b^2 x^4}}+\frac {19 a^2 \sqrt {a-b x^2} \sqrt {a+b x^2} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 \sqrt {b} \sqrt {a^2-b^2 x^4}} \]

output

-1/4*x*(-b*x^2+a)^(3/2)*(b*x^2+a)/(-b^2*x^4+a^2)^(1/2)-9/8*a*x*(b*x^2+a)*( 
-b*x^2+a)^(1/2)/(-b^2*x^4+a^2)^(1/2)+19/8*a^2*arctanh(x*b^(1/2)/(b*x^2+a)^ 
(1/2))*(-b*x^2+a)^(1/2)*(b*x^2+a)^(1/2)/b^(1/2)/(-b^2*x^4+a^2)^(1/2)
 

3.3.5.2 Mathematica [A] (verified)

Time = 2.71 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a-b x^2\right )^{5/2}}{\sqrt {a^2-b^2 x^4}} \, dx=\frac {1}{8} \left (\frac {x \left (-11 a+2 b x^2\right ) \sqrt {a^2-b^2 x^4}}{\sqrt {a-b x^2}}-\frac {19 a^2 \log \left (-a+b x^2\right )}{\sqrt {b}}+\frac {19 a^2 \log \left (a b x-b^2 x^3+\sqrt {b} \sqrt {a-b x^2} \sqrt {a^2-b^2 x^4}\right )}{\sqrt {b}}\right ) \]

input

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

output

((x*(-11*a + 2*b*x^2)*Sqrt[a^2 - b^2*x^4])/Sqrt[a - b*x^2] - (19*a^2*Log[- 
a + b*x^2])/Sqrt[b] + (19*a^2*Log[a*b*x - b^2*x^3 + Sqrt[b]*Sqrt[a - b*x^2 
]*Sqrt[a^2 - b^2*x^4]])/Sqrt[b])/8
 

3.3.5.3 Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.76, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {1396, 318, 27, 299, 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.

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

\(\Big \downarrow \) 1396

\(\displaystyle \frac {\sqrt {a-b x^2} \sqrt {a+b x^2} \int \frac {\left (a-b x^2\right )^2}{\sqrt {b x^2+a}}dx}{\sqrt {a^2-b^2 x^4}}\)

\(\Big \downarrow \) 318

\(\displaystyle \frac {\sqrt {a-b x^2} \sqrt {a+b x^2} \left (\frac {\int \frac {a b \left (5 a-9 b x^2\right )}{\sqrt {b x^2+a}}dx}{4 b}-\frac {1}{4} x \left (a-b x^2\right ) \sqrt {a+b x^2}\right )}{\sqrt {a^2-b^2 x^4}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a-b x^2} \sqrt {a+b x^2} \left (\frac {1}{4} a \int \frac {5 a-9 b x^2}{\sqrt {b x^2+a}}dx-\frac {1}{4} x \left (a-b x^2\right ) \sqrt {a+b x^2}\right )}{\sqrt {a^2-b^2 x^4}}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\sqrt {a-b x^2} \sqrt {a+b x^2} \left (\frac {1}{4} a \left (\frac {19}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx-\frac {9}{2} x \sqrt {a+b x^2}\right )-\frac {1}{4} x \left (a-b x^2\right ) \sqrt {a+b x^2}\right )}{\sqrt {a^2-b^2 x^4}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\sqrt {a-b x^2} \sqrt {a+b x^2} \left (\frac {1}{4} a \left (\frac {19}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}-\frac {9}{2} x \sqrt {a+b x^2}\right )-\frac {1}{4} x \left (a-b x^2\right ) \sqrt {a+b x^2}\right )}{\sqrt {a^2-b^2 x^4}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt {a-b x^2} \sqrt {a+b x^2} \left (\frac {1}{4} a \left (\frac {19 a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}-\frac {9}{2} x \sqrt {a+b x^2}\right )-\frac {1}{4} x \left (a-b x^2\right ) \sqrt {a+b x^2}\right )}{\sqrt {a^2-b^2 x^4}}\)

input

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

output

(Sqrt[a - b*x^2]*Sqrt[a + b*x^2]*(-1/4*(x*(a - b*x^2)*Sqrt[a + b*x^2]) + ( 
a*((-9*x*Sqrt[a + b*x^2])/2 + (19*a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/ 
(2*Sqrt[b])))/4))/Sqrt[a^2 - b^2*x^4]
 

3.3.5.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 318

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S 
imp[1/(b*(2*(p + q) + 1))   Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b 
*c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 
 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G 
tQ[q, 1] && NeQ[2*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[a, b, c, 
d, 2, p, q, x]
 

rule 1396

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_)*((d_) + (e_.)*(x_)^(n_))^(q_.), x 
_Symbol] :> Simp[(a + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d 
+ c*(x^n/e))^FracPart[p])   Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, 
x], x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a* 
e^2, 0] &&  !IntegerQ[p] &&  !(EqQ[q, 1] && EqQ[n, 2])
 

3.3.5.4 Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.62

method	result	size
default	\(\frac {\sqrt {-b^{2} x^{4}+a^{2}}\, \left (2 b^{\frac {3}{2}} x^{3} \sqrt {b \,x^{2}+a}-11 a x \sqrt {b}\, \sqrt {b \,x^{2}+a}+19 \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) a^{2}\right )}{8 \sqrt {-b \,x^{2}+a}\, \sqrt {b \,x^{2}+a}\, \sqrt {b}}\)	\(94\)
risch	\(\frac {x \left (-2 b \,x^{2}+11 a \right ) \sqrt {b \,x^{2}+a}\, \sqrt {\frac {\left (-b \,x^{2}+a \right ) \left (-b^{2} x^{4}+a^{2}\right )}{\left (b \,x^{2}-a \right )^{2}}}\, \left (b \,x^{2}-a \right )}{8 \sqrt {-b \,x^{2}+a}\, \sqrt {-b^{2} x^{4}+a^{2}}}-\frac {19 a^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) \sqrt {\frac {\left (-b \,x^{2}+a \right ) \left (-b^{2} x^{4}+a^{2}\right )}{\left (b \,x^{2}-a \right )^{2}}}\, \left (b \,x^{2}-a \right )}{8 \sqrt {b}\, \sqrt {-b \,x^{2}+a}\, \sqrt {-b^{2} x^{4}+a^{2}}}\)	\(182\)

input

int((-b*x^2+a)^(5/2)/(-b^2*x^4+a^2)^(1/2),x,method=_RETURNVERBOSE)
 

output

1/8*(-b^2*x^4+a^2)^(1/2)*(2*b^(3/2)*x^3*(b*x^2+a)^(1/2)-11*a*x*b^(1/2)*(b* 
x^2+a)^(1/2)+19*ln(b^(1/2)*x+(b*x^2+a)^(1/2))*a^2)/(-b*x^2+a)^(1/2)/(b*x^2 
+a)^(1/2)/b^(1/2)
 

3.3.5.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.74 \[ \int \frac {\left (a-b x^2\right )^{5/2}}{\sqrt {a^2-b^2 x^4}} \, dx=\left [\frac {19 \, {\left (a^{2} b x^{2} - a^{3}\right )} \sqrt {b} \log \left (\frac {2 \, b^{2} x^{4} - a b x^{2} - 2 \, \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {-b x^{2} + a} \sqrt {b} x - a^{2}}{b x^{2} - a}\right ) - 2 \, \sqrt {-b^{2} x^{4} + a^{2}} {\left (2 \, b^{2} x^{3} - 11 \, a b x\right )} \sqrt {-b x^{2} + a}}{16 \, {\left (b^{2} x^{2} - a b\right )}}, \frac {19 \, {\left (a^{2} b x^{2} - a^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b^{2} x^{4} + a^{2}} \sqrt {-b x^{2} + a} \sqrt {-b}}{b^{2} x^{3} - a b x}\right ) - \sqrt {-b^{2} x^{4} + a^{2}} {\left (2 \, b^{2} x^{3} - 11 \, a b x\right )} \sqrt {-b x^{2} + a}}{8 \, {\left (b^{2} x^{2} - a b\right )}}\right ] \]

input

integrate((-b*x^2+a)^(5/2)/(-b^2*x^4+a^2)^(1/2),x, algorithm="fricas")
 

output

[1/16*(19*(a^2*b*x^2 - a^3)*sqrt(b)*log((2*b^2*x^4 - a*b*x^2 - 2*sqrt(-b^2 
*x^4 + a^2)*sqrt(-b*x^2 + a)*sqrt(b)*x - a^2)/(b*x^2 - a)) - 2*sqrt(-b^2*x 
^4 + a^2)*(2*b^2*x^3 - 11*a*b*x)*sqrt(-b*x^2 + a))/(b^2*x^2 - a*b), 1/8*(1 
9*(a^2*b*x^2 - a^3)*sqrt(-b)*arctan(sqrt(-b^2*x^4 + a^2)*sqrt(-b*x^2 + a)* 
sqrt(-b)/(b^2*x^3 - a*b*x)) - sqrt(-b^2*x^4 + a^2)*(2*b^2*x^3 - 11*a*b*x)* 
sqrt(-b*x^2 + a))/(b^2*x^2 - a*b)]
 

3.3.5.6 Sympy [F]

\[ \int \frac {\left (a-b x^2\right )^{5/2}}{\sqrt {a^2-b^2 x^4}} \, dx=\int \frac {\left (a - b x^{2}\right )^{\frac {5}{2}}}{\sqrt {- \left (- a + b x^{2}\right ) \left (a + b x^{2}\right )}}\, dx \]

input

integrate((-b*x**2+a)**(5/2)/(-b**2*x**4+a**2)**(1/2),x)
 

output

Integral((a - b*x**2)**(5/2)/sqrt(-(-a + b*x**2)*(a + b*x**2)), x)
 

3.3.5.7 Maxima [F]

\[ \int \frac {\left (a-b x^2\right )^{5/2}}{\sqrt {a^2-b^2 x^4}} \, dx=\int { \frac {{\left (-b x^{2} + a\right )}^{\frac {5}{2}}}{\sqrt {-b^{2} x^{4} + a^{2}}} \,d x } \]

input

integrate((-b*x^2+a)^(5/2)/(-b^2*x^4+a^2)^(1/2),x, algorithm="maxima")
 

output

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

3.3.5.8 Giac [F]

\[ \int \frac {\left (a-b x^2\right )^{5/2}}{\sqrt {a^2-b^2 x^4}} \, dx=\int { \frac {{\left (-b x^{2} + a\right )}^{\frac {5}{2}}}{\sqrt {-b^{2} x^{4} + a^{2}}} \,d x } \]

input

integrate((-b*x^2+a)^(5/2)/(-b^2*x^4+a^2)^(1/2),x, algorithm="giac")
 

output

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

3.3.5.9 Mupad [F(-1)]

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

input

int((a - b*x^2)^(5/2)/(a^2 - b^2*x^4)^(1/2),x)
 

output

int((a - b*x^2)^(5/2)/(a^2 - b^2*x^4)^(1/2), x)