3.2.99 \(\int \frac {(a+b x^2)^{5/2}}{\sqrt {a^2-b^2 x^4}} \, dx\) [199]

3.2.99.1 Optimal result

Integrand size = 28, antiderivative size = 153 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{\sqrt {a^2-b^2 x^4}} \, dx=-\frac {9 a x \left (a-b x^2\right ) \sqrt {a+b x^2}}{8 \sqrt {a^2-b^2 x^4}}-\frac {x \left (a-b x^2\right ) \left (a+b x^2\right )^{3/2}}{4 \sqrt {a^2-b^2 x^4}}+\frac {19 a^2 \sqrt {a-b x^2} \sqrt {a+b x^2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a-b x^2}}\right )}{8 \sqrt {b} \sqrt {a^2-b^2 x^4}} \]

output

-1/4*x*(-b*x^2+a)*(b*x^2+a)^(3/2)/(-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*arctan(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)
 

3.2.99.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.72 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.64 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{\sqrt {a^2-b^2 x^4}} \, dx=-\frac {\left (11 a x+2 b x^3\right ) \sqrt {a^2-b^2 x^4}}{8 \sqrt {a+b x^2}}+\frac {19 i a^2 \log \left (-2 i \sqrt {b} x+\frac {2 \sqrt {a^2-b^2 x^4}}{\sqrt {a+b x^2}}\right )}{8 \sqrt {b}} \]

input

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

output

-1/8*((11*a*x + 2*b*x^3)*Sqrt[a^2 - b^2*x^4])/Sqrt[a + b*x^2] + (((19*I)/8 
)*a^2*Log[(-2*I)*Sqrt[b]*x + (2*Sqrt[a^2 - b^2*x^4])/Sqrt[a + b*x^2]])/Sqr 
t[b]
 

3.2.99.3 Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.77, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1396, 318, 25, 27, 299, 224, 216}

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

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

\(\Big \downarrow \) 318

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 299

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

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

3.2.99.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 216

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

3.2.99.4 Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.63

method	result	size
default	\(\frac {\sqrt {-b^{2} x^{4}+a^{2}}\, \left (-2 b^{\frac {3}{2}} x^{3} \sqrt {-b \,x^{2}+a}-11 a x \sqrt {-b \,x^{2}+a}\, \sqrt {b}+19 \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {-b \,x^{2}+a}}\right ) a^{2}\right )}{8 \sqrt {b \,x^{2}+a}\, \sqrt {-b \,x^{2}+a}\, \sqrt {b}}\)	\(96\)
risch	\(-\frac {x \left (2 b \,x^{2}+11 a \right ) \sqrt {-b \,x^{2}+a}\, \sqrt {\frac {-b^{2} x^{4}+a^{2}}{b \,x^{2}+a}}\, \sqrt {b \,x^{2}+a}}{8 \sqrt {-b^{2} x^{4}+a^{2}}}+\frac {19 a^{2} \arctan \left (\frac {\sqrt {b}\, x}{\sqrt {-b \,x^{2}+a}}\right ) \sqrt {\frac {-b^{2} x^{4}+a^{2}}{b \,x^{2}+a}}\, \sqrt {b \,x^{2}+a}}{8 \sqrt {b}\, \sqrt {-b^{2} x^{4}+a^{2}}}\)	\(143\)

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*x^2+a 
)^(1/2)*b^(1/2)+19*arctan(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)
 

3.2.99.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.64 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{\sqrt {a^2-b^2 x^4}} \, dx=\left [-\frac {19 \, {\left (a^{2} b x^{2} + a^{3}\right )} \sqrt {-b} \log \left (-\frac {2 \, b^{2} x^{4} + a b x^{2} - 2 \, \sqrt {-b^{2} x^{4} + a^{2}} \sqrt {b x^{2} + a} \sqrt {-b} x - a^{2}}{b x^{2} + a}\right ) + 2 \, \sqrt {-b^{2} x^{4} + a^{2}} {\left (2 \, b^{2} x^{3} + 11 \, a b x\right )} \sqrt {b x^{2} + a}}{16 \, {\left (b^{2} x^{2} + a b\right )}}, -\frac {19 \, {\left (a^{2} b x^{2} + a^{3}\right )} \sqrt {b} \arctan \left (\frac {\sqrt {-b^{2} x^{4} + a^{2}} \sqrt {b x^{2} + a} \sqrt {b}}{b^{2} x^{3} + a b x}\right ) + \sqrt {-b^{2} x^{4} + a^{2}} {\left (2 \, b^{2} x^{3} + 11 \, a b x\right )} \sqrt {b x^{2} + a}}{8 \, {\left (b^{2} x^{2} + a b\right )}}\right ] \]

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 
*(19*(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)]
 

3.2.99.6 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)
 

3.2.99.7 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)
 

3.2.99.8 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)
 

3.2.99.9 Mupad [F(-1)]

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

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)