\(\int (1+b x^2) (-1-b^2 x^4)^{3/2} \, dx\) [287]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (warning: unable to verify)
Fricas [A] (verification not implemented)
Sympy [C] (verification not implemented)
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 22, antiderivative size = 217 \[ \int \left (1+b x^2\right ) \left (-1-b^2 x^4\right )^{3/2} \, dx=-\frac {4 x \sqrt {-1-b^2 x^4}}{15 \left (1+b x^2\right )}-\frac {2}{105} x \left (15+7 b x^2\right ) \sqrt {-1-b^2 x^4}+\frac {1}{63} x \left (9+7 b x^2\right ) \left (-1-b^2 x^4\right )^{3/2}-\frac {4 \left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} E\left (2 \arctan \left (\sqrt {b} x\right )|\frac {1}{2}\right )}{15 \sqrt {b} \sqrt {-1-b^2 x^4}}+\frac {44 \left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{105 \sqrt {b} \sqrt {-1-b^2 x^4}} \] Output:

-4*x*(-b^2*x^4-1)^(1/2)/(15*b*x^2+15)-2/105*x*(7*b*x^2+15)*(-b^2*x^4-1)^(1 
/2)+1/63*x*(7*b*x^2+9)*(-b^2*x^4-1)^(3/2)-4/15*(b*x^2+1)*((b^2*x^4+1)/(b*x 
^2+1)^2)^(1/2)*EllipticE(sin(2*arctan(b^(1/2)*x)),1/2*2^(1/2))/b^(1/2)/(-b 
^2*x^4-1)^(1/2)+44/105*(b*x^2+1)*((b^2*x^4+1)/(b*x^2+1)^2)^(1/2)*InverseJa 
cobiAM(2*arctan(b^(1/2)*x),1/2*2^(1/2))/b^(1/2)/(-b^2*x^4-1)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 7.57 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.35 \[ \int \left (1+b x^2\right ) \left (-1-b^2 x^4\right )^{3/2} \, dx=-\frac {\sqrt {-1-b^2 x^4} \left (3 x \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},\frac {5}{4},-b^2 x^4\right )+b x^3 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3}{4},\frac {7}{4},-b^2 x^4\right )\right )}{3 \sqrt {1+b^2 x^4}} \] Input:

Integrate[(1 + b*x^2)*(-1 - b^2*x^4)^(3/2),x]
 

Output:

-1/3*(Sqrt[-1 - b^2*x^4]*(3*x*Hypergeometric2F1[-3/2, 1/4, 5/4, -(b^2*x^4) 
] + b*x^3*Hypergeometric2F1[-3/2, 3/4, 7/4, -(b^2*x^4)]))/Sqrt[1 + b^2*x^4 
]
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1491, 27, 1491, 27, 1512, 761, 1510}

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.

\(\displaystyle \int \left (b x^2+1\right ) \left (-b^2 x^4-1\right )^{3/2} \, dx\)

\(\Big \downarrow \) 1491

\(\displaystyle \frac {1}{21} \int -2 \left (7 b x^2+9\right ) \sqrt {-b^2 x^4-1}dx+\frac {1}{63} x \left (7 b x^2+9\right ) \left (-b^2 x^4-1\right )^{3/2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{63} x \left (7 b x^2+9\right ) \left (-b^2 x^4-1\right )^{3/2}-\frac {2}{21} \int \left (7 b x^2+9\right ) \sqrt {-b^2 x^4-1}dx\)

\(\Big \downarrow \) 1491

\(\displaystyle \frac {1}{63} x \left (7 b x^2+9\right ) \left (-b^2 x^4-1\right )^{3/2}-\frac {2}{21} \left (\frac {1}{15} \int -\frac {6 \left (7 b x^2+15\right )}{\sqrt {-b^2 x^4-1}}dx+\frac {1}{5} x \sqrt {-b^2 x^4-1} \left (7 b x^2+15\right )\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{63} x \left (7 b x^2+9\right ) \left (-b^2 x^4-1\right )^{3/2}-\frac {2}{21} \left (\frac {1}{5} x \left (7 b x^2+15\right ) \sqrt {-b^2 x^4-1}-\frac {2}{5} \int \frac {7 b x^2+15}{\sqrt {-b^2 x^4-1}}dx\right )\)

\(\Big \downarrow \) 1512

\(\displaystyle \frac {1}{63} x \left (7 b x^2+9\right ) \left (-b^2 x^4-1\right )^{3/2}-\frac {2}{21} \left (\frac {1}{5} x \left (7 b x^2+15\right ) \sqrt {-b^2 x^4-1}-\frac {2}{5} \left (22 \int \frac {1}{\sqrt {-b^2 x^4-1}}dx-7 \int \frac {1-b x^2}{\sqrt {-b^2 x^4-1}}dx\right )\right )\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {1}{63} x \left (7 b x^2+9\right ) \left (-b^2 x^4-1\right )^{3/2}-\frac {2}{21} \left (\frac {1}{5} x \left (7 b x^2+15\right ) \sqrt {-b^2 x^4-1}-\frac {2}{5} \left (\frac {11 \left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{\sqrt {b} \sqrt {-b^2 x^4-1}}-7 \int \frac {1-b x^2}{\sqrt {-b^2 x^4-1}}dx\right )\right )\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {1}{63} x \left (7 b x^2+9\right ) \left (-b^2 x^4-1\right )^{3/2}-\frac {2}{21} \left (\frac {1}{5} x \left (7 b x^2+15\right ) \sqrt {-b^2 x^4-1}-\frac {2}{5} \left (\frac {11 \left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{\sqrt {b} \sqrt {-b^2 x^4-1}}-7 \left (\frac {\left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} E\left (2 \arctan \left (\sqrt {b} x\right )|\frac {1}{2}\right )}{\sqrt {b} \sqrt {-b^2 x^4-1}}+\frac {x \sqrt {-b^2 x^4-1}}{b x^2+1}\right )\right )\right )\)

Input:

Int[(1 + b*x^2)*(-1 - b^2*x^4)^(3/2),x]
 

Output:

(x*(9 + 7*b*x^2)*(-1 - b^2*x^4)^(3/2))/63 - (2*((x*(15 + 7*b*x^2)*Sqrt[-1 
- b^2*x^4])/5 - (2*(-7*((x*Sqrt[-1 - b^2*x^4])/(1 + b*x^2) + ((1 + b*x^2)* 
Sqrt[(1 + b^2*x^4)/(1 + b*x^2)^2]*EllipticE[2*ArcTan[Sqrt[b]*x], 1/2])/(Sq 
rt[b]*Sqrt[-1 - b^2*x^4])) + (11*(1 + b*x^2)*Sqrt[(1 + b^2*x^4)/(1 + b*x^2 
)^2]*EllipticF[2*ArcTan[Sqrt[b]*x], 1/2])/(Sqrt[b]*Sqrt[-1 - b^2*x^4])))/5 
))/21
 

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 761
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1491
Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[x*( 
d*(4*p + 3) + e*(4*p + 1)*x^2)*((a + c*x^4)^p/((4*p + 1)*(4*p + 3))), x] + 
Simp[2*(p/((4*p + 1)*(4*p + 3)))   Int[Simp[2*a*d*(4*p + 3) + (2*a*e*(4*p + 
 1))*x^2, x]*(a + c*x^4)^(p - 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c 
*d^2 + a*e^2, 0] && GtQ[p, 0] && FractionQ[p] && IntegerQ[2*p]
 

rule 1510
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* 
(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

rule 1512
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 2]}, Simp[(e + d*q)/q   Int[1/Sqrt[a + c*x^4], x], x] - Simp[e/q 
 Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a, c 
, d, e}, x] && PosQ[c/a]
 
Maple [A] (warning: unable to verify)

Time = 1.85 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.41

method result size
meijerg \(\frac {{\left (-\operatorname {signum}\left (b^{2} x^{4}+1\right )\right )}^{\frac {3}{2}} x \operatorname {hypergeom}\left (\left [-\frac {3}{2}, \frac {1}{4}\right ], \left [\frac {5}{4}\right ], -b^{2} x^{4}\right )}{\operatorname {signum}\left (b^{2} x^{4}+1\right )^{\frac {3}{2}}}+\frac {b {\left (-\operatorname {signum}\left (b^{2} x^{4}+1\right )\right )}^{\frac {3}{2}} x^{3} \operatorname {hypergeom}\left (\left [-\frac {3}{2}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -b^{2} x^{4}\right )}{3 \operatorname {signum}\left (b^{2} x^{4}+1\right )^{\frac {3}{2}}}\) \(90\)
risch \(\frac {x \left (35 b^{3} x^{6}+45 b^{2} x^{4}+77 b \,x^{2}+135\right ) \left (b^{2} x^{4}+1\right )}{315 \sqrt {-b^{2} x^{4}-1}}+\frac {4 \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )}{7 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}-\frac {4 i \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-i b}, i\right )\right )}{15 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}\) \(171\)
elliptic \(-\frac {b^{3} x^{7} \sqrt {-b^{2} x^{4}-1}}{9}-\frac {b^{2} x^{5} \sqrt {-b^{2} x^{4}-1}}{7}-\frac {11 b \,x^{3} \sqrt {-b^{2} x^{4}-1}}{45}-\frac {3 x \sqrt {-b^{2} x^{4}-1}}{7}+\frac {4 \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )}{7 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}-\frac {4 i \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-i b}, i\right )\right )}{15 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}\) \(196\)
default \(-\frac {b^{2} x^{5} \sqrt {-b^{2} x^{4}-1}}{7}-\frac {3 x \sqrt {-b^{2} x^{4}-1}}{7}+\frac {4 \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )}{7 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}+b \left (-\frac {b^{2} x^{7} \sqrt {-b^{2} x^{4}-1}}{9}-\frac {11 x^{3} \sqrt {-b^{2} x^{4}-1}}{45}-\frac {4 i \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-i b}, i\right )\right )}{15 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}\, b}\right )\) \(201\)

Input:

int((b*x^2+1)*(-b^2*x^4-1)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

(-signum(b^2*x^4+1))^(3/2)/signum(b^2*x^4+1)^(3/2)*x*hypergeom([-3/2,1/4], 
[5/4],-b^2*x^4)+1/3*b*(-signum(b^2*x^4+1))^(3/2)/signum(b^2*x^4+1)^(3/2)*x 
^3*hypergeom([-3/2,3/4],[7/4],-b^2*x^4)
 

Fricas [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.56 \[ \int \left (1+b x^2\right ) \left (-1-b^2 x^4\right )^{3/2} \, dx=-\frac {12 \, \sqrt {-b^{2}} {\left (15 \, b - 7\right )} x \left (-\frac {1}{b^{2}}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {1}{b^{2}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 84 \, \sqrt {-b^{2}} x \left (-\frac {1}{b^{2}}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {1}{b^{2}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (35 \, b^{4} x^{8} + 45 \, b^{3} x^{6} + 77 \, b^{2} x^{4} + 135 \, b x^{2} + 84\right )} \sqrt {-b^{2} x^{4} - 1}}{315 \, b x} \] Input:

integrate((b*x^2+1)*(-b^2*x^4-1)^(3/2),x, algorithm="fricas")
 

Output:

-1/315*(12*sqrt(-b^2)*(15*b - 7)*x*(-1/b^2)^(3/4)*elliptic_f(arcsin((-1/b^ 
2)^(1/4)/x), -1) + 84*sqrt(-b^2)*x*(-1/b^2)^(3/4)*elliptic_e(arcsin((-1/b^ 
2)^(1/4)/x), -1) + (35*b^4*x^8 + 45*b^3*x^6 + 77*b^2*x^4 + 135*b*x^2 + 84) 
*sqrt(-b^2*x^4 - 1))/(b*x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.72 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.72 \[ \int \left (1+b x^2\right ) \left (-1-b^2 x^4\right )^{3/2} \, dx=- \frac {i b^{3} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} - \frac {i b^{2} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} - \frac {i b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} - \frac {i x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:

integrate((b*x**2+1)*(-b**2*x**4-1)**(3/2),x)
 

Output:

-I*b**3*x**7*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), b**2*x**4*exp_polar(I* 
pi))/(4*gamma(11/4)) - I*b**2*x**5*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), b 
**2*x**4*exp_polar(I*pi))/(4*gamma(9/4)) - I*b*x**3*gamma(3/4)*hyper((-1/2 
, 3/4), (7/4,), b**2*x**4*exp_polar(I*pi))/(4*gamma(7/4)) - I*x*gamma(1/4) 
*hyper((-1/2, 1/4), (5/4,), b**2*x**4*exp_polar(I*pi))/(4*gamma(5/4))
 

Maxima [F]

\[ \int \left (1+b x^2\right ) \left (-1-b^2 x^4\right )^{3/2} \, dx=\int { {\left (-b^{2} x^{4} - 1\right )}^{\frac {3}{2}} {\left (b x^{2} + 1\right )} \,d x } \] Input:

integrate((b*x^2+1)*(-b^2*x^4-1)^(3/2),x, algorithm="maxima")
 

Output:

integrate((-b^2*x^4 - 1)^(3/2)*(b*x^2 + 1), x)
 

Giac [F]

\[ \int \left (1+b x^2\right ) \left (-1-b^2 x^4\right )^{3/2} \, dx=\int { {\left (-b^{2} x^{4} - 1\right )}^{\frac {3}{2}} {\left (b x^{2} + 1\right )} \,d x } \] Input:

integrate((b*x^2+1)*(-b^2*x^4-1)^(3/2),x, algorithm="giac")
 

Output:

integrate((-b^2*x^4 - 1)^(3/2)*(b*x^2 + 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (1+b x^2\right ) \left (-1-b^2 x^4\right )^{3/2} \, dx=\int {\left (-b^2\,x^4-1\right )}^{3/2}\,\left (b\,x^2+1\right ) \,d x \] Input:

int((- b^2*x^4 - 1)^(3/2)*(b*x^2 + 1),x)
 

Output:

int((- b^2*x^4 - 1)^(3/2)*(b*x^2 + 1), x)
 

Reduce [F]

\[ \int \left (1+b x^2\right ) \left (-1-b^2 x^4\right )^{3/2} \, dx=\frac {i \left (-35 \sqrt {b^{2} x^{4}+1}\, b^{3} x^{7}-45 \sqrt {b^{2} x^{4}+1}\, b^{2} x^{5}-77 \sqrt {b^{2} x^{4}+1}\, b \,x^{3}-135 \sqrt {b^{2} x^{4}+1}\, x -180 \left (\int \frac {\sqrt {b^{2} x^{4}+1}}{b^{2} x^{4}+1}d x \right )-84 \left (\int \frac {\sqrt {b^{2} x^{4}+1}\, x^{2}}{b^{2} x^{4}+1}d x \right ) b \right )}{315} \] Input:

int((b*x^2+1)*(-b^2*x^4-1)^(3/2),x)
 

Output:

(i*( - 35*sqrt(b**2*x**4 + 1)*b**3*x**7 - 45*sqrt(b**2*x**4 + 1)*b**2*x**5 
 - 77*sqrt(b**2*x**4 + 1)*b*x**3 - 135*sqrt(b**2*x**4 + 1)*x - 180*int(sqr 
t(b**2*x**4 + 1)/(b**2*x**4 + 1),x) - 84*int((sqrt(b**2*x**4 + 1)*x**2)/(b 
**2*x**4 + 1),x)*b))/315