\(\int (d+e x^2)^{3/2} \sqrt {d^2-e^2 x^4} \, dx\) [131]

Optimal result

Integrand size = 28, antiderivative size = 156 \[ \int \left (d+e x^2\right )^{3/2} \sqrt {d^2-e^2 x^4} \, dx=\frac {3 d^2 x \sqrt {d^2-e^2 x^4}}{16 \sqrt {d+e x^2}}+\frac {11 d e x^3 \sqrt {d^2-e^2 x^4}}{24 \sqrt {d+e x^2}}+\frac {e^2 x^5 \sqrt {d^2-e^2 x^4}}{6 \sqrt {d+e x^2}}+\frac {13 d^3 \arctan \left (\frac {\sqrt {e} x \sqrt {d+e x^2}}{\sqrt {d^2-e^2 x^4}}\right )}{16 \sqrt {e}} \] Output:

3/16*d^2*x*(-e^2*x^4+d^2)^(1/2)/(e*x^2+d)^(1/2)+11/24*d*e*x^3*(-e^2*x^4+d^ 
2)^(1/2)/(e*x^2+d)^(1/2)+1/6*e^2*x^5*(-e^2*x^4+d^2)^(1/2)/(e*x^2+d)^(1/2)+ 
13/16*d^3*arctan(e^(1/2)*x*(e*x^2+d)^(1/2)/(-e^2*x^4+d^2)^(1/2))/e^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.89 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.69 \[ \int \left (d+e x^2\right )^{3/2} \sqrt {d^2-e^2 x^4} \, dx=\frac {1}{48} \left (\frac {x \sqrt {d^2-e^2 x^4} \left (9 d^2+22 d e x^2+8 e^2 x^4\right )}{\sqrt {d+e x^2}}+\frac {39 i d^3 \log \left (-2 i \sqrt {e} x+\frac {2 \sqrt {d^2-e^2 x^4}}{\sqrt {d+e x^2}}\right )}{\sqrt {e}}\right ) \] Input:

Integrate[(d + e*x^2)^(3/2)*Sqrt[d^2 - e^2*x^4],x]
 

Output:

((x*Sqrt[d^2 - e^2*x^4]*(9*d^2 + 22*d*e*x^2 + 8*e^2*x^4))/Sqrt[d + e*x^2] 
+ ((39*I)*d^3*Log[(-2*I)*Sqrt[e]*x + (2*Sqrt[d^2 - e^2*x^4])/Sqrt[d + e*x^ 
2]])/Sqrt[e])/48
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.90, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1396, 318, 25, 27, 299, 211, 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.

\(\displaystyle \int \left (d+e x^2\right )^{3/2} \sqrt {d^2-e^2 x^4} \, dx\)

\(\Big \downarrow \) 1396

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

\(\Big \downarrow \) 318

\(\displaystyle \frac {\sqrt {d^2-e^2 x^4} \left (-\frac {\int -d e \sqrt {d-e x^2} \left (11 e x^2+7 d\right )dx}{6 e}-\frac {1}{6} x \left (d+e x^2\right ) \left (d-e x^2\right )^{3/2}\right )}{\sqrt {d-e x^2} \sqrt {d+e x^2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sqrt {d^2-e^2 x^4} \left (\frac {\int d e \sqrt {d-e x^2} \left (11 e x^2+7 d\right )dx}{6 e}-\frac {1}{6} x \left (d-e x^2\right )^{3/2} \left (d+e x^2\right )\right )}{\sqrt {d-e x^2} \sqrt {d+e x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {d^2-e^2 x^4} \left (\frac {1}{6} d \int \sqrt {d-e x^2} \left (11 e x^2+7 d\right )dx-\frac {1}{6} x \left (d-e x^2\right )^{3/2} \left (d+e x^2\right )\right )}{\sqrt {d-e x^2} \sqrt {d+e x^2}}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\sqrt {d^2-e^2 x^4} \left (\frac {1}{6} d \left (\frac {39}{4} d \int \sqrt {d-e x^2}dx-\frac {11}{4} x \left (d-e x^2\right )^{3/2}\right )-\frac {1}{6} x \left (d-e x^2\right )^{3/2} \left (d+e x^2\right )\right )}{\sqrt {d-e x^2} \sqrt {d+e x^2}}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\sqrt {d^2-e^2 x^4} \left (\frac {1}{6} d \left (\frac {39}{4} d \left (\frac {1}{2} d \int \frac {1}{\sqrt {d-e x^2}}dx+\frac {1}{2} x \sqrt {d-e x^2}\right )-\frac {11}{4} x \left (d-e x^2\right )^{3/2}\right )-\frac {1}{6} x \left (d-e x^2\right )^{3/2} \left (d+e x^2\right )\right )}{\sqrt {d-e x^2} \sqrt {d+e x^2}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\sqrt {d^2-e^2 x^4} \left (\frac {1}{6} d \left (\frac {39}{4} d \left (\frac {1}{2} d \int \frac {1}{\frac {e x^2}{d-e x^2}+1}d\frac {x}{\sqrt {d-e x^2}}+\frac {1}{2} x \sqrt {d-e x^2}\right )-\frac {11}{4} x \left (d-e x^2\right )^{3/2}\right )-\frac {1}{6} x \left (d-e x^2\right )^{3/2} \left (d+e x^2\right )\right )}{\sqrt {d-e x^2} \sqrt {d+e x^2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\sqrt {d^2-e^2 x^4} \left (\frac {1}{6} d \left (\frac {39}{4} d \left (\frac {d \arctan \left (\frac {\sqrt {e} x}{\sqrt {d-e x^2}}\right )}{2 \sqrt {e}}+\frac {1}{2} x \sqrt {d-e x^2}\right )-\frac {11}{4} x \left (d-e x^2\right )^{3/2}\right )-\frac {1}{6} x \left (d-e x^2\right )^{3/2} \left (d+e x^2\right )\right )}{\sqrt {d-e x^2} \sqrt {d+e x^2}}\)

Input:

Int[(d + e*x^2)^(3/2)*Sqrt[d^2 - e^2*x^4],x]
 

Output:

(Sqrt[d^2 - e^2*x^4]*(-1/6*(x*(d - e*x^2)^(3/2)*(d + e*x^2)) + (d*((-11*x* 
(d - e*x^2)^(3/2))/4 + (39*d*((x*Sqrt[d - e*x^2])/2 + (d*ArcTan[(Sqrt[e]*x 
)/Sqrt[d - e*x^2]])/(2*Sqrt[e])))/4))/6))/(Sqrt[d - e*x^2]*Sqrt[d + e*x^2] 
)
 

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 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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])
 

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.75

method	result	size
default	\(\frac {\sqrt {-e^{2} x^{4}+d^{2}}\, \left (8 e^{\frac {5}{2}} x^{5} \sqrt {-e \,x^{2}+d}+22 d \,e^{\frac {3}{2}} x^{3} \sqrt {-e \,x^{2}+d}+9 \sqrt {-e \,x^{2}+d}\, \sqrt {e}\, d^{2} x +39 \arctan \left (\frac {\sqrt {e}\, x}{\sqrt {-e \,x^{2}+d}}\right ) d^{3}\right )}{48 \sqrt {e \,x^{2}+d}\, \sqrt {-e \,x^{2}+d}\, \sqrt {e}}\)	\(117\)
risch	\(\frac {x \left (8 e^{2} x^{4}+22 d e \,x^{2}+9 d^{2}\right ) \sqrt {-e \,x^{2}+d}\, \sqrt {\frac {-e^{2} x^{4}+d^{2}}{e \,x^{2}+d}}\, \sqrt {e \,x^{2}+d}}{48 \sqrt {-e^{2} x^{4}+d^{2}}}+\frac {13 d^{3} \arctan \left (\frac {\sqrt {e}\, x}{\sqrt {-e \,x^{2}+d}}\right ) \sqrt {\frac {-e^{2} x^{4}+d^{2}}{e \,x^{2}+d}}\, \sqrt {e \,x^{2}+d}}{16 \sqrt {e}\, \sqrt {-e^{2} x^{4}+d^{2}}}\)	\(154\)

Input:

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

Output:

1/48*(-e^2*x^4+d^2)^(1/2)*(8*e^(5/2)*x^5*(-e*x^2+d)^(1/2)+22*d*e^(3/2)*x^3 
*(-e*x^2+d)^(1/2)+9*(-e*x^2+d)^(1/2)*e^(1/2)*d^2*x+39*arctan(e^(1/2)*x/(-e 
*x^2+d)^(1/2))*d^3)/(e*x^2+d)^(1/2)/(-e*x^2+d)^(1/2)/e^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.77 \[ \int \left (d+e x^2\right )^{3/2} \sqrt {d^2-e^2 x^4} \, dx=\left [-\frac {39 \, {\left (d^{3} e x^{2} + d^{4}\right )} \sqrt {-e} \log \left (-\frac {2 \, e^{2} x^{4} + d e x^{2} - 2 \, \sqrt {-e^{2} x^{4} + d^{2}} \sqrt {e x^{2} + d} \sqrt {-e} x - d^{2}}{e x^{2} + d}\right ) - 2 \, {\left (8 \, e^{3} x^{5} + 22 \, d e^{2} x^{3} + 9 \, d^{2} e x\right )} \sqrt {-e^{2} x^{4} + d^{2}} \sqrt {e x^{2} + d}}{96 \, {\left (e^{2} x^{2} + d e\right )}}, -\frac {39 \, {\left (d^{3} e x^{2} + d^{4}\right )} \sqrt {e} \arctan \left (\frac {\sqrt {-e^{2} x^{4} + d^{2}} \sqrt {e x^{2} + d} \sqrt {e} x}{e^{2} x^{4} - d^{2}}\right ) - {\left (8 \, e^{3} x^{5} + 22 \, d e^{2} x^{3} + 9 \, d^{2} e x\right )} \sqrt {-e^{2} x^{4} + d^{2}} \sqrt {e x^{2} + d}}{48 \, {\left (e^{2} x^{2} + d e\right )}}\right ] \] Input:

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

Output:

[-1/96*(39*(d^3*e*x^2 + d^4)*sqrt(-e)*log(-(2*e^2*x^4 + d*e*x^2 - 2*sqrt(- 
e^2*x^4 + d^2)*sqrt(e*x^2 + d)*sqrt(-e)*x - d^2)/(e*x^2 + d)) - 2*(8*e^3*x 
^5 + 22*d*e^2*x^3 + 9*d^2*e*x)*sqrt(-e^2*x^4 + d^2)*sqrt(e*x^2 + d))/(e^2* 
x^2 + d*e), -1/48*(39*(d^3*e*x^2 + d^4)*sqrt(e)*arctan(sqrt(-e^2*x^4 + d^2 
)*sqrt(e*x^2 + d)*sqrt(e)*x/(e^2*x^4 - d^2)) - (8*e^3*x^5 + 22*d*e^2*x^3 + 
 9*d^2*e*x)*sqrt(-e^2*x^4 + d^2)*sqrt(e*x^2 + d))/(e^2*x^2 + d*e)]
 

Sympy [F]

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

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

Output:

Integral(sqrt(-(-d + e*x**2)*(d + e*x**2))*(d + e*x**2)**(3/2), x)
 

Maxima [F]

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

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

Output:

integrate(sqrt(-e^2*x^4 + d^2)*(e*x^2 + d)^(3/2), x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.44 \[ \int \left (d+e x^2\right )^{3/2} \sqrt {d^2-e^2 x^4} \, dx=-\frac {13 \, d^{3} \log \left ({\left | -\sqrt {-e} x + \sqrt {-e x^{2} + d} \right |}\right )}{16 \, \sqrt {-e}} + \frac {1}{48} \, {\left (2 \, {\left (4 \, e^{2} x^{2} + 11 \, d e\right )} x^{2} + 9 \, d^{2}\right )} \sqrt {-e x^{2} + d} x \] Input:

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

Output:

-13/16*d^3*log(abs(-sqrt(-e)*x + sqrt(-e*x^2 + d)))/sqrt(-e) + 1/48*(2*(4* 
e^2*x^2 + 11*d*e)*x^2 + 9*d^2)*sqrt(-e*x^2 + d)*x
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.47 \[ \int \left (d+e x^2\right )^{3/2} \sqrt {d^2-e^2 x^4} \, dx=\frac {39 \sqrt {e}\, \mathit {asin} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right ) d^{3}+9 \sqrt {-e \,x^{2}+d}\, d^{2} e x +22 \sqrt {-e \,x^{2}+d}\, d \,e^{2} x^{3}+8 \sqrt {-e \,x^{2}+d}\, e^{3} x^{5}}{48 e} \] Input:

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

Output:

(39*sqrt(e)*asin((sqrt(e)*x)/sqrt(d))*d**3 + 9*sqrt(d - e*x**2)*d**2*e*x + 
 22*sqrt(d - e*x**2)*d*e**2*x**3 + 8*sqrt(d - e*x**2)*e**3*x**5)/(48*e)