\(\int \frac {(d+e x^2)^{5/2}}{\sqrt {d^2-e^2 x^4}} \, dx\) [151]

Optimal result

Integrand size = 28, antiderivative size = 116 \[ \int \frac {\left (d+e x^2\right )^{5/2}}{\sqrt {d^2-e^2 x^4}} \, dx=-\frac {11 d x \sqrt {d^2-e^2 x^4}}{8 \sqrt {d+e x^2}}-\frac {e x^3 \sqrt {d^2-e^2 x^4}}{4 \sqrt {d+e x^2}}+\frac {19 d^2 \arctan \left (\frac {\sqrt {e} x \sqrt {d+e x^2}}{\sqrt {d^2-e^2 x^4}}\right )}{8 \sqrt {e}} \] Output:

-11/8*d*x*(-e^2*x^4+d^2)^(1/2)/(e*x^2+d)^(1/2)-1/4*e*x^3*(-e^2*x^4+d^2)^(1 
/2)/(e*x^2+d)^(1/2)+19/8*d^2*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 = 3.02 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.84 \[ \int \frac {\left (d+e x^2\right )^{5/2}}{\sqrt {d^2-e^2 x^4}} \, dx=-\frac {\left (11 d x+2 e x^3\right ) \sqrt {d^2-e^2 x^4}}{8 \sqrt {d+e x^2}}+\frac {19 i d^2 \log \left (-2 i \sqrt {e} x+\frac {2 \sqrt {d^2-e^2 x^4}}{\sqrt {d+e x^2}}\right )}{8 \sqrt {e}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.02, 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.

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

\(\Big \downarrow \) 1396

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

\(\Big \downarrow \) 318

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 299

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

Int[(d + e*x^2)^(5/2)/Sqrt[d^2 - e^2*x^4],x]
 

Output:

(Sqrt[d - e*x^2]*Sqrt[d + e*x^2]*(-1/4*(x*Sqrt[d - e*x^2]*(d + e*x^2)) + ( 
d*((-9*x*Sqrt[d - e*x^2])/2 + (19*d*ArcTan[(Sqrt[e]*x)/Sqrt[d - e*x^2]])/( 
2*Sqrt[e])))/4))/Sqrt[d^2 - e^2*x^4]
 

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])
 

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.83

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

Input:

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

Output:

1/8*(-e^2*x^4+d^2)^(1/2)*(-2*e^(3/2)*x^3*(-e*x^2+d)^(1/2)-11*e^(1/2)*(-e*x 
^2+d)^(1/2)*d*x+19*arctan(e^(1/2)*x/(-e*x^2+d)^(1/2))*d^2)/(e*x^2+d)^(1/2) 
/(-e*x^2+d)^(1/2)/e^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.18 \[ \int \frac {\left (d+e x^2\right )^{5/2}}{\sqrt {d^2-e^2 x^4}} \, dx=\left [-\frac {19 \, {\left (d^{2} e x^{2} + d^{3}\right )} \sqrt {-e} \log \left (-\frac {2 \, e^{2} x^{4} + d e x^{2} - 2 \, \sqrt {-e^{2} x^{4} + d^{2}} \sqrt {e x^{2} + d} \sqrt {-e} x - d^{2}}{e x^{2} + d}\right ) + 2 \, \sqrt {-e^{2} x^{4} + d^{2}} {\left (2 \, e^{2} x^{3} + 11 \, d e x\right )} \sqrt {e x^{2} + d}}{16 \, {\left (e^{2} x^{2} + d e\right )}}, -\frac {19 \, {\left (d^{2} e x^{2} + d^{3}\right )} \sqrt {e} \arctan \left (\frac {\sqrt {-e^{2} x^{4} + d^{2}} \sqrt {e x^{2} + d} \sqrt {e} x}{e^{2} x^{4} - d^{2}}\right ) + \sqrt {-e^{2} x^{4} + d^{2}} {\left (2 \, e^{2} x^{3} + 11 \, d e x\right )} \sqrt {e x^{2} + d}}{8 \, {\left (e^{2} x^{2} + d e\right )}}\right ] \] Input:

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

Output:

[-1/16*(19*(d^2*e*x^2 + d^3)*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*sqrt(-e^ 
2*x^4 + d^2)*(2*e^2*x^3 + 11*d*e*x)*sqrt(e*x^2 + d))/(e^2*x^2 + d*e), -1/8 
*(19*(d^2*e*x^2 + d^3)*sqrt(e)*arctan(sqrt(-e^2*x^4 + d^2)*sqrt(e*x^2 + d) 
*sqrt(e)*x/(e^2*x^4 - d^2)) + sqrt(-e^2*x^4 + d^2)*(2*e^2*x^3 + 11*d*e*x)* 
sqrt(e*x^2 + d))/(e^2*x^2 + d*e)]
 

Sympy [F]

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

integrate((e*x**2+d)**(5/2)/(-e**2*x**4+d**2)**(1/2),x)
 

Output:

Integral((d + e*x**2)**(5/2)/sqrt(-(-d + e*x**2)*(d + e*x**2)), x)
 

Maxima [F]

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

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

Output:

integrate((e*x^2 + d)^(5/2)/sqrt(-e^2*x^4 + d^2), x)
 

Giac [F]

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

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

Output:

integrate((e*x^2 + d)^(5/2)/sqrt(-e^2*x^4 + d^2), x)
 

Mupad [F(-1)]

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

int((d + e*x^2)^(5/2)/(d^2 - e^2*x^4)^(1/2),x)
 

Output:

int((d + e*x^2)^(5/2)/(d^2 - e^2*x^4)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.46 \[ \int \frac {\left (d+e x^2\right )^{5/2}}{\sqrt {d^2-e^2 x^4}} \, dx=\frac {19 \sqrt {e}\, \mathit {asin} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right ) d^{2}-11 \sqrt {-e \,x^{2}+d}\, d e x -2 \sqrt {-e \,x^{2}+d}\, e^{2} x^{3}}{8 e} \] Input:

int((e*x^2+d)^(5/2)/(-e^2*x^4+d^2)^(1/2),x)
 

Output:

(19*sqrt(e)*asin((sqrt(e)*x)/sqrt(d))*d**2 - 11*sqrt(d - e*x**2)*d*e*x - 2 
*sqrt(d - e*x**2)*e**2*x**3)/(8*e)