\(\int \frac {(d^2-e^2 x^2)^{7/2}}{(d+e x)^4} \, dx\) [66]

Optimal result

Integrand size = 24, antiderivative size = 114 \[ \int \frac {\left (d^2-e^2 x^2\right )^{7/2}}{(d+e x)^4} \, dx=\frac {35}{8} d^2 x \sqrt {d^2-e^2 x^2}+\frac {7 (8 d-3 e x) \left (d^2-e^2 x^2\right )^{3/2}}{12 e}+\frac {2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^3}+\frac {35 d^4 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e} \] Output:

35/8*d^2*x*(-e^2*x^2+d^2)^(1/2)+7/12*(-3*e*x+8*d)*(-e^2*x^2+d^2)^(3/2)/e+2 
*(-e^2*x^2+d^2)^(7/2)/e/(e*x+d)^3+35/8*d^4*arctan(e*x/(-e^2*x^2+d^2)^(1/2) 
)/e
 

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.81 \[ \int \frac {\left (d^2-e^2 x^2\right )^{7/2}}{(d+e x)^4} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (160 d^3-81 d^2 e x+32 d e^2 x^2-6 e^3 x^3\right )-210 d^4 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{24 e} \] Input:

Integrate[(d^2 - e^2*x^2)^(7/2)/(d + e*x)^4,x]
 

Output:

(Sqrt[d^2 - e^2*x^2]*(160*d^3 - 81*d^2*e*x + 32*d*e^2*x^2 - 6*e^3*x^3) - 2 
10*d^4*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/(24*e)
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.26, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {465, 464, 469, 455, 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.

\(\displaystyle \int \frac {\left (d^2-e^2 x^2\right )^{7/2}}{(d+e x)^4} \, dx\)

\(\Big \downarrow \) 465

\(\displaystyle 7 \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{(d+e x)^2}dx+\frac {2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^3}\)

\(\Big \downarrow \) 464

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

\(\Big \downarrow \) 469

\(\displaystyle 7 \left (\frac {5}{4} d \int (d-e x) \sqrt {d^2-e^2 x^2}dx+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}\right )+\frac {2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^3}\)

\(\Big \downarrow \) 455

\(\displaystyle 7 \left (\frac {5}{4} d \left (d \int \sqrt {d^2-e^2 x^2}dx+\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e}\right )+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}\right )+\frac {2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^3}\)

\(\Big \downarrow \) 211

\(\displaystyle 7 \left (\frac {5}{4} d \left (d \left (\frac {1}{2} d^2 \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )+\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e}\right )+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}\right )+\frac {2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^3}\)

\(\Big \downarrow \) 224

\(\displaystyle 7 \left (\frac {5}{4} d \left (d \left (\frac {1}{2} d^2 \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )+\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e}\right )+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}\right )+\frac {2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^3}\)

\(\Big \downarrow \) 216

\(\displaystyle 7 \left (\frac {5}{4} d \left (d \left (\frac {d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )+\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e}\right )+\frac {(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}{4 e}\right )+\frac {2 \left (d^2-e^2 x^2\right )^{7/2}}{e (d+e x)^3}\)

Input:

Int[(d^2 - e^2*x^2)^(7/2)/(d + e*x)^4,x]
 

Output:

(2*(d^2 - e^2*x^2)^(7/2))/(e*(d + e*x)^3) + 7*(((d - e*x)*(d^2 - e^2*x^2)^ 
(3/2))/(4*e) + (5*d*((d^2 - e^2*x^2)^(3/2)/(3*e) + d*((x*Sqrt[d^2 - e^2*x^ 
2])/2 + (d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e))))/4)
 

Defintions of rubi rules used

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 455

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

rule 464

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

rule 465

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

rule 469

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

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.73

method	result	size
risch	\(\frac {\left (-6 e^{3} x^{3}+32 d \,e^{2} x^{2}-81 d^{2} e x +160 d^{3}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{24 e}+\frac {35 d^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}}\)	\(83\)
default	\(\frac {\frac {\left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {9}{2}}}{d e \left (x +\frac {d}{e}\right )^{4}}+\frac {5 e \left (\frac {\left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {9}{2}}}{3 d e \left (x +\frac {d}{e}\right )^{3}}+\frac {2 e \left (\frac {\left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {9}{2}}}{5 d e \left (x +\frac {d}{e}\right )^{2}}+\frac {7 e \left (\frac {\left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{7}+d e \left (-\frac {\left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right ) \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{12 e^{2}}+\frac {5 d^{2} \left (-\frac {\left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right ) \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right ) \sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )}{6}\right )\right )}{5 d}\right )}{d}\right )}{d}}{e^{4}}\)	\(403\)

Input:

int((-e^2*x^2+d^2)^(7/2)/(e*x+d)^4,x,method=_RETURNVERBOSE)
 

Output:

1/24*(-6*e^3*x^3+32*d*e^2*x^2-81*d^2*e*x+160*d^3)/e*(-e^2*x^2+d^2)^(1/2)+3 
5/8*d^4/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.73 \[ \int \frac {\left (d^2-e^2 x^2\right )^{7/2}}{(d+e x)^4} \, dx=-\frac {210 \, d^{4} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (6 \, e^{3} x^{3} - 32 \, d e^{2} x^{2} + 81 \, d^{2} e x - 160 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, e} \] Input:

integrate((-e^2*x^2+d^2)^(7/2)/(e*x+d)^4,x, algorithm="fricas")
 

Output:

-1/24*(210*d^4*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (6*e^3*x^3 - 32 
*d*e^2*x^2 + 81*d^2*e*x - 160*d^3)*sqrt(-e^2*x^2 + d^2))/e
 

Sympy [F]

\[ \int \frac {\left (d^2-e^2 x^2\right )^{7/2}}{(d+e x)^4} \, dx=\int \frac {\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}{\left (d + e x\right )^{4}}\, dx \] Input:

integrate((-e**2*x**2+d**2)**(7/2)/(e*x+d)**4,x)
 

Output:

Integral((-(-d + e*x)*(d + e*x))**(7/2)/(d + e*x)**4, x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.37 \[ \int \frac {\left (d^2-e^2 x^2\right )^{7/2}}{(d+e x)^4} \, dx=\frac {35 \, d^{4} \arcsin \left (\frac {e x}{d}\right )}{8 \, e} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}}{4 \, {\left (e^{4} x^{3} + 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x + d^{3} e\right )}} + \frac {7 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{12 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} + \frac {35 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}}{24 \, {\left (e^{2} x + d e\right )}} + \frac {35 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{8 \, e} \] Input:

integrate((-e^2*x^2+d^2)^(7/2)/(e*x+d)^4,x, algorithm="maxima")
 

Output:

35/8*d^4*arcsin(e*x/d)/e + 1/4*(-e^2*x^2 + d^2)^(7/2)/(e^4*x^3 + 3*d*e^3*x 
^2 + 3*d^2*e^2*x + d^3*e) + 7/12*(-e^2*x^2 + d^2)^(5/2)*d/(e^3*x^2 + 2*d*e 
^2*x + d^2*e) + 35/24*(-e^2*x^2 + d^2)^(3/2)*d^2/(e^2*x + d*e) + 35/8*sqrt 
(-e^2*x^2 + d^2)*d^3/e
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.61 \[ \int \frac {\left (d^2-e^2 x^2\right )^{7/2}}{(d+e x)^4} \, dx=\frac {35 \, d^{4} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{8 \, {\left | e \right |}} + \frac {1}{24} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (\frac {160 \, d^{3}}{e} - {\left (81 \, d^{2} + 2 \, {\left (3 \, e^{2} x - 16 \, d e\right )} x\right )} x\right )} \] Input:

integrate((-e^2*x^2+d^2)^(7/2)/(e*x+d)^4,x, algorithm="giac")
 

Output:

35/8*d^4*arcsin(e*x/d)*sgn(d)*sgn(e)/abs(e) + 1/24*sqrt(-e^2*x^2 + d^2)*(1 
60*d^3/e - (81*d^2 + 2*(3*e^2*x - 16*d*e)*x)*x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d^2-e^2 x^2\right )^{7/2}}{(d+e x)^4} \, dx=\int \frac {{\left (d^2-e^2\,x^2\right )}^{7/2}}{{\left (d+e\,x\right )}^4} \,d x \] Input:

int((d^2 - e^2*x^2)^(7/2)/(d + e*x)^4,x)
 

Output:

int((d^2 - e^2*x^2)^(7/2)/(d + e*x)^4, x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.91 \[ \int \frac {\left (d^2-e^2 x^2\right )^{7/2}}{(d+e x)^4} \, dx=\frac {105 \mathit {asin} \left (\frac {e x}{d}\right ) d^{4}+160 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{3}-81 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{2} e x +32 \sqrt {-e^{2} x^{2}+d^{2}}\, d \,e^{2} x^{2}-6 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{3} x^{3}-160 d^{4}}{24 e} \] Input:

int((-e^2*x^2+d^2)^(7/2)/(e*x+d)^4,x)
 

Output:

(105*asin((e*x)/d)*d**4 + 160*sqrt(d**2 - e**2*x**2)*d**3 - 81*sqrt(d**2 - 
 e**2*x**2)*d**2*e*x + 32*sqrt(d**2 - e**2*x**2)*d*e**2*x**2 - 6*sqrt(d**2 
 - e**2*x**2)*e**3*x**3 - 160*d**4)/(24*e)