\(\int \frac {(d+e x)^5}{(d^2-e^2 x^2)^{3/2}} \, dx\) [119]

Optimal result

Integrand size = 24, antiderivative size = 129 \[ \int \frac {(d+e x)^5}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {16 d^3 (d+e x)}{e \sqrt {d^2-e^2 x^2}}+\frac {12 d^2 \sqrt {d^2-e^2 x^2}}{e}+\frac {5}{2} d x \sqrt {d^2-e^2 x^2}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{3 e}-\frac {35 d^3 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \] Output:

16*d^3*(e*x+d)/e/(-e^2*x^2+d^2)^(1/2)+12*d^2*(-e^2*x^2+d^2)^(1/2)/e+5/2*d* 
x*(-e^2*x^2+d^2)^(1/2)-1/3*(-e^2*x^2+d^2)^(3/2)/e-35/2*d^3*arctan(e*x/(-e^ 
2*x^2+d^2)^(1/2))/e
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.84 \[ \int \frac {(d+e x)^5}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {\sqrt {d^2-e^2 x^2} \left (-166 d^3+55 d^2 e x+13 d e^2 x^2+2 e^3 x^3\right )}{6 e (d-e x)}+\frac {35 d^3 \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{2 \sqrt {-e^2}} \] Input:

Integrate[(d + e*x)^5/(d^2 - e^2*x^2)^(3/2),x]
 

Output:

-1/6*(Sqrt[d^2 - e^2*x^2]*(-166*d^3 + 55*d^2*e*x + 13*d*e^2*x^2 + 2*e^3*x^ 
3))/(e*(d - e*x)) + (35*d^3*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(2 
*Sqrt[-e^2])
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {462, 2346, 27, 2346, 27, 455, 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.

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

\(\Big \downarrow \) 462

\(\displaystyle \frac {16 d^3 (d+e x)}{e \sqrt {d^2-e^2 x^2}}-\int \frac {15 d^3+11 e x d^2+5 e^2 x^2 d+e^3 x^3}{\sqrt {d^2-e^2 x^2}}dx\)

\(\Big \downarrow \) 2346

\(\displaystyle \frac {\int -\frac {5 \left (3 d x^2 e^4+7 d^2 x e^3+9 d^3 e^2\right )}{\sqrt {d^2-e^2 x^2}}dx}{3 e^2}+\frac {1}{3} e x^2 \sqrt {d^2-e^2 x^2}+\frac {16 d^3 (d+e x)}{e \sqrt {d^2-e^2 x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {5 \int \frac {3 d x^2 e^4+7 d^2 x e^3+9 d^3 e^2}{\sqrt {d^2-e^2 x^2}}dx}{3 e^2}+\frac {1}{3} e x^2 \sqrt {d^2-e^2 x^2}+\frac {16 d^3 (d+e x)}{e \sqrt {d^2-e^2 x^2}}\)

\(\Big \downarrow \) 2346

\(\displaystyle -\frac {5 \left (-\frac {\int -\frac {7 d^2 e^4 (3 d+2 e x)}{\sqrt {d^2-e^2 x^2}}dx}{2 e^2}-\frac {3}{2} d e^2 x \sqrt {d^2-e^2 x^2}\right )}{3 e^2}+\frac {1}{3} e x^2 \sqrt {d^2-e^2 x^2}+\frac {16 d^3 (d+e x)}{e \sqrt {d^2-e^2 x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {5 \left (\frac {7}{2} d^2 e^2 \int \frac {3 d+2 e x}{\sqrt {d^2-e^2 x^2}}dx-\frac {3}{2} d e^2 x \sqrt {d^2-e^2 x^2}\right )}{3 e^2}+\frac {1}{3} e x^2 \sqrt {d^2-e^2 x^2}+\frac {16 d^3 (d+e x)}{e \sqrt {d^2-e^2 x^2}}\)

\(\Big \downarrow \) 455

\(\displaystyle -\frac {5 \left (\frac {7}{2} d^2 e^2 \left (3 d \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx-\frac {2 \sqrt {d^2-e^2 x^2}}{e}\right )-\frac {3}{2} d e^2 x \sqrt {d^2-e^2 x^2}\right )}{3 e^2}+\frac {1}{3} e x^2 \sqrt {d^2-e^2 x^2}+\frac {16 d^3 (d+e x)}{e \sqrt {d^2-e^2 x^2}}\)

\(\Big \downarrow \) 224

\(\displaystyle -\frac {5 \left (\frac {7}{2} d^2 e^2 \left (3 d \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}-\frac {2 \sqrt {d^2-e^2 x^2}}{e}\right )-\frac {3}{2} d e^2 x \sqrt {d^2-e^2 x^2}\right )}{3 e^2}+\frac {1}{3} e x^2 \sqrt {d^2-e^2 x^2}+\frac {16 d^3 (d+e x)}{e \sqrt {d^2-e^2 x^2}}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {5 \left (\frac {7}{2} d^2 e^2 \left (\frac {3 d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}-\frac {2 \sqrt {d^2-e^2 x^2}}{e}\right )-\frac {3}{2} d e^2 x \sqrt {d^2-e^2 x^2}\right )}{3 e^2}+\frac {1}{3} e x^2 \sqrt {d^2-e^2 x^2}+\frac {16 d^3 (d+e x)}{e \sqrt {d^2-e^2 x^2}}\)

Input:

Int[(d + e*x)^5/(d^2 - e^2*x^2)^(3/2),x]
 

Output:

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

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 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 462

Int[((c_) + (d_.)*(x_))^(n_)/((a_) + (b_.)*(x_)^2)^(3/2), x_Symbol] :> Simp 
[(-2^(n - 1))*d*c^(n - 2)*((c + d*x)/(b*Sqrt[a + b*x^2])), x] + Simp[d^2/b 
  Int[(1/Sqrt[a + b*x^2])*ExpandToSum[(2^(n - 1)*c^(n - 1) - (c + d*x)^(n - 
 1))/(c - d*x), x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 
0] && IGtQ[n, 2]
 

rule 2346

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], 
e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( 
q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1))   Int[(a + b*x^2)^p*ExpandToS 
um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], 
x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.93

method	result	size
risch	\(\frac {\left (2 e^{2} x^{2}+15 d e x +70 d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{6 e}-\frac {35 d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}-\frac {16 d^{3} \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{e^{2} \left (x -\frac {d}{e}\right )}\)	\(120\)
default	\(\frac {d^{3} x}{\sqrt {-e^{2} x^{2}+d^{2}}}+e^{5} \left (-\frac {x^{4}}{3 e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {4 d^{2} \left (-\frac {x^{2}}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {2 d^{2}}{e^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{3 e^{2}}\right )+5 d \,e^{4} \left (-\frac {x^{3}}{2 e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {3 d^{2} \left (\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}\right )}{2 e^{2}}\right )+10 d^{2} e^{3} \left (-\frac {x^{2}}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}+\frac {2 d^{2}}{e^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )+10 d^{3} e^{2} \left (\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}\right )+\frac {5 d^{4}}{e \sqrt {-e^{2} x^{2}+d^{2}}}\)	\(325\)

Input:

int((e*x+d)^5/(-e^2*x^2+d^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/6*(2*e^2*x^2+15*d*e*x+70*d^2)/e*(-e^2*x^2+d^2)^(1/2)-35/2*d^3/(e^2)^(1/2 
)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))-16*d^3/e^2/(x-d/e)*(-(x-d/e)^ 
2*e^2-2*d*e*(x-d/e))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.88 \[ \int \frac {(d+e x)^5}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {166 \, d^{3} e x - 166 \, d^{4} + 210 \, {\left (d^{3} e x - d^{4}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (2 \, e^{3} x^{3} + 13 \, d e^{2} x^{2} + 55 \, d^{2} e x - 166 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, {\left (e^{2} x - d e\right )}} \] Input:

integrate((e*x+d)^5/(-e^2*x^2+d^2)^(3/2),x, algorithm="fricas")
 

Output:

1/6*(166*d^3*e*x - 166*d^4 + 210*(d^3*e*x - d^4)*arctan(-(d - sqrt(-e^2*x^ 
2 + d^2))/(e*x)) + (2*e^3*x^3 + 13*d*e^2*x^2 + 55*d^2*e*x - 166*d^3)*sqrt( 
-e^2*x^2 + d^2))/(e^2*x - d*e)
 

Sympy [F]

\[ \int \frac {(d+e x)^5}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx=\int \frac {\left (d + e x\right )^{5}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((e*x+d)**5/(-e**2*x**2+d**2)**(3/2),x)
 

Output:

Integral((d + e*x)**5/(-(-d + e*x)*(d + e*x))**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.05 \[ \int \frac {(d+e x)^5}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {e^{3} x^{4}}{3 \, \sqrt {-e^{2} x^{2} + d^{2}}} - \frac {5 \, d e^{2} x^{3}}{2 \, \sqrt {-e^{2} x^{2} + d^{2}}} - \frac {34 \, d^{2} e x^{2}}{3 \, \sqrt {-e^{2} x^{2} + d^{2}}} + \frac {37 \, d^{3} x}{2 \, \sqrt {-e^{2} x^{2} + d^{2}}} - \frac {35 \, d^{3} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{2 \, \sqrt {e^{2}}} + \frac {83 \, d^{4}}{3 \, \sqrt {-e^{2} x^{2} + d^{2}} e} \] Input:

integrate((e*x+d)^5/(-e^2*x^2+d^2)^(3/2),x, algorithm="maxima")
 

Output:

-1/3*e^3*x^4/sqrt(-e^2*x^2 + d^2) - 5/2*d*e^2*x^3/sqrt(-e^2*x^2 + d^2) - 3 
4/3*d^2*e*x^2/sqrt(-e^2*x^2 + d^2) + 37/2*d^3*x/sqrt(-e^2*x^2 + d^2) - 35/ 
2*d^3*arcsin(e^2*x/(d*sqrt(e^2)))/sqrt(e^2) + 83/3*d^4/(sqrt(-e^2*x^2 + d^ 
2)*e)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.75 \[ \int \frac {(d+e x)^5}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx=-\frac {35 \, d^{3} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{2 \, {\left | e \right |}} + \frac {1}{6} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left ({\left (2 \, e x + 15 \, d\right )} x + \frac {70 \, d^{2}}{e}\right )} + \frac {32 \, d^{3}}{{\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} - 1\right )} {\left | e \right |}} \] Input:

integrate((e*x+d)^5/(-e^2*x^2+d^2)^(3/2),x, algorithm="giac")
 

Output:

-35/2*d^3*arcsin(e*x/d)*sgn(d)*sgn(e)/abs(e) + 1/6*sqrt(-e^2*x^2 + d^2)*(( 
2*e*x + 15*d)*x + 70*d^2/e) + 32*d^3/(((d*e + sqrt(-e^2*x^2 + d^2)*abs(e)) 
/(e^2*x) - 1)*abs(e))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^5}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^5}{{\left (d^2-e^2\,x^2\right )}^{3/2}} \,d x \] Input:

int((d + e*x)^5/(d^2 - e^2*x^2)^(3/2),x)
 

Output:

int((d + e*x)^5/(d^2 - e^2*x^2)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.55 \[ \int \frac {(d+e x)^5}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx=\frac {-105 \sqrt {-e^{2} x^{2}+d^{2}}\, \mathit {asin} \left (\frac {e x}{d}\right ) d^{3}+105 \mathit {asin} \left (\frac {e x}{d}\right ) d^{4}-105 \mathit {asin} \left (\frac {e x}{d}\right ) d^{3} e x -222 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{3}+55 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{2} e x +13 \sqrt {-e^{2} x^{2}+d^{2}}\, d \,e^{2} x^{2}+2 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{3} x^{3}+222 d^{4}+55 d^{3} e x -68 d^{2} e^{2} x^{2}-15 d \,e^{3} x^{3}-2 e^{4} x^{4}}{6 e \left (\sqrt {-e^{2} x^{2}+d^{2}}-d +e x \right )} \] Input:

int((e*x+d)^5/(-e^2*x^2+d^2)^(3/2),x)
 

Output:

( - 105*sqrt(d**2 - e**2*x**2)*asin((e*x)/d)*d**3 + 105*asin((e*x)/d)*d**4 
 - 105*asin((e*x)/d)*d**3*e*x - 222*sqrt(d**2 - e**2*x**2)*d**3 + 55*sqrt( 
d**2 - e**2*x**2)*d**2*e*x + 13*sqrt(d**2 - e**2*x**2)*d*e**2*x**2 + 2*sqr 
t(d**2 - e**2*x**2)*e**3*x**3 + 222*d**4 + 55*d**3*e*x - 68*d**2*e**2*x**2 
 - 15*d*e**3*x**3 - 2*e**4*x**4)/(6*e*(sqrt(d**2 - e**2*x**2) - d + e*x))