\(\int \frac {(d+e x)^6}{(d^2-e^2 x^2)^{5/2}} \, dx\) [836]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 24, antiderivative size = 143 \[ \int \frac {(d+e x)^6}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {14 (d+e x)^3}{3 e \sqrt {d^2-e^2 x^2}}-\frac {35 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {35 (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}+\frac {35 d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \]

[Out]

2/3*(e*x+d)^5/e/(-e^2*x^2+d^2)^(3/2)+35/2*d^2*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e-14/3*(e*x+d)^3/e/(-e^2*x^2+d^
2)^(1/2)-35/2*d*(-e^2*x^2+d^2)^(1/2)/e-35/6*(e*x+d)*(-e^2*x^2+d^2)^(1/2)/e

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {683, 685, 655, 223, 209} \[ \int \frac {(d+e x)^6}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {35 d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}+\frac {2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {14 (d+e x)^3}{3 e \sqrt {d^2-e^2 x^2}}-\frac {35 \sqrt {d^2-e^2 x^2} (d+e x)}{6 e}-\frac {35 d \sqrt {d^2-e^2 x^2}}{2 e} \]

[In]

Int[(d + e*x)^6/(d^2 - e^2*x^2)^(5/2),x]

[Out]

(2*(d + e*x)^5)/(3*e*(d^2 - e^2*x^2)^(3/2)) - (14*(d + e*x)^3)/(3*e*Sqrt[d^2 - e^2*x^2]) - (35*d*Sqrt[d^2 - e^
2*x^2])/(2*e) - (35*(d + e*x)*Sqrt[d^2 - e^2*x^2])/(6*e) + (35*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e)

Rule 209

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

Rule 223

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 655

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

Rule 683

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*((a + c*x^2)^(p
 + 1)/(c*(p + 1))), x] - Dist[e^2*((m + p)/(c*(p + 1))), Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rule 685

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {7}{3} \int \frac {(d+e x)^4}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx \\ & = \frac {2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {14 (d+e x)^3}{3 e \sqrt {d^2-e^2 x^2}}+\frac {35}{3} \int \frac {(d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx \\ & = \frac {2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {14 (d+e x)^3}{3 e \sqrt {d^2-e^2 x^2}}-\frac {35 (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}+\frac {1}{2} (35 d) \int \frac {d+e x}{\sqrt {d^2-e^2 x^2}} \, dx \\ & = \frac {2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {14 (d+e x)^3}{3 e \sqrt {d^2-e^2 x^2}}-\frac {35 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {35 (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}+\frac {1}{2} \left (35 d^2\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx \\ & = \frac {2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {14 (d+e x)^3}{3 e \sqrt {d^2-e^2 x^2}}-\frac {35 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {35 (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}+\frac {1}{2} \left (35 d^2\right ) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right ) \\ & = \frac {2 (d+e x)^5}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {14 (d+e x)^3}{3 e \sqrt {d^2-e^2 x^2}}-\frac {35 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {35 (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}+\frac {35 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.70 \[ \int \frac {(d+e x)^6}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {\frac {\sqrt {d^2-e^2 x^2} \left (164 d^3-229 d^2 e x+30 d e^2 x^2+3 e^3 x^3\right )}{(d-e x)^2}+210 d^2 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{6 e} \]

[In]

Integrate[(d + e*x)^6/(d^2 - e^2*x^2)^(5/2),x]

[Out]

-1/6*((Sqrt[d^2 - e^2*x^2]*(164*d^3 - 229*d^2*e*x + 30*d*e^2*x^2 + 3*e^3*x^3))/(d - e*x)^2 + 210*d^2*ArcTan[(e
*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/e

Maple [A] (verified)

Time = 2.34 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.09

method result size
risch \(-\frac {\left (e x +12 d \right ) \sqrt {-x^{2} e^{2}+d^{2}}}{2 e}+\frac {35 d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-x^{2} e^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}+\frac {80 d^{2} \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 \left (x -\frac {d}{e}\right ) d e}}{3 e^{2} \left (x -\frac {d}{e}\right )}+\frac {16 d^{3} \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 \left (x -\frac {d}{e}\right ) d e}}{3 e^{3} \left (x -\frac {d}{e}\right )^{2}}\) \(156\)
default \(d^{6} \left (\frac {x}{3 d^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-x^{2} e^{2}+d^{2}}}\right )+e^{6} \left (-\frac {x^{5}}{2 e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {5 d^{2} \left (\frac {x^{3}}{3 e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{e^{2} \sqrt {-x^{2} e^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-x^{2} e^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}\right )}{2 e^{2}}\right )+6 d \,e^{5} \left (-\frac {x^{4}}{e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {4 d^{2} \left (\frac {x^{2}}{e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {2 d^{2}}{3 e^{4} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}\right )}{e^{2}}\right )+\frac {2 d^{5}}{e \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}+15 d^{2} e^{4} \left (\frac {x^{3}}{3 e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{e^{2} \sqrt {-x^{2} e^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-x^{2} e^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}\right )+20 d^{3} e^{3} \left (\frac {x^{2}}{e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {2 d^{2}}{3 e^{4} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}\right )+15 d^{4} e^{2} \left (\frac {x}{2 e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {d^{2} \left (\frac {x}{3 d^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-x^{2} e^{2}+d^{2}}}\right )}{2 e^{2}}\right )\) \(483\)

[In]

int((e*x+d)^6/(-e^2*x^2+d^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*(e*x+12*d)/e*(-e^2*x^2+d^2)^(1/2)+35/2*d^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))+80/3*d^
2/e^2/(x-d/e)*(-(x-d/e)^2*e^2-2*(x-d/e)*d*e)^(1/2)+16/3*d^3/e^3/(x-d/e)^2*(-(x-d/e)^2*e^2-2*(x-d/e)*d*e)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x)^6}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {164 \, d^{2} e^{2} x^{2} - 328 \, d^{3} e x + 164 \, d^{4} + 210 \, {\left (d^{2} e^{2} x^{2} - 2 \, d^{3} e x + d^{4}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (3 \, e^{3} x^{3} + 30 \, d e^{2} x^{2} - 229 \, d^{2} e x + 164 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, {\left (e^{3} x^{2} - 2 \, d e^{2} x + d^{2} e\right )}} \]

[In]

integrate((e*x+d)^6/(-e^2*x^2+d^2)^(5/2),x, algorithm="fricas")

[Out]

-1/6*(164*d^2*e^2*x^2 - 328*d^3*e*x + 164*d^4 + 210*(d^2*e^2*x^2 - 2*d^3*e*x + d^4)*arctan(-(d - sqrt(-e^2*x^2
 + d^2))/(e*x)) + (3*e^3*x^3 + 30*d*e^2*x^2 - 229*d^2*e*x + 164*d^3)*sqrt(-e^2*x^2 + d^2))/(e^3*x^2 - 2*d*e^2*
x + d^2*e)

Sympy [F]

\[ \int \frac {(d+e x)^6}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (d + e x\right )^{6}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((e*x+d)**6/(-e**2*x**2+d**2)**(5/2),x)

[Out]

Integral((d + e*x)**6/(-(-d + e*x)*(d + e*x))**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.46 \[ \int \frac {(d+e x)^6}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {35}{6} \, d^{2} e^{4} x {\left (\frac {3 \, x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} - \frac {2 \, d^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}}\right )} - \frac {e^{4} x^{5}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} - \frac {6 \, d e^{3} x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {44 \, d^{3} e x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {16 \, d^{4} x}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} - \frac {82 \, d^{5}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e} - \frac {61 \, d^{2} x}{6 \, \sqrt {-e^{2} x^{2} + d^{2}}} + \frac {35 \, d^{2} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{2 \, \sqrt {e^{2}}} \]

[In]

integrate((e*x+d)^6/(-e^2*x^2+d^2)^(5/2),x, algorithm="maxima")

[Out]

35/6*d^2*e^4*x*(3*x^2/((-e^2*x^2 + d^2)^(3/2)*e^2) - 2*d^2/((-e^2*x^2 + d^2)^(3/2)*e^4)) - 1/2*e^4*x^5/(-e^2*x
^2 + d^2)^(3/2) - 6*d*e^3*x^4/(-e^2*x^2 + d^2)^(3/2) + 44*d^3*e*x^2/(-e^2*x^2 + d^2)^(3/2) + 16/3*d^4*x/(-e^2*
x^2 + d^2)^(3/2) - 82/3*d^5/((-e^2*x^2 + d^2)^(3/2)*e) - 61/6*d^2*x/sqrt(-e^2*x^2 + d^2) + 35/2*d^2*arcsin(e^2
*x/(d*sqrt(e^2)))/sqrt(e^2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.08 \[ \int \frac {(d+e x)^6}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {35 \, d^{2} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{2 \, {\left | e \right |}} - \frac {1}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (x + \frac {12 \, d}{e}\right )} - \frac {32 \, {\left (4 \, d^{2} - \frac {9 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d^{2}}{e^{2} x} + \frac {3 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d^{2}}{e^{4} x^{2}}\right )}}{3 \, {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} - 1\right )}^{3} {\left | e \right |}} \]

[In]

integrate((e*x+d)^6/(-e^2*x^2+d^2)^(5/2),x, algorithm="giac")

[Out]

35/2*d^2*arcsin(e*x/d)*sgn(d)*sgn(e)/abs(e) - 1/2*sqrt(-e^2*x^2 + d^2)*(x + 12*d/e) - 32/3*(4*d^2 - 9*(d*e + s
qrt(-e^2*x^2 + d^2)*abs(e))*d^2/(e^2*x) + 3*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2*d^2/(e^4*x^2))/(((d*e + sqrt
(-e^2*x^2 + d^2)*abs(e))/(e^2*x) - 1)^3*abs(e))

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^6}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^6}{{\left (d^2-e^2\,x^2\right )}^{5/2}} \,d x \]

[In]

int((d + e*x)^6/(d^2 - e^2*x^2)^(5/2),x)

[Out]

int((d + e*x)^6/(d^2 - e^2*x^2)^(5/2), x)