\(\int \frac {(d+e x)^6}{(d^2-e^2 x^2)^{7/2}} \, dx\) [849]

Optimal result

Integrand size = 24, antiderivative size = 112 \[ \int \frac {(d+e x)^6}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {2 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)^3}{3 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (d+e x)}{e \sqrt {d^2-e^2 x^2}}-\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {683, 667, 223, 209} \[ \int \frac {(d+e x)^6}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}+\frac {2 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)^3}{3 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (d+e x)}{e \sqrt {d^2-e^2 x^2}} \]

[In]

Int[(d + e*x)^6/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(2*(d + e*x)^5)/(5*e*(d^2 - e^2*x^2)^(5/2)) - (2*(d + e*x)^3)/(3*e*(d^2 - e^2*x^2)^(3/2)) + (2*(d + e*x))/(e*S
qrt[d^2 - e^2*x^2]) - ArcTan[(e*x)/Sqrt[d^2 - e^2*x^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 667

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

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

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.77 \[ \int \frac {(d+e x)^6}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {2 \left (\frac {\sqrt {d^2-e^2 x^2} \left (13 d^2-24 d e x+23 e^2 x^2\right )}{(d-e x)^3}+15 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )\right )}{15 e} \]

[In]

Integrate[(d + e*x)^6/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(2*((Sqrt[d^2 - e^2*x^2]*(13*d^2 - 24*d*e*x + 23*e^2*x^2))/(d - e*x)^3 + 15*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2
- e^2*x^2])]))/(15*e)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(579\) vs. \(2(100)=200\).

Time = 2.56 (sec) , antiderivative size = 580, normalized size of antiderivative = 5.18

method	result	size
default	\(d^{6} \left (\frac {x}{5 d^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-x^{2} e^{2}+d^{2}}}}{d^{2}}\right )+e^{6} \left (\frac {x^{5}}{5 e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {\frac {x^{3}}{3 e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{e^{2} \sqrt {-x^{2} e^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-x^{2} e^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}}{e^{2}}\right )+6 d \,e^{5} \left (\frac {x^{4}}{e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {5}{2}}}\right )}{e^{2}}\right )+\frac {6 d^{5}}{5 e \left (-x^{2} e^{2}+d^{2}\right )^{\frac {5}{2}}}+15 d^{2} e^{4} \left (\frac {x^{3}}{2 e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {3 d^{2} \left (\frac {x}{4 e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-x^{2} e^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )}{2 e^{2}}\right )+20 d^{3} e^{3} \left (\frac {x^{2}}{3 e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {5}{2}}}\right )+15 d^{4} e^{2} \left (\frac {x}{4 e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-x^{2} e^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )\)	\(580\)

[In]

int((e*x+d)^6/(-e^2*x^2+d^2)^(7/2),x,method=_RETURNVERBOSE)

[Out]

d^6*(1/5*x/d^2/(-e^2*x^2+d^2)^(5/2)+4/5/d^2*(1/3*x/d^2/(-e^2*x^2+d^2)^(3/2)+2/3*x/d^4/(-e^2*x^2+d^2)^(1/2)))+e
^6*(1/5*x^5/e^2/(-e^2*x^2+d^2)^(5/2)-1/e^2*(1/3*x^3/e^2/(-e^2*x^2+d^2)^(3/2)-1/e^2*(x/e^2/(-e^2*x^2+d^2)^(1/2)
-1/e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2)))))+6*d*e^5*(x^4/e^2/(-e^2*x^2+d^2)^(5/2)-4*d^2/e
^2*(1/3*x^2/e^2/(-e^2*x^2+d^2)^(5/2)-2/15*d^2/e^4/(-e^2*x^2+d^2)^(5/2)))+6/5*d^5/e/(-e^2*x^2+d^2)^(5/2)+15*d^2
*e^4*(1/2*x^3/e^2/(-e^2*x^2+d^2)^(5/2)-3/2*d^2/e^2*(1/4*x/e^2/(-e^2*x^2+d^2)^(5/2)-1/4*d^2/e^2*(1/5*x/d^2/(-e^
2*x^2+d^2)^(5/2)+4/5/d^2*(1/3*x/d^2/(-e^2*x^2+d^2)^(3/2)+2/3*x/d^4/(-e^2*x^2+d^2)^(1/2)))))+20*d^3*e^3*(1/3*x^
2/e^2/(-e^2*x^2+d^2)^(5/2)-2/15*d^2/e^4/(-e^2*x^2+d^2)^(5/2))+15*d^4*e^2*(1/4*x/e^2/(-e^2*x^2+d^2)^(5/2)-1/4*d
^2/e^2*(1/5*x/d^2/(-e^2*x^2+d^2)^(5/2)+4/5/d^2*(1/3*x/d^2/(-e^2*x^2+d^2)^(3/2)+2/3*x/d^4/(-e^2*x^2+d^2)^(1/2))
))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.42 \[ \int \frac {(d+e x)^6}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {2 \, {\left (13 \, e^{3} x^{3} - 39 \, d e^{2} x^{2} + 39 \, d^{2} e x - 13 \, d^{3} + 15 \, {\left (e^{3} x^{3} - 3 \, d e^{2} x^{2} + 3 \, d^{2} e x - d^{3}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (23 \, e^{2} x^{2} - 24 \, d e x + 13 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}\right )}}{15 \, {\left (e^{4} x^{3} - 3 \, d e^{3} x^{2} + 3 \, d^{2} e^{2} x - d^{3} e\right )}} \]

[In]

integrate((e*x+d)^6/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

2/15*(13*e^3*x^3 - 39*d*e^2*x^2 + 39*d^2*e*x - 13*d^3 + 15*(e^3*x^3 - 3*d*e^2*x^2 + 3*d^2*e*x - d^3)*arctan(-(
d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (23*e^2*x^2 - 24*d*e*x + 13*d^2)*sqrt(-e^2*x^2 + d^2))/(e^4*x^3 - 3*d*e^3*x
^2 + 3*d^2*e^2*x - d^3*e)

Sympy [F]

\[ \int \frac {(d+e x)^6}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {\left (d + e x\right )^{6}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]

[In]

integrate((e*x+d)**6/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral((d + e*x)**6/(-(-d + e*x)*(d + e*x))**(7/2), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (100) = 200\).

Time = 0.30 (sec) , antiderivative size = 299, normalized size of antiderivative = 2.67 \[ \int \frac {(d+e x)^6}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {1}{15} \, e^{6} x {\left (\frac {15 \, x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {20 \, d^{2} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {8 \, d^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{6}}\right )} - \frac {1}{3} \, e^{4} x {\left (\frac {3 \, x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} - \frac {2 \, d^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}}\right )} + \frac {6 \, d e^{3} x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {15 \, d^{2} e^{2} x^{3}}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {4 \, d^{3} e x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} - \frac {13 \, d^{4} x}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {26 \, d^{5}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} + \frac {31 \, d^{2} x}{30 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {16 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}}} - \frac {\arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{\sqrt {e^{2}}} \]

[In]

integrate((e*x+d)^6/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

1/15*e^6*x*(15*x^4/((-e^2*x^2 + d^2)^(5/2)*e^2) - 20*d^2*x^2/((-e^2*x^2 + d^2)^(5/2)*e^4) + 8*d^4/((-e^2*x^2 +
d^2)^(5/2)*e^6)) - 1/3*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)) + 6*d*
e^3*x^4/(-e^2*x^2 + d^2)^(5/2) + 15/2*d^2*e^2*x^3/(-e^2*x^2 + d^2)^(5/2) - 4/3*d^3*e*x^2/(-e^2*x^2 + d^2)^(5/2
) - 13/10*d^4*x/(-e^2*x^2 + d^2)^(5/2) + 26/15*d^5/((-e^2*x^2 + d^2)^(5/2)*e) + 31/30*d^2*x/(-e^2*x^2 + d^2)^(
3/2) + 16/15*x/sqrt(-e^2*x^2 + d^2) - arcsin(e^2*x/(d*sqrt(e^2)))/sqrt(e^2)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.61 \[ \int \frac {(d+e x)^6}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {\arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{{\left | e \right |}} - \frac {4 \, {\left (\frac {50 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{e^{2} x} - \frac {100 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{4} x^{2}} + \frac {30 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{e^{6} x^{3}} - \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4}}{e^{8} x^{4}} - 13\right )}}{15 \, {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} - 1\right )}^{5} {\left | e \right |}} \]

[In]

integrate((e*x+d)^6/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-arcsin(e*x/d)*sgn(d)*sgn(e)/abs(e) - 4/15*(50*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) - 100*(d*e + sqrt(-
e^2*x^2 + d^2)*abs(e))^2/(e^4*x^2) + 30*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3/(e^6*x^3) - 15*(d*e + sqrt(-e^2*
x^2 + d^2)*abs(e))^4/(e^8*x^4) - 13)/(((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) - 1)^5*abs(e))

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^6}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^6}{{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \]

[In]

int((d + e*x)^6/(d^2 - e^2*x^2)^(7/2),x)

[Out]

int((d + e*x)^6/(d^2 - e^2*x^2)^(7/2), x)