\(\int \frac {(d+e x)^4}{(d^2-e^2 x^2)^{5/2}} \, dx\) [838]

Optimal result

Integrand size = 24, antiderivative size = 81 \[ \int \frac {(d+e x)^4}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {2 (d+e x)^3}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 (d+e x)}{e \sqrt {d^2-e^2 x^2}}+\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e} \]

[Out]

2/3*(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 = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {683, 667, 223, 209} \[ \int \frac {(d+e x)^4}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}+\frac {2 (d+e x)^3}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 (d+e x)}{e \sqrt {d^2-e^2 x^2}} \]

[In]

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

[Out]

(2*(d + e*x)^3)/(3*e*(d^2 - e^2*x^2)^(3/2)) - (2*(d + e*x))/(e*Sqrt[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)^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)^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)^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)^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 = 0.45 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.91 \[ \int \frac {(d+e x)^4}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {2 \left (-\frac {2 (d-2 e x) \sqrt {d^2-e^2 x^2}}{(d-e x)^2}-3 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )\right )}{3 e} \]

[In]

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

[Out]

(2*((-2*(d - 2*e*x)*Sqrt[d^2 - e^2*x^2])/(d - e*x)^2 - 3*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])]))/(3*
e)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(280\) vs. \(2(73)=146\).

Time = 2.32 (sec) , antiderivative size = 281, normalized size of antiderivative = 3.47

method	result	size
default	\(d^{4} \left (\frac {x}{3 d^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-x^{2} e^{2}+d^{2}}}\right )+e^{4} \left (\frac {x^{3}}{3 e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{e^{2} \sqrt {-x^{2} e^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-x^{2} e^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}\right )+4 d \,e^{3} \left (\frac {x^{2}}{e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {2 d^{2}}{3 e^{4} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}\right )+6 d^{2} e^{2} \left (\frac {x}{2 e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {d^{2} \left (\frac {x}{3 d^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-x^{2} e^{2}+d^{2}}}\right )}{2 e^{2}}\right )+\frac {4 d^{3}}{3 e \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}\)	\(281\)

[In]

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

[Out]

d^4*(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^4*(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))))+4*d*e^3*(x^2/e^2/
(-e^2*x^2+d^2)^(3/2)-2/3*d^2/e^4/(-e^2*x^2+d^2)^(3/2))+6*d^2*e^2*(1/2*x/e^2/(-e^2*x^2+d^2)^(3/2)-1/2*d^2/e^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)))+4/3*d^3/e/(-e^2*x^2+d^2)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.38 \[ \int \frac {(d+e x)^4}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (2 \, e^{2} x^{2} - 4 \, d e x + 2 \, d^{2} + 3 \, {\left (e^{2} x^{2} - 2 \, d e x + d^{2}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - 2 \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (2 \, e x - d\right )}\right )}}{3 \, {\left (e^{3} x^{2} - 2 \, d e^{2} x + d^{2} e\right )}} \]

[In]

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

[Out]

-2/3*(2*e^2*x^2 - 4*d*e*x + 2*d^2 + 3*(e^2*x^2 - 2*d*e*x + d^2)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - 2*
sqrt(-e^2*x^2 + d^2)*(2*e*x - d))/(e^3*x^2 - 2*d*e^2*x + d^2*e)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (73) = 146\).

Time = 0.28 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.88 \[ \int \frac {(d+e x)^4}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {1}{3} \, e^{4} x {\left (\frac {3 \, x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} - \frac {2 \, d^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}}\right )} + \frac {4 \, d e x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {7 \, d^{2} x}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} - \frac {4 \, d^{3}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e} - \frac {5 \, x}{3 \, \sqrt {-e^{2} x^{2} + d^{2}}} + \frac {\arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{\sqrt {e^{2}}} \]

[In]

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

[Out]

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)) + 4*d*e*x^2/(-e^2*x^2 + d^
2)^(3/2) + 7/3*d^2*x/(-e^2*x^2 + d^2)^(3/2) - 4/3*d^3/((-e^2*x^2 + d^2)^(3/2)*e) - 5/3*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 = 0.30 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.06 \[ \int \frac {(d+e x)^4}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {\arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{{\left | e \right |}} + \frac {8 \, {\left (\frac {3 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{e^{2} x} - 1\right )}}{3 \, {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} - 1\right )}^{3} {\left | e \right |}} \]

[In]

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

[Out]

arcsin(e*x/d)*sgn(d)*sgn(e)/abs(e) + 8/3*(3*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) - 1)/(((d*e + sqrt(-e^
2*x^2 + d^2)*abs(e))/(e^2*x) - 1)^3*abs(e))

Mupad [F(-1)]

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

[In]

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

[Out]

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