\(\int \frac {(d+e x)^4}{\sqrt {d^2-e^2 x^2}} \, dx\) [827]

Optimal result

Integrand size = 24, antiderivative size = 149 \[ \int \frac {(d+e x)^4}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {35 d^3 \sqrt {d^2-e^2 x^2}}{8 e}-\frac {35 d^2 (d+e x) \sqrt {d^2-e^2 x^2}}{24 e}-\frac {7 d (d+e x)^2 \sqrt {d^2-e^2 x^2}}{12 e}-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}+\frac {35 d^4 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {685, 655, 223, 209} \[ \int \frac {(d+e x)^4}{\sqrt {d^2-e^2 x^2}} \, dx=\frac {35 d^4 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e}-\frac {35 d^2 (d+e x) \sqrt {d^2-e^2 x^2}}{24 e}-\frac {7 d (d+e x)^2 \sqrt {d^2-e^2 x^2}}{12 e}-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}-\frac {35 d^3 \sqrt {d^2-e^2 x^2}}{8 e} \]

[In]

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

[Out]

(-35*d^3*Sqrt[d^2 - e^2*x^2])/(8*e) - (35*d^2*(d + e*x)*Sqrt[d^2 - e^2*x^2])/(24*e) - (7*d*(d + e*x)^2*Sqrt[d^
2 - e^2*x^2])/(12*e) - ((d + e*x)^3*Sqrt[d^2 - e^2*x^2])/(4*e) + (35*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8
*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 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 {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}+\frac {1}{4} (7 d) \int \frac {(d+e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx \\ & = -\frac {7 d (d+e x)^2 \sqrt {d^2-e^2 x^2}}{12 e}-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}+\frac {1}{12} \left (35 d^2\right ) \int \frac {(d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx \\ & = -\frac {35 d^2 (d+e x) \sqrt {d^2-e^2 x^2}}{24 e}-\frac {7 d (d+e x)^2 \sqrt {d^2-e^2 x^2}}{12 e}-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}+\frac {1}{8} \left (35 d^3\right ) \int \frac {d+e x}{\sqrt {d^2-e^2 x^2}} \, dx \\ & = -\frac {35 d^3 \sqrt {d^2-e^2 x^2}}{8 e}-\frac {35 d^2 (d+e x) \sqrt {d^2-e^2 x^2}}{24 e}-\frac {7 d (d+e x)^2 \sqrt {d^2-e^2 x^2}}{12 e}-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}+\frac {1}{8} \left (35 d^4\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx \\ & = -\frac {35 d^3 \sqrt {d^2-e^2 x^2}}{8 e}-\frac {35 d^2 (d+e x) \sqrt {d^2-e^2 x^2}}{24 e}-\frac {7 d (d+e x)^2 \sqrt {d^2-e^2 x^2}}{12 e}-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}+\frac {1}{8} \left (35 d^4\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 {35 d^3 \sqrt {d^2-e^2 x^2}}{8 e}-\frac {35 d^2 (d+e x) \sqrt {d^2-e^2 x^2}}{24 e}-\frac {7 d (d+e x)^2 \sqrt {d^2-e^2 x^2}}{12 e}-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}+\frac {35 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.62 \[ \int \frac {(d+e x)^4}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {d^2-e^2 x^2} \left (160 d^3+81 d^2 e x+32 d e^2 x^2+6 e^3 x^3\right )+210 d^4 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{24 e} \]

[In]

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

[Out]

-1/24*(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) + 210*d^4*ArcTan[(e*x)/(Sqrt[d^2]
- Sqrt[d^2 - e^2*x^2])])/e

Maple [A] (verified)

Time = 2.35 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.56

method	result	size
risch	\(-\frac {\left (6 e^{3} x^{3}+32 d \,e^{2} x^{2}+81 d^{2} e x +160 d^{3}\right ) \sqrt {-x^{2} e^{2}+d^{2}}}{24 e}+\frac {35 d^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-x^{2} e^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}}\)	\(83\)
default	\(\frac {d^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-x^{2} e^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}+e^{4} \left (-\frac {x^{3} \sqrt {-x^{2} e^{2}+d^{2}}}{4 e^{2}}+\frac {3 d^{2} \left (-\frac {x \sqrt {-x^{2} e^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-x^{2} e^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )}{4 e^{2}}\right )+4 d \,e^{3} \left (-\frac {x^{2} \sqrt {-x^{2} e^{2}+d^{2}}}{3 e^{2}}-\frac {2 d^{2} \sqrt {-x^{2} e^{2}+d^{2}}}{3 e^{4}}\right )+6 d^{2} e^{2} \left (-\frac {x \sqrt {-x^{2} e^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-x^{2} e^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )-\frac {4 d^{3} \sqrt {-x^{2} e^{2}+d^{2}}}{e}\)	\(261\)

[In]

int((e*x+d)^4/(-e^2*x^2+d^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-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)+35/8*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)

none

Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.56 \[ \int \frac {(d+e x)^4}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {210 \, d^{4} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (6 \, e^{3} x^{3} + 32 \, d e^{2} x^{2} + 81 \, d^{2} e x + 160 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, e} \]

[In]

integrate((e*x+d)^4/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-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)*s
qrt(-e^2*x^2 + d^2))/e

Sympy [A] (verification not implemented)

Time = 0.56 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.93 \[ \int \frac {(d+e x)^4}{\sqrt {d^2-e^2 x^2}} \, dx=\begin {cases} \frac {35 d^{4} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{8} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {20 d^{3}}{3 e} - \frac {27 d^{2} x}{8} - \frac {4 d e x^{2}}{3} - \frac {e^{2} x^{3}}{4}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {\begin {cases} d^{4} x & \text {for}\: e = 0 \\\frac {\left (d + e x\right )^{5}}{5 e} & \text {otherwise} \end {cases}}{\sqrt {d^{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate((e*x+d)**4/(-e**2*x**2+d**2)**(1/2),x)

[Out]

Piecewise((35*d**4*Piecewise((log(-2*e**2*x + 2*sqrt(-e**2)*sqrt(d**2 - e**2*x**2))/sqrt(-e**2), Ne(d**2, 0)),
(x*log(x)/sqrt(-e**2*x**2), True))/8 + sqrt(d**2 - e**2*x**2)*(-20*d**3/(3*e) - 27*d**2*x/8 - 4*d*e*x**2/3 -
e**2*x**3/4), Ne(e**2, 0)), (Piecewise((d**4*x, Eq(e, 0)), ((d + e*x)**5/(5*e), True))/sqrt(d**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.74 \[ \int \frac {(d+e x)^4}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {1}{4} \, \sqrt {-e^{2} x^{2} + d^{2}} e^{2} x^{3} - \frac {4}{3} \, \sqrt {-e^{2} x^{2} + d^{2}} d e x^{2} + \frac {35 \, d^{4} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{8 \, \sqrt {e^{2}}} - \frac {27}{8} \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} x - \frac {20 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{3 \, e} \]

[In]

integrate((e*x+d)^4/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

-1/4*sqrt(-e^2*x^2 + d^2)*e^2*x^3 - 4/3*sqrt(-e^2*x^2 + d^2)*d*e*x^2 + 35/8*d^4*arcsin(e^2*x/(d*sqrt(e^2)))/sq
rt(e^2) - 27/8*sqrt(-e^2*x^2 + d^2)*d^2*x - 20/3*sqrt(-e^2*x^2 + d^2)*d^3/e

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.46 \[ \int \frac {(d+e x)^4}{\sqrt {d^2-e^2 x^2}} \, dx=\frac {35 \, d^{4} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{8 \, {\left | e \right |}} - \frac {1}{24} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (\frac {160 \, d^{3}}{e} + {\left (81 \, d^{2} + 2 \, {\left (3 \, e^{2} x + 16 \, d e\right )} x\right )} x\right )} \]

[In]

integrate((e*x+d)^4/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

35/8*d^4*arcsin(e*x/d)*sgn(d)*sgn(e)/abs(e) - 1/24*sqrt(-e^2*x^2 + d^2)*(160*d^3/e + (81*d^2 + 2*(3*e^2*x + 16
*d*e)*x)*x)

Mupad [F(-1)]

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

[In]

int((d + e*x)^4/(d^2 - e^2*x^2)^(1/2),x)

[Out]

int((d + e*x)^4/(d^2 - e^2*x^2)^(1/2), x)