Integrand size = 24, antiderivative size = 149 \[ \int \frac {(d+e x)^4}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {35 d^3 \sqrt {d^2-e^2 x^2}}{8 e}-\frac {35 d^2 (d+e x) \sqrt {d^2-e^2 x^2}}{24 e}-\frac {7 d (d+e x)^2 \sqrt {d^2-e^2 x^2}}{12 e}-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}+\frac {35 d^4 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e} \]
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
Time = 0.56 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.62 \[ \int \frac {(d+e x)^4}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {\sqrt {d^2-e^2 x^2} \left (160 d^3+81 d^2 e x+32 d e^2 x^2+6 e^3 x^3\right )+210 d^4 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{24 e} \]
-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
Time = 0.26 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {469, 469, 469, 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.
\(\displaystyle \int \frac {(d+e x)^4}{\sqrt {d^2-e^2 x^2}} \, dx\) |
\(\Big \downarrow \) 469 |
\(\displaystyle \frac {7}{4} d \int \frac {(d+e x)^3}{\sqrt {d^2-e^2 x^2}}dx-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}\) |
\(\Big \downarrow \) 469 |
\(\displaystyle \frac {7}{4} d \left (\frac {5}{3} d \int \frac {(d+e x)^2}{\sqrt {d^2-e^2 x^2}}dx-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}\right )-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}\) |
\(\Big \downarrow \) 469 |
\(\displaystyle \frac {7}{4} d \left (\frac {5}{3} d \left (\frac {3}{2} d \int \frac {d+e x}{\sqrt {d^2-e^2 x^2}}dx-\frac {(d+e x) \sqrt {d^2-e^2 x^2}}{2 e}\right )-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}\right )-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {7}{4} d \left (\frac {5}{3} d \left (\frac {3}{2} d \left (d \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx-\frac {\sqrt {d^2-e^2 x^2}}{e}\right )-\frac {(d+e x) \sqrt {d^2-e^2 x^2}}{2 e}\right )-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}\right )-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {7}{4} d \left (\frac {5}{3} d \left (\frac {3}{2} d \left (d \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{e}\right )-\frac {(d+e x) \sqrt {d^2-e^2 x^2}}{2 e}\right )-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}\right )-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {7}{4} d \left (\frac {5}{3} d \left (\frac {3}{2} d \left (\frac {d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}-\frac {\sqrt {d^2-e^2 x^2}}{e}\right )-\frac {(d+e x) \sqrt {d^2-e^2 x^2}}{2 e}\right )-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}\right )-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}\) |
-1/4*((d + e*x)^3*Sqrt[d^2 - e^2*x^2])/e + (7*d*(-1/3*((d + e*x)^2*Sqrt[d^ 2 - e^2*x^2])/e + (5*d*(-1/2*((d + e*x)*Sqrt[d^2 - e^2*x^2])/e + (3*d*(-(S qrt[d^2 - e^2*x^2]/e) + (d*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/e))/2))/3))/ 4
3.9.27.3.1 Defintions of rubi rules used
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])
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]
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]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[2*c* ((n + p)/(n + 2*p + 1)) Int[(c + d*x)^(n - 1)*(a + b*x^2)^p, x], x] /; Fr eeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[n, 0] && NeQ[n + 2* p + 1, 0] && IntegerQ[2*p]
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 | \(\frac {d^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-x^{2} e^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}+e^{4} \left (-\frac {x^{3} \sqrt {-x^{2} e^{2}+d^{2}}}{4 e^{2}}+\frac {3 d^{2} \left (-\frac {x \sqrt {-x^{2} e^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-x^{2} e^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )}{4 e^{2}}\right )+4 d \,e^{3} \left (-\frac {x^{2} \sqrt {-x^{2} e^{2}+d^{2}}}{3 e^{2}}-\frac {2 d^{2} \sqrt {-x^{2} e^{2}+d^{2}}}{3 e^{4}}\right )+6 d^{2} e^{2} \left (-\frac {x \sqrt {-x^{2} e^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-x^{2} e^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )-\frac {4 d^{3} \sqrt {-x^{2} e^{2}+d^{2}}}{e}\) | \(261\) |
-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)+3 5/8*d^4/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))
Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.56 \[ \int \frac {(d+e x)^4}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {210 \, d^{4} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (6 \, e^{3} x^{3} + 32 \, d e^{2} x^{2} + 81 \, d^{2} e x + 160 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, e} \]
-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)*sqrt(-e^2*x^2 + d^2))/e
Time = 0.56 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.93 \[ \int \frac {(d+e x)^4}{\sqrt {d^2-e^2 x^2}} \, dx=\begin {cases} \frac {35 d^{4} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{8} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {20 d^{3}}{3 e} - \frac {27 d^{2} x}{8} - \frac {4 d e x^{2}}{3} - \frac {e^{2} x^{3}}{4}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {\begin {cases} d^{4} x & \text {for}\: e = 0 \\\frac {\left (d + e x\right )^{5}}{5 e} & \text {otherwise} \end {cases}}{\sqrt {d^{2}}} & \text {otherwise} \end {cases} \]
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))
Time = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.74 \[ \int \frac {(d+e x)^4}{\sqrt {d^2-e^2 x^2}} \, dx=-\frac {1}{4} \, \sqrt {-e^{2} x^{2} + d^{2}} e^{2} x^{3} - \frac {4}{3} \, \sqrt {-e^{2} x^{2} + d^{2}} d e x^{2} + \frac {35 \, d^{4} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{8 \, \sqrt {e^{2}}} - \frac {27}{8} \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} x - \frac {20 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{3 \, e} \]
-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)))/sqrt(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
Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.46 \[ \int \frac {(d+e x)^4}{\sqrt {d^2-e^2 x^2}} \, dx=\frac {35 \, d^{4} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{8 \, {\left | e \right |}} - \frac {1}{24} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (\frac {160 \, d^{3}}{e} + {\left (81 \, d^{2} + 2 \, {\left (3 \, e^{2} x + 16 \, d e\right )} x\right )} x\right )} \]
35/8*d^4*arcsin(e*x/d)*sgn(d)*sgn(e)/abs(e) - 1/24*sqrt(-e^2*x^2 + d^2)*(1 60*d^3/e + (81*d^2 + 2*(3*e^2*x + 16*d*e)*x)*x)
Timed out. \[ \int \frac {(d+e x)^4}{\sqrt {d^2-e^2 x^2}} \, dx=\int \frac {{\left (d+e\,x\right )}^4}{\sqrt {d^2-e^2\,x^2}} \,d x \]