\(\int \frac {x^3 (d+e x)^2}{(d^2-e^2 x^2)^{7/2}} \, dx\) [46]

Optimal result

Integrand size = 27, antiderivative size = 97 \[ \int \frac {x^3 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {d^2 (d+e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 d (d+e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {5 d+2 e x}{5 d e^4 \sqrt {d^2-e^2 x^2}} \]

[Out]

1/5*d^2*(e*x+d)^2/e^4/(-e^2*x^2+d^2)^(5/2)-4/5*d*(e*x+d)/e^4/(-e^2*x^2+d^2)^(3/2)+1/5*(2*e*x+5*d)/d/e^4/(-e^2*
x^2+d^2)^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1649, 651} \[ \int \frac {x^3 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {d^2 (d+e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 d (d+e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {5 d+2 e x}{5 d e^4 \sqrt {d^2-e^2 x^2}} \]

[In]

Int[(x^3*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(d^2*(d + e*x)^2)/(5*e^4*(d^2 - e^2*x^2)^(5/2)) - (4*d*(d + e*x))/(5*e^4*(d^2 - e^2*x^2)^(3/2)) + (5*d + 2*e*x
)/(5*d*e^4*Sqrt[d^2 - e^2*x^2])

Rule 651

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[((-a)*e + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

Rule 1649

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, Simp[(-d)*f*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2
*a*e*(p + 1))), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)
*Q + f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p
+ 1/2, 0] && GtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {d^2 (d+e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {(d+e x) \left (\frac {2 d^3}{e^3}+\frac {5 d^2 x}{e^2}+\frac {5 d x^2}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d} \\ & = \frac {d^2 (d+e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 d (d+e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {\frac {6 d^3}{e^3}+\frac {15 d^2 x}{e^2}}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2} \\ & = \frac {d^2 (d+e x)^2}{5 e^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {4 d (d+e x)}{5 e^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {5 d+2 e x}{5 d e^4 \sqrt {d^2-e^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.72 \[ \int \frac {x^3 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (2 d^3-4 d^2 e x+d e^2 x^2+2 e^3 x^3\right )}{5 d e^4 (d-e x)^3 (d+e x)} \]

[In]

Integrate[(x^3*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(2*d^3 - 4*d^2*e*x + d*e^2*x^2 + 2*e^3*x^3))/(5*d*e^4*(d - e*x)^3*(d + e*x))

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.67

method	result	size
gosper	\(\frac {\left (-e x +d \right ) \left (e x +d \right )^{3} \left (2 e^{3} x^{3}+d \,e^{2} x^{2}-4 d^{2} e x +2 d^{3}\right )}{5 d \,e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\)	\(65\)
trager	\(\frac {\left (2 e^{3} x^{3}+d \,e^{2} x^{2}-4 d^{2} e x +2 d^{3}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{5 d \,e^{4} \left (-e x +d \right )^{3} \left (e x +d \right )}\)	\(67\)
default	\(e^{2} \left (\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {4 d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )}{e^{2}}\right )+d^{2} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )+2 d e \left (\frac {x^{3}}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {3 d^{2} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )}{2 e^{2}}\right )\)	\(261\)

[In]

int(x^3*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/5*(-e*x+d)*(e*x+d)^3*(2*e^3*x^3+d*e^2*x^2-4*d^2*e*x+2*d^3)/d/e^4/(-e^2*x^2+d^2)^(7/2)

Fricas [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.20 \[ \int \frac {x^3 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {2 \, e^{4} x^{4} - 4 \, d e^{3} x^{3} + 4 \, d^{3} e x - 2 \, d^{4} - {\left (2 \, e^{3} x^{3} + d e^{2} x^{2} - 4 \, d^{2} e x + 2 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{5 \, {\left (d e^{8} x^{4} - 2 \, d^{2} e^{7} x^{3} + 2 \, d^{4} e^{5} x - d^{5} e^{4}\right )}} \]

[In]

integrate(x^3*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/5*(2*e^4*x^4 - 4*d*e^3*x^3 + 4*d^3*e*x - 2*d^4 - (2*e^3*x^3 + d*e^2*x^2 - 4*d^2*e*x + 2*d^3)*sqrt(-e^2*x^2 +
d^2))/(d*e^8*x^4 - 2*d^2*e^7*x^3 + 2*d^4*e^5*x - d^5*e^4)

Sympy [F]

\[ \int \frac {x^3 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {x^{3} \left (d + e x\right )^{2}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]

[In]

integrate(x**3*(e*x+d)**2/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral(x**3*(d + e*x)**2/(-(-d + e*x)*(d + e*x))**(7/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.60 \[ \int \frac {x^3 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {d x^{3}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} - \frac {d^{2} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {3 \, d^{3} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}} + \frac {2 \, d^{4}}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{4}} + \frac {d x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{3}} + \frac {2 \, x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d e^{3}} \]

[In]

integrate(x^3*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

x^4/(-e^2*x^2 + d^2)^(5/2) + d*x^3/((-e^2*x^2 + d^2)^(5/2)*e) - d^2*x^2/((-e^2*x^2 + d^2)^(5/2)*e^2) - 3/5*d^3
*x/((-e^2*x^2 + d^2)^(5/2)*e^3) + 2/5*d^4/((-e^2*x^2 + d^2)^(5/2)*e^4) + 1/5*d*x/((-e^2*x^2 + d^2)^(3/2)*e^3)
+ 2/5*x/(sqrt(-e^2*x^2 + d^2)*d*e^3)

Giac [F]

\[ \int \frac {x^3 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{2} x^{3}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate(x^3*(e*x+d)^2/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

integrate((e*x + d)^2*x^3/(-e^2*x^2 + d^2)^(7/2), x)

Mupad [B] (verification not implemented)

Time = 11.60 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.68 \[ \int \frac {x^3 (d+e x)^2}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (2\,d^3-4\,d^2\,e\,x+d\,e^2\,x^2+2\,e^3\,x^3\right )}{5\,d\,e^4\,\left (d+e\,x\right )\,{\left (d-e\,x\right )}^3} \]

[In]

int((x^3*(d + e*x)^2)/(d^2 - e^2*x^2)^(7/2),x)

[Out]

((d^2 - e^2*x^2)^(1/2)*(2*d^3 + 2*e^3*x^3 + d*e^2*x^2 - 4*d^2*e*x))/(5*d*e^4*(d + e*x)*(d - e*x)^3)