\(\int \frac {(d+e x)^3}{(d^2-e^2 x^2)^{7/2}} \, dx\) [852]

Optimal result

Integrand size = 24, antiderivative size = 103 \[ \int \frac {(d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2}}{5 d e (d-e x)^3}+\frac {2 \sqrt {d^2-e^2 x^2}}{15 d^2 e (d-e x)^2}+\frac {2 \sqrt {d^2-e^2 x^2}}{15 d^3 e (d-e x)} \]

[Out]

1/5*(-e^2*x^2+d^2)^(1/2)/d/e/(-e*x+d)^3+2/15*(-e^2*x^2+d^2)^(1/2)/d^2/e/(-e*x+d)^2+2/15*(-e^2*x^2+d^2)^(1/2)/d
^3/e/(-e*x+d)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {669, 673, 665} \[ \int \frac {(d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {2 \sqrt {d^2-e^2 x^2}}{15 d^2 e (d-e x)^2}+\frac {\sqrt {d^2-e^2 x^2}}{5 d e (d-e x)^3}+\frac {2 \sqrt {d^2-e^2 x^2}}{15 d^3 e (d-e x)} \]

[In]

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

[Out]

Sqrt[d^2 - e^2*x^2]/(5*d*e*(d - e*x)^3) + (2*Sqrt[d^2 - e^2*x^2])/(15*d^2*e*(d - e*x)^2) + (2*Sqrt[d^2 - e^2*x
^2])/(15*d^3*e*(d - e*x))

Rule 665

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

Rule 669

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

Rule 673

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{(d-e x)^3 \sqrt {d^2-e^2 x^2}} \, dx \\ & = \frac {\sqrt {d^2-e^2 x^2}}{5 d e (d-e x)^3}+\frac {2 \int \frac {1}{(d-e x)^2 \sqrt {d^2-e^2 x^2}} \, dx}{5 d} \\ & = \frac {\sqrt {d^2-e^2 x^2}}{5 d e (d-e x)^3}+\frac {2 \sqrt {d^2-e^2 x^2}}{15 d^2 e (d-e x)^2}+\frac {2 \int \frac {1}{(d-e x) \sqrt {d^2-e^2 x^2}} \, dx}{15 d^2} \\ & = \frac {\sqrt {d^2-e^2 x^2}}{5 d e (d-e x)^3}+\frac {2 \sqrt {d^2-e^2 x^2}}{15 d^2 e (d-e x)^2}+\frac {2 \sqrt {d^2-e^2 x^2}}{15 d^3 e (d-e x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.51 \[ \int \frac {(d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (7 d^2-6 d e x+2 e^2 x^2\right )}{15 d^3 e (d-e x)^3} \]

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(7*d^2 - 6*d*e*x + 2*e^2*x^2))/(15*d^3*e*(d - e*x)^3)

Maple [A] (verified)

Time = 2.56 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.49

method	result	size
trager	\(\frac {\left (2 x^{2} e^{2}-6 d e x +7 d^{2}\right ) \sqrt {-x^{2} e^{2}+d^{2}}}{15 d^{3} \left (-e x +d \right )^{3} e}\)	\(50\)
gosper	\(\frac {\left (e x +d \right )^{4} \left (-e x +d \right ) \left (2 x^{2} e^{2}-6 d e x +7 d^{2}\right )}{15 d^{3} e \left (-x^{2} e^{2}+d^{2}\right )^{\frac {7}{2}}}\)	\(55\)
default	\(d^{3} \left (\frac {x}{5 d^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-x^{2} e^{2}+d^{2}}}}{d^{2}}\right )+e^{3} \left (\frac {x^{2}}{3 e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {5}{2}}}\right )+3 d \,e^{2} \left (\frac {x}{4 e^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-x^{2} e^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-x^{2} e^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )+\frac {3 d^{2}}{5 e \left (-x^{2} e^{2}+d^{2}\right )^{\frac {5}{2}}}\)	\(246\)

[In]

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

[Out]

1/15*(2*e^2*x^2-6*d*e*x+7*d^2)/d^3/(-e*x+d)^3/e*(-e^2*x^2+d^2)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.03 \[ \int \frac {(d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {7 \, e^{3} x^{3} - 21 \, d e^{2} x^{2} + 21 \, d^{2} e x - 7 \, d^{3} - {\left (2 \, e^{2} x^{2} - 6 \, d e x + 7 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{3} e^{4} x^{3} - 3 \, d^{4} e^{3} x^{2} + 3 \, d^{5} e^{2} x - d^{6} e\right )}} \]

[In]

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

[Out]

1/15*(7*e^3*x^3 - 21*d*e^2*x^2 + 21*d^2*e*x - 7*d^3 - (2*e^2*x^2 - 6*d*e*x + 7*d^2)*sqrt(-e^2*x^2 + d^2))/(d^3
*e^4*x^3 - 3*d^4*e^3*x^2 + 3*d^5*e^2*x - d^6*e)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.98 \[ \int \frac {(d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {e x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {4 \, d x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {7 \, d^{2}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} + \frac {x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d} + \frac {2 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}} \]

[In]

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

[Out]

1/3*e*x^2/(-e^2*x^2 + d^2)^(5/2) + 4/5*d*x/(-e^2*x^2 + d^2)^(5/2) + 7/15*d^2/((-e^2*x^2 + d^2)^(5/2)*e) + 1/15
*x/((-e^2*x^2 + d^2)^(3/2)*d) + 2/15*x/(sqrt(-e^2*x^2 + d^2)*d^3)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.60 \[ \int \frac {(d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {2 \, {\left (\frac {20 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{e^{2} x} - \frac {40 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{4} x^{2}} + \frac {30 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{e^{6} x^{3}} - \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4}}{e^{8} x^{4}} - 7\right )}}{15 \, d^{3} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} - 1\right )}^{5} {\left | e \right |}} \]

[In]

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

[Out]

-2/15*(20*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) - 40*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2/(e^4*x^2) + 3
0*(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) - 7)/(d
^3*((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) - 1)^5*abs(e))

Mupad [B] (verification not implemented)

Time = 9.87 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.48 \[ \int \frac {(d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (7\,d^2-6\,d\,e\,x+2\,e^2\,x^2\right )}{15\,d^3\,e\,{\left (d-e\,x\right )}^3} \]

[In]

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

[Out]

((d^2 - e^2*x^2)^(1/2)*(7*d^2 + 2*e^2*x^2 - 6*d*e*x))/(15*d^3*e*(d - e*x)^3)