\(\int \frac {1}{(d+e x)^3 (d^2-e^2 x^2)^{7/2}} \, dx\) [857]

Optimal result

Integrand size = 24, antiderivative size = 172 \[ \int \frac {1}{(d+e x)^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {16 x}{165 d^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {1}{11 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8}{99 d^2 e (d+e x)^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8}{99 d^3 e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}+\frac {64 x}{495 d^7 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {128 x}{495 d^9 \sqrt {d^2-e^2 x^2}} \]

[Out]

16/165*x/d^5/(-e^2*x^2+d^2)^(5/2)-1/11/d/e/(e*x+d)^3/(-e^2*x^2+d^2)^(5/2)-8/99/d^2/e/(e*x+d)^2/(-e^2*x^2+d^2)^
(5/2)-8/99/d^3/e/(e*x+d)/(-e^2*x^2+d^2)^(5/2)+64/495*x/d^7/(-e^2*x^2+d^2)^(3/2)+128/495*x/d^9/(-e^2*x^2+d^2)^(
1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {673, 198, 197} \[ \int \frac {1}{(d+e x)^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {8}{99 d^2 e (d+e x)^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {1}{11 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {128 x}{495 d^9 \sqrt {d^2-e^2 x^2}}+\frac {64 x}{495 d^7 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {16 x}{165 d^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8}{99 d^3 e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \]

[In]

Int[1/((d + e*x)^3*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

(16*x)/(165*d^5*(d^2 - e^2*x^2)^(5/2)) - 1/(11*d*e*(d + e*x)^3*(d^2 - e^2*x^2)^(5/2)) - 8/(99*d^2*e*(d + e*x)^
2*(d^2 - e^2*x^2)^(5/2)) - 8/(99*d^3*e*(d + e*x)*(d^2 - e^2*x^2)^(5/2)) + (64*x)/(495*d^7*(d^2 - e^2*x^2)^(3/2
)) + (128*x)/(495*d^9*Sqrt[d^2 - e^2*x^2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -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}& = -\frac {1}{11 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {8 \int \frac {1}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{7/2}} \, dx}{11 d} \\ & = -\frac {1}{11 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8}{99 d^2 e (d+e x)^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {56 \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{7/2}} \, dx}{99 d^2} \\ & = -\frac {1}{11 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8}{99 d^2 e (d+e x)^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8}{99 d^3 e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}+\frac {16 \int \frac {1}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx}{33 d^3} \\ & = \frac {16 x}{165 d^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {1}{11 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8}{99 d^2 e (d+e x)^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8}{99 d^3 e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}+\frac {64 \int \frac {1}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{165 d^5} \\ & = \frac {16 x}{165 d^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {1}{11 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8}{99 d^2 e (d+e x)^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8}{99 d^3 e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}+\frac {64 x}{495 d^7 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {128 \int \frac {1}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{495 d^7} \\ & = \frac {16 x}{165 d^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {1}{11 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8}{99 d^2 e (d+e x)^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {8}{99 d^3 e (d+e x) \left (d^2-e^2 x^2\right )^{5/2}}+\frac {64 x}{495 d^7 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {128 x}{495 d^9 \sqrt {d^2-e^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.73 \[ \int \frac {1}{(d+e x)^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-125 d^8+120 d^7 e x+680 d^6 e^2 x^2+400 d^5 e^3 x^3-720 d^4 e^4 x^4-832 d^3 e^5 x^5+64 d^2 e^6 x^6+384 d e^7 x^7+128 e^8 x^8\right )}{495 d^9 e (d-e x)^3 (d+e x)^6} \]

[In]

Integrate[1/((d + e*x)^3*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-125*d^8 + 120*d^7*e*x + 680*d^6*e^2*x^2 + 400*d^5*e^3*x^3 - 720*d^4*e^4*x^4 - 832*d^3*e
^5*x^5 + 64*d^2*e^6*x^6 + 384*d*e^7*x^7 + 128*e^8*x^8))/(495*d^9*e*(d - e*x)^3*(d + e*x)^6)

Maple [A] (verified)

Time = 2.50 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.70

method	result	size
gosper	\(-\frac {\left (-e x +d \right ) \left (-128 e^{8} x^{8}-384 d \,e^{7} x^{7}-64 d^{2} e^{6} x^{6}+832 d^{3} e^{5} x^{5}+720 d^{4} e^{4} x^{4}-400 d^{5} e^{3} x^{3}-680 d^{6} e^{2} x^{2}-120 d^{7} e x +125 d^{8}\right )}{495 \left (e x +d \right )^{2} d^{9} e \left (-x^{2} e^{2}+d^{2}\right )^{\frac {7}{2}}}\)	\(121\)
trager	\(-\frac {\left (-128 e^{8} x^{8}-384 d \,e^{7} x^{7}-64 d^{2} e^{6} x^{6}+832 d^{3} e^{5} x^{5}+720 d^{4} e^{4} x^{4}-400 d^{5} e^{3} x^{3}-680 d^{6} e^{2} x^{2}-120 d^{7} e x +125 d^{8}\right ) \sqrt {-x^{2} e^{2}+d^{2}}}{495 d^{9} \left (e x +d \right )^{6} \left (-e x +d \right )^{3} e}\)	\(123\)
default	\(\frac {-\frac {1}{11 d e \left (x +\frac {d}{e}\right )^{3} \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}+\frac {8 e \left (-\frac {1}{9 d e \left (x +\frac {d}{e}\right )^{2} \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}+\frac {7 e \left (-\frac {1}{7 d e \left (x +\frac {d}{e}\right ) \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}+\frac {6 e \left (-\frac {-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e}{10 d^{2} e^{2} \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}+\frac {-\frac {2 \left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right )}{15 d^{2} e^{2} \left (-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}-\frac {4 \left (-2 e^{2} \left (x +\frac {d}{e}\right )+2 d e \right )}{15 e^{2} d^{4} \sqrt {-e^{2} \left (x +\frac {d}{e}\right )^{2}+2 d e \left (x +\frac {d}{e}\right )}}}{d^{2}}\right )}{7 d}\right )}{9 d}\right )}{11 d}}{e^{3}}\)	\(327\)

[In]

int(1/(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/495*(-e*x+d)*(-128*e^8*x^8-384*d*e^7*x^7-64*d^2*e^6*x^6+832*d^3*e^5*x^5+720*d^4*e^4*x^4-400*d^5*e^3*x^3-680
*d^6*e^2*x^2-120*d^7*e*x+125*d^8)/(e*x+d)^2/d^9/e/(-e^2*x^2+d^2)^(7/2)

Fricas [A] (verification not implemented)

none

Time = 0.84 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.51 \[ \int \frac {1}{(d+e x)^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {125 \, e^{9} x^{9} + 375 \, d e^{8} x^{8} - 1000 \, d^{3} e^{6} x^{6} - 750 \, d^{4} e^{5} x^{5} + 750 \, d^{5} e^{4} x^{4} + 1000 \, d^{6} e^{3} x^{3} - 375 \, d^{8} e x - 125 \, d^{9} + {\left (128 \, e^{8} x^{8} + 384 \, d e^{7} x^{7} + 64 \, d^{2} e^{6} x^{6} - 832 \, d^{3} e^{5} x^{5} - 720 \, d^{4} e^{4} x^{4} + 400 \, d^{5} e^{3} x^{3} + 680 \, d^{6} e^{2} x^{2} + 120 \, d^{7} e x - 125 \, d^{8}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{495 \, {\left (d^{9} e^{10} x^{9} + 3 \, d^{10} e^{9} x^{8} - 8 \, d^{12} e^{7} x^{6} - 6 \, d^{13} e^{6} x^{5} + 6 \, d^{14} e^{5} x^{4} + 8 \, d^{15} e^{4} x^{3} - 3 \, d^{17} e^{2} x - d^{18} e\right )}} \]

[In]

integrate(1/(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

-1/495*(125*e^9*x^9 + 375*d*e^8*x^8 - 1000*d^3*e^6*x^6 - 750*d^4*e^5*x^5 + 750*d^5*e^4*x^4 + 1000*d^6*e^3*x^3
- 375*d^8*e*x - 125*d^9 + (128*e^8*x^8 + 384*d*e^7*x^7 + 64*d^2*e^6*x^6 - 832*d^3*e^5*x^5 - 720*d^4*e^4*x^4 +
400*d^5*e^3*x^3 + 680*d^6*e^2*x^2 + 120*d^7*e*x - 125*d^8)*sqrt(-e^2*x^2 + d^2))/(d^9*e^10*x^9 + 3*d^10*e^9*x^
8 - 8*d^12*e^7*x^6 - 6*d^13*e^6*x^5 + 6*d^14*e^5*x^4 + 8*d^15*e^4*x^3 - 3*d^17*e^2*x - d^18*e)

Sympy [F]

\[ \int \frac {1}{(d+e x)^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {1}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}} \left (d + e x\right )^{3}}\, dx \]

[In]

integrate(1/(e*x+d)**3/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral(1/((-(-d + e*x)*(d + e*x))**(7/2)*(d + e*x)**3), x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.58 \[ \int \frac {1}{(d+e x)^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {1}{11 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d e^{4} x^{3} + 3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2} e^{3} x^{2} + 3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3} e^{2} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{4} e\right )}} - \frac {8}{99 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2} e^{3} x^{2} + 2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3} e^{2} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{4} e\right )}} - \frac {8}{99 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3} e^{2} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{4} e\right )}} + \frac {16 \, x}{165 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{5}} + \frac {64 \, x}{495 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{7}} + \frac {128 \, x}{495 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{9}} \]

[In]

integrate(1/(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

-1/11/((-e^2*x^2 + d^2)^(5/2)*d*e^4*x^3 + 3*(-e^2*x^2 + d^2)^(5/2)*d^2*e^3*x^2 + 3*(-e^2*x^2 + d^2)^(5/2)*d^3*
e^2*x + (-e^2*x^2 + d^2)^(5/2)*d^4*e) - 8/99/((-e^2*x^2 + d^2)^(5/2)*d^2*e^3*x^2 + 2*(-e^2*x^2 + d^2)^(5/2)*d^
3*e^2*x + (-e^2*x^2 + d^2)^(5/2)*d^4*e) - 8/99/((-e^2*x^2 + d^2)^(5/2)*d^3*e^2*x + (-e^2*x^2 + d^2)^(5/2)*d^4*
e) + 16/165*x/((-e^2*x^2 + d^2)^(5/2)*d^5) + 64/495*x/((-e^2*x^2 + d^2)^(3/2)*d^7) + 128/495*x/(sqrt(-e^2*x^2
+ d^2)*d^9)

Giac [F]

\[ \int \frac {1}{(d+e x)^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int { \frac {1}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} {\left (e x + d\right )}^{3}} \,d x } \]

[In]

integrate(1/(e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

integrate(1/((-e^2*x^2 + d^2)^(7/2)*(e*x + d)^3), x)

Mupad [B] (verification not implemented)

Time = 10.31 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.24 \[ \int \frac {1}{(d+e x)^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (\frac {64\,x}{495\,d^7}+\frac {67}{1584\,d^6\,e}\right )}{{\left (d+e\,x\right )}^2\,{\left (d-e\,x\right )}^2}+\frac {\sqrt {d^2-e^2\,x^2}\,\left (\frac {631\,x}{2640\,d^5}-\frac {113}{528\,d^4\,e}\right )}{{\left (d+e\,x\right )}^3\,{\left (d-e\,x\right )}^3}-\frac {\sqrt {d^2-e^2\,x^2}}{88\,d^4\,e\,{\left (d+e\,x\right )}^6}-\frac {43\,\sqrt {d^2-e^2\,x^2}}{1584\,d^5\,e\,{\left (d+e\,x\right )}^5}-\frac {67\,\sqrt {d^2-e^2\,x^2}}{1584\,d^6\,e\,{\left (d+e\,x\right )}^4}+\frac {128\,x\,\sqrt {d^2-e^2\,x^2}}{495\,d^9\,\left (d+e\,x\right )\,\left (d-e\,x\right )} \]

[In]

int(1/((d^2 - e^2*x^2)^(7/2)*(d + e*x)^3),x)

[Out]

((d^2 - e^2*x^2)^(1/2)*((64*x)/(495*d^7) + 67/(1584*d^6*e)))/((d + e*x)^2*(d - e*x)^2) + ((d^2 - e^2*x^2)^(1/2
)*((631*x)/(2640*d^5) - 113/(528*d^4*e)))/((d + e*x)^3*(d - e*x)^3) - (d^2 - e^2*x^2)^(1/2)/(88*d^4*e*(d + e*x
)^6) - (43*(d^2 - e^2*x^2)^(1/2))/(1584*d^5*e*(d + e*x)^5) - (67*(d^2 - e^2*x^2)^(1/2))/(1584*d^6*e*(d + e*x)^
4) + (128*x*(d^2 - e^2*x^2)^(1/2))/(495*d^9*(d + e*x)*(d - e*x))