3.7.85 \(\int \frac {1}{(d+e x)^3 (d^2-e^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=148 \[ -\frac {2}{21 d^2 e (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {1}{9 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {16 x}{63 d^7 \sqrt {d^2-e^2 x^2}}+\frac {8 x}{63 d^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2}{21 d^3 e (d+e x) \left (d^2-e^2 x^2\right )^{3/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {659, 192, 191} \begin {gather*} \frac {16 x}{63 d^7 \sqrt {d^2-e^2 x^2}}+\frac {8 x}{63 d^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2}{21 d^3 e (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2}{21 d^2 e (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {1}{9 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^3*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

(8*x)/(63*d^5*(d^2 - e^2*x^2)^(3/2)) - 1/(9*d*e*(d + e*x)^3*(d^2 - e^2*x^2)^(3/2)) - 2/(21*d^2*e*(d + e*x)^2*(
d^2 - e^2*x^2)^(3/2)) - 2/(21*d^3*e*(d + e*x)*(d^2 - e^2*x^2)^(3/2)) + (16*x)/(63*d^7*Sqrt[d^2 - e^2*x^2])

Rule 191

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 192

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 659

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*} \int \frac {1}{(d+e x)^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx &=-\frac {1}{9 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 \int \frac {1}{(d+e x)^2 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{3 d}\\ &=-\frac {1}{9 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2}{21 d^2 e (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {10 \int \frac {1}{(d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{21 d^2}\\ &=-\frac {1}{9 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2}{21 d^2 e (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2}{21 d^3 e (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 \int \frac {1}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{21 d^3}\\ &=\frac {8 x}{63 d^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {1}{9 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2}{21 d^2 e (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2}{21 d^3 e (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {16 \int \frac {1}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{63 d^5}\\ &=\frac {8 x}{63 d^5 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {1}{9 d e (d+e x)^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2}{21 d^2 e (d+e x)^2 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2}{21 d^3 e (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {16 x}{63 d^7 \sqrt {d^2-e^2 x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 104, normalized size = 0.70 \begin {gather*} -\frac {\sqrt {d^2-e^2 x^2} \left (19 d^6-6 d^5 e x-66 d^4 e^2 x^2-56 d^3 e^3 x^3+24 d^2 e^4 x^4+48 d e^5 x^5+16 e^6 x^6\right )}{63 d^7 e (d-e x)^2 (d+e x)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^3*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

-1/63*(Sqrt[d^2 - e^2*x^2]*(19*d^6 - 6*d^5*e*x - 66*d^4*e^2*x^2 - 56*d^3*e^3*x^3 + 24*d^2*e^4*x^4 + 48*d*e^5*x
^5 + 16*e^6*x^6))/(d^7*e*(d - e*x)^2*(d + e*x)^5)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.55, size = 104, normalized size = 0.70 \begin {gather*} \frac {\sqrt {d^2-e^2 x^2} \left (-19 d^6+6 d^5 e x+66 d^4 e^2 x^2+56 d^3 e^3 x^3-24 d^2 e^4 x^4-48 d e^5 x^5-16 e^6 x^6\right )}{63 d^7 e (d-e x)^2 (d+e x)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((d + e*x)^3*(d^2 - e^2*x^2)^(5/2)),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-19*d^6 + 6*d^5*e*x + 66*d^4*e^2*x^2 + 56*d^3*e^3*x^3 - 24*d^2*e^4*x^4 - 48*d*e^5*x^5 -
16*e^6*x^6))/(63*d^7*e*(d - e*x)^2*(d + e*x)^5)

________________________________________________________________________________________

fricas [A]  time = 0.54, size = 234, normalized size = 1.58 \begin {gather*} -\frac {19 \, e^{7} x^{7} + 57 \, d e^{6} x^{6} + 19 \, d^{2} e^{5} x^{5} - 95 \, d^{3} e^{4} x^{4} - 95 \, d^{4} e^{3} x^{3} + 19 \, d^{5} e^{2} x^{2} + 57 \, d^{6} e x + 19 \, d^{7} + {\left (16 \, e^{6} x^{6} + 48 \, d e^{5} x^{5} + 24 \, d^{2} e^{4} x^{4} - 56 \, d^{3} e^{3} x^{3} - 66 \, d^{4} e^{2} x^{2} - 6 \, d^{5} e x + 19 \, d^{6}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{63 \, {\left (d^{7} e^{8} x^{7} + 3 \, d^{8} e^{7} x^{6} + d^{9} e^{6} x^{5} - 5 \, d^{10} e^{5} x^{4} - 5 \, d^{11} e^{4} x^{3} + d^{12} e^{3} x^{2} + 3 \, d^{13} e^{2} x + d^{14} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/(-e^2*x^2+d^2)^(5/2),x, algorithm="fricas")

[Out]

-1/63*(19*e^7*x^7 + 57*d*e^6*x^6 + 19*d^2*e^5*x^5 - 95*d^3*e^4*x^4 - 95*d^4*e^3*x^3 + 19*d^5*e^2*x^2 + 57*d^6*
e*x + 19*d^7 + (16*e^6*x^6 + 48*d*e^5*x^5 + 24*d^2*e^4*x^4 - 56*d^3*e^3*x^3 - 66*d^4*e^2*x^2 - 6*d^5*e*x + 19*
d^6)*sqrt(-e^2*x^2 + d^2))/(d^7*e^8*x^7 + 3*d^8*e^7*x^6 + d^9*e^6*x^5 - 5*d^10*e^5*x^4 - 5*d^11*e^4*x^3 + d^12
*e^3*x^2 + 3*d^13*e^2*x + d^14*e)

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/(-e^2*x^2+d^2)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to transpose Error: Bad Argument Value

________________________________________________________________________________________

maple [A]  time = 0.06, size = 99, normalized size = 0.67 \begin {gather*} -\frac {\left (-e x +d \right ) \left (16 e^{6} x^{6}+48 e^{5} x^{5} d +24 e^{4} x^{4} d^{2}-56 e^{3} x^{3} d^{3}-66 e^{2} x^{2} d^{4}-6 x \,d^{5} e +19 d^{6}\right )}{63 \left (e x +d \right )^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d^{7} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/63*(-e*x+d)*(16*e^6*x^6+48*d*e^5*x^5+24*d^2*e^4*x^4-56*d^3*e^3*x^3-66*d^4*e^2*x^2-6*d^5*e*x+19*d^6)/(e*x+d)
^2/d^7/e/(-e^2*x^2+d^2)^(5/2)

________________________________________________________________________________________

maxima [A]  time = 1.46, size = 252, normalized size = 1.70 \begin {gather*} -\frac {1}{9 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d e^{4} x^{3} + 3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} e^{3} x^{2} + 3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3} e^{2} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4} e\right )}} - \frac {2}{21 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} e^{3} x^{2} + 2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3} e^{2} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4} e\right )}} - \frac {2}{21 \, {\left ({\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3} e^{2} x + {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4} e\right )}} + \frac {8 \, x}{63 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{5}} + \frac {16 \, x}{63 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/(-e^2*x^2+d^2)^(5/2),x, algorithm="maxima")

[Out]

-1/9/((-e^2*x^2 + d^2)^(3/2)*d*e^4*x^3 + 3*(-e^2*x^2 + d^2)^(3/2)*d^2*e^3*x^2 + 3*(-e^2*x^2 + d^2)^(3/2)*d^3*e
^2*x + (-e^2*x^2 + d^2)^(3/2)*d^4*e) - 2/21/((-e^2*x^2 + d^2)^(3/2)*d^2*e^3*x^2 + 2*(-e^2*x^2 + d^2)^(3/2)*d^3
*e^2*x + (-e^2*x^2 + d^2)^(3/2)*d^4*e) - 2/21/((-e^2*x^2 + d^2)^(3/2)*d^3*e^2*x + (-e^2*x^2 + d^2)^(3/2)*d^4*e
) + 8/63*x/((-e^2*x^2 + d^2)^(3/2)*d^5) + 16/63*x/(sqrt(-e^2*x^2 + d^2)*d^7)

________________________________________________________________________________________

mupad [B]  time = 0.71, size = 168, normalized size = 1.14 \begin {gather*} \frac {\sqrt {d^2-e^2\,x^2}\,\left (\frac {197\,x}{1008\,d^5}-\frac {155}{1008\,d^4\,e}\right )}{{\left (d+e\,x\right )}^2\,{\left (d-e\,x\right )}^2}-\frac {\sqrt {d^2-e^2\,x^2}}{36\,d^3\,e\,{\left (d+e\,x\right )}^5}-\frac {13\,\sqrt {d^2-e^2\,x^2}}{252\,d^4\,e\,{\left (d+e\,x\right )}^4}-\frac {23\,\sqrt {d^2-e^2\,x^2}}{336\,d^5\,e\,{\left (d+e\,x\right )}^3}+\frac {16\,x\,\sqrt {d^2-e^2\,x^2}}{63\,d^7\,\left (d+e\,x\right )\,\left (d-e\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d^2 - e^2*x^2)^(1/2)*((197*x)/(1008*d^5) - 155/(1008*d^4*e)))/((d + e*x)^2*(d - e*x)^2) - (d^2 - e^2*x^2)^(1
/2)/(36*d^3*e*(d + e*x)^5) - (13*(d^2 - e^2*x^2)^(1/2))/(252*d^4*e*(d + e*x)^4) - (23*(d^2 - e^2*x^2)^(1/2))/(
336*d^5*e*(d + e*x)^3) + (16*x*(d^2 - e^2*x^2)^(1/2))/(63*d^7*(d + e*x)*(d - e*x))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}} \left (d + e x\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**3/(-e**2*x**2+d**2)**(5/2),x)

[Out]

Integral(1/((-(-d + e*x)*(d + e*x))**(5/2)*(d + e*x)**3), x)

________________________________________________________________________________________