3.1.53 \(\int \frac {(d+e x)^2}{x^4 (d^2-e^2 x^2)^{7/2}} \, dx\)

Optimal. Leaf size=209 \[ -\frac {14 e^2 \sqrt {d^2-e^2 x^2}}{3 d^8 x}+\frac {2 e^3 (45 d+53 e x)}{15 d^8 \sqrt {d^2-e^2 x^2}}-\frac {7 e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^8}-\frac {e \sqrt {d^2-e^2 x^2}}{d^7 x^2}-\frac {\sqrt {d^2-e^2 x^2}}{3 d^6 x^3}+\frac {e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}} \]

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Rubi [A]  time = 0.47, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1805, 1807, 807, 266, 63, 208} \begin {gather*} \frac {2 e^3 (45 d+53 e x)}{15 d^8 \sqrt {d^2-e^2 x^2}}+\frac {e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {14 e^2 \sqrt {d^2-e^2 x^2}}{3 d^8 x}-\frac {e \sqrt {d^2-e^2 x^2}}{d^7 x^2}-\frac {\sqrt {d^2-e^2 x^2}}{3 d^6 x^3}-\frac {7 e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2/(x^4*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

(2*e^3*(d + e*x))/(5*d^4*(d^2 - e^2*x^2)^(5/2)) + (e^3*(20*d + 23*e*x))/(15*d^6*(d^2 - e^2*x^2)^(3/2)) + (2*e^
3*(45*d + 53*e*x))/(15*d^8*Sqrt[d^2 - e^2*x^2]) - Sqrt[d^2 - e^2*x^2]/(3*d^6*x^3) - (e*Sqrt[d^2 - e^2*x^2])/(d
^7*x^2) - (14*e^2*Sqrt[d^2 - e^2*x^2])/(3*d^8*x) - (7*e^3*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/d^8

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rubi steps

\begin {align*} \int \frac {(d+e x)^2}{x^4 \left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 d^2-10 d e x-10 e^2 x^2-\frac {10 e^3 x^3}{d}-\frac {8 e^4 x^4}{d^2}}{x^4 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2}\\ &=\frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {15 d^2+30 d e x+45 e^2 x^2+\frac {60 e^3 x^3}{d}+\frac {46 e^4 x^4}{d^2}}{x^4 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^4}\\ &=\frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^3 (45 d+53 e x)}{15 d^8 \sqrt {d^2-e^2 x^2}}-\frac {\int \frac {-15 d^2-30 d e x-60 e^2 x^2-\frac {90 e^3 x^3}{d}}{x^4 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^6}\\ &=\frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^3 (45 d+53 e x)}{15 d^8 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{3 d^6 x^3}+\frac {\int \frac {90 d^3 e+210 d^2 e^2 x+270 d e^3 x^2}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{45 d^8}\\ &=\frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^3 (45 d+53 e x)}{15 d^8 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{3 d^6 x^3}-\frac {e \sqrt {d^2-e^2 x^2}}{d^7 x^2}-\frac {\int \frac {-420 d^4 e^2-630 d^3 e^3 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{90 d^{10}}\\ &=\frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^3 (45 d+53 e x)}{15 d^8 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{3 d^6 x^3}-\frac {e \sqrt {d^2-e^2 x^2}}{d^7 x^2}-\frac {14 e^2 \sqrt {d^2-e^2 x^2}}{3 d^8 x}+\frac {\left (7 e^3\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^7}\\ &=\frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^3 (45 d+53 e x)}{15 d^8 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{3 d^6 x^3}-\frac {e \sqrt {d^2-e^2 x^2}}{d^7 x^2}-\frac {14 e^2 \sqrt {d^2-e^2 x^2}}{3 d^8 x}+\frac {\left (7 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{2 d^7}\\ &=\frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^3 (45 d+53 e x)}{15 d^8 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{3 d^6 x^3}-\frac {e \sqrt {d^2-e^2 x^2}}{d^7 x^2}-\frac {14 e^2 \sqrt {d^2-e^2 x^2}}{3 d^8 x}-\frac {(7 e) \operatorname {Subst}\left (\int \frac {1}{\frac {d^2}{e^2}-\frac {x^2}{e^2}} \, dx,x,\sqrt {d^2-e^2 x^2}\right )}{d^7}\\ &=\frac {2 e^3 (d+e x)}{5 d^4 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {e^3 (20 d+23 e x)}{15 d^6 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 e^3 (45 d+53 e x)}{15 d^8 \sqrt {d^2-e^2 x^2}}-\frac {\sqrt {d^2-e^2 x^2}}{3 d^6 x^3}-\frac {e \sqrt {d^2-e^2 x^2}}{d^7 x^2}-\frac {14 e^2 \sqrt {d^2-e^2 x^2}}{3 d^8 x}-\frac {7 e^3 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^8}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 105, normalized size = 0.50 \begin {gather*} \frac {-5 d^8-55 d^6 e^2 x^2+330 d^4 e^4 x^4-440 d^2 e^6 x^6+6 d^5 e^3 x^3 \, _2F_1\left (-\frac {5}{2},2;-\frac {3}{2};1-\frac {e^2 x^2}{d^2}\right )+176 e^8 x^8}{15 d^8 x^3 \left (d^2-e^2 x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^2/(x^4*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

(-5*d^8 - 55*d^6*e^2*x^2 + 330*d^4*e^4*x^4 - 440*d^2*e^6*x^6 + 176*e^8*x^8 + 6*d^5*e^3*x^3*Hypergeometric2F1[-
5/2, 2, -3/2, 1 - (e^2*x^2)/d^2])/(15*d^8*x^3*(d^2 - e^2*x^2)^(5/2))

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IntegrateAlgebraic [A]  time = 1.05, size = 150, normalized size = 0.72 \begin {gather*} \frac {14 e^3 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^8}+\frac {\sqrt {d^2-e^2 x^2} \left (-5 d^6-5 d^5 e x-40 d^4 e^2 x^2+246 d^3 e^3 x^3-122 d^2 e^4 x^4-247 d e^5 x^5+176 e^6 x^6\right )}{15 d^8 x^3 (d-e x)^3 (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^2/(x^4*(d^2 - e^2*x^2)^(7/2)),x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(-5*d^6 - 5*d^5*e*x - 40*d^4*e^2*x^2 + 246*d^3*e^3*x^3 - 122*d^2*e^4*x^4 - 247*d*e^5*x^5
+ 176*e^6*x^6))/(15*d^8*x^3*(d - e*x)^3*(d + e*x)) + (14*e^3*ArcTanh[(Sqrt[-e^2]*x)/d - Sqrt[d^2 - e^2*x^2]/d]
)/d^8

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fricas [A]  time = 0.47, size = 227, normalized size = 1.09 \begin {gather*} \frac {116 \, e^{7} x^{7} - 232 \, d e^{6} x^{6} + 232 \, d^{3} e^{4} x^{4} - 116 \, d^{4} e^{3} x^{3} + 105 \, {\left (e^{7} x^{7} - 2 \, d e^{6} x^{6} + 2 \, d^{3} e^{4} x^{4} - d^{4} e^{3} x^{3}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (176 \, e^{6} x^{6} - 247 \, d e^{5} x^{5} - 122 \, d^{2} e^{4} x^{4} + 246 \, d^{3} e^{3} x^{3} - 40 \, d^{4} e^{2} x^{2} - 5 \, d^{5} e x - 5 \, d^{6}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{8} e^{4} x^{7} - 2 \, d^{9} e^{3} x^{6} + 2 \, d^{11} e x^{4} - d^{12} x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/x^4/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")

[Out]

1/15*(116*e^7*x^7 - 232*d*e^6*x^6 + 232*d^3*e^4*x^4 - 116*d^4*e^3*x^3 + 105*(e^7*x^7 - 2*d*e^6*x^6 + 2*d^3*e^4
*x^4 - d^4*e^3*x^3)*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - (176*e^6*x^6 - 247*d*e^5*x^5 - 122*d^2*e^4*x^4 + 246*
d^3*e^3*x^3 - 40*d^4*e^2*x^2 - 5*d^5*e*x - 5*d^6)*sqrt(-e^2*x^2 + d^2))/(d^8*e^4*x^7 - 2*d^9*e^3*x^6 + 2*d^11*
e*x^4 - d^12*x^3)

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giac [A]  time = 0.35, size = 325, normalized size = 1.56 \begin {gather*} -\frac {\sqrt {-x^{2} e^{2} + d^{2}} {\left ({\left ({\left ({\left (2 \, x {\left (\frac {53 \, x e^{8}}{d^{8}} + \frac {45 \, e^{7}}{d^{7}}\right )} - \frac {235 \, e^{6}}{d^{6}}\right )} x - \frac {200 \, e^{5}}{d^{5}}\right )} x + \frac {135 \, e^{4}}{d^{4}}\right )} x + \frac {116 \, e^{3}}{d^{3}}\right )}}{15 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} + \frac {x^{3} {\left (\frac {6 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{6}}{x} + \frac {57 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} e^{4}}{x^{2}} + e^{8}\right )} e}{24 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{8}} - \frac {7 \, e^{3} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \, {\left | x \right |}}\right )}{d^{8}} - \frac {{\left (\frac {57 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{16} e^{16}}{x} + \frac {6 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{16} e^{14}}{x^{2}} + \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{3} d^{16} e^{12}}{x^{3}}\right )} e^{\left (-15\right )}}{24 \, d^{24}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/x^4/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")

[Out]

-1/15*sqrt(-x^2*e^2 + d^2)*((((2*x*(53*x*e^8/d^8 + 45*e^7/d^7) - 235*e^6/d^6)*x - 200*e^5/d^5)*x + 135*e^4/d^4
)*x + 116*e^3/d^3)/(x^2*e^2 - d^2)^3 + 1/24*x^3*(6*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^6/x + 57*(d*e + sqrt(-x^2*
e^2 + d^2)*e)^2*e^4/x^2 + e^8)*e/((d*e + sqrt(-x^2*e^2 + d^2)*e)^3*d^8) - 7*e^3*log(1/2*abs(-2*d*e - 2*sqrt(-x
^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^8 - 1/24*(57*(d*e + sqrt(-x^2*e^2 + d^2)*e)*d^16*e^16/x + 6*(d*e + sqrt(-x^2
*e^2 + d^2)*e)^2*d^16*e^14/x^2 + (d*e + sqrt(-x^2*e^2 + d^2)*e)^3*d^16*e^12/x^3)*e^(-15)/d^24

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maple [A]  time = 0.02, size = 249, normalized size = 1.19 \begin {gather*} \frac {22 e^{4} x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d^{4}}+\frac {7 e^{3}}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d^{3}}-\frac {11 e^{2}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d^{2} x}+\frac {88 e^{4} x}{15 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{6}}-\frac {e}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d \,x^{2}}+\frac {7 e^{3}}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{5}}-\frac {7 e^{3} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}\, d^{7}}+\frac {176 e^{4} x}{15 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{8}}-\frac {1}{3 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} x^{3}}+\frac {7 e^{3}}{\sqrt {-e^{2} x^{2}+d^{2}}\, d^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2/x^4/(-e^2*x^2+d^2)^(7/2),x)

[Out]

-1/d*e/x^2/(-e^2*x^2+d^2)^(5/2)+7/5/d^3*e^3/(-e^2*x^2+d^2)^(5/2)+7/3/d^5*e^3/(-e^2*x^2+d^2)^(3/2)+7/d^7*e^3/(-
e^2*x^2+d^2)^(1/2)-7/d^7*e^3/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-11/3*e^2/d^2/x/(-e^2
*x^2+d^2)^(5/2)+22/5*e^4/d^4*x/(-e^2*x^2+d^2)^(5/2)+88/15*e^4/d^6*x/(-e^2*x^2+d^2)^(3/2)+176/15*e^4/d^8*x/(-e^
2*x^2+d^2)^(1/2)-1/3/x^3/(-e^2*x^2+d^2)^(5/2)

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maxima [A]  time = 0.48, size = 243, normalized size = 1.16 \begin {gather*} \frac {22 \, e^{4} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{4}} + \frac {7 \, e^{3}}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3}} + \frac {88 \, e^{4} x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{6}} + \frac {7 \, e^{3}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{5}} - \frac {11 \, e^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2} x} + \frac {176 \, e^{4} x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{8}} - \frac {7 \, e^{3} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{d^{8}} + \frac {7 \, e^{3}}{\sqrt {-e^{2} x^{2} + d^{2}} d^{7}} - \frac {e}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d x^{2}} - \frac {1}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/x^4/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")

[Out]

22/5*e^4*x/((-e^2*x^2 + d^2)^(5/2)*d^4) + 7/5*e^3/((-e^2*x^2 + d^2)^(5/2)*d^3) + 88/15*e^4*x/((-e^2*x^2 + d^2)
^(3/2)*d^6) + 7/3*e^3/((-e^2*x^2 + d^2)^(3/2)*d^5) - 11/3*e^2/((-e^2*x^2 + d^2)^(5/2)*d^2*x) + 176/15*e^4*x/(s
qrt(-e^2*x^2 + d^2)*d^8) - 7*e^3*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d^8 + 7*e^3/(sqrt(-e^2*x^
2 + d^2)*d^7) - e/((-e^2*x^2 + d^2)^(5/2)*d*x^2) - 1/3/((-e^2*x^2 + d^2)^(5/2)*x^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^2}{x^4\,{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^2/(x^4*(d^2 - e^2*x^2)^(7/2)),x)

[Out]

int((d + e*x)^2/(x^4*(d^2 - e^2*x^2)^(7/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{2}}{x^{4} \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2/x**4/(-e**2*x**2+d**2)**(7/2),x)

[Out]

Integral((d + e*x)**2/(x**4*(-(-d + e*x)*(d + e*x))**(7/2)), x)

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