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

Optimal. Leaf size=184 \[ \frac {d+e x}{5 d^2 x^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {16 e \sqrt {d^2-e^2 x^2}}{5 d^8 x}-\frac {7 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^8}-\frac {7 \sqrt {d^2-e^2 x^2}}{2 d^7 x^2}+\frac {35 d+24 e x}{15 d^6 x^2 \sqrt {d^2-e^2 x^2}}+\frac {7 d+6 e x}{15 d^4 x^2 \left (d^2-e^2 x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.16, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {823, 835, 807, 266, 63, 208} \begin {gather*} -\frac {16 e \sqrt {d^2-e^2 x^2}}{5 d^8 x}-\frac {7 \sqrt {d^2-e^2 x^2}}{2 d^7 x^2}+\frac {35 d+24 e x}{15 d^6 x^2 \sqrt {d^2-e^2 x^2}}+\frac {7 d+6 e x}{15 d^4 x^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {d+e x}{5 d^2 x^2 \left (d^2-e^2 x^2\right )^{5/2}}-\frac {7 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^8} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d + e*x)/(5*d^2*x^2*(d^2 - e^2*x^2)^(5/2)) + (7*d + 6*e*x)/(15*d^4*x^2*(d^2 - e^2*x^2)^(3/2)) + (35*d + 24*e*
x)/(15*d^6*x^2*Sqrt[d^2 - e^2*x^2]) - (7*Sqrt[d^2 - e^2*x^2])/(2*d^7*x^2) - (16*e*Sqrt[d^2 - e^2*x^2])/(5*d^8*
x) - (7*e^2*ArcTanh[Sqrt[d^2 - e^2*x^2]/d])/(2*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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 835

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))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rubi steps

\begin {align*} \int \frac {d+e x}{x^3 \left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac {d+e x}{5 d^2 x^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {\int \frac {7 d^3 e^2+6 d^2 e^3 x}{x^3 \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^4 e^2}\\ &=\frac {d+e x}{5 d^2 x^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {7 d+6 e x}{15 d^4 x^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {\int \frac {35 d^5 e^4+24 d^4 e^5 x}{x^3 \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^8 e^4}\\ &=\frac {d+e x}{5 d^2 x^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {7 d+6 e x}{15 d^4 x^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {35 d+24 e x}{15 d^6 x^2 \sqrt {d^2-e^2 x^2}}+\frac {\int \frac {105 d^7 e^6+48 d^6 e^7 x}{x^3 \sqrt {d^2-e^2 x^2}} \, dx}{15 d^{12} e^6}\\ &=\frac {d+e x}{5 d^2 x^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {7 d+6 e x}{15 d^4 x^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {35 d+24 e x}{15 d^6 x^2 \sqrt {d^2-e^2 x^2}}-\frac {7 \sqrt {d^2-e^2 x^2}}{2 d^7 x^2}-\frac {\int \frac {-96 d^8 e^7-105 d^7 e^8 x}{x^2 \sqrt {d^2-e^2 x^2}} \, dx}{30 d^{14} e^6}\\ &=\frac {d+e x}{5 d^2 x^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {7 d+6 e x}{15 d^4 x^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {35 d+24 e x}{15 d^6 x^2 \sqrt {d^2-e^2 x^2}}-\frac {7 \sqrt {d^2-e^2 x^2}}{2 d^7 x^2}-\frac {16 e \sqrt {d^2-e^2 x^2}}{5 d^8 x}+\frac {\left (7 e^2\right ) \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{2 d^7}\\ &=\frac {d+e x}{5 d^2 x^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {7 d+6 e x}{15 d^4 x^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {35 d+24 e x}{15 d^6 x^2 \sqrt {d^2-e^2 x^2}}-\frac {7 \sqrt {d^2-e^2 x^2}}{2 d^7 x^2}-\frac {16 e \sqrt {d^2-e^2 x^2}}{5 d^8 x}+\frac {\left (7 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{4 d^7}\\ &=\frac {d+e x}{5 d^2 x^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {7 d+6 e x}{15 d^4 x^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {35 d+24 e x}{15 d^6 x^2 \sqrt {d^2-e^2 x^2}}-\frac {7 \sqrt {d^2-e^2 x^2}}{2 d^7 x^2}-\frac {16 e \sqrt {d^2-e^2 x^2}}{5 d^8 x}-\frac {7 \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 )}{2 d^7}\\ &=\frac {d+e x}{5 d^2 x^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {7 d+6 e x}{15 d^4 x^2 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {35 d+24 e x}{15 d^6 x^2 \sqrt {d^2-e^2 x^2}}-\frac {7 \sqrt {d^2-e^2 x^2}}{2 d^7 x^2}-\frac {16 e \sqrt {d^2-e^2 x^2}}{5 d^8 x}-\frac {7 e^2 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{2 d^8}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 183, normalized size = 0.99 \begin {gather*} \frac {105 e^2 x^2 (d+e x)^2 (e x-d)^3 \tanh ^{-1}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right )+d \sqrt {1-\frac {e^2 x^2}{d^2}} \left (-15 d^6-15 d^5 e x+176 d^4 e^2 x^2+4 d^3 e^3 x^3-249 d^2 e^4 x^4+9 d e^5 x^5+96 e^6 x^6\right )}{30 d^9 x^2 (d-e x)^2 (d+e x) \sqrt {d^2-e^2 x^2} \sqrt {1-\frac {e^2 x^2}{d^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*Sqrt[1 - (e^2*x^2)/d^2]*(-15*d^6 - 15*d^5*e*x + 176*d^4*e^2*x^2 + 4*d^3*e^3*x^3 - 249*d^2*e^4*x^4 + 9*d*e^5
*x^5 + 96*e^6*x^6) + 105*e^2*x^2*(-d + e*x)^3*(d + e*x)^2*ArcTanh[Sqrt[1 - (e^2*x^2)/d^2]])/(30*d^9*x^2*(d - e
*x)^2*(d + e*x)*Sqrt[d^2 - e^2*x^2]*Sqrt[1 - (e^2*x^2)/d^2])

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IntegrateAlgebraic [A]  time = 0.81, size = 150, normalized size = 0.82 \begin {gather*} \frac {7 e^2 \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 (-15 d^6-15 d^5 e x+176 d^4 e^2 x^2+4 d^3 e^3 x^3-249 d^2 e^4 x^4+9 d e^5 x^5+96 e^6 x^6\right )}{30 d^8 x^2 (d-e x)^3 (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(-15*d^6 - 15*d^5*e*x + 176*d^4*e^2*x^2 + 4*d^3*e^3*x^3 - 249*d^2*e^4*x^4 + 9*d*e^5*x^5 +
96*e^6*x^6))/(30*d^8*x^2*(d - e*x)^3*(d + e*x)^2) + (7*e^2*ArcTanh[(Sqrt[-e^2]*x)/d - Sqrt[d^2 - e^2*x^2]/d])
/d^8

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fricas [A]  time = 0.51, size = 291, normalized size = 1.58 \begin {gather*} \frac {116 \, e^{7} x^{7} - 116 \, d e^{6} x^{6} - 232 \, d^{2} e^{5} x^{5} + 232 \, d^{3} e^{4} x^{4} + 116 \, d^{4} e^{3} x^{3} - 116 \, d^{5} e^{2} x^{2} + 105 \, {\left (e^{7} x^{7} - d e^{6} x^{6} - 2 \, d^{2} e^{5} x^{5} + 2 \, d^{3} e^{4} x^{4} + d^{4} e^{3} x^{3} - d^{5} e^{2} x^{2}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (96 \, e^{6} x^{6} + 9 \, d e^{5} x^{5} - 249 \, d^{2} e^{4} x^{4} + 4 \, d^{3} e^{3} x^{3} + 176 \, d^{4} e^{2} x^{2} - 15 \, d^{5} e x - 15 \, d^{6}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{30 \, {\left (d^{8} e^{5} x^{7} - d^{9} e^{4} x^{6} - 2 \, d^{10} e^{3} x^{5} + 2 \, d^{11} e^{2} x^{4} + d^{12} e x^{3} - d^{13} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(116*e^7*x^7 - 116*d*e^6*x^6 - 232*d^2*e^5*x^5 + 232*d^3*e^4*x^4 + 116*d^4*e^3*x^3 - 116*d^5*e^2*x^2 + 10
5*(e^7*x^7 - d*e^6*x^6 - 2*d^2*e^5*x^5 + 2*d^3*e^4*x^4 + d^4*e^3*x^3 - d^5*e^2*x^2)*log(-(d - sqrt(-e^2*x^2 +
d^2))/x) - (96*e^6*x^6 + 9*d*e^5*x^5 - 249*d^2*e^4*x^4 + 4*d^3*e^3*x^3 + 176*d^4*e^2*x^2 - 15*d^5*e*x - 15*d^6
)*sqrt(-e^2*x^2 + d^2))/(d^8*e^5*x^7 - d^9*e^4*x^6 - 2*d^10*e^3*x^5 + 2*d^11*e^2*x^4 + d^12*e*x^3 - d^13*x^2)

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giac [A]  time = 0.36, size = 260, normalized size = 1.41 \begin {gather*} -\frac {\sqrt {-x^{2} e^{2} + d^{2}} {\left ({\left ({\left (3 \, {\left (x {\left (\frac {11 \, x e^{7}}{d^{8}} + \frac {15 \, e^{6}}{d^{7}}\right )} - \frac {25 \, e^{5}}{d^{6}}\right )} x - \frac {100 \, e^{4}}{d^{5}}\right )} x + \frac {45 \, e^{3}}{d^{4}}\right )} x + \frac {58 \, e^{2}}{d^{3}}\right )}}{15 \, {\left (x^{2} e^{2} - d^{2}\right )}^{3}} - \frac {7 \, e^{2} \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 )}{2 \, d^{8}} + \frac {x^{2} {\left (\frac {4 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{4}}{x} + e^{6}\right )}}{8 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{8}} - \frac {{\left (\frac {4 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} d^{8} e^{8}}{x} + \frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} d^{8} e^{6}}{x^{2}}\right )} e^{\left (-8\right )}}{8 \, d^{16}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*sqrt(-x^2*e^2 + d^2)*(((3*(x*(11*x*e^7/d^8 + 15*e^6/d^7) - 25*e^5/d^6)*x - 100*e^4/d^5)*x + 45*e^3/d^4)*
x + 58*e^2/d^3)/(x^2*e^2 - d^2)^3 - 7/2*e^2*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x))/d^8
+ 1/8*x^2*(4*(d*e + sqrt(-x^2*e^2 + d^2)*e)*e^4/x + e^6)/((d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^8) - 1/8*(4*(d*e
+ sqrt(-x^2*e^2 + d^2)*e)*d^8*e^8/x + (d*e + sqrt(-x^2*e^2 + d^2)*e)^2*d^8*e^6/x^2)*e^(-8)/d^16

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maple [A]  time = 0.02, size = 227, normalized size = 1.23 \begin {gather*} \frac {6 e^{3} x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d^{4}}+\frac {7 e^{2}}{10 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d^{3}}-\frac {e}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d^{2} x}+\frac {8 e^{3} x}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{6}}-\frac {1}{2 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}} d \,x^{2}}+\frac {7 e^{2}}{6 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d^{5}}-\frac {7 e^{2} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{2 \sqrt {d^{2}}\, d^{7}}+\frac {16 e^{3} x}{5 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{8}}+\frac {7 e^{2}}{2 \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)/x^3/(-e^2*x^2+d^2)^(7/2),x)

[Out]

-1/2/d/x^2/(-e^2*x^2+d^2)^(5/2)+7/10*e^2/d^3/(-e^2*x^2+d^2)^(5/2)+7/6*e^2/d^5/(-e^2*x^2+d^2)^(3/2)+7/2*e^2/d^7
/(-e^2*x^2+d^2)^(1/2)-7/2*e^2/d^7/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(-e^2*x^2+d^2)^(1/2))/x)-e/d^2/x/(-e^2*x
^2+d^2)^(5/2)+6/5*e^3/d^4*x/(-e^2*x^2+d^2)^(5/2)+8/5*e^3/d^6*x/(-e^2*x^2+d^2)^(3/2)+16/5*e^3/d^8*x/(-e^2*x^2+d
^2)^(1/2)

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maxima [A]  time = 0.47, size = 221, normalized size = 1.20 \begin {gather*} \frac {6 \, e^{3} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{4}} + \frac {7 \, e^{2}}{10 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3}} + \frac {8 \, e^{3} x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{6}} + \frac {7 \, e^{2}}{6 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{5}} - \frac {e}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2} x} + \frac {16 \, e^{3} x}{5 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{8}} - \frac {7 \, e^{2} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{2 \, d^{8}} + \frac {7 \, e^{2}}{2 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{7}} - \frac {1}{2 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6/5*e^3*x/((-e^2*x^2 + d^2)^(5/2)*d^4) + 7/10*e^2/((-e^2*x^2 + d^2)^(5/2)*d^3) + 8/5*e^3*x/((-e^2*x^2 + d^2)^(
3/2)*d^6) + 7/6*e^2/((-e^2*x^2 + d^2)^(3/2)*d^5) - e/((-e^2*x^2 + d^2)^(5/2)*d^2*x) + 16/5*e^3*x/(sqrt(-e^2*x^
2 + d^2)*d^8) - 7/2*e^2*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d^8 + 7/2*e^2/(sqrt(-e^2*x^2 + d^2
)*d^7) - 1/2/((-e^2*x^2 + d^2)^(5/2)*d*x^2)

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mupad [B]  time = 3.43, size = 181, normalized size = 0.98 \begin {gather*} \frac {161\,e^2}{30\,d^3\,{\left (d^2-e^2\,x^2\right )}^{5/2}}-\frac {1}{2\,d\,x^2\,{\left (d^2-e^2\,x^2\right )}^{5/2}}-\frac {7\,e^2\,\mathrm {atanh}\left (\frac {\sqrt {d^2-e^2\,x^2}}{d}\right )}{2\,d^8}-\frac {49\,e^4\,x^2}{6\,d^5\,{\left (d^2-e^2\,x^2\right )}^{5/2}}+\frac {7\,e^6\,x^4}{2\,d^7\,{\left (d^2-e^2\,x^2\right )}^{5/2}}-\frac {e\,\left (5\,d^6-30\,d^4\,e^2\,x^2+40\,d^2\,e^4\,x^4-16\,e^6\,x^6\right )}{5\,d^8\,x\,{\left (d^2-e^2\,x^2\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(161*e^2)/(30*d^3*(d^2 - e^2*x^2)^(5/2)) - 1/(2*d*x^2*(d^2 - e^2*x^2)^(5/2)) - (7*e^2*atanh((d^2 - e^2*x^2)^(1
/2)/d))/(2*d^8) - (49*e^4*x^2)/(6*d^5*(d^2 - e^2*x^2)^(5/2)) + (7*e^6*x^4)/(2*d^7*(d^2 - e^2*x^2)^(5/2)) - (e*
(5*d^6 - 16*e^6*x^6 - 30*d^4*e^2*x^2 + 40*d^2*e^4*x^4))/(5*d^8*x*(d^2 - e^2*x^2)^(5/2))

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sympy [C]  time = 35.06, size = 2691, normalized size = 14.62

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*Piecewise((-30*I*d**8*sqrt(-1 + e**2*x**2/d**2)/(60*d**15*x**2 - 180*d**13*e**2*x**4 + 180*d**11*e**4*x**6 -
60*d**9*e**6*x**8) + 322*I*d**6*e**2*x**2*sqrt(-1 + e**2*x**2/d**2)/(60*d**15*x**2 - 180*d**13*e**2*x**4 + 18
0*d**11*e**4*x**6 - 60*d**9*e**6*x**8) + 105*d**6*e**2*x**2*log(e**2*x**2/d**2)/(60*d**15*x**2 - 180*d**13*e**
2*x**4 + 180*d**11*e**4*x**6 - 60*d**9*e**6*x**8) - 210*d**6*e**2*x**2*log(e*x/d)/(60*d**15*x**2 - 180*d**13*e
**2*x**4 + 180*d**11*e**4*x**6 - 60*d**9*e**6*x**8) + 210*I*d**6*e**2*x**2*asin(d/(e*x))/(60*d**15*x**2 - 180*
d**13*e**2*x**4 + 180*d**11*e**4*x**6 - 60*d**9*e**6*x**8) - 490*I*d**4*e**4*x**4*sqrt(-1 + e**2*x**2/d**2)/(6
0*d**15*x**2 - 180*d**13*e**2*x**4 + 180*d**11*e**4*x**6 - 60*d**9*e**6*x**8) - 315*d**4*e**4*x**4*log(e**2*x*
*2/d**2)/(60*d**15*x**2 - 180*d**13*e**2*x**4 + 180*d**11*e**4*x**6 - 60*d**9*e**6*x**8) + 630*d**4*e**4*x**4*
log(e*x/d)/(60*d**15*x**2 - 180*d**13*e**2*x**4 + 180*d**11*e**4*x**6 - 60*d**9*e**6*x**8) - 630*I*d**4*e**4*x
**4*asin(d/(e*x))/(60*d**15*x**2 - 180*d**13*e**2*x**4 + 180*d**11*e**4*x**6 - 60*d**9*e**6*x**8) + 210*I*d**2
*e**6*x**6*sqrt(-1 + e**2*x**2/d**2)/(60*d**15*x**2 - 180*d**13*e**2*x**4 + 180*d**11*e**4*x**6 - 60*d**9*e**6
*x**8) + 315*d**2*e**6*x**6*log(e**2*x**2/d**2)/(60*d**15*x**2 - 180*d**13*e**2*x**4 + 180*d**11*e**4*x**6 - 6
0*d**9*e**6*x**8) - 630*d**2*e**6*x**6*log(e*x/d)/(60*d**15*x**2 - 180*d**13*e**2*x**4 + 180*d**11*e**4*x**6 -
60*d**9*e**6*x**8) + 630*I*d**2*e**6*x**6*asin(d/(e*x))/(60*d**15*x**2 - 180*d**13*e**2*x**4 + 180*d**11*e**4
*x**6 - 60*d**9*e**6*x**8) - 105*e**8*x**8*log(e**2*x**2/d**2)/(60*d**15*x**2 - 180*d**13*e**2*x**4 + 180*d**1
1*e**4*x**6 - 60*d**9*e**6*x**8) + 210*e**8*x**8*log(e*x/d)/(60*d**15*x**2 - 180*d**13*e**2*x**4 + 180*d**11*e
**4*x**6 - 60*d**9*e**6*x**8) - 210*I*e**8*x**8*asin(d/(e*x))/(60*d**15*x**2 - 180*d**13*e**2*x**4 + 180*d**11
*e**4*x**6 - 60*d**9*e**6*x**8), Abs(e**2*x**2/d**2) > 1), (30*d**8*sqrt(1 - e**2*x**2/d**2)/(-60*d**15*x**2 +
180*d**13*e**2*x**4 - 180*d**11*e**4*x**6 + 60*d**9*e**6*x**8) - 322*d**6*e**2*x**2*sqrt(1 - e**2*x**2/d**2)/
(-60*d**15*x**2 + 180*d**13*e**2*x**4 - 180*d**11*e**4*x**6 + 60*d**9*e**6*x**8) - 105*d**6*e**2*x**2*log(e**2
*x**2/d**2)/(-60*d**15*x**2 + 180*d**13*e**2*x**4 - 180*d**11*e**4*x**6 + 60*d**9*e**6*x**8) + 210*d**6*e**2*x
**2*log(sqrt(1 - e**2*x**2/d**2) + 1)/(-60*d**15*x**2 + 180*d**13*e**2*x**4 - 180*d**11*e**4*x**6 + 60*d**9*e*
*6*x**8) - 105*I*pi*d**6*e**2*x**2/(-60*d**15*x**2 + 180*d**13*e**2*x**4 - 180*d**11*e**4*x**6 + 60*d**9*e**6*
x**8) + 490*d**4*e**4*x**4*sqrt(1 - e**2*x**2/d**2)/(-60*d**15*x**2 + 180*d**13*e**2*x**4 - 180*d**11*e**4*x**
6 + 60*d**9*e**6*x**8) + 315*d**4*e**4*x**4*log(e**2*x**2/d**2)/(-60*d**15*x**2 + 180*d**13*e**2*x**4 - 180*d*
*11*e**4*x**6 + 60*d**9*e**6*x**8) - 630*d**4*e**4*x**4*log(sqrt(1 - e**2*x**2/d**2) + 1)/(-60*d**15*x**2 + 18
0*d**13*e**2*x**4 - 180*d**11*e**4*x**6 + 60*d**9*e**6*x**8) + 315*I*pi*d**4*e**4*x**4/(-60*d**15*x**2 + 180*d
**13*e**2*x**4 - 180*d**11*e**4*x**6 + 60*d**9*e**6*x**8) - 210*d**2*e**6*x**6*sqrt(1 - e**2*x**2/d**2)/(-60*d
**15*x**2 + 180*d**13*e**2*x**4 - 180*d**11*e**4*x**6 + 60*d**9*e**6*x**8) - 315*d**2*e**6*x**6*log(e**2*x**2/
d**2)/(-60*d**15*x**2 + 180*d**13*e**2*x**4 - 180*d**11*e**4*x**6 + 60*d**9*e**6*x**8) + 630*d**2*e**6*x**6*lo
g(sqrt(1 - e**2*x**2/d**2) + 1)/(-60*d**15*x**2 + 180*d**13*e**2*x**4 - 180*d**11*e**4*x**6 + 60*d**9*e**6*x**
8) - 315*I*pi*d**2*e**6*x**6/(-60*d**15*x**2 + 180*d**13*e**2*x**4 - 180*d**11*e**4*x**6 + 60*d**9*e**6*x**8)
+ 105*e**8*x**8*log(e**2*x**2/d**2)/(-60*d**15*x**2 + 180*d**13*e**2*x**4 - 180*d**11*e**4*x**6 + 60*d**9*e**6
*x**8) - 210*e**8*x**8*log(sqrt(1 - e**2*x**2/d**2) + 1)/(-60*d**15*x**2 + 180*d**13*e**2*x**4 - 180*d**11*e**
4*x**6 + 60*d**9*e**6*x**8) + 105*I*pi*e**8*x**8/(-60*d**15*x**2 + 180*d**13*e**2*x**4 - 180*d**11*e**4*x**6 +
60*d**9*e**6*x**8), True)) + e*Piecewise((5*d**6*e*sqrt(d**2/(e**2*x**2) - 1)/(-5*d**14 + 15*d**12*e**2*x**2
- 15*d**10*e**4*x**4 + 5*d**8*e**6*x**6) - 30*d**4*e**3*x**2*sqrt(d**2/(e**2*x**2) - 1)/(-5*d**14 + 15*d**12*e
**2*x**2 - 15*d**10*e**4*x**4 + 5*d**8*e**6*x**6) + 40*d**2*e**5*x**4*sqrt(d**2/(e**2*x**2) - 1)/(-5*d**14 + 1
5*d**12*e**2*x**2 - 15*d**10*e**4*x**4 + 5*d**8*e**6*x**6) - 16*e**7*x**6*sqrt(d**2/(e**2*x**2) - 1)/(-5*d**14
+ 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4 + 5*d**8*e**6*x**6), Abs(d**2/(e**2*x**2)) > 1), (5*I*d**6*e*sqrt(-
d**2/(e**2*x**2) + 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4 + 5*d**8*e**6*x**6) - 30*I*d**4*e**3
*x**2*sqrt(-d**2/(e**2*x**2) + 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4 + 5*d**8*e**6*x**6) + 40
*I*d**2*e**5*x**4*sqrt(-d**2/(e**2*x**2) + 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4 + 5*d**8*e**
6*x**6) - 16*I*e**7*x**6*sqrt(-d**2/(e**2*x**2) + 1)/(-5*d**14 + 15*d**12*e**2*x**2 - 15*d**10*e**4*x**4 + 5*d
**8*e**6*x**6), True))

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