3.3.61 \(\int \frac {1}{x^3 (d+e x)^2 \sqrt {a+c x^2}} \, dx\)

Optimal. Leaf size=268 \[ \frac {c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{3/2} d^2}-\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^4}+\frac {2 e \sqrt {a+c x^2}}{a d^3 x}+\frac {c e^3 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{d^2 \left (a e^2+c d^2\right )^{3/2}}-\frac {\sqrt {a+c x^2}}{2 a d^2 x^2}+\frac {3 e^3 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{d^4 \sqrt {a e^2+c d^2}}+\frac {e^4 \sqrt {a+c x^2}}{d^3 (d+e x) \left (a e^2+c d^2\right )} \]

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Rubi [A]  time = 0.22, antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {961, 266, 51, 63, 208, 264, 731, 725, 206} \begin {gather*} \frac {c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{3/2} d^2}+\frac {e^4 \sqrt {a+c x^2}}{d^3 (d+e x) \left (a e^2+c d^2\right )}+\frac {3 e^3 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{d^4 \sqrt {a e^2+c d^2}}+\frac {c e^3 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{d^2 \left (a e^2+c d^2\right )^{3/2}}-\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^4}+\frac {2 e \sqrt {a+c x^2}}{a d^3 x}-\frac {\sqrt {a+c x^2}}{2 a d^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(d + e*x)^2*Sqrt[a + c*x^2]),x]

[Out]

-Sqrt[a + c*x^2]/(2*a*d^2*x^2) + (2*e*Sqrt[a + c*x^2])/(a*d^3*x) + (e^4*Sqrt[a + c*x^2])/(d^3*(c*d^2 + a*e^2)*
(d + e*x)) + (c*e^3*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(d^2*(c*d^2 + a*e^2)^(3/2))
+ (3*e^3*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(d^4*Sqrt[c*d^2 + a*e^2]) + (c*ArcTanh[
Sqrt[a + c*x^2]/Sqrt[a]])/(2*a^(3/2)*d^2) - (3*e^2*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/(Sqrt[a]*d^4)

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 264

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

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 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 731

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

Rule 961

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 0])) &&  !(IGtQ[m, 0] || IGtQ[n, 0])

Rubi steps

\begin {align*} \int \frac {1}{x^3 (d+e x)^2 \sqrt {a+c x^2}} \, dx &=\int \left (\frac {1}{d^2 x^3 \sqrt {a+c x^2}}-\frac {2 e}{d^3 x^2 \sqrt {a+c x^2}}+\frac {3 e^2}{d^4 x \sqrt {a+c x^2}}-\frac {e^3}{d^3 (d+e x)^2 \sqrt {a+c x^2}}-\frac {3 e^3}{d^4 (d+e x) \sqrt {a+c x^2}}\right ) \, dx\\ &=\frac {\int \frac {1}{x^3 \sqrt {a+c x^2}} \, dx}{d^2}-\frac {(2 e) \int \frac {1}{x^2 \sqrt {a+c x^2}} \, dx}{d^3}+\frac {\left (3 e^2\right ) \int \frac {1}{x \sqrt {a+c x^2}} \, dx}{d^4}-\frac {\left (3 e^3\right ) \int \frac {1}{(d+e x) \sqrt {a+c x^2}} \, dx}{d^4}-\frac {e^3 \int \frac {1}{(d+e x)^2 \sqrt {a+c x^2}} \, dx}{d^3}\\ &=\frac {2 e \sqrt {a+c x^2}}{a d^3 x}+\frac {e^4 \sqrt {a+c x^2}}{d^3 \left (c d^2+a e^2\right ) (d+e x)}+\frac {\operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+c x}} \, dx,x,x^2\right )}{2 d^2}+\frac {\left (3 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{2 d^4}+\frac {\left (3 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{c d^2+a e^2-x^2} \, dx,x,\frac {a e-c d x}{\sqrt {a+c x^2}}\right )}{d^4}-\frac {\left (c e^3\right ) \int \frac {1}{(d+e x) \sqrt {a+c x^2}} \, dx}{d^2 \left (c d^2+a e^2\right )}\\ &=-\frac {\sqrt {a+c x^2}}{2 a d^2 x^2}+\frac {2 e \sqrt {a+c x^2}}{a d^3 x}+\frac {e^4 \sqrt {a+c x^2}}{d^3 \left (c d^2+a e^2\right ) (d+e x)}+\frac {3 e^3 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^4 \sqrt {c d^2+a e^2}}-\frac {c \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{4 a d^2}+\frac {\left (3 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )}{c d^4}+\frac {\left (c e^3\right ) \operatorname {Subst}\left (\int \frac {1}{c d^2+a e^2-x^2} \, dx,x,\frac {a e-c d x}{\sqrt {a+c x^2}}\right )}{d^2 \left (c d^2+a e^2\right )}\\ &=-\frac {\sqrt {a+c x^2}}{2 a d^2 x^2}+\frac {2 e \sqrt {a+c x^2}}{a d^3 x}+\frac {e^4 \sqrt {a+c x^2}}{d^3 \left (c d^2+a e^2\right ) (d+e x)}+\frac {c e^3 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^2 \left (c d^2+a e^2\right )^{3/2}}+\frac {3 e^3 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^4 \sqrt {c d^2+a e^2}}-\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^4}-\frac {\operatorname {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )}{2 a d^2}\\ &=-\frac {\sqrt {a+c x^2}}{2 a d^2 x^2}+\frac {2 e \sqrt {a+c x^2}}{a d^3 x}+\frac {e^4 \sqrt {a+c x^2}}{d^3 \left (c d^2+a e^2\right ) (d+e x)}+\frac {c e^3 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^2 \left (c d^2+a e^2\right )^{3/2}}+\frac {3 e^3 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d^4 \sqrt {c d^2+a e^2}}+\frac {c \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{3/2} d^2}-\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^4}\\ \end {align*}

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Mathematica [A]  time = 0.45, size = 229, normalized size = 0.85 \begin {gather*} \frac {\frac {\left (c d^2-6 a e^2\right ) \log \left (\sqrt {a} \sqrt {a+c x^2}+a\right )}{a^{3/2}}+\frac {\log (x) \left (6 a e^2-c d^2\right )}{a^{3/2}}+d \sqrt {a+c x^2} \left (\frac {2 e^4}{(d+e x) \left (a e^2+c d^2\right )}-\frac {d-4 e x}{a x^2}\right )+\frac {2 e^3 \left (3 a e^2+4 c d^2\right ) \log \left (\sqrt {a+c x^2} \sqrt {a e^2+c d^2}+a e-c d x\right )}{\left (a e^2+c d^2\right )^{3/2}}-\frac {2 e^3 \left (3 a e^2+4 c d^2\right ) \log (d+e x)}{\left (a e^2+c d^2\right )^{3/2}}}{2 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(d + e*x)^2*Sqrt[a + c*x^2]),x]

[Out]

(d*Sqrt[a + c*x^2]*(-((d - 4*e*x)/(a*x^2)) + (2*e^4)/((c*d^2 + a*e^2)*(d + e*x))) + ((-(c*d^2) + 6*a*e^2)*Log[
x])/a^(3/2) - (2*e^3*(4*c*d^2 + 3*a*e^2)*Log[d + e*x])/(c*d^2 + a*e^2)^(3/2) + ((c*d^2 - 6*a*e^2)*Log[a + Sqrt
[a]*Sqrt[a + c*x^2]])/a^(3/2) + (2*e^3*(4*c*d^2 + 3*a*e^2)*Log[a*e - c*d*x + Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^
2]])/(c*d^2 + a*e^2)^(3/2))/(2*d^4)

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IntegrateAlgebraic [A]  time = 1.78, size = 248, normalized size = 0.93 \begin {gather*} \frac {\left (6 a e^2-c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d^4}-\frac {2 \sqrt {-a e^2-c d^2} \left (3 a e^5+4 c d^2 e^3\right ) \tan ^{-1}\left (\frac {-e \sqrt {a+c x^2}+\sqrt {c} d+\sqrt {c} e x}{\sqrt {-a e^2-c d^2}}\right )}{d^4 \left (a e^2+c d^2\right )^2}+\frac {\sqrt {a+c x^2} \left (-a d^2 e^2+3 a d e^3 x+6 a e^4 x^2-c d^4+3 c d^3 e x+4 c d^2 e^2 x^2\right )}{2 a d^3 x^2 (d+e x) \left (a e^2+c d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^3*(d + e*x)^2*Sqrt[a + c*x^2]),x]

[Out]

(Sqrt[a + c*x^2]*(-(c*d^4) - a*d^2*e^2 + 3*c*d^3*e*x + 3*a*d*e^3*x + 4*c*d^2*e^2*x^2 + 6*a*e^4*x^2))/(2*a*d^3*
(c*d^2 + a*e^2)*x^2*(d + e*x)) - (2*Sqrt[-(c*d^2) - a*e^2]*(4*c*d^2*e^3 + 3*a*e^5)*ArcTan[(Sqrt[c]*d + Sqrt[c]
*e*x - e*Sqrt[a + c*x^2])/Sqrt[-(c*d^2) - a*e^2]])/(d^4*(c*d^2 + a*e^2)^2) + ((-(c*d^2) + 6*a*e^2)*ArcTanh[(Sq
rt[c]*x - Sqrt[a + c*x^2])/Sqrt[a]])/(a^(3/2)*d^4)

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fricas [A]  time = 1.29, size = 1867, normalized size = 6.97

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*((4*a^2*c*d^2*e^4 + 3*a^3*e^6)*x^3 + (4*a^2*c*d^3*e^3 + 3*a^3*d*e^5)*x^2)*sqrt(c*d^2 + a*e^2)*log((2*a
*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + a*c*e^2)*x^2 + 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 +
a))/(e^2*x^2 + 2*d*e*x + d^2)) - ((c^3*d^6*e - 4*a*c^2*d^4*e^3 - 11*a^2*c*d^2*e^5 - 6*a^3*e^7)*x^3 + (c^3*d^7
- 4*a*c^2*d^5*e^2 - 11*a^2*c*d^3*e^4 - 6*a^3*d*e^6)*x^2)*sqrt(a)*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a
)/x^2) - 2*(a*c^2*d^7 + 2*a^2*c*d^5*e^2 + a^3*d^3*e^4 - 2*(2*a*c^2*d^5*e^2 + 5*a^2*c*d^3*e^4 + 3*a^3*d*e^6)*x^
2 - 3*(a*c^2*d^6*e + 2*a^2*c*d^4*e^3 + a^3*d^2*e^5)*x)*sqrt(c*x^2 + a))/((a^2*c^2*d^8*e + 2*a^3*c*d^6*e^3 + a^
4*d^4*e^5)*x^3 + (a^2*c^2*d^9 + 2*a^3*c*d^7*e^2 + a^4*d^5*e^4)*x^2), 1/4*(4*((4*a^2*c*d^2*e^4 + 3*a^3*e^6)*x^3
+ (4*a^2*c*d^3*e^3 + 3*a^3*d*e^5)*x^2)*sqrt(-c*d^2 - a*e^2)*arctan(sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*
x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^2)) - ((c^3*d^6*e - 4*a*c^2*d^4*e^3 - 11*a^2*c*d^2*e^5 - 6
*a^3*e^7)*x^3 + (c^3*d^7 - 4*a*c^2*d^5*e^2 - 11*a^2*c*d^3*e^4 - 6*a^3*d*e^6)*x^2)*sqrt(a)*log(-(c*x^2 - 2*sqrt
(c*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(a*c^2*d^7 + 2*a^2*c*d^5*e^2 + a^3*d^3*e^4 - 2*(2*a*c^2*d^5*e^2 + 5*a^2*c*
d^3*e^4 + 3*a^3*d*e^6)*x^2 - 3*(a*c^2*d^6*e + 2*a^2*c*d^4*e^3 + a^3*d^2*e^5)*x)*sqrt(c*x^2 + a))/((a^2*c^2*d^8
*e + 2*a^3*c*d^6*e^3 + a^4*d^4*e^5)*x^3 + (a^2*c^2*d^9 + 2*a^3*c*d^7*e^2 + a^4*d^5*e^4)*x^2), -1/2*(((c^3*d^6*
e - 4*a*c^2*d^4*e^3 - 11*a^2*c*d^2*e^5 - 6*a^3*e^7)*x^3 + (c^3*d^7 - 4*a*c^2*d^5*e^2 - 11*a^2*c*d^3*e^4 - 6*a^
3*d*e^6)*x^2)*sqrt(-a)*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - ((4*a^2*c*d^2*e^4 + 3*a^3*e^6)*x^3 + (4*a^2*c*d^3*e^
3 + 3*a^3*d*e^5)*x^2)*sqrt(c*d^2 + a*e^2)*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + a*c*e^2)*x^2 +
2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2)) + (a*c^2*d^7 + 2*a^2*c*d^5*e^
2 + a^3*d^3*e^4 - 2*(2*a*c^2*d^5*e^2 + 5*a^2*c*d^3*e^4 + 3*a^3*d*e^6)*x^2 - 3*(a*c^2*d^6*e + 2*a^2*c*d^4*e^3 +
a^3*d^2*e^5)*x)*sqrt(c*x^2 + a))/((a^2*c^2*d^8*e + 2*a^3*c*d^6*e^3 + a^4*d^4*e^5)*x^3 + (a^2*c^2*d^9 + 2*a^3*
c*d^7*e^2 + a^4*d^5*e^4)*x^2), 1/2*(2*((4*a^2*c*d^2*e^4 + 3*a^3*e^6)*x^3 + (4*a^2*c*d^3*e^3 + 3*a^3*d*e^5)*x^2
)*sqrt(-c*d^2 - a*e^2)*arctan(sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2
+ a*c*e^2)*x^2)) - ((c^3*d^6*e - 4*a*c^2*d^4*e^3 - 11*a^2*c*d^2*e^5 - 6*a^3*e^7)*x^3 + (c^3*d^7 - 4*a*c^2*d^5
*e^2 - 11*a^2*c*d^3*e^4 - 6*a^3*d*e^6)*x^2)*sqrt(-a)*arctan(sqrt(-a)/sqrt(c*x^2 + a)) - (a*c^2*d^7 + 2*a^2*c*d
^5*e^2 + a^3*d^3*e^4 - 2*(2*a*c^2*d^5*e^2 + 5*a^2*c*d^3*e^4 + 3*a^3*d*e^6)*x^2 - 3*(a*c^2*d^6*e + 2*a^2*c*d^4*
e^3 + a^3*d^2*e^5)*x)*sqrt(c*x^2 + a))/((a^2*c^2*d^8*e + 2*a^3*c*d^6*e^3 + a^4*d^4*e^5)*x^3 + (a^2*c^2*d^9 + 2
*a^3*c*d^7*e^2 + a^4*d^5*e^4)*x^2)]

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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/x^3/(e*x+d)^2/(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Evaluation time: 0.53Unable to divide, perhaps due to rounding error%%%{%%%{1,[0,0,0,7]%%%},[6,1]%
%%}+%%%{%%{[%%%{-6,[0,0,0,6]%%%},0]:[1,0,%%%{-1,[1,2,0,0]%%%}+%%%{-1,[0,0,1,2]%%%}]%%},[5,1]%%%}+%%%{%%%{12,[1
,2,0,5]%%%}+%%%{15,[0,0,1,7]%%%},[4,1]%%%}+%%%{%%{[%%%{-8,[1,2,0,4]%%%}+%%%{-20,[0,0,1,6]%%%},0]:[1,0,%%%{-1,[
1,2,0,0]%%%}+%%%{-1,[0,0,1,2]%%%}]%%},[3,1]%%%}+%%%{%%%{12,[1,2,1,5]%%%}+%%%{15,[0,0,2,7]%%%},[2,1]%%%}+%%%{%%
{[%%%{-6,[0,0,2,6]%%%},0]:[1,0,%%%{-1,[1,2,0,0]%%%}+%%%{-1,[0,0,1,2]%%%}]%%},[1,1]%%%}+%%%{%%%{1,[0,0,3,7]%%%}
,[0,1]%%%} / %%%{%%{[%%%{-1,[1,2,0,3]%%%}+%%%{-1,[0,0,1,5]%%%},0]:[1,0,%%%{-1,[1,2,0,0]%%%}+%%%{-1,[0,0,1,2]%%
%}]%%},[6,0]%%%}+%%%{%%%{6,[2,4,0,2]%%%}+%%%{12,[1,2,1,4]%%%}+%%%{6,[0,0,2,6]%%%},[5,0]%%%}+%%%{%%{[%%%{-12,[2
,4,0,1]%%%}+%%%{-27,[1,2,1,3]%%%}+%%%{-15,[0,0,2,5]%%%},0]:[1,0,%%%{-1,[1,2,0,0]%%%}+%%%{-1,[0,0,1,2]%%%}]%%},
[4,0]%%%}+%%%{%%%{8,[3,6,0,0]%%%}+%%%{36,[2,4,1,2]%%%}+%%%{48,[1,2,2,4]%%%}+%%%{20,[0,0,3,6]%%%},[3,0]%%%}+%%%
{%%{[%%%{-12,[2,4,1,1]%%%}+%%%{-27,[1,2,2,3]%%%}+%%%{-15,[0,0,3,5]%%%},0]:[1,0,%%%{-1,[1,2,0,0]%%%}+%%%{-1,[0,
0,1,2]%%%}]%%},[2,0]%%%}+%%%{%%%{6,[2,4,2,2]%%%}+%%%{12,[1,2,3,4]%%%}+%%%{6,[0,0,4,6]%%%},[1,0]%%%}+%%%{%%{[%%
%{-1,[1,2,3,3]%%%}+%%%{-1,[0,0,4,5]%%%},0]:[1,0,%%%{-1,[1,2,0,0]%%%}+%%%{-1,[0,0,1,2]%%%}]%%},[0,0]%%%} Error:
Bad Argument Value

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maple [A]  time = 0.01, size = 452, normalized size = 1.69 \begin {gather*} \frac {c \,e^{2} \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, d^{2}}+\frac {\sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, e^{3}}{\left (a \,e^{2}+c \,d^{2}\right ) \left (x +\frac {d}{e}\right ) d^{3}}+\frac {3 e^{2} \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, d^{4}}-\frac {3 e^{2} \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{\sqrt {a}\, d^{4}}+\frac {c \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2 a^{\frac {3}{2}} d^{2}}+\frac {2 \sqrt {c \,x^{2}+a}\, e}{a \,d^{3} x}-\frac {\sqrt {c \,x^{2}+a}}{2 a \,d^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(e*x+d)^2/(c*x^2+a)^(1/2),x)

[Out]

2*e*(c*x^2+a)^(1/2)/a/d^3/x-1/2*(c*x^2+a)^(1/2)/a/d^2/x^2+1/2/d^2*c/a^(3/2)*ln((2*a+2*(c*x^2+a)^(1/2)*a^(1/2))
/x)-3/d^4*e^2/a^(1/2)*ln((2*a+2*(c*x^2+a)^(1/2)*a^(1/2))/x)+3/d^4*e^2/((a*e^2+c*d^2)/e^2)^(1/2)*ln((-2*(x+d/e)
*c*d/e+2*(a*e^2+c*d^2)/e^2+2*((a*e^2+c*d^2)/e^2)^(1/2)*(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(1/2))
/(x+d/e))+1/d^3*e^3/(a*e^2+c*d^2)/(x+d/e)*(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(1/2)+1/d^2*e^2*c/(
a*e^2+c*d^2)/((a*e^2+c*d^2)/e^2)^(1/2)*ln((-2*(x+d/e)*c*d/e+2*(a*e^2+c*d^2)/e^2+2*((a*e^2+c*d^2)/e^2)^(1/2)*(-
2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(1/2))/(x+d/e))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {c x^{2} + a} {\left (e x + d\right )}^{2} x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(c*x^2 + a)*(e*x + d)^2*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{x^3\,\sqrt {c\,x^2+a}\,{\left (d+e\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + c*x^2)^(1/2)*(d + e*x)^2),x)

[Out]

int(1/(x^3*(a + c*x^2)^(1/2)*(d + e*x)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(e*x+d)**2/(c*x**2+a)**(1/2),x)

[Out]

Integral(1/(x**3*sqrt(a + c*x**2)*(d + e*x)**2), x)

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