3.4.39 \(\int \frac {1}{x (d+e x) (a+c x^2)^{3/2}} \, dx\) [339]

Optimal. Leaf size=147 \[ \frac {1}{a d \sqrt {a+c x^2}}-\frac {e (a e+c d x)}{a d \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}+\frac {e^3 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d \left (c d^2+a e^2\right )^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {975, 272, 53, 65, 214, 755, 12, 739, 212} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d}-\frac {e (a e+c d x)}{a d \sqrt {a+c x^2} \left (a e^2+c d^2\right )}+\frac {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 \left (a e^2+c d^2\right )^{3/2}}+\frac {1}{a d \sqrt {a+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 53

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 65

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 212

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

Rule 214

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

Rule 272

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 739

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 755

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(a*e + c*d*x)*
((a + c*x^2)^(p + 1)/(2*a*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^
m*Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[
{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 975

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

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Mathematica [A]
time = 0.75, size = 144, normalized size = 0.98 \begin {gather*} \frac {c (d-e x)}{a \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}-\frac {2 e^3 \tan ^{-1}\left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )}{d \left (-c d^2-a e^2\right )^{3/2}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(362\) vs. \(2(131)=262\).
time = 0.06, size = 363, normalized size = 2.47

method	result	size
default	\(-\frac {\frac {e^{2}}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}+\frac {2 c d e \left (2 c \left (x +\frac {d}{e}\right )-\frac {2 c d}{e}\right )}{\left (a \,e^{2}+c \,d^{2}\right ) \left (\frac {4 c \left (a \,e^{2}+c \,d^{2}\right )}{e^{2}}-\frac {4 c^{2} d^{2}}{e^{2}}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}-\frac {e^{2} \ln \left (\frac {\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}+c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}}{d}+\frac {\frac {1}{a \sqrt {c \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}}{d}\)	\(363\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 302 vs. \(2 (131) = 262\).
time = 2.93, size = 1284, normalized size = 8.73 \begin {gather*} \left [\frac {{\left (a^{2} c x^{2} + a^{3}\right )} \sqrt {c d^{2} + a e^{2}} e^{3} \log \left (-\frac {2 \, c^{2} d^{2} x^{2} - 2 \, a c d x e + a c d^{2} - 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a} + {\left (a c x^{2} + 2 \, a^{2}\right )} e^{2}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) + {\left (c^{3} d^{4} x^{2} + a c^{2} d^{4} + {\left (a^{2} c x^{2} + a^{3}\right )} e^{4} + 2 \, {\left (a c^{2} d^{2} x^{2} + a^{2} c d^{2}\right )} e^{2}\right )} \sqrt {a} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (a c^{2} d^{3} x e - a c^{2} d^{4} + a^{2} c d x e^{3} - a^{2} c d^{2} e^{2}\right )} \sqrt {c x^{2} + a}}{2 \, {\left (a^{2} c^{3} d^{5} x^{2} + a^{3} c^{2} d^{5} + {\left (a^{4} c d x^{2} + a^{5} d\right )} e^{4} + 2 \, {\left (a^{3} c^{2} d^{3} x^{2} + a^{4} c d^{3}\right )} e^{2}\right )}}, -\frac {2 \, {\left (a^{2} c x^{2} + a^{3}\right )} \sqrt {-c d^{2} - a e^{2}} \arctan \left (-\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{c^{2} d^{2} x^{2} + a c d^{2} + {\left (a c x^{2} + a^{2}\right )} e^{2}}\right ) e^{3} - {\left (c^{3} d^{4} x^{2} + a c^{2} d^{4} + {\left (a^{2} c x^{2} + a^{3}\right )} e^{4} + 2 \, {\left (a c^{2} d^{2} x^{2} + a^{2} c d^{2}\right )} e^{2}\right )} \sqrt {a} \log \left (-\frac {c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (a c^{2} d^{3} x e - a c^{2} d^{4} + a^{2} c d x e^{3} - a^{2} c d^{2} e^{2}\right )} \sqrt {c x^{2} + a}}{2 \, {\left (a^{2} c^{3} d^{5} x^{2} + a^{3} c^{2} d^{5} + {\left (a^{4} c d x^{2} + a^{5} d\right )} e^{4} + 2 \, {\left (a^{3} c^{2} d^{3} x^{2} + a^{4} c d^{3}\right )} e^{2}\right )}}, \frac {{\left (a^{2} c x^{2} + a^{3}\right )} \sqrt {c d^{2} + a e^{2}} e^{3} \log \left (-\frac {2 \, c^{2} d^{2} x^{2} - 2 \, a c d x e + a c d^{2} - 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a} + {\left (a c x^{2} + 2 \, a^{2}\right )} e^{2}}{x^{2} e^{2} + 2 \, d x e + d^{2}}\right ) + 2 \, {\left (c^{3} d^{4} x^{2} + a c^{2} d^{4} + {\left (a^{2} c x^{2} + a^{3}\right )} e^{4} + 2 \, {\left (a c^{2} d^{2} x^{2} + a^{2} c d^{2}\right )} e^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) - 2 \, {\left (a c^{2} d^{3} x e - a c^{2} d^{4} + a^{2} c d x e^{3} - a^{2} c d^{2} e^{2}\right )} \sqrt {c x^{2} + a}}{2 \, {\left (a^{2} c^{3} d^{5} x^{2} + a^{3} c^{2} d^{5} + {\left (a^{4} c d x^{2} + a^{5} d\right )} e^{4} + 2 \, {\left (a^{3} c^{2} d^{3} x^{2} + a^{4} c d^{3}\right )} e^{2}\right )}}, -\frac {{\left (a^{2} c x^{2} + a^{3}\right )} \sqrt {-c d^{2} - a e^{2}} \arctan \left (-\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{c^{2} d^{2} x^{2} + a c d^{2} + {\left (a c x^{2} + a^{2}\right )} e^{2}}\right ) e^{3} - {\left (c^{3} d^{4} x^{2} + a c^{2} d^{4} + {\left (a^{2} c x^{2} + a^{3}\right )} e^{4} + 2 \, {\left (a c^{2} d^{2} x^{2} + a^{2} c d^{2}\right )} e^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x^{2} + a}}\right ) + {\left (a c^{2} d^{3} x e - a c^{2} d^{4} + a^{2} c d x e^{3} - a^{2} c d^{2} e^{2}\right )} \sqrt {c x^{2} + a}}{a^{2} c^{3} d^{5} x^{2} + a^{3} c^{2} d^{5} + {\left (a^{4} c d x^{2} + a^{5} d\right )} e^{4} + 2 \, {\left (a^{3} c^{2} d^{3} x^{2} + a^{4} c d^{3}\right )} e^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((a^2*c*x^2 + a^3)*sqrt(c*d^2 + a*e^2)*e^3*log(-(2*c^2*d^2*x^2 - 2*a*c*d*x*e + a*c*d^2 - 2*sqrt(c*d^2 + a
*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a) + (a*c*x^2 + 2*a^2)*e^2)/(x^2*e^2 + 2*d*x*e + d^2)) + (c^3*d^4*x^2 + a*c^2
*d^4 + (a^2*c*x^2 + a^3)*e^4 + 2*(a*c^2*d^2*x^2 + a^2*c*d^2)*e^2)*sqrt(a)*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt
(a) + 2*a)/x^2) - 2*(a*c^2*d^3*x*e - a*c^2*d^4 + a^2*c*d*x*e^3 - a^2*c*d^2*e^2)*sqrt(c*x^2 + a))/(a^2*c^3*d^5*
x^2 + a^3*c^2*d^5 + (a^4*c*d*x^2 + a^5*d)*e^4 + 2*(a^3*c^2*d^3*x^2 + a^4*c*d^3)*e^2), -1/2*(2*(a^2*c*x^2 + a^3
)*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)/(c^2*d^2*x^2 + a*c*d^2 + (a*
c*x^2 + a^2)*e^2))*e^3 - (c^3*d^4*x^2 + a*c^2*d^4 + (a^2*c*x^2 + a^3)*e^4 + 2*(a*c^2*d^2*x^2 + a^2*c*d^2)*e^2)
*sqrt(a)*log(-(c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(a*c^2*d^3*x*e - a*c^2*d^4 + a^2*c*d*x*e^3 -
a^2*c*d^2*e^2)*sqrt(c*x^2 + a))/(a^2*c^3*d^5*x^2 + a^3*c^2*d^5 + (a^4*c*d*x^2 + a^5*d)*e^4 + 2*(a^3*c^2*d^3*x^
2 + a^4*c*d^3)*e^2), 1/2*((a^2*c*x^2 + a^3)*sqrt(c*d^2 + a*e^2)*e^3*log(-(2*c^2*d^2*x^2 - 2*a*c*d*x*e + a*c*d^
2 - 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a) + (a*c*x^2 + 2*a^2)*e^2)/(x^2*e^2 + 2*d*x*e + d^2)) +
2*(c^3*d^4*x^2 + a*c^2*d^4 + (a^2*c*x^2 + a^3)*e^4 + 2*(a*c^2*d^2*x^2 + a^2*c*d^2)*e^2)*sqrt(-a)*arctan(sqrt(-
a)/sqrt(c*x^2 + a)) - 2*(a*c^2*d^3*x*e - a*c^2*d^4 + a^2*c*d*x*e^3 - a^2*c*d^2*e^2)*sqrt(c*x^2 + a))/(a^2*c^3*
d^5*x^2 + a^3*c^2*d^5 + (a^4*c*d*x^2 + a^5*d)*e^4 + 2*(a^3*c^2*d^3*x^2 + a^4*c*d^3)*e^2), -((a^2*c*x^2 + a^3)*
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)/(c^2*d^2*x^2 + a*c*d^2 + (a*c*
x^2 + a^2)*e^2))*e^3 - (c^3*d^4*x^2 + a*c^2*d^4 + (a^2*c*x^2 + a^3)*e^4 + 2*(a*c^2*d^2*x^2 + a^2*c*d^2)*e^2)*s
qrt(-a)*arctan(sqrt(-a)/sqrt(c*x^2 + a)) + (a*c^2*d^3*x*e - a*c^2*d^4 + a^2*c*d*x*e^3 - a^2*c*d^2*e^2)*sqrt(c*
x^2 + a))/(a^2*c^3*d^5*x^2 + a^3*c^2*d^5 + (a^4*c*d*x^2 + a^5*d)*e^4 + 2*(a^3*c^2*d^3*x^2 + a^4*c*d^3)*e^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x\,{\left (c\,x^2+a\right )}^{3/2}\,\left (d+e\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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