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

Optimal result

Integrand size = 22, antiderivative size = 147 \[ \int \frac {1}{x (d+e x) \left (a+c x^2\right )^{3/2}} \, dx=\frac {1}{a d \sqrt {a+c x^2}}-\frac {e (a e+c d x)}{a d \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}+\frac {e^3 \text {arctanh}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{d \left (c d^2+a e^2\right )^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d} \]

[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)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {975, 272, 53, 65, 214, 755, 12, 739, 212} \[ \int \frac {1}{x (d+e x) \left (a+c x^2\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {e^3 \text {arctanh}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{d \left (a e^2+c d^2\right )^{3/2}}-\frac {e (a e+c d x)}{a d \sqrt {a+c x^2} \left (a e^2+c d^2\right )}+\frac {1}{a d \sqrt {a+c x^2}} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x (d+e x) \left (a+c x^2\right )^{3/2}} \, dx=\frac {c (d-e x)}{a \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}-\frac {2 e^3 \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )}{d \left (-c d^2-a e^2\right )^{3/2}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d} \]

[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)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(362\) vs. \(2(131)=262\).

Time = 0.37 (sec) , antiderivative size = 363, normalized size of antiderivative = 2.47

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

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (132) = 264\).

Time = 0.47 (sec) , antiderivative size = 1325, normalized size of antiderivative = 9.01 \[ \int \frac {1}{x (d+e x) \left (a+c x^2\right )^{3/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Sympy [F]

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

[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)

Maxima [F]

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

[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)*(e*x + d)*x), x)

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{x (d+e x) \left (a+c x^2\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]

[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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x (d+e x) \left (a+c x^2\right )^{3/2}} \, dx=\int \frac {1}{x\,{\left (c\,x^2+a\right )}^{3/2}\,\left (d+e\,x\right )} \,d x \]

[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)