\(\int \frac {1}{(d+e x) (a+c x^2)^{3/2}} \, dx\) [573]

Optimal result

Integrand size = 19, antiderivative size = 94 \[ \int \frac {1}{(d+e x) \left (a+c x^2\right )^{3/2}} \, dx=\frac {a e+c d x}{a \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}-\frac {e^2 \text {arctanh}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{\left (c d^2+a e^2\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {755, 12, 739, 212} \[ \int \frac {1}{(d+e x) \left (a+c x^2\right )^{3/2}} \, dx=\frac {a e+c d x}{a \sqrt {a+c x^2} \left (a e^2+c d^2\right )}-\frac {e^2 \text {arctanh}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{3/2}} \]

[In]

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

[Out]

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

Rule 12

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

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

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.11 \[ \int \frac {1}{(d+e x) \left (a+c x^2\right )^{3/2}} \, dx=\frac {a e+c d x}{a \left (c d^2+a e^2\right ) \sqrt {a+c x^2}}+\frac {2 e^2 \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+c x^2}}{\sqrt {-c d^2-a e^2}}\right )}{\left (-c d^2-a e^2\right )^{3/2}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(314\) vs. \(2(86)=172\).

Time = 2.43 (sec) , antiderivative size = 315, normalized size of antiderivative = 3.35

method	result	size
default	\(\frac {\frac {e^{2}}{\left (e^{2} a +c \,d^{2}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}+\frac {2 c d e \left (2 c \left (x +\frac {d}{e}\right )-\frac {2 c d}{e}\right )}{\left (e^{2} a +c \,d^{2}\right ) \left (\frac {4 c \left (e^{2} a +c \,d^{2}\right )}{e^{2}}-\frac {4 c^{2} d^{2}}{e^{2}}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}-\frac {e^{2} \ln \left (\frac {\frac {2 e^{2} a +2 c \,d^{2}}{e^{2}}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}-\frac {2 c d \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (e^{2} a +c \,d^{2}\right ) \sqrt {\frac {e^{2} a +c \,d^{2}}{e^{2}}}}}{e}\)	\(315\)

[In]

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

[Out]

1/e*(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)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (87) = 174\).

Time = 0.33 (sec) , antiderivative size = 456, normalized size of antiderivative = 4.85 \[ \int \frac {1}{(d+e x) \left (a+c x^2\right )^{3/2}} \, dx=\left [\frac {{\left (a c e^{2} x^{2} + a^{2} e^{2}\right )} \sqrt {c d^{2} + a e^{2}} \log \left (\frac {2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} - {\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} - 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, {\left (a c d^{2} e + a^{2} e^{3} + {\left (c^{2} d^{3} + a c d e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{2 \, {\left (a^{2} c^{2} d^{4} + 2 \, a^{3} c d^{2} e^{2} + a^{4} e^{4} + {\left (a c^{3} d^{4} + 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}\right )} x^{2}\right )}}, -\frac {{\left (a c e^{2} x^{2} + a^{2} e^{2}\right )} \sqrt {-c d^{2} - a e^{2}} \arctan \left (\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) - {\left (a c d^{2} e + a^{2} e^{3} + {\left (c^{2} d^{3} + a c d e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{a^{2} c^{2} d^{4} + 2 \, a^{3} c d^{2} e^{2} + a^{4} e^{4} + {\left (a c^{3} d^{4} + 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}\right )} x^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.39 \[ \int \frac {1}{(d+e x) \left (a+c x^2\right )^{3/2}} \, dx=\frac {c d x}{\sqrt {c x^{2} + a} a c d^{2} + \sqrt {c x^{2} + a} a^{2} e^{2}} + \frac {1}{\frac {\sqrt {c x^{2} + a} c d^{2}}{e} + \sqrt {c x^{2} + a} a e} + \frac {\operatorname {arsinh}\left (\frac {c d x}{e \sqrt {\frac {a c}{e^{2}}} {\left | e x + d \right |}} - \frac {a}{\sqrt {\frac {a c}{e^{2}}} {\left | e x + d \right |}}\right )}{{\left (a + \frac {c d^{2}}{e^{2}}\right )}^{\frac {3}{2}} e} \]

[In]

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

[Out]

c*d*x/(sqrt(c*x^2 + a)*a*c*d^2 + sqrt(c*x^2 + a)*a^2*e^2) + 1/(sqrt(c*x^2 + a)*c*d^2/e + sqrt(c*x^2 + a)*a*e)
+ arcsinh(c*d*x/(e*sqrt(a*c/e^2)*abs(e*x + d)) - a/(sqrt(a*c/e^2)*abs(e*x + d)))/((a + c*d^2/e^2)^(3/2)*e)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (87) = 174\).

Time = 0.28 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.91 \[ \int \frac {1}{(d+e x) \left (a+c x^2\right )^{3/2}} \, dx=-\frac {2 \, e^{2} \arctan \left (\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} - a e^{2}}}\right )}{{\left (c d^{2} + a e^{2}\right )} \sqrt {-c d^{2} - a e^{2}}} + \frac {\frac {{\left (c^{2} d^{3} + a c d e^{2}\right )} x}{a c^{2} d^{4} + 2 \, a^{2} c d^{2} e^{2} + a^{3} e^{4}} + \frac {a c d^{2} e + a^{2} e^{3}}{a c^{2} d^{4} + 2 \, a^{2} c d^{2} e^{2} + a^{3} e^{4}}}{\sqrt {c x^{2} + a}} \]

[In]

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

[Out]

-2*e^2*arctan(((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))/((c*d^2 + a*e^2)*sqrt(-c*d^2
- a*e^2)) + ((c^2*d^3 + a*c*d*e^2)*x/(a*c^2*d^4 + 2*a^2*c*d^2*e^2 + a^3*e^4) + (a*c*d^2*e + a^2*e^3)/(a*c^2*d
^4 + 2*a^2*c*d^2*e^2 + a^3*e^4))/sqrt(c*x^2 + a)

Mupad [F(-1)]

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

[In]

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

[Out]

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